Related Rates Triangular Prism

Question

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

(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 1.1 feet deep in ft/min?

(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 in ft^3/min.

Answer 1

Part a)

Let's calculate the volume V(h) in terms of the height h. Using similar triangles we know that 3ft/3ft = h/L, where L = h is the length of the top of the water level at height h. Therefore the cross-sectional triangular area is A = (1/2)h*h ft² = (1/2)h² ft² . Then the volume is just 12A ft³ , or V(h) = 6h² ft³ .

If we want the rate at which the water level is moving (dh/dt) when h = 1.1 ft, simply compute dV/dt by using the chain rule and solve for dh/dt. Then plug in dV/dt = 2 ft³/min and h = 1.1 ft.

dV/dt = (dV/dh)*(dh/dt) = 12h*(dh/dt)

dh/dt = (dV/dt) / (12h) = (2 ft³/min) / (12*1.1 ft²) = 5/33 ft/min.

Part b)

We already have a formula for dV/dt from part a, namely dV/dt = 12h*(dh/dt), so we just need to plug in dh/dt = (3/8)*(1/12) ft/min and h = 2.2 ft, where I'm converting dh/dt from in/min to ft/min.

dV/dt = (12 ft)*(3/(8*12) ft/min)*(2.2 ft) = 33/40 ft³/min.