SalasSV_09_05 - 548 CHAPTER 9 THE CONIC SECTIONS; POLAR...

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Unformatted text preview: 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 9.5 AREA IN POLAR COORDINATES 549 where f (θ ) = 1 [ρ (θ )]2 . 2 Since f is continuous and (1) holds for every partition P of [α , β ], we can conclude that A= Example 1 β α [1, ] r = 1 – cos θ [2, π ] polar axis π 2 f (θ ) d θ = β α 1 [ρ (θ )]2 d θ . 2 Calculate the area enclosed by the cardioid r = 1 − cos θ (Figure 9.5.3) [1, ] Figure 9.5.3 3π 2 SOLUTION The entire curve is traced out as θ increases from 0 to 2π , and 1 − cos θ ≥ 0 on [0, 2π ]. Therefore A= 0 2π 1 (1 2 − cos θ )2 d θ = = ↑ 1 2 1 2 2π 0 2π 0 (1 − 2 cos θ + cos2 θ ) d θ ( 3 − 2 cos θ + 1 cos 2θ ) d θ . 2 2 half-angle formula: cos2 θ = 1 2 + 1 2 cos 2θ 2π Since 0 cos θ d θ = sin θ 2π 0 =0 1 2 2π 0 3 2 2π and 0 cos 2θ d θ = 1 2 sin 2θ 2π 0 = 0, we have A= dθ = 3 4 2π 0 dθ = 3 π . 2 θ=β A slightly more complicated type of region is pictured in Figure 9.5.4. We approach the problem of calculating the area of the region in the same way that we calculated the area between two curves in Section 5.5; that is, we calculate the area out to r = p2 (θ ) and subtract from it the area out to r = p1 (θ ). This gives area of which can be written as (9.5.2) Ω r = ρ1(θ ) r = ρ2(θ ) θ=α = β α 1 [ρ (θ )]2 d θ 22 − β α 1 [ρ (θ )]2 d θ , 21 O polar axis Figure 9.5.4 area of = β α 1 ([ρ2 (θ )]2 2 − [ρ1 (θ )]2 ) d θ . θ= π 3 To find the area between two polar curves, we first determine the curves that serve as outer and inner boundaries of the region and the intervals of θ values over which these boundaries are traced out. Since the polar coordinates of a point are not unique, extra care is needed to determine these intervals of θ values. Example 2 SOLUTION r=1 r = 2 cos θ O polar axis Find the area of the region that consists of all points that lie within the circle r = 2 cos θ but outside the circle r = 1. θ= 5π 3 The region is shown in Figure 9.5.5. Our first step is to find values of θ for the two points where the circles intersect: 2 cos θ = 1, cos θ = 1 , 2 θ = 1π, 5π. 3 3 Figure 9.5.5 550 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS Since the region is symmetric about the polar axis, the area below the polar axis equals the area above the polar axis. Thus A=2 0 π /3 1 ([2 cos θ ]2 2 − [1]2 ) d θ . 1 2 If you carry out the integration, you will see that A = 1 π + 3 θ= π r = 1 – 2 cosθ 1 3 √ 3 ∼ 1. 91. = Example 3 limaçon Find the area A of the region between the inner and outer loops of the r = 1 − 2 cos θ . (Figure 9.5.6) SOLUTION You can verify that r = 0 at θ = π/3 and at θ = 5π/3. The outer loop is formed as θ increases from π/3 to 5π/3. Thus θ= π 5 3 area within outer loop = A1 = 5π/3 π/3 1 [1 2 − 2 cos θ ]2 d θ . Figure 9.5.6 The lower half of the inner loop is formed as θ increases from 0 to π/3, and the upper half as θ increases from 5π/3 to 2π (verify this). Therefore area within inner loop = A2 = 0 π /3 1 [1 2 − 2 cos θ ]2 d θ + 2π 5π/3 1 [1 2 − 2 cos θ ]2 d θ . Now 1 [1 2 − 2 cos θ ]2 d θ = = = 1 2 1 2 1 2 [1 − 4 cos θ + 4 cos2 θ ] d θ [1 − 4 cos θ + 2(1 + cos 2θ )] d θ [3 − 4 cos θ + 2 cos 2θ ] d θ = 1 [3θ − 4 sin θ + sin 2θ ] + C . 2 5π/3 π/3 Therefore A1 = 1 2 3θ − 4 sin θ + sin 2θ = 2π + 3 2 √ 3 and A2 = = 1 2 3θ − 4 sin θ + sin 2θ − 3 4 π /3 0 + 1 2 3θ − 4 sin θ + sin 2θ 2π 5π/3 1 π 2 √ 3+ 1 π 2 √ √ − 3 = π − 3 3. 2 3 4 Thus, A = A 1 − A 2 = 2π + 3 2 √ 3 − (π − 3 2 √ √ 3) = π + 3 3 ∼ 8. 34. = Remark We could have done Example 3 more efficiently by exploiting the symmetry of the region. The region is symmetric about the x-axis. Therefore A=2 π π/3 1 [1 2 − 2 cos θ ]2 d θ − 2 0 π /3 1 [1 2 − 2 cos θ ]2 d θ . 9.5 AREA IN POLAR COORDINATES 551 common to the circle r = 2 sin θ and the limaçon r = − sin θ is indicated in Figure 9.5.7. The θ coordinates of the points of intersection can be found by solving the two equations simultaneously: Example 4 The region 3 2 θ= π 2 2 sin θ = Thus, the area of area of = 0 3 2 − sin θ , sin θ = 1 , 2 and θ = 1π, 5π. 6 6 θ= π 5π/6 1 [2 sin θ ]2 2 5π 6 r = 2 sinθ can be represented as follows: π /6 1 [2 sin θ ]2 d θ 2 [1, ] π 2 θ= π 6 + 5π/6 π/6 13 [ 22 − sin θ ]2 d θ + dθ ; Ω polar axis or, by the symmetry of the region, r= area of =2 0 π /6 1 [2 sin θ ]2 d θ 2 +2 π /2 π/6 3 2 – sin θ 13 [ 22 − sin θ ]2 d θ . As you can verify, the area of EXERCISES 9.5 is 5 π − 4 15 8 √ 3 ∼ 0. 68. = Figure 9.5.7 Calculate the area enclosed by the given curve. Take a > 0. 1. r = a cos θ from θ = −1π 2 −1π 6 to θ = 1 π. 2 18. Outside r = 1 − cos θ , but inside r = 1 + cos θ . 19. Inside r = 4, and to the right of r = 2 sec θ . 20. Inside r = 2, but outside r = 4 cos θ . 21. Inside r = 4, and between the lines θ = 1 π and r = 2 sec θ . 2 22. Inside the inner loop of r = 1 − 2 sin θ . 23. Inside one petal of r = 2 sin 3θ . 24. Outside r = 1 + cos θ , but inside r = 2 − cos θ . 25. Interior to both r = 1 − sin θ and r = sin θ . 26. Inside one petal of r = 5 cos 6θ . 2. r = a cos 3θ from θ = to θ = 1 π . 6 √ 1 3. r = a cos 2θ from θ = − 4 π to θ = 1 π . 4 4. r = a(1 + cos 3θ ) 5. r 2 = a2 sin2 θ . 7. r = tan 2θ 8. r = cos θ , 9. r = 2 cos θ , from θ = − 1 π to θ = 1 π . 3 3 6. r 2 = a2 sin2 2θ . θ = 0, and the rays and the rays θ= 1 π. 8 Calculate the area of the given region. and the rays r = sin θ , r cos θ , θ = 0, θ = 0, θ = 1π. 4 θ = 1π. 4 θ = 1π. 4 27. Outside r = cos 2θ , but inside r = 1. 28. Interior to both r = 2a cos θ and r = 2a sin θ , a > 0. 29. Find the area of the region that is common to the three circles: r = 1, r = 2 cos θ , and r = 2 sin θ . 30. Find the area of the region outside the circle r = a and inside the lemniscate r 2 = 2a2 cos 2θ . 31. Fix a > 0 and let n be a positive integer. Prove that the petal curves r = a cos 2nθ and r = a sin 2nθ all enclose exactly the same area. Find the area. 32. Fix a > 0 and let n be a positive integer. Prove that the petal curves r = a cos ([2n + 1]θ ) and r = a sin ([2n + 1]θ all enclose exactly the same area. Find the area. Centroids in Polar Coordinates Let be the region bounded by the polar curve r = ρ (θ ) between θ = α and θ = β . Since the centroid of a triangle lies on each median, two-thirds of the distance from the vertex to the opposite side (see Exercise 29, Section 6.4), it follows that the x and y coordinates of the centroid of the 10. r = 1 + cos θ , r = cos θ , and the rays θ = 0, θ = 1 π . 2 11. r = a(4 cos θ − sec θ ) 12. r = 1 2 and the rays θ = 0, sec2 1 θ 2 and the vertical line through the origin. Find the area between the curves. 13. r = eθ , the rays 14. r = eθ , the rays 15. r = eθ , the rays 16. r = eθ , the rays 0 ≤ θ ≤ π; r = θ, 0 ≤ θ ≤ π; θ = 0, θ = π . 2π ≤ θ ≤ 3π ; r = θ , 0 ≤ θ ≤ π ; θ = 0, θ = π . 0 ≤ θ ≤ π ; r = eθ/2 , 0 ≤ θ ≤ π ; θ = 2π , θ = 3π . 0 ≤ θ ≤ π ; r = eθ , 2π ≤ θ ≤ 3π ; θ = 0, θ = π . Represent the area by one or more integrals. 17. Outside r = 2, but inside r = 4 sin θ . 552 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS “triangular” region shown in the figure are given approximately by x= 2 ρ (θ ) cos θ 3 36. The region enclosed by r = 2 + sin θ . y = 2 ρ (θ ) sin θ 3 c In Exercises 37 and 38 ,use a graphing utility to draw the polar curve. Then use a CAS to find the area of the region it encloses. 37. r = 2 + cos θ . 38. r = 2 cos 3θ . These approximations improve as the triangle narrows. θ =β y c In Exercises 39 and 40, use a graphing utility to drawn the polar curves. Then use a CAS to find the area of the region inside the first curve and outside the second curve. 39. r = 4 cos 3θ , r = 2. 40. r = 2 cos θ , r = 1 − cos θ . c 41. The curve whose equation in rectangular coordinates is x ρ (θ ) y θ Ω x r = ρ (θ ) y2 = x2 a−x , a+x a > 0, is called a strophoid. (a) Show that the polar equation of this curve has the form r = a cos 2θ sec θ . (b) Use a graphing utility to draw the graph of the curves for a = 1, 2, and 4. (c) Let a = 2. Find the area inside the loop. θ =α 33. Following the approach used in Section 6.4, show that the rectangular coordinates of the centroid of are given by xA = 1 3 β α [ρ (θ )]3 cos θ d θ , . yA = 1 3 β α c 42. The curve whose equation in rectangular coordinates is [ρ (θ )]3 sin θ d θ (x2 + y2 )2 = ax2 y, a > 0, where A is the area of is called a bifolium. (a) Show that the polar equation of this curve has the form r = a sin θ cos2 θ . (b) Use a graphing utility to draw the graph of the curves for a = 1, 2 and 4. (c) Let a = 2. Find the area inside one of the loops. In Exercises 34–36, use the result of Exercise 33 to find the rectangular coordinates of the centroid of the given region. 34. The region enclosed by r = 4 between θ = −α and θ = α , 0 < α < π/2. 35. The region enclosed by the cardioid r = 1 + cos θ . 9.6 CURVES GIVEN PARAMETRICALLY y (x (t), y (t)) So far we have specified curves by equations in rectangular coordinates or by equations in polar coordinates. Here we introduce a more general method. We begin with a pair of functions x = x(t ), y = y(t ) differentiable on the interior of an interval I . At the endpoints of I (if any) we require only continuity. For each number t in I we can interpret (x(t ), y(t )) as the point with x-coordinate x(t ) and y-coordinate y(t ). Then, as t ranges over I , the point (x(t ), y(t )) traces out a path in the xy-plane (Figure 9.6.1). We call such a path a parametrized curve and refer to t as the parameter. x Figure 9.6.1 Example 1 Identify the curve parametrized by the functions x(t ) = t + 1, y(t ) = 2t − 5, t ∈ (−∞, ∞). SOLUTION We can express y(t ) in terms of x(t ) : y(t ) = 2[x(t ) − 1] − 5 = 2x(t ) − 7. ...
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