Let the volume of a metallic hollow sphere be constant. If the inner radius increases at the rate of $\displaystyle 2 \mathrm{~cm} / \mathrm{s}$, find the rate of increase of the outer radius when the radii are $\displaystyle 2$ cm and $\displaystyle 4$ cm respectively.

\[\begin{aligned} & \frac{d \mathrm{r}}{d t}=2 \mathrm{~cm} / \mathrm{s},, \quad\left(\frac{\mathrm{dR}}{\mathrm{dt}}\right)_{\mathrm{R}=4, \mathrm{r}=2}=? \\ & \mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi\left(3 \mathrm{R}^{2} \cdot \frac{d \mathrm{R}}{d t}-3 \mathrm{r}^{2} \frac{d \mathrm{r}}{d t}\right) \\ & \text { When } \mathrm{R}=4 \mathrm{~cm} \text { and } \mathrm{r}=2 \mathrm{~cm}, \\ & 0=\frac{4}{3} \pi\left[3(4)^{2} \cdot \frac{d \mathrm{R}}{d t}-3(2)^{2}(2)\right] \\ & \Rightarrow \frac{d \mathrm{R}}{d t}=\frac{1}{2} \mathrm{~cm} / \mathrm{s} \end{aligned} \]