November 30th, 2008, 09:06 AM  #1 
Newbie Joined: Nov 2008 Posts: 12 Thanks: 0  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 
November 30th, 2008, 09:31 AM  #2 
Guest Joined: Posts: n/a Thanks:  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  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond  Re: Related Rate formula conversion Quote:
 

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at a certain instant each edge of a cube is 5 in long,at a certain instant each edge of a cube,dv/dt= 1.2 x^2 dx/dt cube
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