Surface Area and Volume: Related Rates

Date: 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/