$\begin{array}{1 1} 60\pi\;cm^2/s \\ 80\pi\;cm^2/s \\ 20\pi\;cm^2/s \\ 50\pi\;cm^2/s \end{array} $

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- If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
- $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1:

Given :

$\large\frac{dr}{dt}=$$5cm/s$ and radius=$8cm$

Area of the circle is $\pi r^2$

Differentiating w.r.t $t$ on both sides we get,

$\large\frac{dA}{dt}=$$2\pi r\large\frac{dr}{dt}$

Step 2:

Substituting for $r$ and $\large\frac{dr}{dt}$ we get,

$\large\frac{dA}{dt}$$=2\times \pi\times 8\times 5$

$\quad\;=80\pi\;cm^2/s$

Hence the enclosed area is increasing at the rate of $80\pi\;cm^2/s$

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