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**Unformatted text preview: **546 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS is symmetric about the x-axis: (1)
y 1 if [r , θ ] ∈ C , then [r , −θ ] ∈ C . But this is not easy to see from the polar equation because, in general, if the coordinates of the ﬁrst point satisfy the equation, the coordinates of the second point do not. One way to see that (1) is valid is to note that [r , −θ ] = [−r , π − θ ] and then verify that, if the coordinates of [r , θ ] satisfy the equation, then so do the coordinates of [−r , π − θ ]. But all this is very cumbersome. The easiest way to see that the curve r 2 = sin θ is symmetric about the x-axis is to write it as ( x 2 + y 2 )3 = y 2 . 0.5 0.5 – 0.5 x –1 Figure 9.4.13 The other symmetries of the curve, symmetric about the y-axis and symmetric about the origin, can also be seen from this equation (see Figure 9.4.13). EXERCISES 9.4
Sketch the polar curve. 1. θ = − 1 π . 4 3. r = 4. 5. r = −2 sin θ . 7. r csc θ = 3. 9. r = θ , − 1 π ≤ θ ≤ π . 2 11. r = sin 3θ . 13. r 2 = sin 2θ . 15. r = 4,
2 3 2. r = −3. 4. r = 3 cos θ . 6. θ = 2 π . 3 8. r = 1 − cos θ . 10. r sec θ = −2. 12. r = cos 2θ .
2 37. Show that the point [2, π ] lies both on r 2 = 4 cos θ and on r = 3 + cos θ . 38. Show that the point [2, 1 π ] lies both on r 2 sin θ = 4 and on 2 r = 2 cos 2θ . Sketch the curves and ﬁnd the points at which they intersect. Express your answers in rectangular coordinates. 39. r = sin θ , 40. r = sin θ ,
2 r = − cos θ . r = 2 − sin θ . r = −1. r = 2 cos θ . r = cos θ . r = sin θ . r = 1 + sin θ . 14. r = cos 2θ .
3 π. 4 41. r = cos2 θ , 42. r = 2 sin θ , 0≤θ ≤ 16. r = sin θ . 17. r = 9r . 18. θ = −1π, 4 1 ≤ r < 2. 20. r 2 = 4r .
1 π. 2 1 π. 2 43. r = 1 − cos θ , 44. r = 1 − cos θ , 45. r = sin 2θ , 46. r = 1 − cos θ , 19. r = −1 + sin θ . 21. r = sin 2θ . 22. r = cos 3θ , 23. r = cos 5θ , 24. r = eθ , 27. r = tan θ . 29. r = 2 + sec θ . 31. r = −1 + 2 cos θ . 33. r 2 cos θ = 1; 35. r = sin 1 θ ; 3 36. r 2 = sin 3θ ; [1, π ]. [ 1 , 1 π ]. 22 [1, − 5 π ]. 6 25. r = 2 + sin θ . 0≤θ ≤ 0≤θ ≤ r = sin θ . −π ≤ θ ≤ π (logarithmic spiral). 26. r = cot θ . 28. r = 2 − cos θ . 30. r = 3 − csc θ . 32. r = 1 + 2 sin θ . 34. r 2 = cos 2θ ; [1, 1 π ]. 4 47. Show that the polar equation r = 2a sin θ + 2b cos θ represents a circle. Find the center and radius of the circle and sketch its graph. c 48. a. Use a graphing utility to draw the curves r = 1 + cos (θ − 1 π ) 3 and r = 1 + cos (θ + 1 π ). 6 Determine whether the point lies on the curve. How do your curves compare with the graph of r = 1 + cos θ ? (See Figure 9.4.5) b. In general, what is the relationship between the graph of r = f (θ − α ) and the graph of r = f (θ )? c 49. a. Use a graphing utility to draw the curves
r = 1 + sin θ and r 2 = 4 sin 2θ in the same coordinate system. 9.4 GRAPHING IN POLAR COORDINATES 547 b. Use a CAS to ﬁnd the points of intersection of the two curves. c 50. Repeat Exercise 49 for the pair of equations r = 1 + cos θ and r = 1 + cos ( 1 θ ). 2 Use a graphing utility to draw the equipotential lines for m = 1, 2, 3. (c) Draw the curves r = 2 sin2 θ and r 2 = 2 cos θ in the same coordinate system and estimate the coordinates of their points of intersection. Use four decimal place accuracy. c 51. Repeat Exercise 49 for the pair of equations
r = 2 cos θ and r = 2 sin 2θ . c 55. Use a graphing utility to draw the curves
r = 1 + sin k θ + cos2 (2k θ ) for k = 1, 2, 3, 4, 5. If you were asked to give a name to this family of curves, what would you suggest? c 52. Repeat Exercise 49 for the pair of equations
r=2 and r = 2 sin 3θ . c 53. Repeat Exercise 49 for the pair of equations
r = 1 − 3 cos θ and r = 2 − 5 sin θ . c 54. (a) The electrostatic charge distribution consisting of a charge q (q > 0) at the point [r , 0] and a charge −q at [r , π ] is called a dipole. The lines of force for the dipole are given by the equations
r = k sin θ .
2 c 56. Use a graphing utility to draw the graph r = ecos θ − 2 cos 4θ . What name would you suggest for this curve? c 57. The graphs of r = A cos k θ and r = A sin k θ are called petal curves (see Figure 9.4.10). Use a graphing utility to draw the curves
r = 2 cos k θ and r = 2 sin k θ for k = 3 and k = 5 . Can you predict what the curve will 2 2 look like for k = m/2 for any odd integer m? Use a graphing utility to draw the lines of force for k = 1, 2, 3. (b) The equipotential lines (the set of points with equal electric potential) for the dipole are given by the equations r 2 = m cos θ . c 58. Use a graphing utility to draw the curves
r = 2 cos k θ and r = 2 sin k θ for k = 4 and k = 5 . Can you predict what the curve will 3 3 look like for k = m/3 for any even integer m that is not a multiple of 3? For any odd integer m that is not a multiple of 3? PROJECT 9.4 The Conic Sections in Polar Coordinates
In Section 9.1, we deﬁned a parabola in terms of a focus and directrix, but our deﬁnitions of the ellipse and hyperbola in Section 9.2 were given in terms of two foci; there was no mention of a directrix for either of these conics. In this project, we give a uniﬁed approach to the conic sections that involves a focus and directrix for all three cases. In the plane, let F be a ﬁxed point (the focus) and l a ﬁxed line (the directrix) which does not pass through F . Let e be a positive number (the eccentricity) and consider the set of points P that satisfy (1) distance from P to F = e. distance from P to l
y P r x=d θ
F r cos θ x In the ﬁgure, we have superimposed a polar and rectangular coordinate system and, without loss of generality, we have taken F as the origin and l as the vertical line x = d , d > 0. Problem 1. Show that the set of all points P that satisfy (1) is described by the polar equation: (2) r= ed . 1 + e cos θ Problem 2. Show that: a. If 0 < e < 1, the equation is an ellipse of eccentricity e with right focus at the origin, major axis horizontal: y2 ( x + c )2 +2 =1 a2 a − c2 with a= ed , 1 − e2 c = ea. 548 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS b. If e = 1, the equation represents a parabola with focus at the origin and directrix x = d : y2 = −4 d 2 x− d 2 . Consider the polar equation r = α/(1 + β sin θ ), α , β > 0. Since r= α α = 1 + β sin θ 1 + β cos (θ − π/2) c. If e > 1, the equation represents a hyperbola of eccentricity e with left focus at the origin, transverse axis horizontal: y2 ( x − c )2 −2 =1 a2 c − a2 with a= ed , e2 − 1 c = ea. we can conclude that r = α/(1 + β sin θ ) is simply the conic section r = α/(1 + β cos θ ) rotated π/2 radians in the counterclockwise direction. Problem 4. In terms of rotations, relate the polar equations r= α 1 − β cos θ and r= α , 1 − β sin θ α, β > 0 Problem 3. Identify each of the following conic sections and write the equation in rectangular coordinates. a. r = 8 . 4 + 3 cos θ b. r = 6 . 1 + 2 cos θ c. r = 6 . 2 + 2 cos θ to the conic section r= α . 1 + β cos θ 9.5 AREA IN POLAR COORDINATES
θ =β r = ρ (θ ) Here we develop a technique for calculating the area of a region the boundary of which is given in polar coordinates. As a start, we suppose that α and β are two real numbers with α < β ≤ α + 2π . We take ρ as a function that is continuous on [α , β ] and keeps a constant sign on that interval. We want the area of the polar region generated by the curve
θ =α Γ r = ρ (θ ), α ≤ θ ≤ β. O polar axis Such a region is portrayed in Figure 9.5.1. In the ﬁgure ρ (θ ) remains nonnegative. If ρ (θ ) were negative, the region would appear on the opposite side of the pole. In either case, the area of is given by the formula
(9.5.1) Figure 9.5.1 A= β α 1 [ρ (θ )]2 d θ . 2 β = θn θi θ i –1 α = θo
Ri ri polar axis PROOF We consider the case where ρ (θ ) ≥ 0. We take P = {θ0 , θ1 , . . . , θn } as a partition of [α , β ] and direct our attention to what happens from θi−1 to θi . We set ri = min value of ρ on [θi−1 , θi ] and Ri = max value of ρ on [θi−1 , θi ]. The part of that lies between θi−1 and θi contains a circular sector of radius ri and central angle θi = θi − θi−1 and is contained in a circular sector of radius Ri with the same central angle θi . (See Figure 9.5.2.) Its area Ai must therefore satisfy the inequality
12 r 2i θi ≤ Ai ≤ 1 R2 2i θi . † O Figure 9.5.2 By summing these inequalities from i = 1 to i = n, we can see that the total area A must satisfy the inequality (1) Lf (P ) ≤ A ≤ Uf (P ) † The area of a circular sector of radius r and central angle α is 1 r 2 α . 2 ...

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