If the area of a circle increases at the rate | vedclass

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(A) Let $A$ be the area and $P$ be the perimeter of the circle. We have, $\frac{dA}{dt} = \frac{1}{\sqrt{\pi}}$.Since $A = \pi r^2$, differentiating w

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$2$

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(A) Let $A$ be the area and $P$ be the perimeter of the circle. We have, $\frac{dA}{dt} = \frac{1}{\sqrt{\pi}}$.Since $A = \pi r^2$, differentiating with respect to $t$ gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.Substituting the given rate: $2\pi r \frac{dr}{dt} = \frac{1}{\sqrt{\pi}} \Rightarrow \frac{dr}{dt} = \frac{1}{2\pi r \sqrt{\pi}}$.The perimeter is $P = 2\pi r$. Differentiating with respect to $t$ gives $\frac{dP}{dt} = 2\pi \frac{dr}{dt}$.Substituting $\frac{dr}{dt}$: $\frac{dP}{dt} = 2\pi 	imes \frac{1}{2\pi r \sqrt{\pi}} = \frac{1}{r \sqrt{\pi}}$.Given $P = \sqrt{\pi}$, we have $2\pi r = \sqrt{\pi} \Rightarrow r = \frac{\sqrt{\pi}}{2\pi} = \frac{1}{2\sqrt{\pi}}$.Substituting $r$ into the expression for $\frac{dP}{dt}$: $\frac{dP}{dt} = \frac{1}{(\frac{1}{2\sqrt{\pi}}) \sqrt{\pi}} = \frac{1}{1/2} = 2$ units/sec.

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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