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### Related Rates: The Edges of an Expanding Cube

```
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
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/
```
Associated Topics:
High School Calculus

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