Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.

Let V be the volume of the sphere. Then,

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

$\Rightarrow \frac{d V}{d r}=4 \pi r^{2}$

Let $S$ be the total surface area of sphere. Then,

$S=4 \pi r^{2}$

$\Rightarrow \frac{d S}{d r}=8 \pi r$

$\therefore \frac{d V}{d S}=\frac{d V}{d r} / \frac{d S}{d r}$

$\Rightarrow \frac{d V}{d S}=\frac{4 \pi r^{2}}{8 \pi r}=\frac{r}{2}$

$\Rightarrow\left(\frac{d V}{d S}\right)_{r=2}=\frac{2}{2}$

$=1 \mathrm{~cm}$