A stone is dropped into a quiet lake and waves move in circles with a speed of 4 cm/sec.at the instant when the radius of the circular wave is 10 cm. How fast is the enclosed area increasing?

Hint: Here, we would be using the concept of rate of change of a quantity with respect to the other quantity.

Complete step-by-step answer:
Given, the speed of the stone =4 cm/sec and the radius \[r\] is 10 cm
we know that the speed is given by \[\dfrac{{dr}}{{dt}} = 4cm/\sec \]
also, the area\[A\] of a circle with radius\[r\]is given by \[\pi {r^2}\]
which implies the rate of change of the enclosed area is given by
\[\dfrac{{dA}}{{dt}} = \dfrac{d}{{dt}}\left( {\pi {r^2}} \right) = 2\pi r\dfrac{{dr}}{{dt}}\] (using the chain rule) …… (1)
Putting the value of \[\dfrac{{dr}}{{dt}}\]and \[r\]in equation (1) we get
\[\dfrac{{dA}}{{dt}} = 2\pi \times 10 \times 4\]
\[ \Rightarrow \dfrac{{dA}}{{dt}} = 80\pi \]
Hence, the enclosed area is increasing at a rate of \[80\pi c{m^2}/\sec \].

Note: We can conclude that the area is increasing as the rate of area \[\dfrac{{dA}}{{dt}}\] is positive.