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Related Rates

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 
    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


   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/   
Associated Topics:
High School Calculus

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