Question

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

Answer 1

Let $r$ be the radius

Given, Waves move in a circle at speed of $5\text{cm/s}$

i.e., Radius of circle increasing at a rate of $5\text{cm/s}$

$\frac{dr}{dt}=5\text{cm/s}$

Area of circle, $A=\pi r^2$

Now,

$\frac{dA}{dt}=\frac{d\left(\pi r^2\right)}{dt}$

$=\pi \frac{d\left(r^2\right)}{dt}$

$=2\pi r\frac{dr}{dt}$

$=10\pi r[\because \frac{dr}{dt}=5]$

When $r=8\text{cm}$

$\frac{dA}{dt}=10\pi \times 8=80\pi {\text{cm}}^2/\text{s}$

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