Question

If the area of a circle increases at the rate of $\frac{1}{\sqrt{\pi}} s q$. 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,
$\frac{d A}{d t}=\frac{1}{\sqrt{\pi}}$
$ \Rightarrow \frac{d}{d t}\left(\pi r^{2}\right)=\frac{1}{\sqrt{\pi}} $
$\Rightarrow 2 \pi r \frac{d r}{d t}=\frac{1}{\sqrt{\pi}} $
$\Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r \sqrt{\pi}}$
Now, $P=2 \pi r$
$\Rightarrow \frac{d P}{d r}=2 \pi \times \frac{d r}{d t}=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