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Thread: Meridians onf surface of revolution.

  1. #1
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    Meridians onf surface of revolution.

    Hi Guys,

    I have a homework questions that I am stuck on, it's the last in the set and I can't do it!

    It reads:

    Prove that the meridians are geodesics when the curve: (x(u),0,u) is revolved around the z-axis.

    I have an idea of using the Geodesic Curvature formula etc but do not know where to start.

    If anyone could give me a hand, please let me know!

    Cheers.

    Patrick.
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  2. #2
    MHF Contributor
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    Re: Meridians onf surface of revolution.

    Consider a constant speed curve $\displaystyle \alpha : I \rightarrow \mathbb{R}^3$ along a meridian of the revolved surface S,
    where $\displaystyle I$ is some open interval of $\displaystyle \mathbb{R}$.
    To show $\displaystyle \alpha$ is a geodesic, you want to show that $\displaystyle \alpha''$ is orthogonal to S.

    Since it has constant speed, you can say something about the relationship between $\displaystyle \alpha'$ and $\displaystyle \alpha''$.
    Consider the plane P that defines the meridian by cutting through S.
    Since $\displaystyle \alpha$ stays in P, so do all of its derivatives.
    At any point along $\displaystyle \alpha$, P is spanned by $\displaystyle \alpha'$ and the normal to S at that point.
    Put it all together, and you deduce something about $\displaystyle \alpha''$ that proves that $\displaystyle \alpha$ is a geodesic.
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