Polar Curve Review Answer Key

Version 001 – Polar Curve Review This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider the polar curve r = 2 sin(3θ) for 0 ≤ θ ≤ π. a) Find the area of region inside the curve. 1 1 4 1 2. correct 2 1 3. 5 1 4. 3 1. 5. 0 Explanation: 1. 4π 2. 3π x = 2 sin 3θ cos θ dx = −2 sin 3θ sin θ + 6 cos 3θ cos θ dθ 3. π correct 4. 5π y = 2 sin 3θ sin θ dy = 2 sin 3θ cos θ + 6 cos 3θ sin θ dθ 5. 2π Explanation: π dy dx At θ = , = −2 and = −4, so 4 dθ dθ dy −2 1 = = . dx −4 2 Alternate Solution: Z 1 π A= 4 sin2 3θ dθ 2 0 Z π = (1 − cos 6θ) dθ 0 =θ− 1 sin 6θ 6 π 0 =π Alternate Solution 1: Z π 3 3 A= 4 sin2 3θ dθ = · · · = π 2 0 Alternate Solution 2: Z π 6 6 A= 4 sin2 3θ dθ = · · · = π 2 0 002 (part 2 of 2) 10.0 points b) Find the slope of the curve at the point π where θ = . 4
2 x2 + y 2 = 6x2 y − 2y 3  
dy dy dy 2 2 2 x +y 2x + 2y = 6x2 − 6y 2 dx dx dx + 12xy π At θ = , x = 1 and y = 1, so 4   dy dy dy 4 2+2 =6 + 12 − 6 dx dx dx 8+8 dy dy 1 = 12 ⇒ = dx dx 2 003 10.0 points The length of the path described by the parametric equations x = cos3 t and y = sin3 t, for π 0 ≤ t ≤ , is given by 2 Z πp 2 1. 9 cos4 t sin2 t + 9 sin4 t cos2 t dt 0 correct Version 001 – Polar Curve Review 2. π 2 Z p 0 3. π 2 Z p 0 4. π 2 Z p 0 5. π 2 Z p 0 2 −3 cos2 t sin t + 3 sin2 t cos t dt 3 cos2 t + 3 sin2 t dt 9 cos4 t + 9 sin4 t dt π r = 2 cos θ and r = cos θ intersect at θ = . 2 Because of symmetry with respect to the xaxis, cos6 t + sin6 t dt Explanation: Basic Concept: Path length is s Z t2   2   2 dx dy L= + dt dt dt t1 A 1 = 2 2 A=3 Z π 2 0 Z π 2 Z π 2 cos2 θ dθ 0 cos2 θ dθ 0 005 Solution: x = cos3 t dx = 3 cos2 t (− sin t) dt 1 4 cos θ dθ − 2 2 10.0 points The shaded region in y = sin3 t dy = 3 sin2 t (cos t) dt L= π 2 Z 0 p 9 cos4 t sin2 t + 9 sin4 t cos2 t dt 004 10.0 points Which of the following is equal to the area of the region inside the polar curve r = 2 cos θ and outside the polar curve r = cos θ? 3 1. 2 2. 3 Z π 2 cos2 θ dθ 0 Z π 2 2 cos θ dθ correct 0 3. 3 4. 3 Z π 2 cos θ dθ Z0 π 5. 3 0 π/2 1 2. I = 2 Z 3. I = Z 1 4. I = 2 Z π 1 2 Z π 5 sin2 θ dθ 0 π sin θ dθ 0 cos2 θ dθ 0 Z lies inside the polar curve r = 3 sin θ and outside the polar curve r = 2 sin θ. Which of the following integrals gives the area of this region? Z π 1. I = 5 sin2 θ dθ π cos θ dθ 0 Explanation: 5. I = 5 sin2 θ dθ correct 0 0 sin θ dθ Version 001 – Polar Curve Review 1 6. I = 2 Z 3 On the other hand, the graph of π/2 sin θ dθ r = 4 + sin θ 0 Explanation: As the graphs show, the polar curves intersect when is the cardioid similar to the one shown in 3 sin θ = 2 sin θ , i.e. at θ = 0, π. Thus the area of the shaded region is given by Z o 1 πn 2 2 I = (3 sin θ) −(2 sin θ) dθ . 2 0 Consequently, 1 I = 2 Z π 5 sin2 θ dθ . 0 so in this case we can take θ0 = 0 and θ1 = 2π. Thus the area of the region enclosed by the graph is given by the integral keywords: area, polar cooordinates, definite integral, circle, 006 10.0 points 1 A = 2 Z 2π (4 + sin θ)2 dθ . 0 Now (4 + sin θ)2 = 16 + 2 sin θ + sin2 θ Find the area of the region enclosed by the graph of the polar function r = 4 + sin θ . 1. area = 35 π 2 2. area = 18π 3. area = 33 π correct 2 4. area = 17π = 33 1 + 2 sin θ − cos 2θ , 2 2 since sin2 θ = But then, 1 A = 2 Z 2π  1 1 − cos 2θ . 2   33 1 + 2 sin θ − cos 2θ dθ 2 2 0 i2π 1 h 33 1 = θ − 2 cos θ − sin 2θ . 2 2 4 0 Consequently, 5. area = 35π Explanation: The area of a region bounded by the graph of the polar function r = f (θ) and the rays θ = θ0 , θ1 is given by the integral Z 1 θ1 A = f (θ)2 dθ . 2 θ0 area = A = 33 π . 2 keywords: polar graph, area, cardioid Version 001 – Polar Curve Review 007 10.0 points Thus Find the area of one loop of the graph of the polar function r = 3 cos 4θ . 1. area = 4 1 A = 2 Z 21 π 32 3. area = 19 π 32 4. area = 11 π 16 5. area = 5 π 8 (3 cos 4θ)2 dθ −π/8 9 = 4 9 π correct 16 2. area = π/8 Z π/8 ( 1 + cos 8θ) dθ −π/8 = iπ/8 9h 1 θ + sin 8θ . 4 8 −π/8 Consequently, area = A = 9 π . 16 keywords: polar graph, area, cardioid, Explanation: The area of a region bounded by the graph of the polar function r = f (θ) and the rays θ = θ0 , θ1 is given by the integral Z 1 θ1 A = f (θ)2 dθ . 2 θ0 008 10.0 points Find the area of the shaded region in Now the graph of r = 3 cos 4θ is the ‘rose’ specified by the graphs of the circles r = cos θ , 1 4 1 2. area = π 4 1 3. area = π 2  1π 4. area = −1 4 2 1. area = so the values of θ0 and θ1 needed to find the area of one loop will be consecutive solutions of cos 4θ = 0; for example, we can use π θ0 = − , 8 θ1 = π . 8 r = sin θ . 5 Version 001 – Polar Curve Review  1π 5. area = +1 2 4  1π 6. area = + 1 correct 4 2 Explanation: The area of a region bounded by the graph of polar function r = f (θ) between the rays θ = θ0 , θ1 is given by the integral Z 1 θ1 A = f (θ)2 dθ . 2 θ0 Consequently, the shaded region has  1π +1 . 4 2 area = keywords: definite integral, area between curves, polar area, circle, 009 10.0 points Now the graph of r = cos θ is a circle centered on the x-axis, while that of r = sin θ is a circle centered on the y-axis; in addition, both pass through the origin and have the same radius. So the circles intersect also at θ = π/4, as the figure shows. Find the area of the region lying outside the polar curve The area of the shaded region can thus be computed as the difference Z 1 π I1 − I2 = sin2 θ dθ 2 π/4 and inside the polar curve r = 6 sin θ. 1 − 2 Z π/2 cos2 θ dθ r = 2(1 + sin θ) 1. area = 2π 2. area = 4π correct π/4 of the area of the respective shaded regions in 3. area = π 4. area = 1 π 2 5. area = 3π Explanation: The area of a region bounded by the graph of the polar function r = f (θ) and the rays θ = θ0 , θ1 is given by the integral Now π 1 I1 = (1 − cos 2θ) dθ 4 π/4 iπ 1h 1 1  3π 1  = θ − sin 2θ = + , 4 2 4 4 2 π/4 while Z 1 π/2 I2 = (1 + cos 2θ) dθ 4 π/4 Z = iπ/2 1h 1 1π 1 θ + sin 2θ = − . 4 2 4 4 2 π/4 1 A = 2 Z θ1 f (θ)2 dθ . θ0 the graph of r = 2(1 + sin θ) is a cardioid, while the graph of r = 6 sin θ 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 001 – Polar Curve Review The values of θ0 , θ1 will be the solutions of 2(1 + sin θ) = 6 sin θ , i.e., θ0 = π/6 , θ1 = 5π/6, as the graph confirms. Thus the area of the shaded region is given by 1 A = 2 Z 5π/6 π/6 n o (6 sin θ)2 − 4(1 + sin θ)2 dθ . Now (6 sin θ)2 − 4(1 + sin θ)2 = 4(9 sin2 θ − (1 + 2 sin θ + sin2 θ)) = 4(8 sin2 θ − 2 sin θ − 1) = 4(3 − 2 sin θ − 4 cos 2θ) . But then A = 2 Z 5π/6 π/6 n o 3 − 2 sin θ − 4 cos 2θ dθ h = 2 3θ + 2 cos θ − 2 sin 2θ i5π/6 π/6 . Consequently, area = A = 4π . keywords: area, polar curves, polar graphs, circle, cardioid 6