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 1.7 feet deep?

(b) If the water is rising at a rate of 3/8 inch per minute when h = 2.1, determine the rate at which water is being pumped into the trough.

Answer 1

Hello!

PART A

We are given the rate of the volume (dV/dt) of water being added to the trough and asked to find how fast the water height is rising (dh/dt)

dV/dt = 2 ft³ / min

We need an expression relating volume (V) and the height of the trough h. This will be the filled area of the front or back sides of the trough, times the trough depth of 12 ft.

To find the area of the front or back side, we use the triangle area formula A = 1/2 b * h

To find the area of the trough face, we can use similar triangles. The length top side of the trough is 3 ft long, and it's height is 3 ft high. The length of the top of the water is b, and the height is h. As these are similar triangles, we can relate b and h as

3 / 3 = b / h

Solving for b in terms of h

b = h

We can this place this in our triangle area formula

A = 1 / 2 (h) * h = 1/2 h² ft²

and we now multiply by the depth of 12 ft to get the volume of water

V = 12 * A = 12 / 2 * h² = 6 h² ft³

V = 6 h²

We now take the derivative of both sides with respect to time,

(dV / dt) = 6 * 2 * h * (dh/dt) = 12 h * (dh/dt)

Plugging in for h = 1.7 ft and the given dV/dt = 2 ft³ / min, we have

2 = 12 * 1.7 * (dh/dt)

(dh/dt) = 0.0784 ft / min

The water level is rising at a rate of 0.0784 ft / min

PART B

We use our previous derived expression,

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

with (dh/dt) = (3/8) inch / min, and h = 2.1 ft, and solve for (dV/dt)

(dV / dt) = 12 ft * 2.1 ft * (3/8 inch / min)

Convert inches to feet in order to evaluate the expression,

(dV / dt) = 12 ft * 2.1 ft * (3/8 inch / min) * (1 ft / 12 inch)

(dV / dt) = 0.7875 ft / min