First thing is that your diagram is wrong. This is simply a 'v' shaped trough. That is, if you take a vertical section through the trough at right angles to its length, what you see is a v shape, an isosceles triangle.

Next, pretty much all of these related rates problems use the function of a function rule.

For differentiation purposes the rule is usually written as , but in 'practical' examples, it doesn't matter what letters are used.

In this example the variables are, , the volume of water, , the height of the water and , time.

You are given the rate of change of volume,

What you are required to calculate is the rate of change of height, (when ).

You now have to figure out a suitable function of a function relationship linking these quantities.

It is :

In order to make use of this, you need to calculate , and in order to calculate that, you need a relationship between and .

Now, when the height of the water in the trough is ft., the cross-sectional area of the water will be (use similar triangles to work out the 'width' of the water, then it's simply base times height divided by 2), in which case

So, use this to calculate put the whole lot together and calculate