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Math Help - Differentiation Problem

  1. #1
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    Differentiation Problem

    So here's the problem:

    A curve is represented parametrically by x = (t-1), y = t. Find dy/dx in terms of t and show that
    <br />
\frac {d^2 y}{dx^2} = <br />
\frac {-3(t^2 +1)}{16t(t^2- 1)^3}<br />

    I tried to work it out, and I got

    dy/dx = <br />
\frac{3t^2}{4t^3-4t}<br />
(not sure if it's correct). And about the second derivative, it got too complicated.

    Thanks for any help.
    Last edited by mr fantastic; March 28th 2010 at 01:51 PM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by fdylan View Post
    First of all I'm not sure whether it's in the correct Section.

    So here's the problem:

    A curve is represented parametrically by x = (t-1), y = t. Find dy/dx in terms of t and show that
    <br />
\frac {d^2 y}{dx^2} = <br />
\frac {-3(t^2 +1)}{16t(t^2- 1)^3}<br />

    I tried to work it out, and I got

    dy/dx = <br />
\frac{3t^2}{4t^3-4t}<br />
(not sure if it's correct). And about the second derivative, it got too complicated.

    Thanks for any help.
    x = (t^2-1)^2 \: \rightarrow \: \frac{dx}{dt} = 4t(t^2-1) = 4t^3-4t

    \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{4t(t^2-1)} (this will be used later in the chain rule)


    y = t^3 \: \rightarrow \: \frac{dy}{dt} = 3t^2


    The chain rule says that \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}

    \frac{dy}{dx} = \frac{3t^2}{2t(t^2-1)} = \frac{3t}{4t^2-4} I have expanded the brackets to make the next step easier

    No problem with your original answer although you can cancel out a t.


    For the second part use the quotient rule.

    u = 3t \: \rightarrow u' = 3

    v = 4t^2-4 \: \rightarrow v' = 8t
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  3. #3
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    Many Thanks. Cutting out 't' makes it a lot easier I guess.
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  4. #4
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    Quote Originally Posted by fdylan View Post
    for the x part, I worked it out different.
    <br />
x = (t^2 -1)^2<br />
x = (t^2 -1)(t^2 -1)<br />
x = t^4 -2t^2 +1<br />
\frac{dx}{dt} = 4t^3 -4t<br /> <br />
\frac{dy}{dx} = \frac{3t^2}{4t^3 -4t}<br />
    It still works, I just used the chain rule.

    \frac{3t^2}{4t^3-4t} = \frac{3t^2}{4t(t^2-1)} = \frac{3t}{4(t^2-1)}

    t cancels
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