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# A stone is dropped into a quiet lake and waves move in circles at the speed of $5 \;cm/s$. At the instant when the radius of the circular wave is $8 \;cm$, how fast is the enclosed area increasing?

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

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## 1 Answer

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Toolbox:
• 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 $80\pi\;cm^2/s$
answered Jul 5, 2013

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