Question

A rock is thrown into a still pond. The circular ripple move outward from the point of impact of the rock so that the area of the circle formed by a ripple increases at the rate of 100 square feet per minute. Find the rate at which the radius is changing at the instant the radius is 5 feet.

Answer 1

Answer:

$\approx 31.42 \text{ Square feet per minute}$

Step-by-step explanation:

Area of the Pond A= $\pi r^2$

The increase in the area of the pond as it moves outward is: $\frac{dA}{dt}$

$\frac{dA}{dt}=\frac{d}{dt}\pi r^2\\\frac{dA}{dt}= 2\pi r \frac{dr}{dt}$

Since the area of the circle formed increases at the rate of 100 square feet per minute.

$\frac{dA}{dt}$ = 100 Square feet per minute

When the radius, r = 5 feet, we want to determine the rate at which the radius is changing, $\frac{dr}{dt}$.

$\frac{dA}{dt}= 2\pi r \frac{dr}{dt}\\100=2*5*\pi \frac{dr}{dt}\\\frac{dr}{dt} =\frac{100}{10 \pi} =10 \pi \approx 31.42 \text{ Square feet per minute}$