Air is pumped into a spherical ball which expands at a rate of 8 cm cubed per second (cm^3/s). Find the exact rate of increase of the radius of the ball when the radius is 2cm.

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} .

For the last part there was some mistake in the previous post, it is

t=\frac{4\pi}{3}

and leads to \frac{1}{3}(\frac{6}{\pi})^{1/3} (\frac{3}{4\pi})^{2/3}
=\frac{1}{2\pi} .