Question

A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the area at the instant the radius is 10ft?

Answer 1

The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
$\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}$

The rate of change of the radius is given as
$\frac{dr}{dt} =-2 \, \frac{ft}{s}$
Therefore
$\frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s}$

When r = 10 ft, obtain
$\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s}$

Answer: - 40π ft²/s (or - 127.5 ft²/s)