A pebble is dropped in a calm pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of .5 feet per second.

Question

When the radius is 4 feet, at what rate is the total area A of the disturbed water changing?

Answer 1

The area of the disturbed water is

A(r) = π r^2,

where r is the radius of the outer ripple. (This is because the disturbed water is in the shape of a disk of radius r.)

So, dA / dr = 2 π r is the derivative of A with respect to r.

So, by the chain rule, the derivative of A with respect the time t is

dA / dt = (dA / dr) · (dr / dt).

Since you know the value of r (it's given), you know the value of dA / dr = 2 π r. You're also given the value of dr / dt. You should do the rest.