A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s.

The area of a circle $(A)$ with radius $(r)$ is given by $A=\pi r^{2}$.

Therefore, the rate of change of area (A) with respect to time (t) is given by,

$\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right)=\frac{d}{d r}\left(\pi r^{2}\right) \frac{d r}{d t}=2 \pi r \frac{d r}{d t}$ [By chain rule]

It is given that $\frac{d r}{d t}=5 \mathrm{~cm} / \mathrm{s}$.

Thus, when r = 8 cm,

$\frac{d A}{d t}=2 \pi(8)(5)=80 \pi$

Hence, when the radius of the circular wave is $8 \mathrm{~cm}$, the enclosed area is increasing at the rate of $80 \pi \mathrm{cm}^{2} / \mathrm{s}$.