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Math Help - Length of Curves

  1. #1
    Member VitaX's Avatar
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    Length of Curves

    Find the length of the curve

     x = \frac{y^3}{3} + \frac{1}{4y} from y=1 to y=3

    g'(y) = y^2 + \frac{y^{-2}}{4}

    L = \int \sqrt{1 + \left(y^2 + \frac{y^{-2}}{4}\right)^2}dy

    L = \int \sqrt{1 + y^4 + \frac{1}{2} + \frac{y^{-4}}{16}}dy

    L = \int \sqrt{1 + y^4 + \frac{3}{2} + \frac{y^{-4}}{16}}dy

    It says it in the book that 1 + \left(\frac{dx}{dy}\right)^2 is a perfect square but I'm not seeing that. Any help?

    Edit: I think I found my error. :|
    Last edited by VitaX; January 18th 2010 at 06:05 PM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    The problem starts here

    g'(y) = y^2 - 1/(4y^2)

    g' (y) ^2 = y^4 -1/2 + 1/(16y^2)

    1 + g' (y)^2 = y^4 +1/2 + 1/(16y^2) = [y^2 +1/(4y^2)]^2
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