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 March 22nd, 2009, 06:30 AM #1 Newbie   Joined: Jan 2009 Posts: 6 Thanks: 0 determine dx/dt If y^3 = 9 x^2, determine {dx}/{dt} when x = 3 and {dy}/{dt} = -3 . I did the following step, but the answer I have is wrong. Anyone point out what I did wrong please. y^3=9x^2 3y^2dy/dt=18xdx/dt 3y^2(-3)=18(3)dx/dt -9y^2=54dx/dt (-9y^2)/54=dx/dt That's where I was up to, but the answer was wrong. I can't finish it up to this point.
 March 22nd, 2009, 07:50 AM #2 Newbie   Joined: Mar 2009 Posts: 4 Thanks: 0 Re: determine dx/dt I think {dy}/{dt} = -3 means that y = -3t Then -81t^2dt = 18xdx and dx/dt = (-81t^2)/18x Using techniques for differentiation of an implicit function, namely Z(x,t)

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### dy/dt means y=3

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