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Math Help - Stuck on Parametric Arc Length

  1. #1
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    Stuck on Parametric Arc Length

    Hey I am not sure how to finish this problem...

    x= e^t +e^-t
    y=5-2t
    0\leq{t}\leq{5}

    \frac{dx}{dt}=e^t-e^{-t}
    \frac{dy}{dt}=-2
    (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t}-e^{-2t}+4
    A= \int_0^5 \! \sqrt{(e^{2t}-e^{-2t}+4)}  \, dt

    Do I have to complete the square to simplify?
    Thanks
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  2. #2
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    (e^{t}-e^{-t})^2 does not simplify to e^{2t}-e^{-2t}
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  3. #3
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    Quote Originally Posted by mmattson07 View Post
    Hey I am not sure how to finish this problem...

    x= e^t +e^-t
    y=5-2t
    0\leq{t}\leq{5}

    \frac{dx}{dt}=e^t-e^{-t}
    \frac{dy}{dt}=-2
    (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t}{\color{red}+e^{-2t}+2}
    ... which factorises, to square factors that cancel the square root... and then you can substitute u = e^t
    Last edited by mr fantastic; December 10th 2009 at 06:14 PM. Reason: Fixed the color code in latex
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  4. #4
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    Quote Originally Posted by mmattson07 View Post
    Hey I am not sure how to finish this problem...

    x= e^t +e^-t
    y=5-2t
    0\leq{t}\leq{5}

    \frac{dx}{dt}=e^t-e^{-t}
    \frac{dy}{dt}=-2
    (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t} -2e^{t}e^{-t} +e^{-2t}+4
    A= \int_0^5 \! \sqrt{(e^{2t}-e^{-2t}+4)}  \, dt

    Do I have to complete the square to simplify?
    Thanks
    Something times its reciprocal is simply 1.
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  5. #5
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    So (\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}-2e^te^{-t}+e^{-2t}+4

    and

    A= \int_0^5 \! \sqrt{(e^{2t}+e^{-2t}+2)}  \, dt

    I still am unsure how this simplifies...
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  6. #6
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    Quote Originally Posted by mmattson07 View Post
    So (\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}-2e^te^{-t}+e^{-2t}+4

    and

    A= \int_0^5 \! \sqrt{(e^{2t}+e^{-2t}+2)} \, dt

    I still am unsure how this simplifies...
    e^{2t} + e^{-2t} + 2 = (e^t + e^{-t})^2.
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  7. #7
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    Thank you...don't know how I didn't see that.
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