Surface area of a balloon (spherical), when air is blown into it, increases at a rate of $\displaystyle 5 \mathrm{~mm}^{2} / \mathrm{s}$. When the radius of the balloon is $\displaystyle 8$ mm , find the rate at which the volume of the balloon is increasing.

\[\begin{aligned} & \frac{d S}{d t}=5 \mathrm{~mm}^{2} / \mathrm{s}, \quad\left(\frac{d V}{d t}\right)_{r=8}=? \\ & S=4 \pi r^{2} \Rightarrow \frac{d S}{d t}=8 \pi r \cdot \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{5}{8 \pi r} \\ & V=\frac{4}{3} \pi r^{3} \Rightarrow \frac{d V}{d t}=4 \pi r^{2} \cdot \frac{d r}{d t} \Rightarrow \frac{d V}{d t}=\frac{5}{2} r \\ & \Rightarrow\left(\frac{d V}{d t}\right)_{r=8}=20 \mathrm{~mm}^{3} / \mathrm{s} \end{aligned} \]