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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
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Let $A$ and $P$ are area and perimeter of circle. We have,
$\frac{d A}{d t}=\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d}{d t}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi}}$
$\Rightarrow \quad 2 \pi r \frac{d r}{d t}=\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r \sqrt{\pi}}$
Now, $P=2 \pi r$
$\begin{aligned}
\Rightarrow & \frac{d P}{d r}=2 \pi \times \frac{d r}{d t}=2 \pi \times \frac{1}{2 \pi r \sqrt{\pi}}=\frac{1}{r \sqrt{\pi}} \\
= & \frac{1}{\frac{1}{2 \sqrt{\pi}} \times \sqrt{\pi}}\left[\because P=\sqrt{\pi} \Rightarrow 2 \pi r=\sqrt{\pi} \Rightarrow r=\frac{1}{2 \sqrt{\pi}}\right] \\
& =2 \text { unit/sec. }
\end{aligned}$
$\frac{d A}{d t}=\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d}{d t}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi}}$
$\Rightarrow \quad 2 \pi r \frac{d r}{d t}=\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r \sqrt{\pi}}$
Now, $P=2 \pi r$
$\begin{aligned}
\Rightarrow & \frac{d P}{d r}=2 \pi \times \frac{d r}{d t}=2 \pi \times \frac{1}{2 \pi r \sqrt{\pi}}=\frac{1}{r \sqrt{\pi}} \\
= & \frac{1}{\frac{1}{2 \sqrt{\pi}} \times \sqrt{\pi}}\left[\because P=\sqrt{\pi} \Rightarrow 2 \pi r=\sqrt{\pi} \Rightarrow r=\frac{1}{2 \sqrt{\pi}}\right] \\
& =2 \text { unit/sec. }
\end{aligned}$
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