Answer:
[tex]\approx 31.42 \text{ Square feet per minute}[/tex]
Step-by-step explanation:
Area of the Pond A= [tex]\pi r^2[/tex]
The increase in the area of the pond as it moves outward is: [tex]\frac{dA}{dt}[/tex]
[tex]\frac{dA}{dt}=\frac{d}{dt}\pi r^2\\\frac{dA}{dt}= 2\pi r \frac{dr}{dt}[/tex]
Since the area of the circle formed increases at the rate of 100 square feet per minute.
[tex]\frac{dA}{dt}[/tex] = 100 Square feet per minute
When the radius, r = 5 feet, we want to determine the rate at which the radius is changing, [tex]\frac{dr}{dt}[/tex].
[tex]\frac{dA}{dt}= 2\pi r \frac{dr}{dt}\\100=2*5*\pi \frac{dr}{dt}\\\frac{dr}{dt} =\frac{100}{10 \pi} =10 \pi \approx 31.42 \text{ Square feet per minute}[/tex]
