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Math Help - Integral for the length of a curve.

  1. #1
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    Integral for the length of a curve.

    Any help with the following problem would be greatly appreciated.

    Set up, but do not evaluate, an integral for the length of the curve.

    x^2/a^2 + y^2/b^2 = 1

    Thanks.
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  2. #2
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    There are many methods , the choice depends on what you know about it .

    we have  L = \int_a^b \sqrt{ 1 + y'^2 }~dx~,~ y = f(x)

    or parametric form :

     L = \int_{t_1}^{t_2} \sqrt{ x'^2(t) + y'^2(t) }~dt


    Make y as the subject , we have

     y = (b/a) \sqrt{ a^2 - x^2 } and

     y' = (b/a) \cdot \frac{ -2x}{2 \sqrt{a^2-x^2}}  = \frac{ -bx}{a \sqrt{ a^2 - x^2 } }

    Therefore ,  L = \int_{-a}^a \sqrt{ 1 + \left ( \frac{ -bx}{a \sqrt{ a^2 - x^2} } \right )^2 } ~dx

    But  L indicates the length of the semi-ellipse but not the whole ellipse , for the whole ellipse you need to multiply  L by  2 .
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  3. #3
    MHF Contributor

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    As simplependulum said, you can also do this with parametric equations.

    Standard parametric equations for the ellipse \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 are x= a cos(t) and y= b sin(t). Then x'= -a sin(t) and y'= b cos(t) so the arclength is \int_{t= 0}^{2\pi}\sqrt{a^2 sin^2 t+ b^2 cos^2 t} dt

    By the way, the reason the problem said "Set up, but do not evaluate" the integral is that "ellipitic integrals" [b]cannot[b] be integrated analytically.
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