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

If water is being pumped into the trough at 2 cubic feet per minute, how fast (in ft/min) is the water level rising when his 1.6 feet deep?

Answer 1

The trough, I believe, is a triangular prism where the two bases are isosceles triangles. I was just making sure when I asked for the figure. Anyway, this is how it goes.

Volume of the prism, as you know perhaps, is:

V =(Area of the triangle)•h

V = [(1/2) b•a]•h

Where: b = the base of the triangle

a = the altitude of the triangle

h = the height (or length) of the prism

In your case:

b = 3 ft.

a = 3 ft.

Since b = a, we can write down the formula as:

V = (1/2) a²h

Since h = 12 ft.

V = (1/2) a²(12)

V = 6a²

Derive both sides of the equation with respect to time (t):

dV/dt = 12a • da/dt

Given:

dV/dt = 2 ft³/min

a = 1.6 ft

da/dt = ?

Substitute:

2 = 12 (1.6) (da/dt)

da/dt ≈ 0.104 ft./min.