calculus
posted by Tori .
A hollow cone has height 5 feet and base diameter 4 feet. The vertex of the cone is pointed down so that it can serve as a container. Water is poured into the cone at the rate 3/2 cubic feet per second. At what rate (in feet per second) is the depth of the water increasing when the depth of the water is 3 feet?

Find the area of the surface of the water when water is 3 feet from the tip (A =pi r^2) where r = (2/5)(3)
dh (A) = dV
dh/dt = (1/A)(dV/dt)
but dV/dt is given, 3/2 ft^3/s 
25/24pi....I think is the answer

r = 6/5
pi r^2 = pi (36/25)
dh/dt = 25/(36pi) * 3/2
= 25 /24pi
agree