A spherical medicine ball when dropped in water dissolves in such a way that the rate of decrease of volume at any instant is proportional to its surface area. Calculate the rate of decrease of its radius.

Let ' $\displaystyle V$ ' and ' $\displaystyle S$ ' be the volume and surface area of the spherical medicine ball with radius ' $\displaystyle r$ '. \[\frac{d V}{d t}=-k S, k>0 \] $\displaystyle V=\frac{4}{3} \pi r^{3} \Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \Rightarrow-k S=4 \pi r^{2} \frac{d r}{d t}$ \[\Rightarrow-k\left(4 \pi r^{2}\right)=4 \pi r^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=-k \] ∴ Radius decreases at a constant rate.