AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form - FRQs - Exam Style Questions
No-Calc Question
The graphs of the polar curves \(r=4\) and \(r=3+2\cos\theta\) are shown in the figure above. The curves intersect at \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\).
(a) Let \(R\) be the shaded region that is inside the graph of \(r=4\) and also outside the graph of \(r=3+2\cos\theta\), 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 \(r=3+2\cos\theta\) at \(\theta=\dfrac{\pi}{2}\).
(c) A particle moves along the portion of the curve \(r=3+2\cos\theta\) for \(0<\theta<\dfrac{\pi}{2}\). 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 \(\theta\) changes with respect to time at the instant when the position of the particle corresponds to \(\theta=\dfrac{\pi}{3}\). Indicate units of measure.
Most-appropriate topic codes (CED):
• TOPIC 9.8 (BC): Area of a Polar Region — part (a)
• TOPIC 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• TOPIC 4.5:Solving Related Rates Problems (Chain Rule) — part (c)
• TOPIC 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• TOPIC 4.5:Solving Related Rates Problems (Chain Rule) — part (c)
▶️ Answer/Explanation
(a) Area of \(R\)
Inside the circle \(r=4\) and outside \(r=3+2\cos\theta\) between the intersection angles \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\):
\[ \boxed{\text{Area}(R)=\frac12\int_{\pi/3}^{5\pi/3}\!\Big(4^{2}-\big(3+2\cos\theta\big)^{2}\Big)\,d\theta }. \]
(b) Slope \(\dfrac{dy}{dx}\) at \(\theta=\dfrac{\pi}{2}\)
For polar \(x=r\cos\theta,\ y=r\sin\theta\): \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} =\frac{r’\sin\theta+r\cos\theta}{r’\cos\theta-r\sin\theta}\), with \(r’=dr/d\theta\).
Here \(r=3+2\cos\theta\Rightarrow r’=-2\sin\theta\). At \(\theta=\dfrac{\pi}{2}\): \(\sin=\!1,\ \cos=\!0,\ r=3\).
Numerator \(=r’\sin\theta+r\cos\theta=(-2)(1)+3(0)=-2\).
Denominator \(=r’\cos\theta-r\sin\theta=(-2)(0)-3(1)=-3\).
\[ \boxed{\left.\frac{dy}{dx}\right|_{\theta=\pi/2}=\frac{-2}{-3}=\frac{2}{3}}. \]
(c) Angular rate \(\dfrac{d\theta}{dt}\) at \(\theta=\dfrac{\pi}{3}\)
Given the radial distance increases at \(3\) units/s: \(\displaystyle \frac{dr}{dt}=3\). Chain rule: \(\displaystyle \frac{dr}{dt}=\frac{dr}{d\theta}\cdot\frac{d\theta}{dt}\). Since \(r’=dr/d\theta=-2\sin\theta\), \[ \frac{d\theta}{dt}=\frac{\frac{dr}{dt}}{\frac{dr}{d\theta}} =\frac{3}{-2\sin\theta}. \] Evaluate at \(\theta=\dfrac{\pi}{3}\) (\(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)): \[ \boxed{\frac{d\theta}{dt}=\frac{3}{-2\cdot(\sqrt3/2)}=-\frac{3}{\sqrt3}=-\sqrt3\ \text{radians/second}}. \] (Negative sign indicates \(\theta\) is decreasing at that instant.)
Inside the circle \(r=4\) and outside \(r=3+2\cos\theta\) between the intersection angles \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\):
\[ \boxed{\text{Area}(R)=\frac12\int_{\pi/3}^{5\pi/3}\!\Big(4^{2}-\big(3+2\cos\theta\big)^{2}\Big)\,d\theta }. \]
(b) Slope \(\dfrac{dy}{dx}\) at \(\theta=\dfrac{\pi}{2}\)
For polar \(x=r\cos\theta,\ y=r\sin\theta\): \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} =\frac{r’\sin\theta+r\cos\theta}{r’\cos\theta-r\sin\theta}\), with \(r’=dr/d\theta\).
Here \(r=3+2\cos\theta\Rightarrow r’=-2\sin\theta\). At \(\theta=\dfrac{\pi}{2}\): \(\sin=\!1,\ \cos=\!0,\ r=3\).
Numerator \(=r’\sin\theta+r\cos\theta=(-2)(1)+3(0)=-2\).
Denominator \(=r’\cos\theta-r\sin\theta=(-2)(0)-3(1)=-3\).
\[ \boxed{\left.\frac{dy}{dx}\right|_{\theta=\pi/2}=\frac{-2}{-3}=\frac{2}{3}}. \]
(c) Angular rate \(\dfrac{d\theta}{dt}\) at \(\theta=\dfrac{\pi}{3}\)
Given the radial distance increases at \(3\) units/s: \(\displaystyle \frac{dr}{dt}=3\). Chain rule: \(\displaystyle \frac{dr}{dt}=\frac{dr}{d\theta}\cdot\frac{d\theta}{dt}\). Since \(r’=dr/d\theta=-2\sin\theta\), \[ \frac{d\theta}{dt}=\frac{\frac{dr}{dt}}{\frac{dr}{d\theta}} =\frac{3}{-2\sin\theta}. \] Evaluate at \(\theta=\dfrac{\pi}{3}\) (\(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)): \[ \boxed{\frac{d\theta}{dt}=\frac{3}{-2\cdot(\sqrt3/2)}=-\frac{3}{\sqrt3}=-\sqrt3\ \text{radians/second}}. \] (Negative sign indicates \(\theta\) is decreasing at that instant.)