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Truly Bizarre - The unit tangent and unit normal vectors aren't orthogonal!

  1. Sep 24, 2012 #1
    OK, this looks like a differential geometry problem, which it is, but at the end of the day I am trying to figure out why the unit normal and unit tangent vectors to a curve aren't orthogonal, so even if you don't know about DG, please respond.

    Joseph-2.png

    Obviously the two choices for E_1 and E_2 are the unit normal and unit tangent vectors to the curve.

    Using Mathematica...

    Or by hand...


    [itex]\alpha(t) = {Cos(t), 2Sin(t)}[/itex]
    [itex]\alpha'(t) = {-Sin(t), 2Cos(t)}[/itex]
    [itex]\alpha''(t) = {-Cos(t), -2Sin(t)}[/itex]

    However, graphically, the unit tangent and unit normal vectors are far from perpendicular on this curve!

    642.png

    Here is my mathematica code

     
  2. jcsd
  3. Sep 24, 2012 #2

    Dick

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    The unit normal isn't alphaprimeprime[t] / Norm[alphaprimeprime[t]]. It's the derivative of the UNIT tangent vector divided by the norm of the derivative of the UNIT tangent vector. alphaprime isn't UNIT.
     
    Last edited: Sep 24, 2012
  4. Sep 24, 2012 #3
    Thanks, and that would seem to do the trick, but mathematica is hating me right now. Any ideas what's going on? Here is the newly defined unit normal vector


    failure.png


    Implementation

     
  5. Sep 24, 2012 #4

    Dick

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    Not really, sorry. It looks right. But I haven't done Mathematica since other people stopped paying for it for me. It's really expensive and unfree. So I can't test that in any detail.
     
  6. Sep 24, 2012 #5
    Well, at least theoretically, would you agree that my choice of E1 and E2 provide a frame field on the curve?
     
  7. Sep 24, 2012 #6

    Dick

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    Sure. The unit tangent T points along your curve. T' must be perpendicular to that. Just differentiate T.T=1. It's gotta work, right?
     
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