Question

A receptacle is in the shape of an inverted square pyramid 10 inches in height and with a 6 x 6 square base. The volume of such a pyramid is given by
(1/3)(area of the base)(height)

Suppose that the receptacle is being filled with water at the rate of .2 cubic inches per second. How fast is water rising when it is 2 inches deep?

Answer 1

V = 1/3s^2h; where V is the volume, s is the side length of the square base, and h is the height.
s/h = 6/10 = 3/5
s = 3h/5
V = 1/3(3h/5)^2 h = 1/3(9h^2/25)h = 3h^3/25

dV/dt = 9h^2/25 dh/dt = 0.2 = 1/5
dh/dt = 5/(9h^2)

When h = 2
dh/dt = 5/(9(2)^2) = 5/(9 * 4) = 5/36 = 0.1389 inches per seconds.