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Math Help - Finding arc lengths

  1. #1
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    Finding arc lengths

    find the exact arc length of the parametric curve:

    x=t^2cost, y=t^2sint, 0<=t<=2

    how do i go about solving the above question?
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  2. #2
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     \displaystyle L = \int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2 + \bigg(\frac{dy}{dt}\bigg)^2} \, dt

    for  \displaystyle\alpha \leq t \leq \beta (so in your case  \alpha = 0 \, , \beta = 2

    So, find  \frac{dx}{dt} and  \frac{dy}{dx} , then plug them into that equation, simplify the integrand as much as possible, and then integrate it with the limits.
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  3. #3
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Jonny-123 View Post
    find the exact arc length of the parametric curve:

    x=t^2cost, y=t^2sint, 0<=t<=2

    how do i go about solving the above question?
    Also see this picture



    This is easier to calculate the integral \int\limits_0^2 {t\sqrt {{t^2} + 4} dt}, using this substitution \sqrt {{t^2} + 4}  = u.

    \int\limits_0^2 {t\sqrt {{t^2} + 4} dt}  = \left\{ \begin{gathered}\sqrt {{t^2} + 4}  = u, \hfill \\tdt = udu, \hfill \\0 \leqslant t \leqslant 2, \hfill \\2 \leqslant u \leqslant 2\sqrt 2  \hfill \\ \end{gathered}  \right\} = \int\limits_2^{2\sqrt 2 } {{u^2}du}  = \left. {\frac{{{u^3}}}{3}} \right|_2^{2\sqrt 2 } =

    = \frac{1}{3}\left( {8\sqrt 8  - 8} \right) = \frac{8}{3}\left( {2\sqrt 2  - 1} \right) \approx 4.8758{\text{ }}\left( {{\text{linear units}}} \right).
    Last edited by DeMath; August 6th 2009 at 10:29 AM.
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