Question

A stone is dropped into a quiet lake and waves move in circles at the speed of $5 cm/s$. At the instant when the radius of the circular wave is $8 cm$, how fast is the enclosed area increasing?

Answer 1

The correct answer $80π cm^2 /s.$
The area of a circle $(A)$ with radius $(r)$ is given by $A=πr^2.$
Therefore, the rate of change of area $(A)$ with respect to time $(t)$ is given by,
$\frac{dA}{dt}=\frac{d}{dt}(πr^2)=\frac{d}{dr}(πr^2)\frac{dr}{dt}=2πr^2 \frac{dr}{dt} ...$[By chain rule]
It is given that $\frac{dr}{dt}=5cm/s$
Thus, when $r = 8 cm$
$\frac{dA}{dt}=2π(8)(5)=80π$
Hence, when the radius of the circular wave is $8 cm$, the enclosed area is increasing at the rate of $80π cm^2 /s.$