AP Calculus BC Fun 4 A Justify Conclusions About the Behavior of a Function Based on the Behavior of Its Derivatives Topic 5 4 Questions

C It can be shown that $\frac{d x}{d \theta}=4 \sin \theta \cos ^{2} \theta-2 \sin ^{3} \theta$ for curve C. For $0 \leq \theta \leq \frac{\pi}{2}$, find the value of $\theta$ that

corresponds to the point on curve C that is farthest from the y-axis. Justify your answer.

Scoring Notes for Part C

- P5 is earned for considering $\frac{d x}{d \theta}=0. \mathbf{P 5}$ is not earned by just presenting $\theta=0.955317$.

A response that discusses the sign of $\frac{d x}{d \theta}$ changing or uses the phrase "critical points of $x(\theta)$ " also

earns P5.

- The value $\theta=0.955317$ might be presented as $\arccos \left(\frac{1}{\sqrt{3}}\right), \arcsin \left(\sqrt{\frac{2}{3}}\right)$, or $\arctan (\sqrt{2})$.

- To earn P6 using a candidates test, a response must make a global argument by correctly evaluating

$x(\theta)$ at $\theta=0, \theta=0.955317$, and $\theta=\frac{\pi}{2}$. The evaluations must be correct to the first digit after

the decimal, rounded or truncated.

- Alternate justifications:

○ $\frac{d x}{d \theta}>0$ for $0<\theta<0.955$, and $\frac{d x}{d \theta}<0$ for $0.955<\theta<\frac{\pi}{2}$. Therefore, $\theta=0.955$ is the

location of the absolute maximum for $x(\theta)$ on the interval $0 \leq \theta \leq \frac{\pi}{2}$.

○ Because $\frac{d x}{d \theta}$ changes sign from positive to negative at $\theta=0.955$ (this might be presented as

" $\frac{d x}{d \theta}>0$ for $\theta<0.955$, and $\frac{d x}{d \theta}<0$ for $\theta>0.955$ "), it is the location of a relative maximum

for $x(\theta)$. And because $\theta=0.955$ is the only critical point of $x(\theta)$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$,

it is the location of the absolute maximum for $x(\theta)$ on the interval.

○ Because $\left.\frac{d x}{d \theta}\right|_{\theta=0.955}=0$ and $\left.\frac{d^{2} x}{d \theta^{2}}\right|_{\theta=0.955}<0, \theta=0.955$ is the location of a relative

maximum for $x(\theta)$. And because $\theta=0.955$ is the only critical point of $x(\theta)$ in the interval

$0 \leq \theta \leq \frac{\pi}{2}$, it is the location of the absolute maximum for $x(\theta)$ on the interval.

- A response that presents only a local argument (such as a First Derivative Test or a Second

Derivative Test) or an incorrect global argument does not earn P6 but is eligible for P7 with 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, unless an earlier

point was not earned due to inappropriate rounding.