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Math Help - Arc Length

  1. #1
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    Arc Length

    x = 1 + 3t^2
    y = 4 + 2t^3
    0<= t >= 1

    k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2)
    then i mess up somewhere when im solving that
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  2. #2
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    Quote Originally Posted by amiv4 View Post
    x = 1 + 3t^2
    y = 4 + 2t^3
    0<= t >= 1

    k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2)
    then i mess up somewhere when im solving that
    L = \int_{0}^{1} 6t \sqrt{1 + t^2} \, dt.

    Make the substitution u = 1 + t^2.
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