Home / AP Calculus BC: 9.1 Defining and Differentiating Parametric Equations – Exam Style questions with Answer- FRQ

AP Calculus BC: 9.1 Defining and Differentiating Parametric Equations – Exam Style questions with Answer- FRQ

Question

(a) Topic-9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

(b) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

(c) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

(d) Topic-9.1 Defining and Differentiating Parametric Equations

Let $S$ be the region bounded by the graph of the polar curve $r(\theta) = 3\sqrt{\theta}\sin(\theta^2)$ for $0 \leq \theta \leq \sqrt{\pi}$, as shown in the figure above.

(a) Find the area of $S$.

(b) What is the average distance from the origin to a point on the polar curve $r(\theta) = 3\sqrt{\theta}\sin(\theta^2)$ for $0 \leq \theta \leq \sqrt{\pi}$?

(c) There is a line through the origin with positive slope $m$ that divides the region $S$ into two regions with equal areas. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of $m$.

(d) For $k > 0$, let $A(k)$ be the area of the portion of region $S$ that is also inside the circle $r = k \cos \theta$. Find
\(\lim_{k \to \infty}\) A(k)

▶️Answer/Explanation

\(\textbf{2(a)}\)
\(
r(\theta) = 3\sqrt{\theta} \sin(\theta^2) \quad \text{for } 0 \leq \theta \leq \sqrt{\pi}.
\)
\(
\text{Area} = \frac{1}{2} \int_{0}^{\sqrt{\pi}} \left(r(\theta)\right)^2 \, d\theta \approx 3.534291735.
\)
\(\textbf{2(b)}\)
The distance from the origin to a point on the curve is $r$.
\(
r_{\text{avg}} = \frac{1}{\sqrt{\pi} – 0} \int_{0}^{\sqrt{\pi}} r(\theta) \, d\theta \approx 1.57993277.
\)
\(\textbf{2(c)}\)
Since the slope of the line is $m$, then $m = \tan \theta$. So $\theta = \tan^{-1} m$ is where the line intersects $r(\theta)$.
So, the area from $0$ to $\tan^{-1} m$ is equal to the area from $\tan^{-1} m$ to $\sqrt{\pi}$:
\(
\frac{1}{2} \int_{0}^{\tan^{-1} m} \left(r(\theta)\right)^2 \, d\theta = \frac{1}{2} \int_{\tan^{-1} m}^{\sqrt{\pi}} \left(r(\theta)\right)^2 \, d\theta.
\)
\(\textbf{2(d)}\)
As $k \to \infty$, $r = k \cos \theta$ is a circle that gets bigger, and $\theta \to \frac{\pi}{2}$.
\(
\lim_{k \to \infty} A(k) = \frac{1}{2} \int_{0}^{\pi/2} \left(r(\theta)\right)^2 \, d\theta \approx 3.324470722.
\)

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