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 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. $V=x^{3}, \;\ \frac{dV}{dt}=3x^{2}\cdot\frac{dx}{dt}$ 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. $S=6x^{2}$
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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