Version 1 Polar Curve Review rittenhouse (RittBCblock 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. BC 1993 FR 4 1 (part 1 of 1. points Consider the polar curve r = sin(3θ for θ π. a Find the area of region inside the curve. 1. 4π. 3π 3. π correct 4. 5π 5. π 1. 1 4. 1 correct 3. 1 5 4. 1 3 5. x = sin3θcosθ dx = sin3θsinθ +6cos3θcosθ dθ y = sin3θsinθ = sin3θcosθ +6cos3θsinθ dθ At θ = π 4, dx = and = 4, so dθ dθ dx = 4 = 1. Alternate Solution: A = 1 = 4sin 3θ dθ (1 cos6θ dθ = θ 1 6 sin6θ π Alternate Solution 1: A = 3 3 Alternate Solution : A = 6 6 = π 4sin 3θ dθ = = π 4sin 3θ dθ = = π (part of 1. points b Find the slope of the curve at the point where θ = π 4. ( x +y = 6x y y 3 ( x +y ( x+y = 6x dx dx 6y dx +1xy At θ = π, x = 1 and y = 1, so 4 ( 4 + = 6 dx dx +1 6 dx 8+8 dx = 1 dx = 1 BC 1997 MCnc 15 3 1. points The length of the path described by the parametric equations x = cos 3 t and y = sin 3 t, for t π, is given by 1. 9cos 4 tsin t+9sin 4 tcos t dt correct
Version 1 Polar Curve Review rittenhouse (RittBCblock. 3. 4. 5. 3cos tsint+3sin tcost dt 3cos t+3sin t dt 9cos 4 t+9sin 4 t dt cos 6 t+sin 6 t dt Basic Concept: Path length is t (dx L = + dt Solution: t 1 ( dt dt x = cos 3 t dx dt = 3cos t ( sint r = cosθ andr = cosθ intersect atθ = π. Because of symmetry with respect to the x- axis, A = 1 A = 3 4cos θ dθ 1 cos θ dθ CalC11d1b 5 1. points The shaded region in cos θ dθ L = y = sin 3 t dt = 3sin t (cost 9cos 4 tsin t+9sin 4 tcos t dt BC 1997 MCnc 1 4 1. points Which of the following is equal to the area of the region inside the polar curve r = cosθ and outside the polar curve r = cosθ? 1. 3. 3 3. 3 4. 3 5. 3 cos θ dθ cos θ dθ correct cosθ dθ cos θ dθ cosθ dθ lies inside the polar curve r = 3sinθ and outside the polar curve r = sinθ. Which of the following integrals gives the area of this region? 1. I =. I = 1 3. I = 4. I = 1 5. I = 1 / 5sin θdθ sinθdθ 5sin θdθ 5sin θdθ correct sinθdθ
Version 1 Polar Curve Review rittenhouse (RittBCblock 3 6. I = 1 / sinθdθ As the graphs show, the polar curves intersect when 3sinθ = sinθ, i.e. at θ =, π. Thus the area of the shaded region is given by I = 1 { (3sinθ (sinθ } dθ. I = 1 5sin θdθ. keywords: area, polar cooordinates, definite integral, circle, CalC11d6a 6 1. points Find the area of the region enclosed by the graph of the polar function 1. area = 35 π. area = 18π r = 4+sinθ. 3. area = 33 π correct 4. area = 17π 5. area = 35π of the polar function r = f(θ and the rays θ = θ, θ 1 is given by the integral A = 1 θ f(θ dθ. On the other hand, the graph of r = 4+sinθ is the cardioid similar to the one shown in sointhiscasewecantakeθ = andθ 1 = π. Thus the area of the region enclosed by the graph is given by the integral Now since A = 1 π (4+sinθ dθ. (4+sinθ = 16+sinθ +sin θ But then, A = 1 = 33 +sinθ 1 cosθ, sin θ = 1 π ( 1 cosθ. ( 33 +sinθ 1 cosθ dθ = 1 [ 33 θ cosθ 1 ] π 4 sinθ. area = A = 33 π. keywords: polar graph, area, cardioid CalC11d7a
Version 1 Polar Curve Review rittenhouse (RittBCblock 4 7 1. points Thus Find the area of one loop of the graph of the polar function r = 3cos4θ. 1. area = 9 16 π correct. area = 1 3 π 3. area = 19 3 π 4. area = 11 16 π 5. area = 5 8 π of the polar function r = f(θ and the rays θ = θ, θ 1 is given by the integral A = 1 = 9 4 /8 π/8 /8 π/8 = 9 4 (3cos4θ dθ (1+cos8θdθ [θ + 1 8 sin8θ ] π/8 π/8. area = A = 9 16 π. keywords: polar graph, area, cardioid, CalC11d11a 8 1. points Find the area of the shaded region in A = 1 θ f(θ dθ. Now the graph of is the rose r = 3cos4θ specified by the graphs of the circles r = cosθ, r = sinθ. 1. area = 1 4 so the values of θ and θ 1 needed to find the area of one loop will be consecutive solutions of cos4θ = ; for example, we can use θ = π 8, θ 1 = π 8.. area = 1 4 π 3. area = 1 π 4. area = 1 ( π 4 1
Version 1 Polar Curve Review rittenhouse (RittBCblock 5 5. area = 1 ( π 4 +1 6. area = 1 ( π 4 +1 correct of polar function r = f(θ between the rays θ = θ, θ 1 is given by the integral A = 1 θ f(θ dθ. Now the graph of r = cosθ is a circle centeredonthex-axis,whilethatofr = sinθ isa circle centered on the y-axis; in addition, both pass throughthe originandhave the same radius. So the circles intersect also at θ =, as the figure shows. The area of the shaded region can thus be computed as the difference I 1 I = 1 1 sin θdθ / cos θdθ of the area of the respective shaded regions in the shaded region has area = 1 4 ( π +1 keywords: definite integral, area between curves, polar area, circle, CalC11d16a 9 1. points Findtheareaoftheregionlyingoutsidethe polar curve r = (1+sinθ and inside the polar curve r = 6sinθ. 1. area = π. area = 4π correct 3. area = π. 4. area = 1 π 5. area = 3π of the polar function r = f(θ and the rays θ = θ, θ 1 is given by the integral Now I 1 = 1 4 = 1 4 (1 cosθdθ [ θ 1 ] π sinθ = 1 ( 3π 4 4 + 1, A = 1 the graph of θ f(θ dθ. while I = 1 4 = 1 4 / [θ+ 1 sinθ ] π/ (1+cosθdθ = 1 4( π 4 1. r = (1+sinθ is a cardioid, while the graph of r = 6sinθ is a circle passing through the origin, having center on the y-axis and diameter 6, so the region between them is the shaded crescent in
Version 1 Polar Curve Review rittenhouse (RittBCblock 6 The values of θ, θ 1 will be the solutions of (1+sinθ = 6sinθ, i.e., θ = π/6, θ 1 = 5π/6, as the graph confirms. Thus the area of the shaded region is given by A = 1 Now 5π/6 π/6 (6sinθ 4(1+sinθ But then A = { (6sinθ 4(1+sinθ } dθ. = 4(9sin θ (1+sinθ+sin θ = 4(8sin θ sinθ 1 5π/6 π/6 = 4(3 sinθ 4cosθ. { } 3 sinθ 4cosθ dθ [ ] 5π/6 = 3θ +cosθ sinθ. π/6 area = A = 4π. keywords: area, polar curves, polar graphs, circle, cardioid