AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form - Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Detailed solution
Parametric form: \(x(\theta)=r\cos\theta=\theta^2\cos\theta\), \(y(\theta)=r\sin\theta=\theta^2\sin\theta\).
Differentiate: \(x'(\theta)=2\theta\cos\theta-\theta^2\sin\theta\), \(y'(\theta)=2\theta\sin\theta+\theta^2\cos\theta\).
Evaluate at \( \theta=\pi \): \(x'(\pi)=-2\pi\), \(y'(\pi)=-\pi^2\).
Slope \( \dfrac{dy}{dx}=\dfrac{y’}{x’}=\dfrac{-\pi^2}{-2\pi}=\dfrac{\pi}{2}\).
✅ Correct Answer: (B) \( \dfrac{\pi}{2} \)
No-Calc Question
Let \(f\) be differentiable with \(f(\pi)=2\) and \(f'(\pi)=-3\). What is the slope of the line tangent to the polar curve \(r=f(\theta)\) at \(\theta=\pi\)?
(A) \(-3\)
(B) \(-2\)
(C) \(-\dfrac{3}{2}\)
(D) \(-\dfrac{2}{3}\)
(B) \(-2\)
(C) \(-\dfrac{3}{2}\)
(D) \(-\dfrac{2}{3}\)
▶️ Answer/Explanation
Detailed solution
For polar \(x=r\cos\theta,\ y=r\sin\theta\): \(\displaystyle \dfrac{dy}{dx}=\dfrac{r’ \sin\theta + r \cos\theta}{r’ \cos\theta – r \sin\theta}\).
At \(\theta=\pi\): \(r=2,\ r’=-3,\ \sin\pi=0,\ \cos\pi=-1\). Then \(\displaystyle \dfrac{dy}{dx}=\dfrac{(-3) x 0 + 2 x (-1)}{(-3) x (-1) – 2 x 0}=\dfrac{-2}{3}.\)
✅ Answer: (D)