Surface Area and Volume: Related RatesDate: 02/22/99 at 19:30:53 From: Jack Cronin Subject: Calculus - Related Rates - A Word Problem. The surface area of a cube is changing at a rate of 8 inches squared per second. How fast is the volume changing when the surface area is 60 square inches? I'm pretty sure I have to use the chain rule for this one. I know that S = 6e^2 and V = e^3, and I have to find dV/dt, but I'm not sure how to go about it. Thanks. -Jack Date: 02/22/99 at 22:18:56 From: Doctor Jaffee Subject: Re: Calculus - Related Rates - A Word Problem. It looks to me as if you are off to a good start. Here is what you need to be thinking about now. Saying that the surface area of a cube is changing at a rate of 8 inches squared per second is the same as saying dS/dt = 8. Furthermore, you should be able to determine the value of e, the length of an edge of the cube when the surface area is 60. So, if you work with the formula S = 6e^2 and take the derivative of both sides with respect to t using implicit differentiation (which is just an application of the chain rule, as you predicted) and then replace e with its value when the surface area is 60, you should be able to get a numerical value for de/dt. Next, using the other formula V = e^3, take the derivative of both sides with respect to t using implicit differentiation again; then substitute the values of e and de/dt that you get when S = 60 and you will have your solution. I hope this is enough explanation to help you. If not, write back and I or one of the other Math Doctors will give you some more help. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
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