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Math Help - parametric derivative

  1. #1
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    parametric derivative

    If x=t^3, y=2t^2 for t>0, then what does d^2y/dx^2 equal?
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  2. #2
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    \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}
    \frac{d^2y}{dx^2} = \frac{d}{dt} \left ( \frac{dy}{dx} \right ) \cdot \frac{dt}{dx}
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  3. #3
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    This is what the question is asking.
    \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}
    Step 1. Express y' = \frac{{dy}}{{dx}} in terms of t.

    Step 2. find \frac{{dy'}}{{dt}}, i.e. the derivative of step 1 with respect to t.

    Step 3. divide \frac{{dy'}}{{dt}} by  \frac{{dx}}{{dt}}.
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  4. #4
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    So is this what I should do?
     y=\frac{{2x}}{{t}} <br />
    Step 1: <br />
y'=\frac{{2}}{{t}}<br />
    Step 2: <br />
\frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}}<br />
    Step 3: <br />
\frac{{dx}}{{dt}}=3t^2 <br />
    so the answer would be  \frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}}<br />
.
    Is that right?
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  5. #5
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    Quote Originally Posted by chaddy View Post
    So is this what I should do?
     y=\frac{{2x}}{{t}} <br />
    Step 1: <br />
y'=\frac{{2}}{{t}}<br />
No, this is incorrect, you cannot treat t as a constant when you know it is a variable
    Step 2: <br />
\frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}}<br />
    Step 3: <br />
\frac{{dx}}{{dt}}=3t^2 <br />
    so the answer would be  \frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}}<br />
.
    Is that right?

    first find \frac{dx}{dt} and \frac{dy}{dt}

    you should get \frac{dx}{dt} = 3t^2 and \frac{dy}{dt} = 4t

    then you use \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}

    giving \frac{dy}{dx} = 4t \cdot \frac{1}{3t^2}

    want to finish this off ?
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  6. #6
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    Okay, so <br />
\frac{dy}{dx} = \frac{4}{3t}<br />
    Step 2: <br />
\frac{{dy'}}{{dt}}=\frac{{-4}}{{3t^2}}<br />
    Step 3: <br />
\frac{{dx}}{{dt}}=3t^2<br />
    so the answer would be <br />
\frac{{d^2y}}{{dx^2}}=\frac{{-4}}{{9t^4}}<br />
.
    Is that right?
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  7. #7
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    Yeah that looks good to me.
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  8. #8
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    Quote Originally Posted by chaddy View Post
    Okay, so <br />
\frac{dy}{dx} = \frac{4}{3t}<br />
    Step 2: <br />
\frac{{dy'}}{{dt}}=\frac{{-4}}{{3t^2}}<br />
    Step 3: <br />
\frac{{dx}}{{dt}}=3t^2<br />
    so the answer would be <br />
\frac{{d^2y}}{{dx^2}}=\frac{{-4}}{{9t^4}}<br />
.
    Is that right?
    By the way .... since y = 2t^2 you can express the result as a function of y ......
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