Question

A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 3 feet/minute. Find the rate at which the area is changing at the instance when the radius is 9 feet.

Answer 1

The rate at which the area is changing when the radius is 9 feet is 54π feet^2/minute, which is approximately 169.56 feet^2/minute.

Step-by-step explanation:

To find the rate at which the area is changing at the instance when the radius is 9 feet, we can use the formula for the area of a circle, which is A = πr^2. We want to find dA/dt, the rate of change of the area with respect to time. We know that dr/dt, the rate at which the radius is changing, is 3 feet/minute. We can differentiate the area function with respect to time.

A = πr^2

dA/dt = 2πr * dr/dt

Substituting dr/dt = 3 and r = 9 into the equation, we get:

dA/dt = 2π * 9 * 3 = 54π feet^2/minute

So, the rate at which the area is changing when the radius is 9 feet is 54π feet^2/minute. Since π is approximately equal to 3.14, the rate can be approximated as 169.56 feet^2/minute.