Subject: Story Problem using differentiation Name: Don Who are you: Student I need help with a related rate problem, using differentiation to solve it. The problem is: The cross section of a 5-meter trough is an isosceles trapezoid with a 2-meter lower base, a 3-meter upper base and an altitude of 2 meters. Water is running into the trough at a rate of 1 cubic meter per minute. How fast is the water level rising when the water is 1 meter deep? Hi Don. If we call the water level height h and the volume v and time t, then you are seeking dh/dt when h is 1 meter. You will need to figure out the formula for the volume of the trough, given a constant length (5) and a cross-section as you described and the height h. I will assume you can do that geometry and algebra and are looking for help with the differentiation. So let's go from the point where you have V = f(h) If you now take the derivative of both sides of the equation with respect to time t, then dv/dt = df/dt. The left side is easy: you are given that dv/dt is 1 cubic meter per minute. But df/dt is trickier, because f is not given in terms of t, it is given in terms of h. This should remind you of the chain rule. Let's take an easy example. Say g(x) = 8x2 + x and x = 2p. Then dg/dp = (dg/dx)(dx/dp) = (16x+1)(dx/dp) = (16x+1)(2) = 32x+2 = 64p+2 Now yours is a little simpler. You are actually trying to find dh/dt so you will get 1 = df/dt = (df/dh)(dh/dt) which becomes: dh/dt = 1 / (df/dh) So if you now take the derivate of f(h) with respect to h, the right hand side is solved and you have an expression for dh/dt in terms of h. At this point, you simply put in the value of h that you are given (1 meter) and compute dh/dt's final value. Hope this helps, Stephen La Rocque.