Related Rates--yeah...?

Question

A trough is 14 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12 ft3/min, how fast is the water level rising when the water is 5/12 ft deep?

Note: rate is 12 feet cubed! (for clarification)

Answer 1

There are at least two ways of doing this problem. One is the formal use of the machinery of calculus.

You want to find dH/dt when what you are given is dV/dt, so you need to find H as a function of V.

As it happens, it is easy to find the volume of water V as a function of the height H so to get H as a function of V, just invert the function V(H).

Then dH/dt = (dH/dV)(dV/dt) where dV/dt is a given constant.

So, what is the relationship between volume and height? Since the water level is always going to be horizontal, the volume of water is going to be a triangular prism with the triangular cross-section geometrically similar to that of the trough.

For a triangular prism:

volume(t) = (1/2) x height(t) x base area(t)
base area(t) = width(t) x length(t)

In this case:

length(t) = 14 feet ( a constant)
width(t) = height(t) x (width of trough) / (height of trough)

Substituting through we get V(t) on the left and a function in H(t) on the right.

First solve for V(t) when H(t) = 5/12

Knowing V(t) and dV/dt, you can compute the corresponding t

Then solve for H(t) in terms of V(t).

Then solve for dH/dV and compute dH/dt for the computed value of t.