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Unit speed curves

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Let ##\alpha(s)## and ##\beta(s)## be two unit speed curves and assume that ##\kappa_{\alpha}(s)=\kappa_{\beta}(s)## and ##\tau_{\alpha}(s)=\tau_{\beta}(s)##, where ##\kappa## and ##\tau## are respectively the curvature and torsion. Let ##J(s) = T_{\alpha}(s)\dot\ T_{\beta}(s)+N_{\alpha}(s) \dot\ N_{\beta}(s) +B_{\alpha}(s) \dot\ B_{\beta}(s).##

    Show that:

    ##J(0)=3## and ##J(s)=3## implies that the Frenet frames of ##\alpha## and ##\beta## agree at ##s##

    ##J'(s) = 0## and ##\alpha(s) = \beta(s)## for all ##s##.


    2. Relevant equations

    Frenet frames


    3. The attempt at a solution

    For the first question, I know that the Frenet frame vectors $T,B,N$ are unit vectors but how can I formally prove the given statement?

    I know for the second statement that ##J'(s) = 0## everywhere implies that the Frenet frames agree and since ##\alpha## and ##\beta## are unit speed, they are equal to the integral of their tangent vector. Also, how can I formally prove this second part?
     
  2. jcsd
  3. Oct 1, 2013 #2

    Dick

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    For any two vectors v.w=|v||w|cos(θ). If v and w are unit vectors then that's less than one unless the vectors are parallel. So?
     
    Last edited: Oct 1, 2013
  4. Oct 1, 2013 #3
    Dick -

    If they are unit vectors and are parallel then there dot product is 1.
     
  5. Oct 1, 2013 #4

    Dick

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    The point is the dot product is ONLY 1 if they are parallel, otherwise it's less. You have three of them summing to 3. So?
     
  6. Oct 1, 2013 #5
    Since they sum to 3 they are parallel hence their dot product equals 1.

    Is it really that simple or am I missing something?
     
  7. Oct 1, 2013 #6

    Dick

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    No, I think it's really that simple.
     
  8. Oct 2, 2013 #7
    Thanks Dick!

    For D(s), is it essentially the same thing?
     
  9. Oct 2, 2013 #8

    Dick

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    What's D(s)? Look, if J(s)=3 then all three Frenet frame vectors must be equal in ##\alpha## and ##\beta##, yes? Am I missing something?
     
    Last edited: Oct 2, 2013
  10. Oct 2, 2013 #9
    Opps, I meant J(s), sorry about that. But yes they will be equal, thanks again!
     
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