



 
Hi David. Perhaps if I help you interpret the word problem down to the level of equations, you can take it from there. This is a volume problem, so we first have to come up with an expression relating the water level (h) to the volume (v). The volume is the length of the trough (12 ft) times the area of the triangle that is filled. The triangle's shape is defined by the trough shape, so the base (which is really the top surface of the water) is 3 ft when the water depth is 3 ft. The base and the depth match. So really, if you know the depth (h) of the water, you also know the base. And the area of a triangle is one half the base times the height, so the area of the triangle is (1/2) h^{2}. This makes the volume 12 x (1/2) h^{2}. Let's call the volume v. Thus, v = 6 h^{2}. Now take the derivative of both sides with respect to t. Notice that you'll need the chain rule here. You will be left with an equation that has four variables: h, v, dh/dt and dv/dt. These last two are the rate of change of the depth of the water and the rate of change of the volume in the trough, respectively. For part (a) of the question, you are given h and you can use the original equation v = 6 h^{2} to find v. You are also given dv/dt, so you have three quantities to put into the differential equation and you are asked to solve for the unknown variable dh/dt. For part (b), you solve it similarly but this time you are given dh/dt and asked for the corresponding dv/dt. Cheers,  


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