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February 1st, 2016, 05:18 PM   #11
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At $t=-1$, $x=\sqrt{(-1)^2+3}=2$. $\dfrac{dx}{dt}=\dfrac{t}{\sqrt{t^2+3}}$. At $t=-1$, $\dfrac{dx}{dt}=-\dfrac12$.

So we have $\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=(3x^2-10x)\cdot\dfrac{-1}{2}=-\dfrac12(3(2)^2-10(2))=4$.
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Last edited by greg1313; February 2nd, 2016 at 06:52 AM.
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February 2nd, 2016, 04:06 AM   #12
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At $t=-1, \ \dfrac{dx}{dt}=-\dfrac12$, not $-\dfrac14$.

Also, $\dfrac{dy}{dx} = 3x^2 - 10x$, not $3x^2 - 5x$.
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February 2nd, 2016, 06:03 AM   #13
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Thanks. I've edited my post.
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February 2nd, 2016, 06:50 AM   #14
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Your final answer now has the wrong sign.
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February 2nd, 2016, 06:53 AM   #15
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Ok. Edited.
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February 2nd, 2016, 09:41 AM   #16
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So my answer was right , I just did not simplify
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