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 rage of 4 feet per second. Find the rate at which the area is changing at the instant the radius is 8 feet
We know that the area of a circle is given by $A = \pi r^2$. Taking the derivative of both sides of this equation with respect to time $t$, we have $\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$.
Given that $\frac{dr}{dt} = 4$ ft/s and $r = 8$ ft, we can substitute those values into the equation above to find $\frac{dA}{dt}$.
$\frac{dA}{dt} = 2\pi(8)(4) = 64\pi$ ft$^2$/s.
Therefore, the rate at which the area is changing at the instant the radius is 8 feet is $64\pi$ ft$^2$/s.