AP Calculus BC 9.7 Polar Derivatives Review

[Maximum number: 3]

Curve C is defined by the polar equation $r(\theta)=2 \sin ^{2} \theta$ for $0 \leq \theta \leq \pi$ . Curve C and the semicircle $r=\frac{1}{2}$ for $0 \leq \theta \leq \pi$ are shown in the x y-plane.

(Note: Your calculator should be in radian mode.)

(a)

Find the rate of change of r with respect to $\theta$ at the point on curve C where $\theta=1.3$ . Show the setup for your calculations.

[ 1 ]

} \\ \hline &

$\left.\frac{d r}{d \theta}\right\|_{\theta=1.3}=1.031003$
The rate of change of r with respect to $\theta$ at the point on curve C where $\theta=1.3$ is 1.031.

& Answer with setup Point 1 (P1) \\ \hline \end{tabular}

Scoring Notes for Part A

- An exact answer of $\left.\frac{d r}{d \theta}\right|_{\theta=1.3}=4 \sin (1.3) \cos (1.3)$ earns P1.

- To earn P1, a response must indicate differentiation of r and provide the correct answer. A reported

answer should be accurate to three places after the decimal point, rounded or truncated. An

inappropriately rounded answer does not earn the point.

○ Examples of responses with correct communication include $\left.\frac{d r}{d \theta}\right|_{\theta=1.3}=1.031$ and

$r^{\prime}(1.3)=1.031$ .

○ Responses with incorrect communication, such as $r^{\prime}(\theta)=1.031, r^{\prime}=1.031$ , or $\frac{d r}{d \theta}=1.031$ ,

are sufficient to earn P1.

(b)

A particle travels along curve C so that $\frac{d \theta}{d t}=15$ for all times t. Find the rate at which the particle's distance from the origin changes with respect to time when the particle is at the point where $\theta=1.3$ . Show the setup for your calculations.

[ 2 ]

D A particle travels along curve C so that $\frac{d \theta}{d t}=15$ for all times t. Find the rate at which the particle's

distance from the origin changes with respect to time when the particle is at the point where $\theta=1.3$ .

Show the setup for your calculations.

$\left.\frac{d r}{d t}\right\|_{\theta=1.3}=\left.\left(\frac{d r}{d \theta} \cdot \frac{d \theta}{d t}\right)\right\|_{\theta=1.3}=1.031003 \cdot 15=15.465041$

$\frac{d r}{d \theta}$	$\frac{d \theta}{d t}$

Point 8 (P8)

Answer

Point 9 (P9)

The particle's distance from the origin changes at a rate of

15.465.

Scoring Notes for

AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form Questions

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)