how fast is the water level rising when the water is 3 inches deep?

Question

A trough is 8 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is filled with water at a rate of 10ft.^3/min. how fast is the water level rising when the water is 3 inches deep?

If you could show how to do it and what the answer is, it would be greatly appreciatedd!!

Answer 1

We need to find the volume of the water at any given time. Volume of a prism is basically the area of the base times the length of the trough. The area of the base is basically the area of a triangle. The area of a right triangle is A = 0.5bh. Since this is an isosceles triangle, we can split it down the middle to get two congruent right triangles, so the area of each will be A = 0.5bh, and the area of the whole thing would simply be A = bh.

The base and the height are constantly changing, so we need to express the base and height in terms of variables. If we set our height to be some unknown value h, then through similar triangles we can write the base in terms of h:

height of the whole thing / base of the whole thing = height at any time / base at any time

1/4 = h/base
base = 4h

So now in terms of our variable h, the volume of water at any given height (depth) of water h is:

V = area of base * length
V = (4h)(h) * 8
V = 32h^2

We have a changing rate of the volume (dV) and a height (3 inches, but we need to convert that to 0.25 feet since the volume formula is in feet). So we need to take a derivative and plug in numbers to find the changing height:

V = 32h^2
dV = 64h dh
10 = (64)(.25) dh
10 = 16 dh
dh = 0.625 ft/min, or 7.5 in/min

So the height is changing at a rate of 7.5 in/min at that very instant in time.

Answer 2

A related rates problem, not much different from one last night. Your trough has a volume of 0.5 base x height x length. The trick is express base in terms of height, which, from your statement, is 4h = b. Then we have, V = 28 h^2. Then, we can determine dV/dh as dV/dh = 56h. We know dV/dt=14, and we can convert dV/dh to dV/dt x dt/dh= 56h. Then 14 dt/dh = 56h, or dt/dh = 56/14 h= 4h. We really want dh/dt, which is the reciprocal, or dh/dt = 1/(4h) For h of 0.5 ft, we have 0.5 ft/minute