Related Rates: Trough

Question

A trough is 9 feet long, and its cross section is in the shape of an isosceles right triangle with hypotenuse 2 feet, as shown above. Water begins flowing into the empty trough at the rate of 2 cubic feet per minute. At what rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow?

Answer 1

First let's find the Volume of the trough, I assume it's an inverted triangle where the sides are equal since the diagram is not shown. All units match so I will work with numbers only.

Volume of a triangle prism is given by:

V = (b * h )/2 * l

Length or l is a constant at 9, so we can sub it in the equation.

V = 9/2 * (b * h)

Substitute out base

Base and height of the filled trough at any time is related by trigonometry. We're interested in height so we will substitute out base with height.

This part would be a lot easier with a diagram, but unfortunately I can't upload one. Using properties of isosceles right triangle and tan:

tan(45 degrees) = h/(b/2)

b = 2h

V = 9/2 *(2h * h) = 9h^2

Rate

Take derivative of volume with respect to time to find equation for rate of filling the trough.

dV/dt = 2 * 9 *h dh/dt = 18h dh/dt

re-arranging for dh/dt

dh/dt = dV/dt /(18h)

Subbing in dV/dt = 2

dh/dt = 2/(18 h) = 1/(9h)

Find h at 2 min

V = 9h^2

V = t * rate = 2 * 2 = 4

4 = 9h^2

re-arranging for h:

h = sqrt(4/9) = 2/3

Subbing in h to rate equation

dh/dt = 1/(9h) = 1/(9 * 2/3) = 1/6

The rate of change for height is 1/6 cubic feet/min at 2 min.