Water Trough Related Rate ProblemDate: 10/21/2003 at 22:29:24 From: Ricardo Subject: Calculus and Derivatives A storage tank is 20 ft long and its ends are isosceles triangles having bases and altitudes of 3 ft. Water is poured into the tank at a rate of 4 (ft)^3/min. How fast is the water level rising when the water in the tank is 6 in. deep? I can't get the answer that was provided to us. Here is how I set the problem up: The volume of a tank of that description would be the area of a triangle times its length: V(h) = (1/2)bhL. I was given its base, height and length, which is 3, 3 and 20, respectively. I am also given dV/dt = 4 ft^3/min. I first take the derivative of the equation, with respect to the height: V(h) = bL/2 dh/dt Since I'm asked to find the rate the water level is rising, I set up the rates in this way: 4 = bL/2 dh/dt 8/bL = dh/dt 8/(3 * 20) = dh/dt 8/60 = dh/dt Thus, 2/15 ft/min = dh/dt, which is approximately .133 ft/min. My teacher gave us an answer of .04 ft/min. Did I do the problem correctly? Date: 10/22/2003 at 09:53:25 From: Doctor Luis Subject: Re: Calculus and Derivatives Hi Ricardo, Your last steps are not quite correct. Let me show you why. You wrote V = (1/2)bh * L, which is correct. However, when you took the derivative dV/dt, you forgot that the base becomes wider as h increases. You treated it as a constant equal to the base of the tank, but it's actually smaller and it depends on the height. To solve these related rates problems, take note of all the quantities that are changing before you start applying a time derivative. Take a look at this diagram: The proportion h/b = H/B allows you to eliminate b from the formula for V, leaving you with h and some constants. At that point, you can find dV/dt in terms of h and dh/dt, and then you can solve for dh/dt and evaluate it at h = 6 in = 0.5 ft, knowing that dV/dt = 4 ft^3/min. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ Date: 10/22/2003 at 12:35:43 From: Ricardo Subject: Thank you (Calculus and Derivatives) Thank you so much for your reply. When thinking in terms of "what variables are changing over time", it all came together. I've since been able to complete two additional Calculus (rates of change) word problems. Thanks again! |
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