Question:
The volume of metal in a hollow sphere is constant. If the inner radius is increasing at the rate of 1 cm/sec, find the rate of increase of the outer radius when the radii are 4 cm and 8 cm respectively.
Solution:
Let $r_{1}$ be the inner radius and $r_{2}$ be the outer radius and $V$ be the volume of the hollow sphere at any time $t .$ Then,
$V=\frac{4}{3} \pi\left(r_{1}^{3}-r_{2}^{3}\right)$
$\Rightarrow \frac{d V}{d t}=4 \pi\left(r_{1}^{2} \frac{d r_{1}}{d t}-r_{2}^{2} \frac{d r_{2}}{d t}\right)$
$\Rightarrow r_{1}^{2} \frac{d r_{1}}{d t}=r_{2}^{2} \frac{d r_{2}}{d t}$ $\left[\because \frac{d V}{d t}=0\right]$
$\Rightarrow(4)^{2} \times 1=(8)^{2} \frac{d r_{2}}{d t}$
$\Rightarrow \frac{d r_{2}}{d t}=\frac{1}{4} \mathrm{~cm} / \mathrm{sec}$