Question

If the area of a circle increases at the rate of $\frac{1}{\sqrt{\pi }}sq$. units/sec, then the rate (in units/sec) at which the perimeter of the circle changes, when perimeter is $\sqrt{\pi }$ units, is

• A. $2$
• B. $4$
• C. $\frac{1}{\sqrt{\pi }}$
• D. $\sqrt{\pi }$

Answer 1

Answer: (A)

Let $A$ and $P$ are area and perimeter of circle.
We have,
$\frac{dA}{dt}=\frac{1}{\sqrt{\pi }}$
$\Rightarrow \frac{d}{dt}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi }}$
$\Rightarrow 2\pi r\frac{dr}{dt}=\frac{1}{\sqrt{\pi }}$
$\Rightarrow \frac{dr}{dt}=\frac{1}{2\pi r\sqrt{\pi }}$
Now, $P=2\pi r$
$\Rightarrow \frac{dP}{dr}=2\pi \times \frac{dr}{dt}=2\pi \times \frac{1}{2\pi r\sqrt{\pi }}=\frac{1}{r\sqrt{\pi }}$
$=\frac{1}{\frac{1}{2\sqrt{\pi }}\times \sqrt{\pi }}\left[\because P=\sqrt{\pi }\Rightarrow 2\pi r=\sqrt{\pi }\Rightarrow r=\frac{1}{2\sqrt{\pi }}\right]$
$=2$ unit / sec