Question

If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

Answer 1

Let the radius of circle = r and area of the circle,$A=\pi r^2\therefore \frac{d}{dt}A=\frac{d}{dt}\pi r^2\Rightarrow \frac{dA}{dt}=2\pi r.\frac{dr}{dt}$
Since the area of a circle increases at a uniform rate, then
$\frac{dA}{dt}=k$
Where, k is a constant.
from Eqs. (i) and (ii) , $2\pi r.\frac{dr}{dt}=k\Rightarrow \frac{dr}{dt}=\frac{k}{2\pi r}=\frac{k}{2\pi }.\left(\frac{1}{r}\right)$
Let the perimeter $P=2\pi r\frac{dP}{dt}=\frac{d}{dt}.2\pi r\Rightarrow \frac{dP}{dt}=2\pi .\frac{dr}{dt}=2\pi .\frac{k}{2\pi }.\frac{1}{r}=\frac{k}{r}\Rightarrow \frac{dP}{dt}\propto \frac{1}{r}$