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havj0y · Dec 13, 2011, 06:26 PM

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.

Aurora2000 · Dec 14, 2011, 02:40 AM

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 $\frac{1}{3}(\frac{6}{\pi})^{-1/3}2^{-2/9}$ .

Aurora2000 · Dec 14, 2011, 02:47 AM

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

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

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