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(a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) 11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2+y2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.5 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2(θ) − 6r cos(θ) + 9 + r2 sin2(θ) = 9 r2 cos2(θ) + sin2(θ) − 6r cos(θ) = 0 Subtract 9 from both sides. Since cos2(θ) + sin2(θ) = 1 Factor. r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our ﬁnal answer.6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r sin(θ) = −r cos(θ). Rearranging, we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the
origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4. In this equation, the variable r is free,7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our ﬁnal answer θ = − π 4 here.8 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x2 and get r sin(θ) = (r cos(θ))2, or r2 cos2(θ) − r sin(θ) = 0. Factoring, we get r(r cos2(θ) − sin(θ)) = 0 so that either r = 0 or r cos2(θ) = sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos2(θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos2(θ) = 0, then cos(θ) = 0, and for the equation r cos2(θ) = sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeﬀ’s. 2 into r = 6 cos(θ), we recover the point r = 0, so we aren’t losing anything 6Note that when we substitute θ = π by disregarding r = 0. 7See Section 8.1. 8We could take it to be any of θ = − π 4 + πk for integers k. 928 Applications of Trigonometry information by dividing both sides of r cos2(θ) = sin(θ
) by cos2(θ). Doing so, we get r = sin(θ) cos2(θ), or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our ﬁnal answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We could solve r2 = x2 + y2 for r to get r = ± x2 + y2 + πk for and solving tan(θ) = y integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y2, r cos(θ) = x, r sin(θ) = y and/or tan(θ) = y x requires the arctangent function to get θ = arctan y x present themselves. x (a) Starting with r = −3, we can square both sides to get r2 = (−3)2 or r2 = 9. We may now substitute r2 = x2 + y2 to get the equation x2 + y2 = 9. As we have seen,9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r2 = 9 might be satisﬁed by more points than r = −3. On the surface, this appears to be the case since r2 = 9 is equivalent to r = ±3, not just r = −3. However, any point with polar coordinates (3, θ) can be represented as (−3, θ + π), which means any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent10 representation which satisﬁes r = −3. (b) We take the tangent of both sides the equation θ = 4π Since tan(θ) = y if, geometrically, the equations θ = 4π 3 and y = x The same argument presented in number 1b applies equally well here so we are done. 3 to get tan(θ) = tan
4π 3. 3. Of course, we pause a moment to wonder 3 generate the same set of points.11 x, we get y 3 or c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r2 = r − r cos(θ). We now have an r2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r2 + r cos(θ) and squaring both sides yields r2 = r2 + r cos(θ)2. Substituting r2 = x2 + y2 and r cos(θ) = x gives x2 + y2 = x2 + y2 + x2. Once again, we have performed some 9Exercise 5.3.1 in Section 5.3, for instance... 10Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3, π) are diﬀerent, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = −3 represent diﬀerent relations since they correspond to diﬀerent sets of ordered pairs. Since polar coordinates were deﬁned geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we ﬁrst deﬁned relations as sets of points in the plane in Section 1.2. Back then, a point in the plane was identiﬁed with a unique ordered pair given by its Cartesian coordinates. 11In addition to taking the tangent of both sides of an equation (There are inﬁnitely many solutions to tan(θ) = √ √ 3 is only one of them
!), we also went from y x = 3, in which x cannot be 0, to y = x √ 3, 3 in which we assume and θ = 4π x can be 0. 11.4 Polar Coordinates 929 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1 − cos(θ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates (r, θ) which satisfy r2 = r2 + r cos(θ)2 but do not satisfy r = r2 + r cos(θ)? Suppose (r, θ) satisﬁes r2 = r2 + r cos(θ)2. Then r = ± (r)2 + r cos(θ). If we have that r = (r)2+r cos(θ), we are done. What if r = − (r)2 + r cos(θ) = −(r)2−r cos(θ)? We claim that the coordinates (−r, θ + π), which determine the same point as (r, θ), satisfy r = r2 + r cos(θ). We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to see if we get a true statement. −r − −(r)2 − r cos(θ) (r)2 + r cos(θ) (r)2 + r cos(θ)? = (−r)2 + (−r cos(θ + π))? = (r)2 − r cos(θ + π)? = (r)2 − r(− cos(θ)) = (r)2 + r cos(θ) Since r = −(r)2 − r cos(θ) Since cos(θ + π) = − cos(θ) Since both sides worked out to be equal, (−r, θ + π) satisﬁes r = r2 + r
cos(θ) which means that any point (r, θ) which satisﬁes r2 = r2 + r cos(θ)2 has a representation which satisﬁes r = r2 + r cos(θ), and we are done. In practice, much of the pedantic veriﬁcation of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r2 = 9 happens without a second thought. Your instructor will ultimately decide how much, if any, justiﬁcation is warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things in rectangular coordinates, such as y = x2, can turn ugly in polar coordinates, and vice-versa. In the next section, we devote our attention to graphing equations like the ones given in Example 11.4.3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates. If nothing else, number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. 930 Applications of Trigonometry 11.4.1 Exercises In Exercises 1 - 16, plot the point given in polar coordinates and then give three diﬀerent expressions (c) r > 0 and θ ≥ 2π for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π, (b) r > 0 and θ ≤ 0 2, 1. π 3 5. 12, − 7π 6 9. (−20, 3π) 2. 5, 7π 4 6. 3, − 10. −4, 5π 4 5π 4 3. 1 3, 3π 2 √ 7. 2 2, −π 11. −1, 2π 3 13. −3, − 11π 6 14. −2.5, − π 4 15. √ − 5, − 4π 3 4. 8. 5 2 7 2, 5π 6, − 13π 6 −3, 12. π 2 16. (−π, −π) In Exercises 17 - 36, convert the point from polar coordinates into rectangular coordinates. 18. 2, π 3 22. −4, 5π 6 7π 6 19. 11,
− 23. 9, 7π 2 20. (−20, 3π) 24. −5, − 9π 4 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 17. 5, 7π 4 21. 3 5, π 2 25. 42, 13π 6 29. −3, arctan 31. 33. 2, π − arctan −1, π + arctan 4 3 1 2 3 4 34. 2 3 35. (π, arctan(π)) 36. 13, arctan 30. 5, arctan − 4 3 32. − 1 2, π − arctan (5), π + arctan 2 √ 2 12 5 In Exercises 37 - 56, convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 37. (0, 5) 41. (−3, 0) 38. (3, √ 3) 39. (7, −7) √ 42. − √ 2 2, 43. −4, −4 √ 3 √ 3) 40. (−3, − √ 44. 3 4, − 1 4 11.4 Polar Coordinates 931 45. − √ 3 3 10 3 10, − √ 46. − √ 5, − 5 47. (6, 8) √ √ 5) 5, 2 48. ( 49. (−8, 1) √ 50. (−2 √ 10, 6 10) 53. (24, −7) 54. (12, −9) 51. (−5, −12) √ 2 4 √ 6 4, 55. 52. − √ 5 15 √ 5 2 15, − 56. − √ 65 5 2, √ 65 5 In Exercises 57 - 76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #63. In Exercises 60 - 63, you need to solve for θ 57. x = 6 61. y = −x 58. x = −3 √ 3 62. y = x 59. y = 7 60. y = 0 63. y = 2x 64. x2 + y2 = 25 65. x2 + y2 = 117 66
. y = 4x − 19 67. x = 3y + 1 68. y = −3x2 69. 4x = y2 70. x2 + y2 − 2y = 0 71. x2 − 4x + y2 = 0 72. x2 + y2 = x 73. y2 = 7y − x2 75. x2 + (y − 3)2 = 9 74. (x + 2)2 + y2 = 4 76. 4x2 + 4 y − 2 1 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78. r = −3 82. θ = π 79. r = 83. θ = √ 2 3π 2 80. θ = π 4 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88. r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) √ 91. r = − 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = − csc(θ) cot(θ) 94. r2 = sin(2θ) 95. r = 1 − 2 cos(θ) 96. r = 1 + sin(θ) 97. Convert the origin (0, 0) into polar coordinates in four diﬀerent ways. 98. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 932 Applications of Trigonometry 11.4.2 Answers 1. 2,, π 3 2, − 5π 3 −2,, 4π 3 7π 3 2, 2. 5, 7π 4 π 4,, −5, 5, 3π 4 15π 4 5, − 3. 1 3 1 3 4. 5 2 5 2,,, 3π,, 5π 6 7π 7π 2,, 11π 6 17π 6 y y 2 1 1 −1 1 2 x −1 1 2 3 x −1 −2 −3 −2 −1
−1 1 2 3 x −2 −3 11.4 Polar Coordinates 933 5. 12, − 12, −, −12,, 12, 7π 6 19π 6 11π 6 17π 6 6. 3, − 3, −, −3,, 3, 5π 4 13π 4 7π 4 11π 4 √ √ 7. 2 2 √ 2, −π, −2 √ 2, −3π, 2 2, 0 2, 3π 8. 7 2 7 2, −, − 13π 23π 6 5π 6 y 6 3 −12 −9 −6 −3 x y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 3 2 1 y −3 −2 −1 1 2 3 x −1 −2 −3 4 3 2 1 y −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 934 Applications of Trigonometry 9. (−20, 3π), (−20, π) (20, −2π), (20, 4π) 10. −4, 4, − 5π 4 7π 4 11. −1, 1, − 2π 3 π 3 13π 4,, −4, 4, 9π 4 8π 3,, 1, −1, 11π 3 12. −3, 3, −,, π 2 π 2 5π 2 −3, 3, 7π 2 10 x 20 −20 −10 y y 1 −1 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 2 1 y −2 −1 1 2 x −1 −2 y 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 935 13. −3, − 11π 6, 3, − 5π 6 π 6 −3, 19π 6, 3,,, 14. −2.5, − 2.5, − π 4 5π 4 −2.5, 2.5, 7π 4 11π 4 √ 15. − 5, − √ 5, − π 3 4π 3,, √ √ − 2π 3 5, 5, 11π 3 16. (−π, −π), (−π, π) (
π, −2π), (π, 2π3 −2 −1 −1 −2 −3 2 1 −2 −1 −1 −2 2 1 −2 −1 −1 −2 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 936 Applications of Trigonometry √ 5 2 2, − √ 2 5 2 17. 18. 1, √ 3 19. − √ 11 2 3, 11 2 22. 2 √ 3, −2 23. (0, −9) 20. (20, 0) 24 28. √ 10, 3 √ 10 √ 26 52 √ 5 26 52, − 32. 36. (5, 12) √ 2 3, 40. 44. 1 2, 11π 6 7π 6 4 3 48. (5, arctan (2)) 26. (117, 0) 30. (3, −4 34. √ 2 3, 38. 42. 2, √ 46. 3π 4 10, π 6 5π 4 √ 5 6 5 5 √ 12 5, 27. 31 35. π√ 1+π2, √ π2 1+π2 7π 4 √ 7 2, 39. 43. 8, 4π 3 47. 10, arctan 21. 0, 3 5 √ 25. 21 3, 21 29. − 9 5, − 12 5 33. 4 5, 3 5 37. 5, π 2 41. (3, π) 4π 3, 3 5 √ 51. 13, π + arctan 25, 2π − arctan 1 8 12 5 7 24 45. 49. 53. 55. 65, π − arctan 50. (20, π − arctan(3)), π + arctan (2) 52. 1 3 54. 15, 2π − arctan 3 4 √ 2 2, π 3 56. √ 13, π − arctan(2) 57. r = 6 sec(θ) 58. r = −3 sec(θ) 59. r = 7 csc(θ) 60. θ = 0 61. θ = 3π 4 √ 65. r = 117 62. θ = π 3 63. θ = arct
an(2) 64. r = 5 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = − sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 937 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = sin(θ) 77. x2 + y2 = 49 78. x2 + y2 = 9 79. x2 + y2 = 2 80. y = x √ 81. y = − 3x 83. x = 0 82. y = 0 84. x2 + y2 = 4x or (x − 2)2 + y2 = 4 85. 5x2 + 5y2 = x or x − 2 1 10 + y2 = 1 100 86. x2 + y2 = 3y or x2 + y − 2 3 2 = 9 4 87. x2 + y2 = −2y or x2 + (y + 1)2 = 1 88. x = 7 89. y = 1 12 √ 91. y = − 5 93. y2 = −x 90. x = −2 92. x2 = 2y 94. x2 + y22 = 2xy 95. x2 + 2x + y22 = x2 + y2 96. x2 + y2 + y2 97. Any point of the form (0, θ) will work, e.g. (0, π), (0, −117), 0, = x2 + y2 23π 4 and (0, 0). 938 Applications of Trigonometry 11.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has inﬁnitely many diﬀerent representations in polar coordinates, our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. We state it
below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a point P (r, θ) is on the graph of an equation if and only if there is a representation of P, say (r, θ), such that r and θ satisfy the equation. Our ﬁrst example focuses on some of the more structurally simple polar equations. Example 11.5.1. Graph the following polar equations. 1. r = 4 2. r = −3 √ 2 3. θ = 5π 4 4. θ = − 3π 2 Solution. In each of these equations, only one of the variables r and θ is present making the other variable free.1 This makes these graphs easier to visualize than others. 1. In the equation r = 4, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation (4, θ), for any choice of θ. Graphically this translates into tracing out all of the points 4 units away from the origin. This is exactly the deﬁnition of circle, centered at the origin, with a radius of 44 x 4 In r = 4, θ is free The graph of r = 4 −4 2. Once again we have θ being free in the equation r = −3 2, θ) gives us a circle of radius 3 form (−3 √ √ 2 centered at the origin. 2. Plotting all of the points of the √ 1See the discussion in Example 11.4.3 number 2a. 11.5 Graphs of Polar Equations 939 4 x 4 In r = −3 √ 2, θ is free −4 The graph of r = −3 √ 2 3. In the equation θ = 5π. 4, r is free, so we plot all of the points with polar representation r, 5π What we ﬁnd is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. 4 y r < 0 θ = 5π 4 y 4 r = 0 x −4 x 4 r > 0 In θ = 5π 4, r is free The graph of θ = 5π 4 −4 4. As in the previous example, the variable r
is free in the equation θ = − 3π 2. Plotting r, − 3π 2 for various values of r shows us that we are tracing out the y-axis. 940 Applications of Trigonometry y r > 0 r = 0 θ = − 3π 2 y 4 x −4 x 4 r < 0 −4 In θ = − 3π 2, r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following result clear. Theorem 11.8. Graphs of Constant r and θ: Suppose a and α are constants, a = 0. The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius \|a\|. The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cos(θ). A reasonable way to start is to treat θ as the independent variable, r as the dependent variable, evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.2 We generate the table below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π r = 6 cos(θ) 6 √ 3 2 3 √ √ (r, θ) (6, 0) 2, π 4 0, π 2 2, 3π 4 (−6, π) √ 2, 5π 4 0, 3π 2 √ 2, 7π 4 (6, 2π) 0 2 −3 √ −3 −6 √ −3 √ 3 2 6 2 −3 0 3 y 3 −3 3 x 6 2For a review of these concepts and this process, see Sections 1.4 and 1.6. 11.5 Graphs of Polar Equations 941 Despite having nine ordered pairs, we get only four distinct points on the graph. For this reason, we employ a slightly diﬀerent strategy. We graph one cycle of r = 6 cos(θ) on the θr-plane3 and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π 2, r ranges from 6 to 0. In
the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the ﬁgure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. r 6 3 −3 −6 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x Next, we repeat the process as θ ranges from π 2 to π. Here, the r values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. r 6 3 −3 −6 y θ runs from π 2 to π π 2 π 3π 2 2π θ x r < 0 so we plot here As θ ranges from π to 3π 2, the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the \|r\| for these values of θ match the r values for θ in 3The graph looks exactly like y = 6 cos(x) in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r, θ) on the xy-plane. 942 Applications of Trigonometry, we have that the curve begins to retrace itself at this point. Proceeding further, we ﬁnd 2 ≤ θ ≤ 2π, we retrace the portion of the curve in Quadrant IV that we ﬁrst traced 2 ≤ θ ≤ π. The reader is invited to verify that plotting any range of θ outside the interval 0, π 2 that when 3π out as π [0
, π] results in retracting some portion of the curve.4 We present the ﬁnal graph below. r 6 3 −3 −6 π 2 π θ y 3 −3 3 x 6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r2 = 16 cos(2θ) Solution. 1. We ﬁrst plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. To help us visualize what is going on graphically, we divide up [0, 2π] into the usual four subintervals 0, π 2, π, 2 π, 3π 2, r decreases from 2 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. 2, 2π, and proceed as we did above. As θ ranges from 0 to π and 3π, π r 6 4 2 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x 4The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of Polar Equations 943 Next, as θ runs from π oﬀ, we gradually pull the graph away from the origin until we reach the negative x-axis. 2 to π, we see that r increases from 2 to 4. Picking up where we left θ runs from π 2 to 3π 2 2π θ, we see that r increases from 4 to 6. On the xy-plane, the curve Over the interval π, 3π 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis 3π 2 2π θ θ runs from π to 3π 2 Finally, as θ takes on values from 3π xy
-plane pulls in from the negative y-axis to ﬁnish where we started. 2 to 2π, r decreases from 6 back to 4. The graph on the r 6 4 2 y x π 2 π 3π 2 2π θ θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions of the curve, so we are ﬁnished. 944 Applications of Trigonometry sin(θ) in the θr-plane 3π 2 2π y 2 −4 x 4 −6 r = 4 − 2 sin(θ) in the xy-plane. 2. The ﬁrst thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0, 2π] is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our ﬁrst task is to determine when this happens. Setting r = 0 we get 2 + 4 cos(θ) = 0, or cos(θ) = − 1 2. Solving for θ in [0, 2π] gives θ = 2π 3. Since these values of θ are important geometrically, we break the interval [0, 2π] into six subintervals: 0, π 2, 2π. As 3, π, π, 4π 2 θ ranges from 0 to π 2, r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. 3 and θ = 4π and 3π, 2π, 4π 3, 3π, π 2, 2π 3 3 2 r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x On the interval π will eventually cross through) the origin. Not only do we reach the origin when θ = 2π theorem from Calculus5 states that the curve hugs the line θ = 2π, r decreases from 2 to 0,
which means the graph is heading into (and 3, a 3 as it approaches the origin. 2, 2π 3 5The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 945 r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x 3, π, r ranges from 0 to −2. Since r ≤ 0, the curve passes through the On the interval 2π origin in the xy-plane, following the line θ = 2π 3 and continues upwards through Quadrant IV towards the positive x-axis.6 Since \|r\| is increasing from 0 to 2, the curve pulls away from the origin to ﬁnish at a point on the positive x-axis. r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x Next, as θ progresses from π to 4π graph in the ﬁrst quadrant, heading into the origin along the line θ = 4π 3. 3, r ranges from −2 to 0. Since r ≤ 0, we continue our 6Recall that one way to visualize plotting polar coordinates (r, θ) with r < 0 is to start the rotation from the left 3 and π radians from the negative x-axis in side of the pole - in this case, the negative x-axis. Rotating between 2π this case determines the region between the line θ = 2π 3 and the x-axis in Quadrant IV. 946 Applications of Trigonometry r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x On the interval 4π line θ = 4π 3, 3π 2, r returns to positive values and increases from 0 to 2. We hug the 3 as we move through the origin and head towards the negative y-axis. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x As we round out the interval, we ﬁnd that as θ runs through 3π to 6, and we end up back where we started, 6 units from the origin on the
positive x-axis. 2 to 2π, r increases from 2 out y r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ x θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 947 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our ﬁnal graph is below. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 2 θ = 4π 3 −2 2 6 x r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As usual, we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane, which in this case, occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely 0, π, π 4, r 4 increases from 0 to 5. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4. 4, π. As θ ranges from 0 to π and 3π, π 2, 3π 3π 4 π θ x r 5 −5 Next, we see that r decreases from 5 to 0 as θ runs through π heading negative as θ crosses π as the curve heads to the origin. 2. Hence, we draw the curve hugging the line θ = π, and furthermore, r is 2 (the y-axis) 4, π 2 948 Applications of Trigonometry y π 4 π 2 3π 4 π θ x r 5 −5 As θ runs from π pulls away from the negative y-axis into Quadrant IV. 2 to 3π 4, r becomes negative and ranges from 0 to −5. Since r ≤ 0, the curve y π 4 π 2 3π 4 π θ x r 5 −5 For 3π 4
≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. y π 4 π 2 3π 4 π θ x r 5 −5 11.5 Graphs of Polar Equations 949 Even though we have ﬁnished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we ﬁnd that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the ﬁrst. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. r 5 π 1 5π 4 2 3π 2 3 7π 4 4 2π θ 4 1 −5 We have the ﬁnal graph below 3π 4 π 5π 4 3π 2 7π 4 2π θ −5 x 5 −5 −5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. Graphing r2 = 16 cos(2θ) is complicated by the r2, so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). How do we sketch such a curve? First oﬀ, we sketch a fundamental period cos(2θ) is of r = cos(2θ) which we have dotted in the ﬁgure below. When cos(2θ) < 0,. On the intervals which remain, undeﬁned, so we don’t have any values on the interval π cos(2θ) ranges from 0 to 1 as well.7 From cos(2θ) ranges from 0 to 1, inclusive. Hence, cos(2θ) ranges continuously from 0 to ±4, respectively. Below we this, we know r = ±4 cos(2θ) on the θr plane and use them to sketch the graph both r = 4 corresponding pieces of the curve r2 = 16 cos(2θ) in the xy-plane. As we have seen in earlier cos(2θ) and r = −4 4, 3π 4 7Owing to
the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is deﬁned. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4, which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ), θ = 3π 4 π 4 π 2 3π 4 4 3 π θ r = 4 cos(2θ) and r = −4cos(2θ) As we plot points corresponding to values of θ outside of the interval [0, π], we ﬁnd ourselves retracing parts of the curve,8 so our ﬁnal answer is below. r 4 θ = 3π 3π 4 π θ −4 x 4 −4 r = ±4 cos(2θ) in the θr-plane −4 r2 = 16 cos(2θ) in the xy-plane A few remarks are in order. First, there is no relation, in general, between the period of the function f (θ) and the length of the interval required to sketch the complete graph of r = f (θ) in the xyplane. As we saw on page 941, despite the fact that the period of f (θ) = 6 cos(θ) is 2π, we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 11.5.2, number 3, the period of f (θ) = 5 sin(2θ) is π, but in order to obtain the complete graph of r = 5 sin(2θ), we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8In this case, we could have generated the entire graph by using just the plot r = 4cos(2θ), but graphed over the interval [0, 2π
] in the θr-plane. We leave the details to the reader. 9Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 951 coordinate system, but also prepare you for what is needed in Calculus. Second, the symmetry seen in the examples is also a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. In our next example, we are given the task of ﬁnding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 938, in order for a point P to be on the graph of a polar equation, it must have a representation P (r, θ) which satisﬁes the equation. What complicates matters in polar coordinates is that any given point has inﬁnitely many representations. As a result, if a point P is on the graph of two diﬀerent polar equations, it is entirely possible that the representation P (r, θ) which satisﬁes one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to ﬁnd our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 − 2 sin(θ) 2. r = 2 and r = 3 cos(θ) 3. r = 3 and r = 6 cos(2θ) Solution. 4. r = 3 sin θ 2 and r = 3 cos θ 2 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and ﬁnd it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 − 2 sin(θ) is a special kind of
lima¸con called a ‘cardioid.’10 y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the ﬁrst quadrant, one in the second quadrant, and the origin. Our next task is to ﬁnd polar representations of these points. In 10Presumably, the name is derived from its resemblance to a stylized human heart. 952 Applications of Trigonometry 2. From this, we get θ = π 6 into r = 2 sin(θ), we get r = 2 sin π order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P (r, θ) which satisﬁes r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly diﬀerent) representation P (r, θ) which satisﬁes r = 2 sin(θ). We ﬁrst try to see if we can ﬁnd any points which have a single representation P (r, θ) that satisﬁes both r = 2 sin(θ) and r = 2 − 2 sin(θ). Assuming such a pair (r, θ) exists, then equating11 the expressions for r gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 1 6 + 2πk = 1, which for integers k. Plugging θ = π is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). Hence, 1, π is one 6 representation for the point of intersection in the ﬁrst quadrant. For the point of intersection, so this is in the second quadrant, we try θ = 5π our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π, and as the reader can verify, we are
at the origin exactly when θ = πk for integers k. On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determined the three points of intersection to be 1, π 6 6. Both equations give us the point 1, 5π 2, and more generally, when θ = π 6 + 2πk or θ = 5π = 2 1 2 and the origin., 1, 5π 6 6 6 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a circle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ) is also a circle - but this one is centered at the point with rectangular coordinates 3 2, 0 and has a radius of 3 2. y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ) We have two intersection points to ﬁnd, one in Quadrant I and one in Quadrant IV. Proceeding as above, we ﬁrst determine if any of the intersection points P have a representation (r, θ) which satisﬁes both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 3. To solve this equation, we need the arccosine function. We get 11We are really using the technique of substitution to solve the system of equations r = 2 sin(θ) r = 2 − 2 sin(θ) 11.5 Graphs of Polar Equations 953 θ = arccos 2 + 2πk or θ = 2π − arccos 2 3 3 2, arccos 2 as one representation for our answer in Quadrant I, and 2, 2π − arccos 2 3 3 as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraically and geometrically. + 2πk for integers k.
From these solutions, we get 3. Proceeding as above, we ﬁrst graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12 y 6 3 −6 −3 3 x 6 −3 −6 r = 3 and r = 6 cos(2θ), 3, 5π 6, 3, 7π 6 and 3, 11π 6 6 + πk or θ = 5π 2. Solving, we get θ = π It appears as if there are eight points of intersection - two in each quadrant. We ﬁrst look to see if there any points P (r, θ) with a representation that satisﬁes both r = 3 and r = 6 cos(2θ). For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 6 + πk for integers k. Out of all of these solutions, we obtain just four distinct points represented by 3, π. To determine the coordinates of the remaining four 6 points, we have to consider how the representations of the points of intersection can diﬀer. We know from Section 11.4 that if (r, θ) and (r, θ) represent the same point and r = 0, then either r = r or r = −r. If r = r, then θ = θ+2πk, so one possibility is that an intersection point P has a representation (r, θ) which satisﬁes r = 3 and another representation (r, θ+2πk) for some integer, k which satisﬁes r = 6 cos(2θ). At this point,13 if we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if, by equating the resulting expressions for r, we get any more solutions for θ. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k,
however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r, we have that θ = θ + (2k + 1)π for integers k. We look to see if we can ﬁnd points P which have a representation (r, θ) that satisﬁes r = 3 and another, 12See Example 11.5.2 number 3. 13The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of eﬃciency. 954 Applications of Trigonometry (−r, θ + (2k + 1)π), that satisﬁes r = 6 cos(2θ). To do this, we substitute14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). Since cos(2(θ + (2k + 1)π)) = cos(2θ + (2k + 1)(2π)) = cos(2θ) for all integers k, the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 1 3 + πk. From these solutions, we obtain15 the remaining four intersection points with representations −3, π 3, which we can readily check graphically. 3 + πk or θ = 2π 2. We get θ = π and −3, 5π 3, −3, 2π 3, −3, 4π 3. Using the techniques
4. As usual, we begin by graphing r = 3 sin θ 2 presented in Example 11.5.2, we ﬁnd that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! and r = 3 cos θ 2 y 3 −3 3 x −3 r = 3 sin θ 2 and r = 3 cos θ 2 appear to determine the same curve in the xy-plane and r = 3 cos θ 2 To verify this incredible claim,16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (r, θ) which satisﬁes both r = 3 sin θ. Equating these two expressions for r gives 2. While normally we discourage dividing by a variable = 3 cos θ the equation 3 sin θ 2 2 expression (in case it could be 0), we note here that if 3 cos θ = 0, then for our equation 2 = 0 as well. Since no angles have both cosine and sine equal to zero, to hold, 3 sin θ 2 to get we are safe to divide both sides of the equation 3 sin θ 2 tan θ 2 + 2πk for integers k. From these solutions, however, we 2 = 1 which gives θ = π by 3 cos θ 2 = 3 cos θ 2 14Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. 15We obtain these representations by substituting the values for θ into r = 6 cos(2θ), once again, for illustration purposes. Again, in the interests of eﬃciency, we could ‘plug’ these values for θ into r = 3 (where there is no θ) and represents the same point get the list of points: 3, π as −3, π, we still get the same set of solutions.. While it is not true that 3, π and 3, 5π, 3, 4π, 3, 2π 3 3 3 3 3 16A quick sketch of r = 3 sin θ and r = 3 cos θ in the θr-plane will convince you that, viewed
as functions of r, 3 2 2 these are two diﬀerent animals. 11.5 Graphs of Polar Equations 955 √ 3 2 2, π 2 2 + πk = 3 cos θ 2 [θ + 2πk] = 3 cos θ get only one intersection point which can be represented by. We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation (r, θ) which satisﬁes r = 3 sin θ and the same point P has a diﬀerent 2. Substituting representation (r, θ + 2πk) for some integer k which satisﬁes r = 3 cos θ 2 into the latter, we get r = 3 cos 1 2 + πk. Using the sum formula for, since sin (πk) = ±3 cos θ cosine, we expand 3 cos θ 2 sin(πk) = 0 for all integers k, and cos (πk) = ±1 for all integers k. If k is an even integer, we get the same equation r = 3 cos θ as before. If k is odd, we get r = −3 cos θ. This 2 2 = −1. Solving, latter expression for r leads to the equation 3 sin θ 2 √ we get θ = − π 2, − π. Next, we assume P has a representation (r, θ) which satisﬁes r = 3 sin θ and a representation 2 for some integer k. Substituting (−r) for (−r, θ + (2k + 1)π) which satisﬁes r = 3 cos θ 2 r and (θ + (2k + 1)π) in for θ into r = 3 cos θ 2 [θ + (2k + 1)π]. Once gives −r = 3 cos 1 2 again, we use the sum formula for cosine to get 2 + 2πk for integers k, which gives the intersection point cos(πk) − 3 sin θ 2, or tan θ 2 3 = −3 cos θ 2 2 2 2 cos 1 2 [θ + (2k + 1)π] = cos 2 θ 2 + (2k+1)π cos (2k+1)
π 2 = cos θ 2 = ± sin θ 2 − sin θ 2 sin (2k+1)π 2 = 0 and sin (2k+1)π (2k+1)π where the last equality is true since cos 2 2 2 [θ + (2k + 1)π] can be rewritten as r = ±3 sin θ Hence, −r = 3 cos 1 (2k+1)π = 1, and the equation −r = 3 cos 1 = sin π then sin 2 2 reduces to −r = −3 sin θ, or r = 3 sin θ 2 2 What this means is that if a polar representation (r, θ) for the point P satisﬁes r = 3 sin( θ then the representation (−r, θ + π) for P automatically satisﬁes r = 3 cos θ 2 equations r = 3 sin( θ = ±1 for integers k.. If we choose k = 0, 2 [θ + (2k + 1)π] in this case which is the other equation under consideration! 2 ),. Hence the 2 ) determine the same set of points in the plane. 2 ) and r = 3 cos( θ 2 Our work in Example 11.5.3 justiﬁes the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To ﬁnd the points of intersection of the graphs of two polar equations E1 and E2: Sketch the graphs of E1 and E2. Check to see if the curves intersect at the origin (pole). Solve for pairs (r, θ) which satisfy both E1 and E2. Substitute (θ + 2πk) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 956 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. Example 11.5.4. Sketch
the region in the xy-plane described by the following sets. 1. (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 3. (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 4. (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 3 ≤ θ ≤ 4π 3 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π 6 ≤ θ ≤ π 2 Solution. Our ﬁrst step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π 2, we are tracing out the ‘leaf’ of the rose which lies in the ﬁrst quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0, π 2. Hence, the region we seek is the leaf itself. r 5 −5 π 4 π 2 3π 4 π θ y x (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π 6, so the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π 6. In other words, we are looking at the points outside or on the circle (since r ≥ 3
) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below and r = 6 cos(2θ) (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 11.5 Graphs of Polar Equations 957 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given values 2π 3. Since the values r takes on in this interval are non-positive, the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense, and we are looking for all of the points between the pole r = 0 and the lima¸con as θ ranges over the interval 2π r. In other words, we shade in the inner loop of the lima¸con. 3 to 4π 3, 4π 3 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 θ = 4π 3 x (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 4. We have two regions described here connected with the union symbol ‘∪.’ We shade each in turn and ﬁnd our ﬁnal answer by combining the two. In Example 11.5.3, number 1, we found that the curves r = 2 sin(θ) and r = 2 − 2 sin(θ) intersect when θ = π 6. Hence, for the, we are shading the region between the origin ﬁrst region, (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π 6, which is the angle of intersection of the two curves. For the second region, (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π, θ picks up 6 ≤ θ ≤ π where it left oﬀ at π 2. In this case, however
, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2. Putting these two regions together gives us our ﬁnal answer. 6 and continues to sin(θ) and r = 2 sin(θ) (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π ∪ 6 ≤ θ ≤ π 2 958 Applications of Trigonometry 11.5.1 Exercises In Exercises 1 - 20, plot the graph of the polar equation by hand. Carefully label your graphs. 1. Circle: r = 6 sin(θ) 2. Circle: r = 2 cos(θ) 3. Rose: r = 2 sin(2θ) 4. Rose: r = 4 cos(2θ) 5. Rose: r = 5 sin(3θ) 6. Rose: r = cos(5θ) 7. Rose: r = sin(4θ) 8. Rose: r = 3 cos(4θ) 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ) 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) 16. Lima¸con: r = 3 − 5 cos(θ) 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) In Exercises 21 - 30, ﬁnd the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 21. r
= 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 − cos(θ) 23. r = 1 − 2 sin(θ) and r = 2 24. r = 1 − 2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 27. r2 = 4 cos(2θ) and r = 2 √ 3 sin(θ) √ 26. r = 3 cos(θ) and r = sin(θ) 28. r2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In Exercises 31 - 40, sketch the region in the xy-plane described by the given set. 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π} 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 11.5 Graphs of Polar Equations 959 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 4 3 ≤ θ ≤ π 3 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 38. (r, θ) \| 1 ≤ r ≤ 2 sin(2θ), 13π √ 12 ≤ θ ≤ 17π 12 39. (r, θ) \| 0 ≤ r ≤ 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 3 sin(θ), 0 ≤ θ ≤ π 6 ∪ (r, θ) \| 0 ≤ r ≤ 1
, π 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the left half of the circle r = 6 sin(θ). 44. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. 45. The region inside the top half of the cardioid r = 3 − 3 cos(θ) 46. The region inside the cardioid r = 2 − 2 sin(θ) which lies in Quadrants I and IV. 47. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. The region inside the circle r = 5 but outside the circle r = 3. 49. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to your understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difﬁcult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The ﬁrst thing you must do is switch the MODE of your calculator to POL, which stands for “polar”. 960 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu,
we ﬁnd three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more about the θ-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [θmin, θmax] using the θstep. For most graphs, a θstep of 0.1 is ﬁne. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this. You will need to experiment with the settings in order to get a nice graph. Exercises 51 - 60 give you some curves to graph using your calculator. Notice that some of them have explicit bounds on θ and others do not. 11.5 Graphs of Polar Equations 961 51. r = θ, 0 ≤ θ ≤ 12π 52. r = ln(θ), 1 ≤ θ ≤ 12π 53. r = e.1θ, 0 ≤ θ ≤ 12π 54. r = θ3 − θ, −1.2 ≤ θ ≤ 1.2 55. r = sin(5θ) − 3 cos(θ) 57. r = arctan(θ), −π ≤ θ ≤ π 59. r = 1 2 − cos(θ) 56. r = sin3 θ 2 + cos2 θ 3 58. r = 1 1 − cos(θ) 60. r = 1 2 − 3 cos(θ) 61. How many petals does the polar
rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section, it’s clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26. In Exercises 62 - 64, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves. 62. Show that if f is even17 then the graph of r = f (θ) is symmetric about the x-axis. (a) Show that f (θ) = 2 + 4 cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis. (See Example 11.5.2 number 2.) (b) Show that f (θ) = 3 sin θ 2 is not even, yet the graph of r = 3 sin θ 2 is symmetric about the x-axis. (See Example 11.5.3 number 4.) 63. Show that if f is odd18 then the graph of r = f (θ) is symmetric about the origin. (a) Show that f (θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 sin(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f (θ) = 3 cos θ 2 is not odd, yet the graph of r = 3 cos θ 2 is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric about the y-axis. (a) For f (θ) = 4 − 2 sin(
θ), show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17Recall that this means f (−θ) = f (θ) for θ in the domain of f. 18Recall that this means f (−θ) = −f (θ) for θ in the domain of f. 962 Applications of Trigonometry (b) For f (θ) = 5 sin(2θ), show that f π − π 4 = f π 4, yet the graph of r = 5 sin(2θ) is symmetric about the y-axis. (See Example 11.5.2 number 3.) In Section 1.7, we discussed transformations of graphs. classmates explore transformations of polar graphs. In Exercise 65 we have you and your 65. For Exercises 65a and 65b below, let f (θ) = cos(θ) and g(θ) = 2 − sin(θ). (a) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of. Repeat this process for g(θ). In general, how do you think the graph of r = f (θ + α) compares with the graph of r = f (θ)? and r = f θ − 3π 4, r = f θ + 3π 4 (b) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of r = 2f (θ), r = 1 2 f (θ), r = −f (θ) and r = −3f (θ). Repeat this process for g(θ). In general, how do you think the graph of r = k · f (θ) compares with the graph of r = f (θ)? (Does it matter if k > 0 or k < 0?) 66. In light of Exercises 62 - 64, how would the graph of r = f (−θ) compare with the graph of r = f (θ) for a generic function f? What about the graphs of r = −f (θ) and r = f (θ)? What
about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research. 11.5 Graphs of Polar Equations 963 11.5.2 Answers 1. Circle: r = 6 sin(θ) y −6 6 −6 3. Rose: r = 2 sin(2θ) y −2 2 −2 2. Circle: r = 2 cos(θ) y 2 6 x −2 2 x −2 4. Rose: r = 4 cos(2θ) y 4 θ = 3π 4 θ = π 4 2 x −4 4 x 5. Rose: r = 5 sin(3θ) y 5 θ = 2π 3 θ = π 3 −5 5 x −4 6. Rose: r = cos(5θ) y 1 θ = 7π 10 θ = 3π 10 θ = 9π 10 −1 θ = π 10 1 x −5 −1 964 Applications of Trigonometry 7. Rose: r = sin(4θ) y 1 θ = 3π 4 θ = π 4 −1 1 x 8. Rose: r = 3 cos(4θ) y θ = 5π 8 3 θ = 3π 8 θ = 7π 8 −3 θ = π 8 3 x −1 −3 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) y 6 3 y 10 5 −6 −3 3 6 x −10 −5 5 10 x −3 −6 −5 −10 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) y 4 2 y 2 1 −4 −2 2 4 x −2 −1 1 2 x −2 −4 −1 −2 11.5 Graphs of Polar Equations 965 13. Lima¸con: r
= 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ = 5π 6 −3 −1 1 3 x −3 −1 1 3 x −1 −3 θ = 5π 3 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) y √ 2 3 + 4 θ = 5π 6 √ 2 3 √ −2 3 − 4 θ = 7π 6 √ −2 3 √ −1 −3 16. Lima¸con: r = 3 − 5 cos(θ) y 8 3 θ = arccos 3 5 −8 −2 8 x −3 −8 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) y 8 θ = π − arcsin 3 5 θ = arcsin 3 5 y 9 5 −8 −3 3 −2 −8 8 x −9 θ = π + arcsin −2 2 7 2 9 θ = 2π − arcsin 2 7 x −9 966 Applications of Trigonometry 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) y 1 y 2 θ = 3π 4 θ = π 4 −1 1 x −2 2 x −1 −2 3 2, π 3, 3 2, 5π 3, pole √ 2 + 2, √ 2 − 2 2, 7π 4 2, 3π 4, pole 21. r = 3 cos(θ) and r = 1 + cos(θ) y 3 2 1 −3 −2 −1 1 2 3 −1 −2 −3 x 22. r = 1 + sin(θ) and r = 1 − cos(θ) y 2 1 −2 −1 1 2 x −1 −2 11.5 Graphs of Polar Equations 967 2, 7π 6, 2, 11π 6 1,, π 2 1, 3π 2, (−1, 0) √ 3, π 6, pole 23. r = 1 − 2 sin(θ) and r = 2 y 3 1 −3 −1
1 3 x −1 −3 24. r = 1 − 2 cos(θ) and r = 1 y 3 1 −3 −1 1 3 x −1 −3 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) y 4 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 968 Applications of Trigonometry √ 3 10 10, arctan(3), pole √ 2,, π 6 √, 5π 6 2, √, 7π 6 2, √ 2, 11π 6 1,, π 12 1,, 5π 12 1, 13π 12, 1, 17π 12 26. r = 3 cos(θ) and r = sin(θ) y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 27. r2 = 4 cos(2θ) and r = √ 2 y 2 −2 2 x −2 28. r2 = 2 sin(2θ) and r = 1 y √ 2 1 −1 11.5 Graphs of Polar Equations 969 29. r = 4 cos(2θ) and r = 2 y 4 2, 2,, π 6 11π 6, 7π 6 2, 2, 5π 6, −2,, π 3, −2,, 2π 3 −2, 4π 3, −2, 5π 3 −4 4 x −4 30. r = 2 sin(2θ) and r = 1 y 2 −2 2 x −2, π 12, 17π 12 1, 1, 1, −1,, 19π 12, 5π 12 1,, 7π 12 −1, −1, 23π 12, 13π 12 −1,, 11π 12 970 Applications of Trigonometry 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π3 −2 −1 1 2 3 −1 −2 −3 x x −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 33. (r, θ) \| 0 ≤ r ≤ 3 cos
(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 3 −2 −1 1 2 3 −1 −2 −3 x −2 2 x −2 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 y 4 y 3 1 −4 4 x −3 −1 1 3 x −4 −1 −3 11.5 Graphs of Polar Equations 971 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − 3 −2 −1 1 2 3 −1 −2 −3 x 38. (r, θ) \| 1 ≤ r ≤ y 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 √ − 2 −1 √ 2 1 −r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 39. (r, θ) \| 0 ≤ r ≤ 2 3 sin(θ), 3 −2 −1 1 2 3 x −1 −2 −3 −4 972 Applications of Trigonometry ∪ (r, θ) \| 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 y −2 2 −2 2 x 41. {(r, θ) \| 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 42. (r, θ) \| 0 ≤ r ≤ 5, π ≤ θ ≤ 3π 2 43. (r, θ) \| 0 ≤ r ≤ 6 sin(θ), π 44. (r, θ) \| 4 cos(θ) ≤ r ≤ 0 45. {(r, θ) \| 0 ≤ r ≤ 3 − 3 cos(θ), 0 ≤ θ ≤ π} 46. (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 0 ≤ θ ≤ π
2 or (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 47. (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 or (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), − ≤ 5π 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 2 ≤ θ ≤ 2π ∪ (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π 48. {(r, θ) \| 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 49. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 50. (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ θ ≤ π (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 5π 2 2 ≤ θ ≤ 0 ∪ {(r, θ) \| sin(θ) ≤ r ≤ 3 cos(θ), 0 ≤ θ ≤ arctan(3)} ∪ (r, θ) \| 0 ≤ r ≤ 3, π 12 ≤ θ ≤ 5π 12 ∪ 11.6 Hooked on Conics Again 973 11.6 Hooked on Conics Again In this section, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our ﬁrst task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More speciﬁcally, it is the hyperbola obtained by rotating the graph of x2 − y2 = 4 counter-clockwise through a 45◦ angle. Armed with polar coordinates, we can generalize the process of rotating axes as shown below. 11.6.1 Rotation of Axes Consider the x- and y-axes below along with the dashed x- and y-axes obtained by rotating the xand y-axes counter-clockwise through an angle θ and
consider the point P (x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to ﬁnd rectangular coordinates based on the x- and y-axes. That is, we wish to determine P (x, y). While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive x-axis and whose terminal side contains the point P. y y P (x, y) = P (x, y) x θ φ θ x We relate P (x, y) and P (x, y) by converting them to polar coordinates. Converting P (x, y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). To convert the point P (x, y) into polar coordinates, we ﬁrst match the polar axis with the positive x-axis, choose the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). Using the sum formulas for sine and cosine, we have x = r cos(θ + φ) = r cos(θ) cos(φ) − r sin(θ) sin(φ) = (r cos(φ)) cos(θ) − (r sin(φ)) sin(θ) = x cos(θ) − y sin(θ) Sum formula for cosine Since x = r cos(φ) and y = r sin(φ) 974 Applications of Trigonometry Similarly, using the sum formula for sine we get y = x sin(θ) + y cos(θ). These equations enable us to easily convert points with xy-coordinates back into xy-coordinates. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y.1 If we want equations which enable us to convert points with xy-coordinates into xy-coordinates, we need to solve the system x cos(θ) − y sin(θ) = x x sin(θ) + y cos(θ) = y for
x and y. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = cos(θ) − sin(θ) cos(θ) sin(θ Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2(θ) + sin2(θ) = 1, the determinant of A is not zero so A is invertible and X = A−1X. Using the formula given in Equation 8.2 with det(A) = 1, we ﬁnd so that A−1 = cos(θ) − sin(θ) sin(θ) cos(θ) X = A−1X x y x y = = cos(θ) − sin(θ) sin(θ) cos(θ) x y x cos(θ) + y sin(θ) −x sin(θ) + y cos(θ) From which we get x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ). To summarize, Theorem 11.9. Rotation of Axes: Suppose the positive x and y axes are rotated counterclockwise through an angle θ to produce the axes x and y, respectively. Then the coordinates P (x, y) and P (x, y) are related by the following systems of equations x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) and x = x cos(θ) + y sin(θ) y = −x sin(θ) + y cos(θ) We put the formulas in Theorem 11.9 to good use in the following example. 1Sound familiar? In Section 11.4, the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates, and they make it easy to convert equations from rectangular coordinates into polar coordinates. 2We could, of course, interchange the roles of x and x, y and y and replace φ
with −φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. Let P (x, y) = (2, −4) and ﬁnd P (x, y). Check your answer algebraically and graphically. 2. Convert the equation 21x2 + 10xy √ 3 + 31y2 = 144 to an equation in x and y and graph. Solution. √ 3, Theorem 11.9 gives x = x cos(θ) + y sin(θ) = 2 cos π √ 1. If P (x, y) = (2, −4) then x = 2 and y = −4. Using these values for x and y along with which simpliﬁes 3. Similarly, y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which √ 3 3 3. To check our answer 3. Hence P (x, y to x = 1 − 2 √ gives y = − algebraically, we use the formulas in Theorem 11.9 to convert P (x, y) = 1 − 2 back into x and y coordinates. We get + (−4) sin π 3 3 − 2 = −2 − 3, −2 − 3, −2 − √ √ 3 x = x cos(θ) − y sin(θ) √ − (−2 − 3) cos ) sin π 3 = ( Similarly, using y = x sin(θ) + y cos(θ), we obtain y = −4 as required. To check our answer graphically, we sketch in the x-axis and y-axis to see if the new coordinates P (x, y) = 3 ≈ (−2.46, −3.73) seem reasonable. Our graph is below. 1 − 2 3, −x, y) = (2, −4) P (x, y) ≈ (−2.46, −3.73) 2. To convert the
equation 21x2 +10xy we substitute x = x cos π 3 − y sin π 3 √ 3+31y2 = 144 to an equation in the variables x and y, = x 2 + y and y = x sin π 3 + y cos 976 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start by verifying that x2 = (x)2 4 − √ xy 2 3 + 3(y)2 4, √ (x)2 4 3 − xy 2 − xy = √ 3, y2 = 3(x)2 4 + √ xy 2 3 + (y)2 4 To our surprise and delight, the equation 21x2 + 10xy 3 + 31y2 = 144 in xy-coordinates 4 + (y)2 reduces to 36(x)2 + 16(y)2 = 144, or (x)2 9 = 1 in xy-coordinates. The latter is an ellipse centered at (0, 0) with vertices along the y-axis with (xy-coordinates) (0, ±3) and whose minor axis has endpoints with (xy-coordinates) (±2, 0). We graph it below. (y) 21x2 + 10xy √ 3 + 31y2 = 144 √ The elimination of the troublesome ‘xy’ term from the equation 21x2 + 10xy 3 + 31y2 = 144 in Example 11.6.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7. It is natural to wonder if we can always do this. That is, given an equation of the form Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, with B = 0, is there an angle θ so that if we rotate the x and yaxes counter-clockwise through that angle θ, the equation in the rotated variables x and y contains no xy term? To explore this conjecture, we make the usual substitutions x = x cos(θ) − y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax2 +
Bxy + Cy2 + Dx + Ey + F = 0 and set the coeﬃcient of the xy term equal to 0. Terms containing xy in this expression will come from the ﬁrst three terms of the equation: Ax2, Bxy and Cy2. We leave it to the reader to verify that x2 = (x)2 cos2(θ) − 2xy cos(θ) sin(θ) + (y)2 sin(θ) xy = (x)2 cos(θ) sin(θ) + xy cos2(θ) − sin2(θ) − (y)2 cos(θ) sin(θ) y2 = (x)2 sin2(θ) + 2xy cos(θ) sin(θ) + (y)2 cos2(θ) 11.6 Hooked on Conics Again 977 The contribution to the xy-term from Ax2 is −2A cos(θ) sin(θ), from Bxy it is B cos2(θ) − sin2(θ), and from Cy2 it is 2C cos(θ) sin(θ). Equating the xy-term to 0, we get −2A cos(θ) sin(θ) + B cos2(θ) − sin2(θ) + 2C cos(θ) sin(θ) = 0 −A sin(2θ) + B cos(2θ) + C sin(2θ) = 0 Double Angle Identities From this, we get B cos(2θ) = (A − C) sin(2θ), and our goal is to solve for θ in terms of the coeﬃcients A, B and C. Since we are assuming B = 0, we can divide both sides of this equation by B. To solve for θ we would like to divide both sides of the equation by sin(2θ), provided of course that we have assurances that sin(2θ) = 0. If sin(2θ) = 0, then we would have B cos(2θ) = 0, and since B = 0, this would force cos(2θ) = 0. Since no angle θ can have both sin(2
θ) = 0 and cos(2θ) = 0, we can safely assume3 sin(2θ) = 0. We get cos(2θ) B. We have just proved the following theorem. B, or cot(2θ) = A−C sin(2θ) = A−C Theorem 11.10. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any xy terms by rotating the xand y- axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C B. We put Theorem 11.10 to good use in the following example. Example 11.6.2. Graph the following equations. 1. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 2. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. 1. Since the equation 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 is already given to us in the form required by Theorem 11.10, we identify A = 5, B = 26 and C = 5 so that cot(2θ) = A−C 26 = 0. This means cot(2θ) = 0 which gives θ = π 2 k for integers k. √ √ 2 + y 2 2 We choose θ = π 2. The reader should verify that 4 so that our rotation equations are x = x 4 + π and y = x B = 5−5 2 − y 2 + 16y √ √ 2 2 2 √ √ x2 = (x)2 2 − xy + (y)2 2, xy = (x)2 2 − (y)2 2, y2 = (x)2 2 + xy + (y)2 2 Making the other substitutions, we get that 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 2 + 16y reduces to 18(x)2 − 8(y)2 + 32y − 104 = 0, or (x)2 9 = 1. The latter is the equation of a hyperbola centered at the
xy-coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in xy-coordinates) and asymptotes y = ± 3 4 − (y−2)2 2 x + 2. We graph it below. √ √ 3The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 978 Applications of Trigonometry 24 2. From 16x2 + 24xy + 9y2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that cot(2θ) = 7 24. Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to ﬁnd cos(θ) and sin(θ), which means we have two If we use the arccotangent function immediately, after the usual calculations we options.. To get cos(θ) and sin(θ) from this, we would need to use half angle 2 arccot 7 get θ = 1 identities. Alternatively, we can start with cot(2θ) = 7 24, use a double angle identity, and then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 7 24, we have tan(2θ) = 24 7, which 7. Using the double angle identity for tangent, we have gives 24 tan2(θ) + 14 tan(θ) − 24 = 0. Factoring, we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which gives tan(θ) = − 4 4. While either of these values of tan(θ) satisﬁes the equation. To 4, since this produces an acute angle,4 θ = arctan 3 cot(2θ) = 7. ﬁnd the rotation equations, we need cos(θ) = cos arctan 3 4 Using the techniques developed in Section 10.6 we get cos(θ) = 4 5. Our rotation 5 + 4y 5 − 3y equations are x = x cos(θ) − y sin(θ)
= 4x 5. As usual, we now substitute these quantities into 16x2 + 24xy + 9y2 + 15x − 20y = 0 and simplify. As a ﬁrst step, the reader can verify 3 or tan(θ) = 3 24, we choose tan(θ) = 3 5 and y = x sin(θ) + y cos(θ) = 3x and sin(θ) = sin arctan 3 4 5 and sin(θ) = 3 1−tan2(θ) = 24 2 tan(θ) 4 x2 = 16(x)2 25 − 24xy 25 + 9(y)2 25, xy = 12(x)2 25 + 7xy 25 − 12(y)2 25, y2 = 9(x)2 25 + 24xy 25 + 16(y)2 25 Once the dust settles, we get 25(x)2 − 25y = 0, or y = (x)2, whose graph is a parabola opening along the positive y-axis with vertex (0, 0). We graph this equation below = arctan 3 4 x 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 16x2 + 24xy + 9y2 + 15x − 20y = 0 4As usual, there are inﬁnitely many solutions to tan(θ) = 3. The 4 reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C B and what this means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 4. We choose the acute angle θ = arctan 3 3 in our ﬁnal graph. (Hint: 3 = −1.) − 4 3 4 11.6 Hooked on Conics Again 979 We note that even though the coeﬃcients of x2 and y2 were both positive numbers in parts 1 and 2 of Example 11.6.2, the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola. Whereas in Chapter 7, we could easily pick out
which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coeﬃcients, Example 11.6.2 demonstrates that all bets are oﬀ when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coeﬃcients of the quadratic terms. We have the following theorem. Theorem 11.11. Suppose the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 describes a non-degenerate conic section.a If B2 − 4AC > 0 then the graph of the equation is a hyperbola. If B2 − 4AC = 0 then the graph of the equation is a parabola. If B2 − 4AC < 0 then the graph of the equation is an ellipse or circle. aRecall that this means its graph is either a circle, parabola, ellipse or hyperbola. See page 497. As you may expect, the quantity B2 −4AC mentioned in Theorem 11.11 is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’, we will at least work through the proof of Theorem 11.11 mechanically to show that it is true.5 First note that if the coeﬃcient B = 0 in the equation Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, Theorem 11.11 reduces to the result presented in Exercise 34 in Section 7.5, so we proceed here under the assumption that B = 0. We rotate the xy-axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C to produce an equation with no xy-term in accordance with B Theorem 11.10: A(x)2 + C(y)2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 34 in Section 7.5 once more using the product AC. Our goal is to ﬁnd the product AC in terms of the coeﬃcients A, B and C in the original equation.
To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coeﬃcient A on (x)2 and the coeﬃcient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ) + C sin2(θ) C = A sin2(θ) − B cos(θ) sin(θ) + C cos2(θ) In order to make use of the condition cot(2θ) = A−C the power reduction formulas. After some regrouping, we get B, we rewrite our formulas for A and C using 2A = [(A + C) + (A − C) cos(2θ)] + B sin(2θ) 2C = [(A + C) − (A − C) cos(2θ)] − B sin(2θ) Next, we try to make sense of the product (2A)(2C) = {[(A + C) + (A − C) cos(2θ)] + B sin(2θ)} {[(A + C) − (A − C) cos(2θ)] − B sin(2θ)} 5We hope that someday you get to see why this works the way it does. 980 Applications of Trigonometry We break this product into pieces. First, we use the diﬀerence of squares to multiply the ‘ﬁrst’ quantities in each factor to get [(A + C) + (A − C) cos(2θ)] [(A + C) − (A − C) cos(2θ)] = (A + C)2 − (A − C)2 cos2(2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C) + (A − C) cos(2θ)] +B sin(2θ) [(A + C) − (
A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B2 sin2(2θ). Putting all of this together yields 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) From cot(2θ) = A−C sin(2θ) = A−C twice along with the Pythagorean Identity cos2(2θ) = 1 − sin2(2θ) to get B, we get cos(2θ) B, or (A−C) sin(2θ) = B cos(2θ). We use this substitution 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 1 − sin2(2θ) − 2B cos(2θ)B cos(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + (A − C)2 sin2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [(A − C) sin(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [B cos(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + B2 cos2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C
)2 − (A − C)2 − B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) + sin2(2θ) = (A + C)2 − (A − C)2 − B2 = A2 + 2AC + C2 − A2 − 2AC + C2 − B2 = 4AC − B2 Hence, B2 − 4AC = −4AC, so the quantity B2 − 4AC has the opposite sign of AC. The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. 1. 21x2 + 10xy √ 3 + 31y2 = 144 2. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 3. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. This is a straightforward application of Theorem 11.11. 11.6 Hooked on Conics Again 981 1. We have A = 21, B = 10 3)2 − 4(21)(31) = −2304 < 0. Theorem 11.11 predicts the graph is an ellipse, which checks with our work from Example 11.6.1 number 2. 3 and C = 31 so B2 − 4AC = (10 √ √ 2. Here, A = 5, B = 26 and C = 5, so B2 − 4AC = 262 − 4(5)(5) = 576 > 0. Theorem 11.11 classiﬁes the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1. 3. Finally, we have A = 16, B = 24 and C = 9 which gives 242 − 4(16)(9) = 0. Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2. 11.6.2 The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more uniﬁed perspective. We have our ‘new
’ deﬁnition below. Deﬁnition 11.1. Given a ﬁxed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricity of the conic section. We have seen the notions of focus and directrix before in the deﬁnition of a parabola, Deﬁnition 7.3. There, a parabola is deﬁned as the set of points equidistant from the focus and directrix, giving an eccentricity e = 1 according to Deﬁnition 11.1. We have also seen the concept of eccentricity before. It was introduced for ellipses in Deﬁnition 7.5 in Section 7.4, and later extended to hyperbolas in Exercise 31 in Section 7.5. There, e was also deﬁned as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the ‘old’ ideas of focus, directrix and eccentricity with the ‘new’ ones presented in Deﬁnition 11.1 is to derive equations for the conic sections using Deﬁnition 11.1 and compare these parameters with what we know from Chapter 7. We begin by assuming the conic section has eccentricity e, a focus F at the origin and that the directrix is the vertical line x = −d as in the ﬁgure below. y d r cos(θ) P (r, θ) r θ x = −d O = F x 982 Applications of Trigonometry Using a polar coordinate representation P (r, θ) for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = r d + r cos(θ) so that r = e(d + r cos(θ)). Solving this equation for r, yields r = ed 1 − e cos(θ) At this
point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e = 1, the usual conversion process outlined in Section 11.4 gives6 1 − e22 e2d2 x − e2d 1 − e2 2 1 − e2 e2d2 + y2 = 1 1−e2, so the major axis has length 2ed 1−e2 and the minor axis has length 2ed√ e2d 1−e2, 0 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at with major axis along the x-axis. Using the notation from Section 7.4, we have a2 = e2d2 (1−e2)2 and b2 = e2d2 1−e2. Moreover, we ﬁnd that one focus is (0, 0) and working through the formula given in Deﬁnition 7.5 gives the eccentricity 1−e2, 0 to be e, as required. If e > 1, then the equation generates a hyperbola with center whose transverse axis lies along the x-axis. Since such hyperbolas have the form (x−h)2 b2 = 1, we need to take the opposite reciprocal of the coeﬃcient of y2 to ﬁnd b2. We get7 a2 = e2d2 (e2−1)2 and b2 = − e2d2 e2−1 and the conjugate axis has length 2ed√. e2−1 Additionally, we verify that one focus is at (0, 0), and the formula given in Exercise 31 in Section ed 1−e cos(θ) reduces to 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r =. This is a parabola with vertex r = − d 2, the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented in De�
��nition 11.1 correspond with the ‘old’ deﬁnitions given in Chapter 7. 1−cos(θ) which gives the rectangular equation y2 = 2d x + d 2, 0 opening to the right. In the language of Section 7.3, 4p = 2d so p = d e2−1, so the transverse axis has length 2ed a2 − y2 (1−e2)2 = e2d2 1−e2 = e2d2 e2d d 2 ed Before we summarize our ﬁndings, we note that in order to arrive at our general equation of a conic 1−e cos(θ), we assumed that the directrix was the line x = −d for d > 0. We could have just as r = easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases 1+e sin(θ), respectively. The key thing to 1−e sin(θ) and r = we obtain the forms r = remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is (0, 0) and the directrix also tells us which way the parabola opens. We have established the following theorem. 1+e cos(θ), r = ed ed ed 6Turn r = e(d + r cos(θ)) into r = e(d + x) and square both sides to get r2 = e2(d + x)2. Replace r2 with x2 + y2, expand (d + x)2, combine like terms, complete the square on x and clean things up. 7Since e > 1 in this case, 1 − e2 < 0. Hence, we rewrite 1 − e22 = e2 − 12 to help simplify things later on. 11.6 Hooked on Conics Again 983 Theorem 11.12. Suppose e and d are positive numbers. Then the graph of r = ed 1−e cos(θ) is the graph of a conic section with directrix x = −d. the graph of r = ed 1+e cos(θ) is the graph of
a conic section with directrix x = d. the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose minor axis has length 2ed√ 1−e2 If e = 1, the graph is a parabola whose focal diameter is 2d. If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1. We test out Theorem 11.12 in the next example. Example 11.6.4. Sketch the graphs of the following equations. 1. r = 4 1 − sin(θ) Solution. 2. r = 12 3 − cos(θ) 3. r = 6 1 + 2 sin(θ) 4 1. From r = 1−sin(θ), we ﬁrst note e = 1 which means we have a parabola on our hands. Since ed = 4, we have d = 4 and considering the form of the equation, this puts the directrix at y = −4. Since the focus is at (0, 0), we know that the vertex is located at the point (in rectangular coordinates) (0, −2) and must open upwards. With d = 4, we have a focal diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below. 4 12 2. We ﬁrst rewrite r = 4 1−(1/3) cos(θ). 3−cos(θ) in the form found in Theorem 11.12, namely r = Since e = 1 3 satisﬁes 0 < e < 1, we know that the graph of this equation is an ellipse. Since ed = 4, we have d = 12 and, based on the form of the equation, the direct
rix is x = −12. This means that the ellipse has its major axis along the x-axis. We can ﬁnd the vertices of the ellipse by ﬁnding the points of the ellipse which lie on the x-axis. We ﬁnd r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3, 0) and (6, 0), so these are our 2, 0.8 vertices. The center of the ellipse is the midpoint of the vertices, which in this case is 3 2, 0 and this allows us to ﬁnd the We know one focus is (0, 0), which is 3 1−e2 = (2)(4) 2 from the center 3 8As a quick check, we have from Theorem 11.12 the major axis should have length 2ed 1−(1/3)2 = 9. 984 Applications of Trigonometry other focus (3, 0), even though we are not asked to do so. Finally, we know from Theorem 2ed√ 3 which means the endpoints 1−e2 = 11.12 that the length of the minor axis is of the minor axis are 3 1−(1/3)2 = 6 4√ 2. We now have everything we need to graph r = 12 √ √ 3−cos(θ). 2, ±3 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 −3 y = −4 r = 4 1−sin(θ) y 4 3 2 1 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 x = −12 r = 12 3−cos(θ) 3. From r = 6 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get d = 3, and from the form of the equation, we know the directrix is y = 3. This means the transverse axis of the hyperbola lies along the y-axis, so we can ﬁnd the vertices by looking where the hyperbola intersects the y-axis. We ﬁnd r π = −6. These two 2 points correspond to the rectangular points (0, 2) and
(0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 3. Putting this together with the location of the vertices, we get that √ 3 the asymptotes of the hyperbola have slopes ± 2 3. Since the center of the hyperbola √ 2 = 2 and r 3π 2 = (2)(6) √ 22−1 = ± 2ed√ = 4 e2−1 √ 3 is (0, 4), the asymptotes are y = ± √ 3 3 x + 4. We graph the hyperbola below5 −4 −3 −2 −+2 sin(θ) 11.6 Hooked on Conics Again 985 In light of Section 11.6.1, the reader may wonder what the rotated form of the conic sections would look like in polar form. We know from Exercise 65 in Section 11.5 that replacing θ with (θ − φ) in an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. For instance, to graph r = 1−sin(θ), which we obtained in Example 11.6.4 number 1, counter-clockwise by π all we need to do is rotate the graph of r = 4 1−sin(θ− π 4 ) 4 radians, as shown below4 −3 −2 −1 −1 −2 −3 r = 4 1−sin(θ− π 4 ) Using rotations, we can greatly simplify the form of the conic sections presented in Theorem 11.12, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π 2. Since rotations do not aﬀect lengths, all of the formulas for lengths Theorem 11.12 remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0. We conclude this section with the statement of the following theorem. Theorem 11.13. Given constants > 0,
e ≥ 0 and φ, the graph of the equation r = 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0, 0). If e = 0, the graph is a circle centered at (0, 0) with radius. If e = 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e. minor axis has length 2ed√ 1−e2 – If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose – If e = 1, the graph is a parabola whose focal diameter is 2d. – If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1. 986 Applications of Trigonometry 11.6.3 Exercises Graph the following equations. 1. x2 + 2xy + y2 − x √ √ 2 − 6 = 0 2. 7x2 − 4xy √ 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2y = 0 2 + y √ 5. 13x2 − 34xy √ 3 + 47y2 − 64 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 Graph the following equations. 9. r = 11. r = 2 1 − cos(θ) 3 2 − cos(θ) 13. r = 4 1 + 3 cos(θ) 15. r = 2 1 + sin(θ − π 3 ) √ √ 4. x2 + 2 6. x2 − 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 3xy − y2 + 8 = 0 8. 8x2 + 12xy + 17y2 − 20 = 0 10. r = 12. r = 3 2 + sin(θ) 2 1 + sin(θ) 14. r = 2 1 − 2 sin(θ) 16. r = 6 3 − cos θ +
π 4 The matrix A(θ) = cos(θ) − sin(θ) cos(θ) sin(θ) is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A π 4. 18. Discuss with your classmates how to use A(θ) to rotate points in the plane. 19. Using the even / odd identities for cosine and sine, show A(θ)−1 = A(−θ). Interpret this geometrically. 11.6 Hooked on Conics Again 987 11.6.4 Answers 1. x2 + 2xy + y2 − becomes (x)2 = −(y − 3) after rotating counter-clockwise through x2 + 2xy + y2 − x √ 2 + y √ 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2 − 6 = 0 √ 2y = 0 becomes (x)2 + (y+2)2 counter-clockwise through θ = π 4. 4 = 1 after rotating. 7x2 − 4xy 3 + 3y2 − 2x − 2y 3 − 5 = 0 9 + (y)2 = 1 after rotating becomes (x−2)2 counter-clockwise through √ 7x2 − 4xy 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 √ 4. x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 becomes(x)2 = y + 4 after rotating counter-clockwise through 5x2 + 6xy + 5y2 − 4 √ 2x + 4 √ 2y = 0 √ x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 988 Applications of Trigonometry √ 5. 13x2 − 34xy 3 + 47y2 − 64 = 0 becomes (y)2 − (x)2 counter-clockwise through θ = π 6. 16 = 1 after rotating √ 6. x2 − 2 3xy − y2 + 8 = 0 4 − (y)2 becomes (
x)2 counter-clockwise through θ = π 3 4 = 1 after rotating 13x2 − 34xy √ 3 + 47y2 − 64 = 0 √ x2 − 2 3xy − y2 + 8 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 becomes (y)2 = x after rotating counter-clockwise through θ = arctan 1 2 y. y 8. 8x2 + 12xy + 17y2 − 20 = 0 becomes (x)2 + (y)2 4 = 1 after rotating counter-clockwise through θ = arctan(2) y x y x θ = arctan 1 2 x θ = arctan(2) x x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 8x2 + 12xy + 17y2 − 20 = 0 11.6 Hooked on Conics Again 989 2 9. r = 1−cos(θ) is a parabola directrix x = −2, vertex (−1, 0) focus (0, 0), focal diameter 4 10. r = 3 2+sin(θ) = 3 2 1+ 1 2 sin(θ) is an ellipse directrix y = 3, vertices (0, 1), (0, −3) center (0, −2), foci (0, 0), (0, −2) minor axis length 4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 11. r = 3 2−cos(θ) = 3 2 1− 1 2 cos(θ) is an ellipse directrix x = −3, vertices (−1, 0), (3, 0) center (1, 0), foci (0, 0), (2, 0) minor axis length 2 √ 3 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 12. r = 2 1+sin(θ) is a parabola directrix y = 2, vertex (0, 1) focus (0, 0), focal diameter 4 y 4 3 2 1 −4 −3 −
2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis length 2 3, vertices (1, 0), (2, 0) √ 2 2, 0, foci (0, 0), (3, 0) 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices 0, − 2 3 center 0, − 4, foci (0, 0), 0, − 8 3 3 √ conjugate axis length 2 3 3, (0, −24 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 15. r = 2 3 ) is 1+sin(θ− π 2 the parabola r = 1+sin(θ) rotated through 1 −2 −3 −4 is the ellipse 16. r = 6 3−cos(θ+ π 4 ) 6 3−cos(θ) = r = 3 cos(θ) rotated through φ = − π 4 14 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 φ = − π 4 x 11.7 Polar Form of Complex Numbers 991 11.7 Polar Form of Complex Numbers √ In this section, we return to our study of complex numbers which were ﬁrst introduced in Section 3.4. Recall that a complex number is a number of the form z = a + bi where a and b are real −1. The number a is called the real part of numbers and i is the imaginary unit deﬁned by i = z, denoted Re(z), while the real number b is called the imaginary part of z, denoted Im(z). From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Re(z) and Im(z) are well-deﬁned.1 To start oﬀ this section, we associate each complex number z = a + bi with the point (a, b) on
the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Imaginary Axis (3, 0) ←→ z = 3 0 1 2 3 4 Real Axis (−4, 2) ←→ z = −4 + 2i 4i 3i 2i i −4 −3 −2 −1 −i −2i −3i (0, −3) ←→ z = −3i −4i The Complex Plane Since the ordered pair (a, b) gives the rectangular coordinates associated with the complex number z = a + bi, the expression z = a + bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates (r, θ). Although it is not as straightforward as the deﬁnitions of Re(z) and Im(z), we can still give r and θ special names in relation to z. Deﬁnition 11.2. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Re(z) and b = Im(z). Let (r, θ) be a polar representation of the point with rectangular coordinates (a, b) where r ≥ 0. The modulus of z, denoted \|z\|, is deﬁned by \|z\| = r. The angle θ is an argument of z. The set of all arguments of z is denoted arg(z). If z = 0 and −π < θ ≤ π, then θ is the principal argument of z, written θ = Arg(z). 1‘Well-deﬁned’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. 992 Applications of Trigonometry Some remarks about Deﬁnition 11.2 are in order. We know from Section 11.4 that every point in the plane has inﬁnitely many polar coordinate representations (r, θ)
which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-deﬁned. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, \|z\| = 0 is well-deﬁned. If z = 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r ≥ 0 in our deﬁnition so this pins down the value of \|z\| to one and only one number. Thus the modulus is well-deﬁned in this case, too.2 Even with the requirement r ≥ 0, there are inﬁnitely many angles θ which can be used in a polar representation of a point (r, θ). If z = 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval (−π, π], and this angle is what we call the principal argument of z, Arg(z). In fact, the set arg(z) of all arguments of z can be described using set-builder notation as arg(z) = {Arg(z) + 2πk \| k is an integer}. Note that since arg(z) is a set, we will write ‘θ ∈ arg(z)’ to mean ‘θ is in3 the set of arguments of z’. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as (0, θ) for any angle θ. In this case, we have arg(0) = (−∞, ∞) and since there is no one value of θ which lies (−π, π], we leave Arg(0) undeﬁned.4 It is time for an example. Example 11.7.1. For each of the following complex numbers ﬁnd Re(z), Im(z), \|z\|, arg(z)
and Arg(z). Plot z in the complex plane. 1. z = √ 3 − i Solution. 2. z = −2 + 4i 3. z = 3i 4. z = −117 √ √ √ 1. For z = 3 − i = 3 + (−1)i, we have Re(z) = 3 and Im(z) = −1. To ﬁnd \|z\|, arg(z) and Arg(z), we need to ﬁnd a polar representation (r, θ) with r ≥ 0 for the point P ( 3, −1) associated with z. We know r2 = ( 3)2 + (−1)2 = 4, so r = ±2. Since we require r ≥ 0, we choose r = 2, so \|z\| = 2. Next, we ﬁnd a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tan(θ) = −1√ 6 + 2πk 3 6 + 2πk \| k is an integer. Of these values, only θ = − π for integers k. Hence, arg(z) = − π satisﬁes the requirement that −π < θ ≤ π, hence Arg(z) = − π 6. 3, so. The complex number z = −2 + 4i has Re(z) = −2, Im(z) = 4, and is associated with the point P (−2, 4). Our next task is to ﬁnd a polar representation (r, θ) for P where r ≥ 0. 5. To ﬁnd θ, we get Running through the usual calculations gives r = 2 tan(θ) = −2, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We ﬁnd θ = π + arctan(−2) + 2πk, or, more succinctly θ = π − arctan(2) + 2πk for integers k. Hence arg(z) = {π − arctan(2) + 2πk \| k is an integer}. Only θ = π − arctan(2) satisﬁes the requirement −
π < θ ≤ π, so Arg(z) = π − arctan(2). 5, so \|z\| = 2 √ √ 2In case you’re wondering, the use of the absolute value notation \|z\| for modulus will be explained shortly. 3Recall the symbol being used here, ‘∈,’ is the mathematical symbol which denotes membership in a set. 4If we had Calculus, we would regard Arg(0) as an ‘indeterminate form.’ But we don’t, so we won’t. 11.7 Polar Form of Complex Numbers 993 3. We rewrite z = 3i as z = 0 + 3i to ﬁnd Re(z) = 0 and Im(z) = 3. The point in the plane which corresponds to z is (0, 3) and while we could go through the usual calculations to ﬁnd the required polar form of this point, we can almost ‘see’ the answer. The point (0, 3) lies 3 units away from the origin on the positive y-axis. Hence, r = \|z\| = 3 and θ = π 2 + 2πk for integers k. We get arg(z) = π 2 + 2πk \| k is an integer and Arg(z) = π 2. 4. As in the previous problem, we write z = −117 = −117 + 0i so Re(z) = −117 and Im(z) = 0. The number z = −117 corresponds to the point (−117, 0), and this is another instance where we can determine the polar form ‘by eye’. The point (−117, 0) is 117 units away from the origin along the negative x-axis. Hence, r = \|z\| = 117 and θ = π + 2π = (2k + 1)πk for integers k. We have arg(z) = {(2k + 1)π \| k is an integer}. Only one of these values, θ = π, just barely lies in the interval (−π, π] which means and Arg(z) = π. We plot z along with the other numbers in this example below. Imaginary Axis z = 3i 4i 3i 2i i z = −2 + 4i z = −117 −
117 −2 −1 − Real Axis Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 11.14. Properties of the Modulus: Let z and w be complex numbers. \|z\| is the distance from z to 0 in the complex plane \|z\| ≥ 0 and \|z\| = 0 if and only if z = 0 \|z\| = Re(z)2 + Im(z)2 Product Rule: \|zw\| = \|z\|\|w\| Power Rule: \|zn\| = \|z\|n for all natural numbers, n Quotient Rule: z w = \|z\| \|w\|, provided w = 0 To prove the ﬁrst three properties in Theorem 11.14, suppose z = a + bi where a and b are real numbers. To determine \|z\|, we ﬁnd a polar representation (r, θ) with r ≥ 0 for the point (a, b). From √ Section 11.4, we know r2 = a2 + b2 so that r = ± a2 + b2. Since we require r ≥ 0, then it must be a2 + b2. Using the distance formula, we ﬁnd the distance a2 + b2, which means \|z\| = that r = √ √ 994 Applications of Trigonometry √ a2 + b2, establishing the ﬁrst property.5 For the second property, note from (0, 0) to (a, b) is also that since \|z\| is a distance, \|z\| ≥ 0. Furthermore, \|z\| = 0 if and only if the distance from z to 0 is 0, and the latter happens if and only if z = 0, which is what we were asked to show.6 For the third property, we note that since a = Re(z) and b = Im(z), z = To prove the product rule, suppose z = a + bi and w = c + di for real numbers a, b, c and d. Then zw = (a + bi)(c + di). After the usual arithmetic7 we get zw = (ac − bd) + (ad + bc)i. Therefore, Re(z)2 + Im(z)2. a2 + b2 = �
� \|zwac − bd)2 + (ad + bc)2 a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 Expand a2c2 + a2d2 + b2c2 + b2d2 a2 (c2 + d2) + b2 (c2 + d2) (a2 + b2) (c2 + d2) a2 + b2 c2 + d2 Factor Factor √ Rearrange terms = = \|z\|\|w\| Product Rule for Radicals Deﬁnition of \|z\| and \|w\| Hence \|zw\| = \|z\|\|w\| as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction8 to prove the power rule. Let P (n) be the statement \|zn\| = \|z\|n. Then P (1) is true since z1 = \|z\|k for some k ≥ 1. Our job = \|z\|k+1. As is customary with induction proofs, is to show that P (k + 1) is true, namely we ﬁrst try to reduce the problem in such a way as to use the Induction Hypothesis. = \|z\| = \|z\|1. Next, assume P (k) is true. That is, assume zk+1 zk zk+1 Properties of Exponents \|z\| Product Rule zkz = zk = = \|z\|k\|z\| = \|z\|k+1 Properties of Exponents Induction Hypothesis Hence, P (k + 1) is true, which means \|zn\| = \|z\|n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w = 0 (so \|w\| = 0) and we get (z) 1 w Product Rule. = = \|z\| z w 1 w 5Since the absolute value \|x\| of a real number x can be viewed as the distance from x to 0 on the number line, this ﬁrst property justiﬁes the notation \|z\| for modulus. We leave it to the reader to show that if z is real, then the deﬁnition of modulus coincides with absolute
value so the notation \|z\| is unambiguous. 6This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is a2 + b2 = 0 if and only if a2 + b2 = 0, which is true if and only if a = b = 0. √ true. We know \|z\| = 0 if and only if The latter happens if and only if z = a + bi = 0. There. 7See Example 3.4.1 in Section 3.4 for a review of complex number arithmetic. 8See Section 9.3 for a review of this technique. 11.7 Polar Form of Complex Numbers 995 1 Hence, the proof really boils down to showing w Next, we characterize the argument of a complex number in terms of its real and imaginary parts. \|w\|. This is left as an exercise. = 1 Theorem 11.15. Properties of the Argument: Let z be a complex number. If Re(z) = 0 and θ ∈ arg(z), then tan(θ) = Im(z) Re(z). 2 + 2πk \| k is an integer. 2 + 2πk \| k is an integer. If Re(z) = 0 and Im(z) > 0, then arg(z) = π If Re(z) = 0 and Im(z) < 0, then arg(z) = − π If Re(z) = Im(z) = 0, then z = 0 and arg(z) = (−∞, ∞). To prove Theorem 11.15, suppose z = a + bi for real numbers a and b. By deﬁnition, a = Re(z) and b = Im(z), so the point associated with z is (a, b) = (Re(z), Im(z)). From Section 11.4, we know that if (r, θ) is a polar representation for (Re(z), Im(z)), then tan(θ) = Im(z) Re(z), provided Re(z) = 0. If Re(z) = 0 and Im(z) > 0, then z lies on the positive imaginary axis. Since we take r > 0, we have that θ is coterminal with π 2, and the result follows. If Re(z
) = 0 and Im(z) < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with − π 2. The last property in the theorem was already discussed in the remarks following Deﬁnition 11.2. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the ﬁgure below. Imaginary Axis bi (a, b) ←→ z = a + bi ←→ (r, θ ∈ arg(z) 0 a Real Axis Polar coordinates, (r, θ) associated with z = a + bi with r ≥ 0. We know from Theorem 11.7 that a = r cos(θ) and b = r sin(θ). Making these substitutions for a and b gives z = a + bi = r cos(θ) + r sin(θ)i = r [cos(θ) + i sin(θ)]. The expression ‘cos(θ) + i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Since r = \|z\| and θ ∈ arg(z), we get Deﬁnition 11.3. A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). The expression: is called a polar form for z. \|z\|cis(θ) = \|z\| [cos(θ) + i sin(θ)] 996 Applications of Trigonometry Since there are inﬁnitely many choices for θ ∈ arg(z), there inﬁnitely many polar forms for z, so we used the indeﬁnite article ‘a’ in Deﬁnition 11.3. It is time for an example. Example 11.7.2. 1. Find the rectangular form of the following complex numbers. Find Re(z) and Im(z). (a) z = 4cis 2π 3 (b) z = 2cis − 3π 4 (c) z = 3cis(0) (d) z = cis π 2 2. Use the results from Example 11.7.1 to ﬁnd a polar form
of the following complex numbers. (a) z = √ 3 − i Solution. (b) z = −2 + 4i (c) z = 3i (d) z = −117 1. The key to this problem is to write out cis(θ) as cos(θ) + i sin(θ). (a) By deﬁnition, z = 4cis 2π √ 3 z = −2 + 2i + i sin 2π = 4 cos 2π √ 3 3 3, so that Re(z) = −2 and Im(z) = 2 = 2 cos − 3π 4. After some simplifying, we get 3. + i sin − 3π 4. From this, we ﬁnd (b) Expanding, we get z = 2cis − 3π √ 4 2 = Im(z). 2, so Re(zc) We get z = 3cis(0) = 3 [cos(0) + i sin(0)] = 3. Writing 3 = 3 + 0i, we get Re(z) = 3 and (d) Lastly, we have z = cis π 2 Im(z) = 0, which makes sense seeing as 3 is a real number. = cos π 2 = i. Since i = 0 + 1i, we get Re(z) = 0 and Im(z) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. + i sin π 2 2. To write a polar form of a complex number z, we need two pieces of information: the modulus \|z\| and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 11.7.1 to ﬁnd what we need. (a) For z = √ 3 − i, \|z\| = 2 and θ = − π 6, so z = 2cis − π 6. We can check our answer by √ converting it back to rectangular form to see that it simpliﬁes to z = 3 − i. √ √ (b) For z = −2 + 4i, \|z\| = 2 5 and θ = π − arctan(2). Hence, z = 2 5cis(π − arctan(2)). It
is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, \|z\| = 3 and θ = π 2. In this case, z = 3cis π 2 geometrically. Head out 3 units from 0 along the positive real axis. Rotating π counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i.. This can be checked 2 radians (d) Last but not least, for z = −117, \|z\| = 117 and θ = π. We get z = 117cis(π). As with the previous problem, our answer is easily checked geometrically. 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = \|z\|cis(α) and w = \|w\|cis(β). Then Product Rule: zw = \|z\|\|w\|cis(α + β) Power Rule (DeMoivre’s Theorem) : zn = \|z\|ncis(nθ) for every natural number n Quotient Rule: z w = \|z\| \|w\| cis(α − β), provided \|w\| = 0 The proof of Theorem 11.16 requires a healthy mix of deﬁnition, arithmetic and identities. We ﬁrst start with the product rule. zw = [\|z\|cis(α)] [\|w\|cis(β)] = \|z\|\|w\| [cos(α) + i sin(α)] [cos(β) + i sin(β)] We now focus on the quantity in brackets on the right hand side of the equation. [cos(α) + i sin(α)] [cos(β) + i sin(β)] = cos(α) cos(β) + i cos(α) sin(β) + i sin(α) cos(β) + i2 sin(α) sin(β) = cos(α) cos(β) + i2 sin(α) sin(β) Rearranging terms + i sin(α) cos(β) + i cos(α) sin
(β) = (cos(α) cos(β) − sin(α) sin(β)) Since i2 = −1 + i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α + β) + i sin(α + β) Sum identities = cis(α + β) Deﬁnition of ‘cis’ Putting this together with our earlier work, we get zw = \|z\|\|w\|cis(α + β), as required. Moving right along, we next take aim at the Power Rule, better known as DeMoivre’s Theorem.9 We proceed by induction on n. Let P (n) be the sentence zn = \|z\|ncis(nθ). Then P (1) is true, since z1 = z = \|z\|cis(θ) = \|z\|1cis(1 · θ). We now assume P (k) is true, that is, we assume zk = \|z\|kcis(kθ) for some k ≥ 1. Our goal is to show that P (k + 1) is true, or that zk+1 = \|z\|k+1cis((k + 1)θ). We have zk+1 = zkz = \|z\|kcis(kθ) (\|z\|cis(θ)) = \|z\|k\|z\| cis(kθ + θ) = \|z\|k+1cis((k + 1)θ) Properties of Exponents Induction Hypothesis Product Rule 9Compare this proof with the proof of the Power Rule in Theorem 11.14. 998 Applications of Trigonometry Hence, assuming P (k) is true, we have that P (k + 1) is true, so by the Principle of Mathematical Induction, zn = \|z\|ncis(nθ) for all natural numbers n. The last property in Theorem 11.16 to prove is the quotient rule. Assuming \|w\| = 0 we have z w = = \|z\|cis(α) \|w\|cis(β) \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) Next, we multiply both the numer
ator and denominator of the right hand side by (cos(β) − i sin(β)) which is the complex conjugate of (cos(β) + i sin(β)) to get z w = \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) · cos(β) − i sin(β) cos(β) − i sin(β) If we let the numerator be N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] and simplify we get N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] = cos(α) cos(β) − i cos(α) sin(β) + i sin(α) cos(β) − i2 sin(α) sin(β) Expand = [cos(α) cos(β) + sin(α) sin(β)] + i [sin(α) cos(β) − cos(α) sin(β)] Rearrange and Factor = cos(α − β) + i sin(α − β) = cis(α − β) Diﬀerence Identities Deﬁnition of ‘cis’ If we call the denominator D then we get D = [cos(β) + i sin(β)] [cos(β) − i sin(β)] = cos2(β) − i cos(β) sin(β) + i cos(β) sin(β) − i2 sin2(β) Expand = cos2(β) − i2 sin2(β) Simplify = cos2(β) + sin2(β) Again, i2 = −1 Pythagorean Identity = 1 Putting it all together, we get · cos(β) − i sin(β) cos(β) − i sin(β) z w = = = \|z\| \|w\| \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) cis(α − β) 1 \|z\| \|w\| cis(α − β) and we are done. The next example makes good use of Theorem 11.16. 11.7 Polar Form of Complex Numbers 999 Example 11.7.3. Let z = 2 √
3 + 2i and w = −1 + i √ 3. Use Theorem 11.16 to ﬁnd the following. 1. zw 2. w5 Write your ﬁnal answers in rectangular form. 3. z w √ √ √ 3)2 + (2)2 = Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 3+2i, √ 3 3. Since 3, we ﬁnd \|z\| = (2 z lies in Quadrant I, we have θ = π 16 = 4. If θ ∈ arg(z), we know tan(θ) = Im(z) 6 + 2πk for integers k. Hence, z = 4cis π 3)2 = 2. For an argument θ of w, we have tan(θ) = 3 + 2πk for integers k and w = 2cis 2π = 8cis π 2cis 2π √ 3 After simplifying, we get zw = −4 3 + 4i. = 3. For w = −1 + i √ 3 −1 = − we have \|w\| = w lies in Quadrant II, θ = 2π 1. We get zw = 4cis π 6. We can now proceed. + i sin 5π 6. = 8 cos 5π 6 = 8cis 5π 6 Re(z) = 2 (−1)2 + ( 6 + 2π 3. Since √ √ √ √ 6 2 3 3 2. We use DeMoivre’s Theorem which yields w5 = 2cis 2π 3 + i sin 4π 3 is coterminal with 4π Since 10π 3 = 25cis 5 · 2π 3 = −16 − 16i = 32cis 10π √ 3 5 3. 3. 3. Last, but not least, we have = 4 2 is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating π 2 radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that z 6 − 2π = 2cis − π 2. Since − π 3
w = −2i. 3, we get w5 = 32 cos 4π z 4cis( π 6 ) 2cis( 2π 3 ) w 2 cis π = Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out zw = (2 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of complex numbers. Consider how we computed w5 above and compare that to using the Binomial Theorem, Theorem 9.4, to accomplish the same feat by 3)5. Division is tricky in the best of times, and we saved ourselves a lot of expanding (−1 + i time and eﬀort using Theorem 11.16 to ﬁnd and simplify z w using their polar forms as opposed to starting with 2, rationalizing the denominator, and so forth. 3 + 2i)(−1 + i √ √ √ √ 3+2i √ 3 −1+i There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11.16. Take the product rule, for instance. If z = \|z\|cis(α) and w = \|w\|cis(β), the formula zw = \|z\|\|w\|cis(α + β) can be viewed geometrically as a two step process. The multiplication of \|z\| by \|w\| can be interpreted as magnifying10 the distance \|z\| from z to 0, by the factor \|w\|. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise.11 Focusing on z and w from Example 10Assuming \|w\| > 1. 11Assuming β > 0. 1000 Applications of Trigonometry 11.7.3, we can arrive at the product zw by plotting z, doubling its distance from 0 (since \|w\| = 2), and rotating 2π 3 radians counter-clock
wise. The sequence of diagrams below attempts to describe this process geometrically. Imaginary Axis Imaginary Axis 6i 5i 4i 3i 2i i z\|w\| = 8cis π 6 z = 4cis π 6 zw = 8cis π 6 + 2π 3 6i 5i 4i 3i 2i i z\|w\| = 8cis Real Axis −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 Real Axis Multiplying z by \|w\| = 2. Rotating counter-clockwise by Arg(w) = 2π 3 radians. Visualizing zw for z = 4cis π 6 and w = 2cis 2π 3. We may also visualize division similarly. Here, the formula z \|w\| cis(α − β) may be interpreted as shrinking12 the distance from 0 to z by the factor \|w\|, followed up by a clockwise 13 rotation of β radians. In the case of z and w from Example 11.7.3, we arrive at z w by ﬁrst halving the distance from 0 to z, then rotating clockwise 2π w = \|z\| 3 radians. Imaginary Axis Imaginary Axis 3i 2i i z = 4cis π 6 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis i −i −2i 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis zw = 2cis π 6 2π 3 Dividing z by \|w\| = 2. Rotating clockwise by Arg(w) = 2π 3 radians. Visualizing z w for z = 4cis π 6 and w = 2cis 2π 3. Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. Deﬁnition 11.4. Let z and w be complex numbers. If there is a natural number n such that wn = z, then w is an nth root of z. Unlike Deﬁnition 5.4 in Section 5.3, we do not specify one particular prinicpal nth root, hence the use of the indeﬁnite article ‘an’ as in ‘an nth root
of z’. Using this deﬁnition, both 4 and −4 are 12Again, assuming \|w\| > 1. 13Again, assuming β > 0. 11.7 Polar Form of Complex Numbers 1001 √ √ 16 means the principal square root of 16 as in square roots of 16, while 16 = 4. Suppose we wish to ﬁnd all complex third (cube) roots of 8. Algebraically, we are trying to solve w3 = 8. We √ know that there is only one real solution to this equation, namely w = 3 8 = 2, but if we take the time to rewrite this equation as w3 − 8 = 0 and factor, we get (w − 2) w2 + 2w + 4 = 0. The quadratic factor gives two more cube roots w = −1 ± i 3, for a total of three cube roots of 8. In accordance with Theorem 3.14, since the degree of p(w) = w3 − 8 is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we get z = 8cis(0). If we let w = \|w\|cis(α) be a polar form of w, the equation w3 = 8 becomes √ w3 = 8 (\|w\|cis(α))3 = 8cis(0) \|w\|3cis(3α) = 8cis(0) DeMoivre’s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates \|w\|3, 3α, while the complex number on the right hand side corresponds to the point with polar coordinates (8, 0). Since \|w\| ≥ 0, so is \|w\|3, which means \|w\|3, 3α and (8, 0) are two polar representations corresponding to the same complex number, both with positive r values. From Section 11.4, we know \|w\|3 = 8 and 3α = 0 + 2πk for integers k. Since \|w\|
is a real number, √ we solve \|w\|3 = 8 by extracting the principal cube root to get \|w\| = 3 8 = 2. As for α, we get α = 2πk for integers k. This produces three distinct points with polar coordinates corresponding to 3 k = 0, 1 and 2: speciﬁcally (2, 0), 2, 2π. These correspond to the complex numbers 3 w0 = 2cis(0), w1 = 2cis 2π, respectively. Writing these out in rectangular form 3 yields w0 = 2, w1 = −1 + i 3. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to ﬁnd, for example, all of the ﬁfth roots of 32. (Try using Chapter 3 techniques on that!) If we start with a generic complex number in polar form z = \|z\|cis(θ) and solve wn = z in the same manner as above, we arrive at the following theorem. and w2 = 2cis 4π √ and 2, 4π 3 3 and w2 = −1 − i √ 3 Theorem 11.17. The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct nth roots, which we denote by w0, w1,..., wn − 1, and they are given by the formula √ wk = n rcis θ n + 2π n k The proof of Theorem 11.17 breaks into to two parts: ﬁrst, showing that each wk is an nth root, and second, showing that the set {wk \| k = 0, 1,..., (n − 1)} consists of n diﬀerent complex numbers. To show wk is an nth root of z, we use DeMoivre’s Theorem to show (wk)n = z. 1002 Applications of Trigonometry √ (wk)n = n √ = ( n rcis θ n + 2π r)n cis n · θ n kn n + 2π n k DeMoivre’s Theorem = rcis (θ + 2πk) Since k is a whole
number, cos(θ + 2πk) = cos(θ) and sin(θ + 2πk) = sin(θ). Hence, it follows that cis(θ + 2πk) = cis(θ), so (wk)n = rcis(θ) = z, as required. To show that the formula in Theorem 11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note √ that the modulus of each of the wk is the same, namely n r. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments diﬀer by an integer multiple of 2π. Suppose k and j are whole numbers between 0 and (n − 1), inclusive, with k = j. Since k and j are diﬀerent, let’s assume for the sake of argument that k > j. Then θ. For this to be an integer multiple of 2π, (k − j) must be a multiple of n. But because of the restrictions on k and j. (Think this through.) Hence, (k − j) is a positive number less than n, so it cannot be a multiple of n. As a result, wk and wj are diﬀerent complex numbers, and we are done. By Theorem 3.14, we know there at most n distinct solutions to wn = z, and we have just found all of them. We illustrate Theorem 11.17 in the next example. n j = 2π n k − θ k−j n n + 2π n + 2π Example 11.7.4. Use Theorem 11.17 to ﬁnd the following: 1. both square roots of z = −2 + 2i √ 3 2. the four fourth roots of z = −16 3. the three cube roots of z = √ √ 2 + i 2 4. the ﬁve ﬁfth roots of z = 1. Solution. √ 1. We start by writing z = −2 + 2i 3 = 4cis 2π 3 θ = 2π in Theorem 11.17, we’ll call them w0 and w1. We get w0 =. To use Theorem
11.17, we identify r = 4, 3 and n = 2. We know that z has two square roots, and in keeping with the notation = 2cis π 3. In rectangular form, the two square roots of 3. We can check our answers by squaring them and 2 + 2π 2 (1) 3 and w1 = −1 − i = 2cis 4π 3 √ (2π/3) √ 2 + 2π 2 (0) and w1 = (2π/3) 4cis 4cis √ √ z are w0 = 1 + i showing that we get z = −2 + 2i √ 3. 2. Proceeding as above, we get z = −16 = 16cis(π). With r = 16, θ = π and n = 4, we get the √ √ 4 (1) = 4 (0) = 2cis π 16cis π 16cis π, w1 = 4 4 + 2π 4 + 2π four fourth roots of z to be w0 = 4 √ √ 4. and w3 = 4, w2 = 4 16cis π 2cis 3π 4 (3) = 2cis 7π 4 (2) = 2cis 5π 16cis −i 2, w2 = − Converting these to rectangular form gives w0 = 2 √ and w3 = 4 + 2π √ 2+i 2, w1 = − 4 + 2π 2 − i 2+i √ 2. 11.7 Polar Form of Complex Numbers 1003 √ √ 3. For z = 2cis π 12 2, we have z = 2cis π √ 4, w1 = 3 2cis 9π 12 4 and n = 3 the usual computations 2+i √ 2cis 17π yield w0 = 3 If we were 12 to convert these to rectangular form, we would need to use either the Sum and Diﬀerence Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w0 and w2. Since we are not explicitly told to do so, we leave this as a good, but messy, exercise.. With r = 2, θ = π √ = 3 √ and w2 =
3 2cis 3π 4. 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 2π 4. To ﬁnd the ﬁve ﬁfth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. and. The situation here is even graver than in the previous example, since we have 5. At this stage, we √ Since 5 w4 = cis 8π not developed any identities to help us determine the cosine or sine of 2π could approximate our answers using a calculator, and we leave this as an exercise., w3 = cis 6π, w2 = cis 4π 5 5 5 5 Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Essentially, Theorem 11.17 says that to ﬁnd the nth roots of a complex number, we ﬁrst take the nth root of the modulus and divide the argument by n. This gives the ﬁrst root w0. Each succeessive root is found by adding 2π n to the argument, which amounts to rotating w0 by 2π n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 11.7.4 below. Imaginary Axis w1 2i i w0 −2 −1 0 1 2 Real Axis w2 −i −2i w3 The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications.14 For now, the following exercises will have to suﬃce. 14For more on this, see the beautifully written epilogue to Section 3.4 found on page 294. 1004 Applications of Trigonometry 11.7.1 Exercises In Exercises 1 - 20, ﬁnd a polar representation for the complex number z and then identify Re(z), Im(z), \|z\|, arg(z) and Arg(
z). 2. z = 5 + 5i √ 3 3. z = 6i 4. z = −3 √ √ 2 2+3i 1. z = 9 + 9i 5. z = −6 √ 3 + 6i 9. z = −5i 13. z = 3 + 4i 6. z = −2 7. z = − √ √ 2 − 2i 2 11. z = 6 10. z = 2 √ 14. z = 2 + i 15. z = −7 + 24i 16. z = −2 + 6i √ 3 2 − 1 2 i 8. z = −3 − 3i √ 12. z = i 3 7 17. z = −12 − 5i 18. z = −5 − 2i 19. z = 4 − 2i 20. z = 1 − 3i In Exercises 21 - 40, ﬁnd the rectangular form of the given complex number. Use whatever identities are necessary to ﬁnd the exact values. 21. z = 6cis(0) 22. z = 2cis π 6 3π 4 3π 2 √ 26. z = 6cis √ 30. z = 13cis √ 23. z = 7 2cis π 4 24. z = 3cis 27. z = 9cis (π) 28. z = 3cis π 2 4π 3 31. z = 1 2 cis 32. z = 12cis − π 3 7π 4 7π 8 34. z = 2cis 10cis arctan √ 3 arctan − √ 1 3 2 5 12 4 3 √ 36. z = 37. z = 15cis (arctan (−2)) 38. z = 39. z = 50cis π − arctan 7 24 40. z = 1 2 cis π + arctan For Exercises 41 - 52, use z = − your answers in polar form using the principal argument. i and w = 3 + 2 2 − 3i √ 3 3 3 2 √ √ 2 to compute the quantity. Express 41. zw 45. w3 42. z w 46. z5w2 43. w z 47. z3w2 44. z4 48. z2 w 25. z = 4cis 2π 3 29
. z = 7cis − 3π 4 33. z = 8cis π 12 35. z = 5cis arctan 11.7 Polar Form of Complex Numbers 1005 49. w z2 50. z3 w2 51. w2 z3 52. 6 w z In Exercises 53 - 64, use DeMoivre’s Theorem to ﬁnd the indicated power of the given complex number. Express your ﬁnal answers in rectangular form. 53. −2 + 2i √ 33 √ 54. (− 3 − i)3 57 + 61 58. 62. (2 + 2i)5 55. (−3 + 3i)4 59)5 63. ( √ 56. ( 3 + i)4 4 √ 3 3 − 1 3 i 60. 64. (1 − i)8 In Exercises 65 - 76, ﬁnd the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. 65. the two square roots of z = 4i 66. the two square roots of z = −25i 67. the two square roots of z = 1 + i √ 3 68. the two square roots of 5 2 − 5 √ 3 2 i 69. the three cube roots of z = 64 70. the three cube roots of z = −125 71. the three cube roots of z = i 72. the three cube roots of z = −8i 73. the four fourth roots of z = 16 74. the four fourth roots of z = −81 75. the six sixth roots of z = 64 76. the six sixth roots of z = −729 77. Use the Sum and Diﬀerence Identities in Theorem 10.16 or the Half Angle Identities in 2 in rectangular form. (See 2 + i Theorem 10.19 to express the three cube roots of z = Example 11.7.4, number 3.) √ √ 78. Use a calculator to approximate the ﬁve ﬁfth roots of 1. (See Example 11.7.4, number 4.) 79. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors. In Exercise 28 in Section 8.7, we showed you how to
factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i)). By multiplying the ﬁrst two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that p(x) = (x2 − 2x + 2)(x2 + 2x + 2). Use the 12 complex 12th roots of 4096 to factor p(x) = x12 − 4096 into a product of linear and irreducible quadratic factors. 1006 Applications of Trigonometry 80. Complete the proof of Theorem 11.14 by showing that if w = 0 than 1 w = 1 \|w\|. 81. Recall from Section 3.4 that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a − bi. (a) Prove that \|z\| = \|z\|. √ (b) Prove that \|z\| = (c) Show that Re(z) = zz z + z 2 and Im(z) = z − z 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). Interpret this result geometrically. (e) Is it always true that Arg (z) = −Arg(z)? 82. Given any natural number n ≥ 2, the n complex nth roots of the number z = 1 are called the In the following exercises, assume that n is a ﬁxed, but arbitrary, nth Roots of Unity. natural number such that n ≥ 2. (a) Show that w = 1 is an nth root of unity. (b) Show that if both wj and wk are nth roots of unity then so is their product wjwk. (c) Show that if wj is an nth root of unity then there exists another nth root of unity wj such that wj
wj = 1. Hint: If wj = cis(θ) let wj = cis(2π − θ). You’ll need to verify that wj = cis(2π − θ) is indeed an nth root of unity. 83. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula deﬁnes eit = cos(t) + i sin(t). (a) Use Theorem 11.16 to show that eixeiy = ei(x+y) for all real numbers x and y. (b) Use Theorem 11.16 to show that eixn = ei(nx) for any real number x and any natural number n. (c) Use Theorem 11.16 to show that eix eiy = ei(x−y) for all real numbers x and y. (d) If z = rcis(θ) is the polar form of z, show that z = reit where θ = t radians. (e) Show that eiπ + 1 = 0. (This famous equation relates the ﬁve most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that cos(t) = eit + e−it 2 and that sin(t) = eit − e−it 2i for all real numbers t. 11.7 Polar Form of Complex Numbers 1007 11.7.2 Answers √ 2cis π 4, Re(z) = 9, 1. z = 9 + 9i = 9 4 + 2πk \| k is an integer and Arg(z) = π arg(z) = π 4. √ √, Re(z) = 5, 3 = 10cis π 3, 3 3 + 2πk \| k is an integer and Arg(z) = π 3. arg(z) = π 2. z = 5 + 5i Im(z) = 9, Im(z) = 5 √ \|z\| = 9 2 \|z\| = 10 3. z = 6i = 6cis π 2 arg(z) = π, Re(z) = 0, \|z\| = 6 2 + 2πk \| k is an integer and Arg(
z) = π 2. Im(z) = 6, 4. z = −3 √ √ 2 + 3i 2 = 6cis 3π 4, Re(z) = −3 √ 2, Im(z) = 3 √ 2, \|z\| = 6 arg(z) = 3π √ 5. z = −6 arg(z) = 5π 4 + 2πk \| k is an integer and Arg(z) = 3π 4., Re(z) = −6 6 + 2πk \| k is an integer and Arg(z) = 5π 6. 3 + 6i = 12cis 5π 6 √ 3, Im(z) = 6, \|z\| = 12 6. z = −2 = 2cis (π), Re(z) = −2, Im(z) = 0, \|z\| = 2 arg(z) = {(2k + 1)π \| k is an integer} and Arg(z) = π. √ 3 7. z = − 2 − 1 arg(z) = 7π 8. z = −3 − 3i = 3 arg(z) = 5π √ 3 2, 6, Re(z) = − 2 i = cis 7π Im(z) = − 1 2, 6 + 2πk \| k is an integer and Arg(z) = − 5π 6., Re(z) = −3, Im(z) = −3, 4 + 2πk \| k is an integer and Arg(z) = − 3π 4. 2cis 5π 4 √ \|z\| = 1 √ \|z\| = 3 2 9. z = −5i = 5cis 3π 2 arg(z) = 3π √, Re(z) = 0, Im(z) = −5, \|z\| = 5 2 + 2πk \| k is an integer and Arg(z) = − π 2. √, Re(z) = 2 4 + 2πk \| k is an integer and Arg(z) = − π 4. 2 = 4cis 7π 4 √ 2, 10. z = 2 2 − 2i arg(z) = 7π √ Im(z) = −2 2, \|z\| = 4 11. z = 6 = 6
cis (0), Re(z) = 6, Im(z) = 0, \|z\| = 6 arg(z) = {2πk \| k is an integer} and Arg(z) = 0. √ Im(z) = 3, Re(z) = 0, √ 12. z = i 3 7, 7cis π 2 √ \|z\| = 3 7 2 + 2πk \| k is an integer and Arg(z) = π 2. √ 7 = 3 arg(z) = π 13. z = 3 + 4i = 5cis arctan 4 3, Re(z) = 3, Im(z) = 4, \|z\| = 5 arg(z) = arctan 4 3 + 2πk \| k is an integer and Arg(z) = arctan 4 3. 1008 14. z = √ 2 + i = √ arctan √ 2 2, Re(z) = √ 2, Im(z) = 1, \|z\| = arg(z) = arctan + 2πk \| k is an integer and Arg(z) = arctan 3cis √ 2 2 √ 3 √ 2 2. Applications of Trigonometry 15. z = −7 + 24i = 25cis π − arctan 24 7, Re(z) = −7, Im(z) = 24, \|z\| = 25 arg(z) = π − arctan 24 7 + 2πk \| k is an integer and Arg(z) = π − arctan 24 7. 16. z = −2 + 6i = 2 √ 10cis (π − arctan (3)), Re(z) = −2, Im(z) = 6, √ \|z\| = 2 10 arg(z) = {π − arctan (3) + 2πk \| k is an integer} and Arg(z) = π − arctan (3). 17. z = −12 − 5i = 13cis π + arctan 5 12, Re(z) = −12, Im(z) = −5, \|z\| = 13 arg(z) = π + arctan 5 12 + 2πk \| k is an integer and Arg(z)
= arctan 5 12 18. z = −5 − 2i = √ 29cis π + arctan 2 5, Re(z) = −5, Im(z) = −2, arg(z) = π + arctan 2 5 5cis arctan − 1 2 19. z = 4 − 2i = 2 √ + 2πk \| k is an integer and Arg(z) = arctan 2 5 √, Re(z) = 4, Im(z) = −2, \|z\| = 2 arg(z) = arctan − 1 2 + 2πk \| k is an integer and Arg(z) = arctan − 1 2 − π. √ \|z\| = 29 − π. 5 = − arctan 1 2. 20. z = 1 − 3i = √ 10cis (arctan (−3)), Re(z) = 1, Im(z) = −3, \|z\| = √ 10 arg(z) = {arctan (−3) + 2πk \| k is an integer} and Arg(z) = arctan (−3) = − arctan(3). 21. z = 6cis(0) = 6 √ 23. z = 7 25. z = 4cis 2π 3 2cis π 4 = −2 + 2i = 7 + 7i √ 3 27. z = 9cis (π) = −9 √ 3 + i = 22. z = 2cis π 6 24. z = 3cis π 2 6cis 3π 4 26. z = √ = 3i √ = − √ 3 + i 3 28. z = 3cis 4π 3 √ 30. z = = − 3 2 − 3i √ √ 3 2 13 13cis 3π 2 = −i 32. z = 12cis − π 3 = 6 − 6i √ 3 29. z = 7cis − 3π 4 31. z = 1 2 cis 7π 33. z = 8cis π 12 35. z = 5cis arctan + 4i 2 − √ 3 = 3 + 4i √ 37. z = 15cis (arctan (−2)) = 3 5 − 6i �
� 5 √ 38. z = 34. z = 2cis 7π 8 √ 36 10cis arctan 1 3 √ 3cis arctan − 2 = 1 − i √ 2 39. z = 50cis π − arctan 7 24 = −48 + 14i 40. z = 1 2 cis π + arctan 5 12 = − 6 13 − 5i 26 11.7 Polar Form of Complex Numbers 1009 In Exercises 41 - 52, we have that z = − 3 we get the following. √ 3 2 + 3 2 i = 3cis 5π 6 41. zw = 18cis 7π 12 44. z4 = 81cis − 2π 3 47. z3w2 = 972cis(0) 50. z3 w2 = 3 4 cis(π) √ 53. −2 + 2i 33 = 64 and w = 3 √ √ 2 − 3i 2 = 6cis − π 4 so 43. w z = 2cis 11π 46. z5w2 = 8748cis − π 3 12 42. z w = 1 2 cis − 11π 12 45. w3 = 216cis − 3π 4 48. z2 w = 3 2 cis − π 12 49. w z2 = 2 3 cis π 12 51. w2 z3 = 4 √ 3 cis(π) 54. (− 3 − i)3 = −8i 52. w z 6 = 64cis − π 2 55. (−3 + 3i)4 = −324 √ 56. ( 3 + i)4 = −8 + 8i √ 3 57. 5 2 + 5 2 i3 = − 125 4 + 125 58. 59. 3 2 − 3 2 i3 = − 27 4 − 27 4 i 60. 62. (2 + 2i)5 = −128 − 128i 63. ( 3 − i)5 = −16 3 − 16i 81 − 8i √ √ 3 81 √ 2 2 + 4 √ 2 2 i 61. = −1 64. (1 − i)8 = 16 65. Since z = 4i = 4cis π 2 w0 = 2cis π 4 = √ 2 + i 2 we have √ 66. Since z = −25i = 25
cis 3π 2 we have w0 = 5cis 3π w1 = 2cis 5π 4 √ = − √ 2 − i 2 w1 = 5cis 7π 67. Since z = 1 + i √ 3 = 2cis π 3 we have w0 = √ 2cis w1 = √ 2cis 7π 68. Since z = 5 2 − 5 3 2 i = 5cis 5π 3 we have √ w0 = √ 5cis 5π 6 = − √ 15 2 + √ 5 2 i w1 = √ 5cis 11π 6 = √ 15 2 − √ 5 2 i 69. Since z = 64 = 64cis (0) we have w0 = 4cis (0) = 4 w1 = 4cis 2π 3 = −2 + 2i √ 3 w2 = 4cis 4π 3 = −2 − 2i √ 3 1010 Applications of Trigonometry 70. Since z = −125 = 125cis (π) we have w0 = 5cis w1 = 5cis (π) = −5 w2 = 5cis 5π 71. Since z = i = cis π 2 we have w0 = cis w1 = cis 5π w2 = cis 3π 2 = −i 72. Since z = −8i = 8cis 3π 2 we have w0 = 2cis π 2 = 2i w1 = 2cis 7π 6 √ = − 3 − i w2 = cis 11π 6 = √ 3 − i 73. Since z = 16 = 16cis (0) we have w0 = 2cis (0) = 2 w2 = 2cis (π) = −2 74. Since z = −81 = 81cis (π) we have w1 = 2cis π w3 = 2cis 3π = 2i = −2i 2 2 w0 = 3cis π 4 = 3 2 2 + 3 √ w2 = 3cis 5π w1 = 3cis 3π 4 w3 = 3cis 7π 75. Since z = 64 = 64cis(0) we have w0 = 2cis(0) = 2 w3 = 2cis (π) = −2 √
w1 = 2cis π w4 = 2cis − 2π 3 3 = 1 + 3i = −1 − √ w2 = 2cis 2π 3i w5 = 2cis − π = −1 + √ = 1 − 3 3i 3 √ 3i 76. Since z = −729 = 729cis(π) we have w0 = 3cis w1 = 3cis π 2 = 3i w2 = 3cis 5π w3 = 3cis 7π w4 = 3cis − 3π 2 = −3i w5 = 3cis − 11π 77. Note: In the answers for w0 and w2 the ﬁrst rectangular form comes from applying the 3 − π appropriate Sum or Diﬀerence Identity ( π 4, respectively) and the second comes from using the Half-Angle Identities. √ √ = 3 2 √ w0 = 3 4 and 17π 12 = 2π 3 + 3π 12 = 2cis π 12 2+ 2 2− 2 6− 6 4 + i − √ 2 √ w1 = 3 2cis 3π 4 √ w2 = 3 2cis 17π 12 √ − √ 6 √ = 3 2 2− 4 √ √ 3 2− 2 + i √ 3 √ 2+ 2 11.7 Polar Form of Complex Numbers 1011 5 78. w0 = cis(0) = 1 w1 = cis 2π w2 = cis 4π w3 = cis 6π w4 = cis 8π ≈ 0.309 + 0.951i ≈ −0.809 + 0.588i ≈ −0.809 − 0.588i ≈ 0.309 − 0.951i 5 5 5 79. p(x) = x12 −4096 = (x−2)(x+2)(x2 +4)(x2 −2x+4)(x2 +2x+4)(x2 −2 √ 3x+4)(x2 +2 √ 3+4) 1012 Applications of Trigonometry 11.8 Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suﬃce; that is, real
numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suﬃce. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. (Foreshadowing the use of bearings in the exercises, perhaps?) To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors.1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt2 the ‘arrow’ notation, so the symbol v is read as ‘the vector v’. Below is a typical vector v with endpoints P (1, 2) and Q (4, 6). The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct −−→ v completely from P and Q, we write v = P Q, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) Q (4, 6) P (1, 2) −−→ P Q v = While it is true that P and Q completely determine v, it is important to note that since vectors are deﬁned in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as v is considered to be the same vector as v, regardless of its initial point. In the case of our vector v above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v = 3, 4, where the ﬁrst number, 3, is called the x-component of v and the second number, 4, is called the y-component of v. If we wanted to reconstruct v = 3, 4 with initial point P (−2, 3), then we would ﬁnd the terminal point of v
by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q(1, 7), as seen below. 1The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’ 2Other textbook authors use bold vectors such as v. We ﬁnd that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the ‘arrow’ notation. 3If this idea of ‘over’ and ‘up’ seems familiar, it should. The slope of the line segment containing v is 4 3. 11.8 Vectors 1013 Q (1, 7) up 4 P (−2, 3) over 3 v = 3, 4 with initial point P (−2, 3). The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our deﬁnition below. Deﬁnition 11.5. Suppose v is represented by a directed line segment with initial point P (x0, y0) and terminal point Q (x1, y1). The component form of v is given by −−→ P Q = x1 − x0, y1 − y0 v = Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, v1, v2 = v 1 and v2 = v 2. (Again, think about this before reading on.) We now set about deﬁning operations on vectors. Suppose we are given two vectors v and w. The sum, or resultant vector v + w is obtained as follows. First, plot v. Next, plot w so that its initial point is the terminal point of v. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. It is helpful to think of the vector v + w as the ‘net result’ of moving along v then moving along w. 2 if and only if v1 = v 1, v v + w w v v, w, and v + w Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 11.8.1. A plane leaves an airport with an airspeed5 of 175 miles per hour at
a bearing of N40◦E. A 35 mile per hour wind is blowing at a bearing of S60◦E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. 4If necessary, review page 905 and Section 11.3. 5That is, the speed of the plane relative to the air around it. If there were no wind, plane’s airspeed would be the same as its speed as observed from the ground. How does wind aﬀect this? Keep reading! 1014 Applications of Trigonometry Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook.6 We let v denote the plane’s velocity and w denote the wind’s velocity in the diagram below. The ‘true’ speed and bearing is found by analyzing the resultant vector, v + w. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 175, the magnitude of w, which is 35, and the magnitude of v + w, which we’ll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100◦. N 40◦ w 60◦ v v + w N 35 100◦ 175 α c 40◦ E E 60◦ 31850 − 12250 cos(100◦) ≈ 184, which means the From the Law of Cosines, we determine c = true speed of the plane is (approximately) 184 miles per hour. To determine the true bearing of the plane, we need to determine the angle α. Using the Law of Cosines once more,7 we ﬁnd cos(α) = c2+29400 so that α ≈ 11◦. Given the geometry of the situation, we add α to the given 40◦ and ﬁnd the true bearing of the plane to be (approximately) N51◦E. 350c Our next step is to deﬁne addition of vectors component-wise to match the geometric action.8 Deﬁnition 11.6. Suppose v = v1, v2 and w
= w1, w2. The vector v + w is deﬁned by v + w = v1 + w1, v2 + w2 Example 11.8.2. Let v = 3, 4 and suppose w = and interpret this sum geometrically. −−→ P Q where P (−3, 7) and Q(−2, 5). Find v + w Solution. Before can add the vectors using Deﬁnition 11.6, we need to write w in component form. Using Deﬁnition 11.5, we get w = −2 − (−3), 5 − 7 = 1, −2. Thus 6See Section 10.1.1, for instance. 7Or, since our given angle, 100◦, is obtuse, we could use the Law of Sines without any ambiguity here. 8Adding vectors ‘component-wise’ should seem hauntingly familiar. Compare this with how matrix addition was deﬁned in section 8.3. In fact, in more advanced courses such as Linear Algebra, vectors are deﬁned as 1 × n or n × 1 matrices, depending on the situation. 11.8 Vectors 1015 v + w = 3, 4 + 1, −2 = 3 + 1, 4 + (−2) = 4, 2 To visualize this sum, we draw v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w with its initial point at (3, 4). Moving one to the right and two down, we ﬁnd the terminal point of w to be (4, 2). We see that the vector v + w has initial point (0, 0) and terminal point (4, 2) so its component form is 4, 2, as required In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our deﬁnition of vectors to include a ‘zero vector’, 0 = 0, 0. Geometrically, 0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of 0, since after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of 0
should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undeﬁned, but this minor hiccup in the natural ﬂow of things is worth the beneﬁts we reap by including 0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition Commutative Property: For all vectors v and w, v + w = w + v. Associative Property: For all vectors u, v and w, (u + v) + w = u + (v + w). Identity Property: The vector 0 acts as the additive identity for vector addition. That is, for all vectors v. Inverse Property: Every vector v has a unique additive inverse, denoted −v. That is, for every vector v, there is a vector −v so that v + (−v) = (−v) + v = 0. 1016 Applications of Trigonometry The properties in Theorem 11.18 are easily veriﬁed using the deﬁnition of vector addition.9 For the commutative property, we note that if v = v1, v2 and w = w1, w2 then v + w = v1, v2 + w1, w2 = v1 + w1, v2 + w2 = w1 + v1, w2 + v2 = w + v Geometrically, we can ‘see’ the commutative property by realizing that the sums v + w and w + v are the same directed diagonal determined by the parallelogram below Demonstrating the commutative property of vector addition. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector v = v1, v2, suppose we wish to ﬁnd a vector w = w1, w2 so that v + w = 0. By the deﬁnition of vector addition, we have v1 + w1, v2 + w2 = 0, 0, and hence, v1 + w1 = 0 and v2 + w2 = 0. We get w1 = −v1 and w2 = −v2 so that w = −v1, −v
2. Hence, v has an additive inverse, and moreover, it is unique and can be obtained by the formula −v = −v1, −v2. Geometrically, the vectors v = v1, v2 and −v = −v1, −v2 have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum v + (−v) results in starting at the initial point of v and ending back at the initial point of v, or in other words, the net result of moving v then −v is not moving at all. v −v Using the additive inverse of a vector, we can deﬁne the diﬀerence of two vectors, v − w = v + (− w). If v = v1, v2 and w = w1, w2 then 9The interested reader is encouraged to compare Theorem 11.18 and the ensuing discussion with Theorem 8.3 in Section 8.3 and the discussion there. 11.8 Vectors 1017 v − w = v + (− w) = v1, v2 + −w1, −w2 = v1 + (−w1), v2 + (−w2) = v1 − w1, v2 − w2 In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v − w geometrically, we note w + (v − w) = w + (v + (− w)) Deﬁnition of Vector Subtraction = w + ((− w) + v) Commutativity of Vector Addition = ( w + (− w)) + v Associativity of Vector Addition = 0 + v = v Deﬁnition of Additive Inverse Deﬁnition of Additive Identity This means that the ‘net result’ of moving along w then moving along v − w is just v itself. From the diagram below, we see that v − w may be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. It is also worth mentioning that in the parallelogram determined by the vectors v and w, the vector v − w is one of the diagonals – the other being v + w Next, we discuss scalar multiplication – that is, taking a real number times a vector.
We deﬁne scalar multiplication for vectors in the same way we deﬁned it for matrices in Section 8.3. Deﬁnition 11.7. If k is a real number and v = v1, v2, we deﬁne kv by kv = k v1, v2 = kv1, kv2 Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. 1018 Applications of Trigonometry v 2v 1 2 v −2v Note that, by deﬁnition 11.7, (−1)v = (−1) v1, v2 = (−1)v1, (−1)v2 = −v1, −v2 = −v. This, and other properties of scalar multiplication are summarized below. Theorem 11.19. Properties of Scalar Multiplication Associative Property: For every vector v and scalars k and r, (kr)v = k(rv). Identity Property: For all vectors v, 1v = v. Additive Inverse Property: For all vectors v, −v = (−1)v. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k + r)v = kv + rv Distributive Property of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k, k(v + w) = kv + k w Zero Product Property: If v is vector and k is a scalar, then kv = 0 if and only if k = 0 or v = 0 The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the deﬁnition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let v = v1, v2. If k and r are scalars then (kr)v = (kr) v1, v2 = (kr)v1, (kr)v2 Deﬁnition of Scalar Multiplication = k(rv1), k(rv2) Associative Property of Real Number Multiplication = k rv

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