Ask Me Help Desk - Calculus 3

The ball expands at $8 cm^3/s$ , thus put $t$ the time and
$Vol(t)$ the volume at time $t$ : you have that $Vol(t)=8t$ .

Using the volume formula $Vol=\frac{4}{3}\pi r^3$ where $r$ is the radius,
you have that at time $t$ the radius is

$r(t)=(\frac{3}{4\pi}Vol(t))^{1/3}=(\frac{3}{4\pi}8t)^{1/3}=(\frac{6}{\pi})^{1/3}t^{1/3}$

The rate of expansion of radius is given by the derivative

$r'(t)= \frac{1}{3}(\frac{6}{\pi})^{1/3} t^{-2/3}$

and to find which time the radius has length 2cm impose
$r(t)=(\frac{6}{\pi})^{1/3}t^{1/3}=2$
which leads to $t=(\frac{\pi}{6}) 2^{1/3}$ .

Thus the answer is $\frac{1}{3}(\frac{6}{\pi})^{1/3} ((\frac{\pi}{6}) 2^{1/3})^{-2/3}$
which simplifies to $\frac{1}{3}(\frac{6}{\pi})^{-1/3}2^{-2/9}$ .