Question

A lonely guy throws a stone into a still pond causing a circular ripple to spread. If the radius
of the circular ripple spreads at the rate 1.5 ft/sec, how fast is the enclosed area increasing at the end of 2 seconds?

Answer 1

Are of circle =πr²..... A = πr²

Differentiating it with respect to time..

$\LARGE\frac{dA}{dt}$ =2πr $\LARGE\frac{dr}{dt}$

If the radius is increasing at a constant rate of $\LARGE\frac{1.5 ft}{sec}$

then after 2 seconds radius = 3 fts.

We know that $\LARGE\frac{dr}{dt}$ = $\LARGE\frac{1.5ft}{sec}$

and so $\LARGE\frac{dA}{dt}$ = [2]π[3][2]

$=\LARGE12πft2sec$ .