# At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

Mar 25, 2015

We know the volume of a sphere relative to its radius is given by:
$V \left(r\right) = \frac{4}{3} \pi {r}^{3}$

We are given that
Surface area at the time in question is $4 {m}^{2}$
which implies since $S = 4 \pi {r}^{2}$ that the radius at that time is
$r = \frac{1}{\sqrt{\pi}}$

We are asked for $\frac{\mathrm{dV}}{\mathrm{dt}}$

$\frac{d V}{\mathrm{dt}} = \frac{d V}{\mathrm{dr}} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$

$\frac{d V}{\mathrm{dr}} = 4 \pi {r}^{2}$ (using the derivative ofour formula for the Volume)
and we are told that $\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{6} \frac{m}{\sec}$

So at the time indicated
$\frac{d V}{\mathrm{dt}} = 4 \pi {r}^{2} \cdot \frac{1}{6}$

and with $r = \frac{1}{\sqrt{\pi}}$

$\frac{d V}{\mathrm{dt}} = \frac{2}{3} \frac{m}{\sec}$