Question 5

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?

Answer

The area of a circle (*A*) with radius (*r*) is given by A = πr^{2}.

Therefore, the rate of change of area (*A)* with respect to time (*t)* is given by,

\begin{align} \frac{dA}{dt}=\frac{d}{dt}(\pi r^2)=\frac{d}{dt}(\pi r^2).\frac{dr}{dt}=2\pi r\frac{dr}{dt}\;\;\;[By\; Chain \;Rule]\end{align}

It is given that

\begin{align} \frac{dr}{dt}=5\; cm/s\end{align}

Thus, when *r* = 8 cm,

\begin{align} \frac{dA}{dt}=2\pi(8)(5)=80\pi\end{align}

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm^{2}/s.

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