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November 30th, 2008, 09:06 AM   #1
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Related Rate formula conversion

At a certain instant each edge of a cube is 5 in. long and the volume is is increasing at the rate of 2 in^3/min. How fast is the surface area of the cube increasing?

What i've done so far is taken the derivative of the volume of a cube formula and plugged in the numbers into the derivative to end up with dV/dt = 150 in^3/min.
I'm not sure how I'm supposed to import this number into the surface area equation and that's what I need help with.

Thanks for your help
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November 30th, 2008, 09:31 AM   #2
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Re: Related Rate formula conversion

We are given dV/dt=2 and x=5 at a certain instant.



In this equation, we are given dV/dt and x.

Solve this for dx/dt and then plug it all into the derivative of the surface area formula.

 
November 30th, 2008, 09:42 AM   #3
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Re: Related Rate formula conversion

Quote:
Originally Posted by pranavpuck
...the volume is is increasing at the rate of 2 in^3/min.
dV/dt = 2 in^3/min, not 150.
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