A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet.

Question

(a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.6 feet deep?

(b) If the water is rising at a rate of 3/8 inch per minute when h = 2.2, determine the rate at which water is being pumped into the trough.

Answer 1

If base is equal to the height, by similar triangles:

Volume v = base x height x length/2 = bhl/2 = h²12/2 = 6h².

(a) dv/dt = 12hdh/dt = 2 ft³/min

dh/dt = 2/(12h)

when h = 1.6, dh/dt = 2/(12•1.6) ≈ 0.1 ft³/min

(b) Assuming h is in inches. Need to convert length to inches also.

dv/dt = (12•12)h (dh/dt) = 144(2.2)(3/8) = 118.8 inch³/min = 0.825 ft³/min