Question

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

• A. $8\pi c{m}^2/s$
• B. $80\pi c{m}^2/s$
• C. $6\pi c{m}^2/s$
• D. $\frac{8}{3}\pi c{m}^2/s$

Answer 1

The correct option is B $80\pi c{m}^2/s$

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

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

$\frac{dA}{dt}=\frac{d}{dt}\left(\pi r^2\right)=\frac{d}{dt}\left(\pi r^2\right).\frac{dr}{dt}=2\pi r\frac{dr}{dt}\left[ByChainRule\right]$

It is given that: $\frac{dr}{dt}=5cm$

Thus, when $r=8cm$,

$\Rightarrow \frac{dA}{dt}=2\pi \left(8\right)\left(5\right)=80\pi$

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