Solutions

a) We have $A(t)=\pi r(t)^2$, so $A'(t)=2\pi r(t)r'(t)$. Therefore, the rate of change of the area when the radius is $6$ is $48\pi$.

b) Let $V(t)$ be the function representing the volume of the sphere at time $t$. Then

$$V(t)=\frac{4}{3}\pi r(t)^3$$

and

$$V'(t)=4\pi r(t)^2r'(t).$$

The rate of change of the volume when the radius is $6$ is therefore $426^2\pi$.

c) We're missing the rate of change for the radius $r$ from the first method. Isolating $r'(t)$ in $A'(t)$, we get

$$A'(t)=5=2\pi r(t)r'(t)=2\pi(3)r'(t)\implies r'(t)=\frac{5}{6\pi}.$$

Thus,

$$V'(t)=4\pi r(t)^2r'(t)=4\pi(3)^2\left(\frac{5}{6\pi}\right)=30.$$

With the second method, we have $A(t)=\pi r(t)^2$. This implies $r(t)=\sqrt{\frac{A(t)}{\pi}}$. Hence,

$$\begin{align*} V(t)&=\frac{4}{3}\pi r(t)^3 \\ &=\frac{4}{3}\pi\left(\frac{A(t)}{\pi}\right)^{3/2} \\ &=\frac{4}{3\sqrt{\pi}}\sqrt{A(t)^3} \end{align*}$$

and

$$\begin{align*} V'(t)&=\frac{4}{3\sqrt{\pi}}\left(\frac{1}{2\sqrt{A(t)^3}}\right)3A(t)^2A'(t) \\ &=\frac{4}{3\sqrt{\pi}}\cdot\frac{3}{2}\sqrt{A(t)}A'(t). \end{align*}$$

Consequently, when the rate of change of the area is $5$, the rate of change of the volume is

$$\begin{align*} V'(t)&=\frac{2}{\sqrt{\pi}}\sqrt{\pi(3)^2}\cdot 5 \\ &=30. \end{align*}$$