So, Allison, the given data are: dr/dt = 4 (units of ft s^-1) . == equation 1
You also know an equation for the area of a circle: A = .pi. r^2 . == equation 2
So you can put these together: dA/dr = 2 .pi. r . (differentiation of equation 2 with respect to r) == equation 3
Multiply equation s 1 and 3 together: dA/dr dr/dt = 4 x 2 .pi.r = dA/dt = 8 .pi. r == equation 4
Since after 12 s the ripple will have r = 4 ft/s x 12 s = 48 ft,
then dA/dt at that time will have value 8 .pi. 48 = 384 .pi. ft^2/s (unless I've done my multiplication wrong).
By the way, you would also get the same result by considering the problem *geometrically* as the area covered by the ripple acting for a small amount of time at t=12 sec, when it has a circumference of 2 .pi. r = 96 .pi. ft. Then the small "slice" of circumference (annulus), unwrapped to make it straight, is length 96 .pi. ft, width 4 ft/s, or 384 .pi. ft^2/s.
Somewhat more interesting, perhaps, would be your finding if you tried to measure the speed of travel of a single crest of the ripple. You'd find it's travelling at 8 ft/s, or twice as fast as the ripple as a whole is moving! If you get interested in why, look up "deep water wave propagation".