find the average rate of change for the function r = 3 cos θ on the interval π/3, π/2. then estimate the…

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. For the function $r = 3\cos\theta$ on the interval $[\frac{\pi}{3},\frac{\pi}{2}]$, we have $a=\frac{\pi}{3}$, $b = \frac{\pi}{2}$, $f(\theta)=3\cos\theta$.

Step2: Calculate $f(\frac{\pi}{3})$ and $f(\frac{\pi}{2})$

$f(\frac{\pi}{3})=3\cos\frac{\pi}{3}=3\times\frac{1}{2}=\frac{3}{2}$, $f(\frac{\pi}{2})=3\cos\frac{\pi}{2}=0$.

Step3: Compute the average rate of change

The average rate of change is $\frac{f(\frac{\pi}{2})-f(\frac{\pi}{3})}{\frac{\pi}{2}-\frac{\pi}{3}}=\frac{0 - \frac{3}{2}}{\frac{\pi}{2}-\frac{\pi}{3}}=\frac{-\frac{3}{2}}{\frac{3\pi - 2\pi}{6}}=\frac{-\frac{3}{2}}{\frac{\pi}{6}}=-\frac{9}{\pi}$.

Step4: Find the halfway - point of the interval

The halfway - point of the interval $[\frac{\pi}{3},\frac{\pi}{2}]$ is $\theta=\frac{\frac{\pi}{3}+\frac{\pi}{2}}{2}=\frac{\frac{2\pi + 3\pi}{6}}{2}=\frac{5\pi}{12}$.

Step5: Estimate $f(\frac{5\pi}{12})$

We know that $\cos(A + B)=\cos A\cos B-\sin A\sin B$, and $\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$. So $\cos\frac{5\pi}{12}=\cos(\frac{\pi}{4}+\frac{\pi}{6})=\cos\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{4}\sin\frac{\pi}{6}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$. Then $f(\frac{5\pi}{12})=3\cos\frac{5\pi}{12}=\frac{3(\sqrt{6}-\sqrt{2})}{4}$.

Answer:

Average rate of change: $-\frac{9}{\pi}$; $f(\frac{5\pi}{12})=\frac{3(\sqrt{6}-\sqrt{2})}{4}$