How fast is the water level rising when the water is 4 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 being filled with water at a rate of 15 ft³/min, how fast is the water level rising when the water is 4 inches deep?

Answer 1

The volume of water in the trough forms the shape of a triangular prism and that volume is calculated by taking the volume of the triangular base times the length. So V_water = (Area_triangle)(Length) or V_water = 8(Area_triangle)

The cross section of the trough looks like this:

So cutting it down the middle, and adding the water I would get this:

So, using similar triangle, we can say 2/1 = x/h or x = 2h. This gives a relationship between x and h.

The volume of water at any time t can be determined by remembering the volume equation we discussed above V_water = 8(Area_triangle) = 8(1/2bh) but b = 2x so, plugging that in, we get:

V_water = 8(1/2)(2x)(h) but, then again, since x = 2h, we get V_water = 8(1/2)(2(2h)(h) = 16h² so:

V(h) = 16h²

Taking the derivative as a function of time (dV/dt) we get dV/dt = 32h(dh/dt) and the solving for dh/dt we get:

dh/dt = (dV/dt)/(32h)

We know that dV/dt = 15 ft³/min. We are looking for dh/dt when h = 4 inches but the other units are all in feet so we must consider h in feet. 4 inches = 4/12 ft = 1/3 ft.

So dh/dt = 15/(32(1/3) = 45/32 ft/min