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### 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

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

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

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/
```
Associated Topics:
High School Calculus

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