A stone is dropped into a quiet lake and waves move in circles at a speed of $4 \text{ cm}$ per second. At the instant, when the radius of the circular wave is $10 \text{ cm}$, how fast is the enclosed area increasing?

$80\pi \text{ cm}^2/\text{s}$

The correct answer is Option (3) → $80\pi \text{ cm}^2/\text{s}$ ##

The area $A$ of a circle with radius $r$ is given by $A = \pi r^2$. Therefore, the rate of change of area $A$ with respect to time $t$ is

$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \frac{d}{dr}(\pi r^2) \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \quad \text{(By Chain Rule)}$

It is given that

$\frac{dr}{dt} = 4 \text{ cm/s}$

Therefore, when $r = 10 \text{ cm}$,

$\frac{dA}{dt} = 2\pi(10)(4) = 80\pi$

Thus, the enclosed area is increasing at the rate of $80\pi \text{ cm}^2/\text{s}$, when $r = 10 \text{ cm}$.