The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is
[tex] \frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt} [/tex]
The rate of change of the radius is given as
[tex] \frac{dr}{dt} =-2 \, \frac{ft}{s} [/tex]
Therefore
[tex] \frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s} [/tex]
When r = 10 ft, obtain
[tex] \frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s} [/tex]
Answer: - 40π ft²/s (or - 127.5 ft²/s)
