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March 22nd, 2009, 06:30 AM   #1
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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.
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March 22nd, 2009, 07:50 AM   #2
keb
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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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