Date: 03/11/98 at 01:19:10 From: Jen Armstrong Subject: Calculus The volume of a cube is increasing at the rate of 1200 cm cubed/min at the instant its edges are 20 cm long. At what rate are the edges changing at the instant?
Date: 03/11/98 at 08:49:19 From: Doctor Jerry Subject: Re: Calculus Hi Jen, Delay with specific numbers as long as possible. That's my best advice on these problems. So, V = x^3, where I'm thinking of both V and x as functions of time t. Holding off on any numbers for a bit longer, differentiate, to relate dV/dt and dx/dt: dV/dt = 3x^2*dx/dt, by the chain rule. OK, it is given that dV/dt = 1200. So, 1200 = 3x^2*dx/dt. So, if we know x, we can calculate dx/dt. -Doctor Jerry, The Math Forum Check out our web site http://mathforum.org/dr.math/
Date: 03/11/98 at 12:30:14 From: Doctor Sam Subject: Re: Calculus Jen, 1. The first thing to do in any related rates problem is to identify the rate that you want, in this case "the rate of change of the edge." If we call the edge of the cube x, that tells us we want to find dx/dt. 2. The second step is to find an equation involving that variable; in this case, V = x^3 gives the volume of the cube. 3. The third step is to differentiate with respect to time. A good idea at this step is to look at your equation and ask yourself, "Which things are changing?" In this case, both V and x are changing, so you need to treat them as functions of t. In some problems, there will be quantities that do not change (the length of a ladder in a sliding ladder problem, for example), and these quantities do not need to be named with variables since they are constant and their derivatives will be zero. So, differentiating gives dV/dt = 3x^2 dx/dt. You must remember to use the Chain Rule in doing all these problems, since this in not like differentiating y = x^3, which gives dy/dx; rather, we have dV/dt = (dV/dx)(dx/dt). 4. The next step often involves the numerical information in the problem. Here, you want to find dx/dt; but the equation we have, dV/dt = 3x^2(dx/dt) involves three different quanitites: dV/dt, dx/dt, and x . . . so you must find enough data in the statement of the problem to calculate the two quantities dV/dt and x. The problem states that "the volume of a cube is increasing at the rate of 1200 cm cubed/min at the instant that the edges are 20 cm." (Look for the key word "rate" to spot a derivative.) The first part, "the volume is increasing at the rate of," gives dV/dt = 1200 cm^3/min, and the second part, "the edges are 20 cm," gives x = 20. Now substitute to get 1200 = 3(20)^2 dx/dt, which can be solved to give dx/dt = 1 cm/min. I hope that helps. -Doctor Sam, The Math Forum Check out our web site http://mathforum.org/dr.math/
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