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Plane curves question

  1. Mar 9, 2007 #1
    1. The problem statement, all variables and given/known data

    I want to show that if the chhord length ||f(s)-f(t)|| depends only on |s-t| then the f is part of a line or circlle. f may not be regular or unit speed.

    2. Relevant equations



    3. The attempt at a solution

    I'm trying to differentiete it and taking ||f'(t)||, but I really need help, because it's not working for me.
     
  2. jcsd
  3. Mar 9, 2007 #2

    Dick

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    You should be able to show |f'(t)| is a constant. For the rest of the problem I would think about trying to express the curvature of f(t) in terms of things like |f(s)-f(t)| as s->t and hence argue that it is also a constant.
     
  4. Mar 10, 2007 #3
    How do I show |f'(t)| is a constant? I get to the part: |f'(t)| = lim(dt->0) |a(dt)/dt| where a is the function ||f(s)-f(t)|| =a(|s-t|)... and then i'm stuck
     
  5. Mar 10, 2007 #4

    Dick

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    Divide both sides of your last expression by |s-t| and let s->t. I would conclude |f'(t)|=a'(0) for an appropriately defined a. Or this "|f'(t)| = lim(dt->0) |a(dt)/dt|". That looks to me like the definition of a'(0).
     
    Last edited: Mar 10, 2007
  6. Mar 10, 2007 #5
    Yes, but I think the 'appropriately defined a' is the catch. a can be any differentiable function, adn although i can show that |f'(t)| is a constant for certain a, how can I show it for all a?
     
  7. Mar 10, 2007 #6

    Dick

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    There's no 'catch'. You've defined a(|s-t|)=|f(s)-f(t)|. That defines a(x) for non-negative x. The only 'appropriately defined' case is whether you bother to define a(x) for negative x or keep referring to one sided derivatives. It's really nothing.
     
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