Answer
$\frac{dr}{dt}=\frac{1}{ \sqrt \pi} \frac{mi}{h}$
Work Step by Step
Data
Let $A$ denote the area.
Rate of increase in area= $\frac{dA}{dt}=6 \frac{mi^2} {h}$
$A=9$ $ mi^2$
Solution:
Let $r$ denote radius.
$A=\pi r^2$ ............... eq (1)
Taking derivative with respect to time
$\frac{dA}{dt}=\frac{d(\pi r^2)}{dt}=\pi\frac{d(r^2)}{dt}=\pi\frac{d(r^2)}{dr} \frac{dr}{dt}$
$\frac{dA}{dt}=2\pi r \frac{dr}{dt}$......... eq(2)
Putting $A=9$ in equation (1)
$\pi r^2=9 $ $ \Longrightarrow$ $r=\frac{3}{ \sqrt \pi} $
Putting $\frac{dA}{dt}=6 $ and $r=\frac{3}{ \sqrt \pi} $ in equation (2)
$ 6=2\pi \frac{3}{ \sqrt \pi} \frac{dr}{dt} \Longrightarrow$
$\frac{dr}{dt}=\frac{\sqrt \pi}{\pi}= \frac{1}{\sqrt \pi} \frac{mi}{h}$
