Question

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 1

The correct answer \(80π cm^2 /s.\)
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,
\(\frac{dA}{dt}=\frac{d}{dt}(πr^2)=\frac{d}{dr}(πr^2)\frac{dr}{dt}=2πr^2 \frac{dr}{dt} ...\)[By chain rule]
It is given that \(\frac{dr}{dt}=5cm/s\)
Thus, when \(r = 8 cm\)
\(\frac{dA}{dt}=2π(8)(5)=80π\)
Hence, when the radius of the circular wave is \(8 cm\), the enclosed area is increasing at the rate of \(80π cm^2 /s.\)