Question

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?

Answer 1

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.