calculus related rates

Question

An inverted pyramid is being filled with water at a constant rate of 75 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 5 cm, and the height is 13 cm.

Find the rate at which the water level is rising when the water level is 6 cm. 

Answer 1

By similar triangles,

5/13 = S/H where S is the measure of the side of the square and H is the height of the water

5 H = 13 S

(5/13)H = S

(5/13) (dH/dt) = dS/dt

V = (1/3)(S^2)H

Product rule says:

dV/dt = (1/3) [ ( S^2 * dH/dt + 2*S*H* dS/dt)]

= (1/3) [ (5/13H)^2 * dH/dt + 2 * (5/13 H) * H * (5/13) (dH/dt)] <-- substitutes S = (5/13)H

and dS/dt = (5/13)(dH/dt)

(1/3) [ (25/169) H^2 * dH/dt + 50/169 H^2 dH/dt]

(1/3) [ (75/169) H^2 * dH/dt]

25/169 * H^2 * dH/dt

75 = (25/169) * 6^2 * dH/dt

75 = (25/169)*36 * dH/dt

75*169/(25*36) = dH/dt

14 and 1/12 = dH/dt