Home / AP Calculus BC: 9.9 Finding the Area of the Region Bounded by Two Polar Curves – Exam Style questions with Answer- FRQ

AP Calculus BC: 9.9 Finding the Area of the Region Bounded by Two Polar Curves – Exam Style questions with Answer- FRQ

Question

The graphs of the polar curves r = 4 and r =  3 + 2 cos θ are shown in the figure above. The curves intersect at \(\)  and \(\).

(a) Let R be the shaded region that is inside the graph of r = 4 and also outside the graph of r = 3 + 2 cos θ, as shown in the figure above. Write an expression involving an integral for the area of R.
(b) Find the slope of the line tangent to the graph of r = 3 + 2 cos θ at \(\theta =\frac{\pi }{2}.\).
(c) A particle moves along the portion of the curve r = 3 + 2 cos θ for \(0<\theta <\frac{\pi }{2}.\) The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 3 units per second. Find the rate at which the angle θ changes with respect to time at the instant when the position of the particle corresponds to \(\theta =\frac{\pi }{3}.\) Indicate units of measure. 

Answer/Explanation

Ans:

(a) \(Area = \frac{1}{2}\int_{\pi /3}^{5\pi /3}\left ( 4^{2}-(3+2cos\theta )^{2} \right )d\theta \)

(b) \(\frac{dr}{d\theta }=-sin\theta \Rightarrow \frac{dr}{d\theta }|_{\theta =\pi /2}=-2\)

\(r\left ( \frac{\pi }{2} \right )=3+2cos\left ( \frac{\pi }{2} \right )=3\)

\(y = r sin\theta \Rightarrow \frac{dy}{d\theta }=\frac{dr}{d\theta }sin\theta +rcos\theta \)

\(x = r cos\theta \Rightarrow \frac{dy}{d\theta }=\frac{dr}{d\theta }cos\theta +rsin\theta \)

\(\frac{dy}{dx}|_{\theta =\pi /2}=\frac{dy/d\theta }{dx/d\theta }|_{\theta =\pi /2}=\frac{-2sin\left ( \frac{\pi }{2} \right )+3cos\left ( \frac{\pi }{2} \right )}{-2cos\left ( \frac{\pi }{2} \right )-3sin\left ( \frac{\pi }{2} \right )}=\frac{2}{3}\)

The slope of the line tangent to the graph of r = 3 + 2cos θ at \(\theta =\frac{\pi }{2}is \frac{2}{3}.\)

– OR – 

\(y = r sin \theta =(3+2cos\theta )sin\theta \Rightarrow \frac{dy}{d\theta }=3 cos\theta +2cos^{2}\theta -2sin^{2}\theta \)

\(x = r cos \theta =(3+2cos\theta )cos\theta \Rightarrow \frac{dx}{d\theta }=-3 sin\theta -4sin\theta cos\theta \)

\(\frac{dy}{dx}|_{\theta =\pi /2}=\frac{dy/d\theta }{dx/d\theta }|_{\theta =\pi /2}=\frac{3cos\left ( \frac{\pi }{2} \right )+2cos^{2}\left ( \frac{\pi }{2} \right )-2sin^{2}\left ( \frac{\pi }{2} \right )}{-3sin\left ( \frac{\pi }{2} \right )-4sin\left ( \frac{\pi }{2} \right )cos\left ( \frac{\pi }{2} \right )}=\frac{2}{3}\)

The slope of the line tangent to the graph of r = 3 + 2cos θ at \(\theta =\frac{\pi }{2}\) is \(\frac{2}{3}.\)

(c) \(\frac{dr}{dt}=\frac{dr}{d\theta }\cdot \frac{d\theta }{dt}=-2sin\theta \cdot \frac{d\theta }{dt}\Rightarrow \frac{dr}{dt}\cdot \frac{1}{-2sin\theta }\)

\(\frac{d\theta }{dt}|_{\theta =\pi /3}=3\cdot \frac{1}{-2sin\left ( \frac{\pi }{3} \right )}=\frac{3}{-\sqrt{3}}=-\sqrt{-3}\) radians per second 

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