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Prove that any curve can be parameterized by arc length

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that any curve [itex]\Gamma[/itex] can be parameterized by arc length.

    2. Relevant equations

    Hint: If η is any parameterization (of [itex]\Gamma[/itex] I am guessing), let [itex]h(s) = \int^{s}_{a} \left| \eta ' (t) \right| dt[/itex] and consider [itex]\gamma = \eta \circ h^{-1}.[/itex]

    3. The attempt at a solution

    Given the hint, we have [itex]\gamma = \eta \circ h^{-1} (t) = (x(h^{-1} (t), y(h^{-1} (t))[/itex] and thus [itex] \left| \gamma ' (t) \right| = \sqrt{ [x' (h^{-1} (t))]^{2} + [y' (h^{-1} (t))]^{2}}[/itex] but what good is this if I can't find the inverse of h(t)?

    I know I have to show there exists a parameterization [itex]\gamma : [a, b] \rightarrow \Re^{2}[/itex] of [itex]\Gamma[/itex] so that [itex]\left| \gamma ' (t) \right| = 1[/itex] for all t. Perhaps this hint will reveal this parameterization, but I can't see how. Any help would be appreciated.
     
  2. jcsd
  3. Mar 26, 2013 #2
    Your expression for ## |\gamma'(t)| ## is incorrect. Fix that, then apply the rule for differentiating inverse functions.
     
  4. Mar 28, 2013 #3
    What exactly is wrong with my expression?

    If gamma is a parameterization, then |γ'| = sqrt ( x'(t)^2 + y'(t)^2), is it not?
     
  5. Mar 28, 2013 #4
    The equation is correct in #3, but not in #1.
     
  6. Mar 28, 2013 #5
    But the parametrization is a function of h^-1 which is a function of t. Would I need to use chain rule?
     
  7. Mar 28, 2013 #6
    Yes, you have to use the chain rule. Which you tried, but you either did not do it correctly, or did not write the result correctly.
     
  8. Mar 28, 2013 #7
    Oh I forgot because h is a function of s, not t. All this time I thought it was a function of t. Alright I'm plugging away I will be back.
     
  9. Mar 28, 2013 #8
    No, that was not the problem. The problem is that the chain rule should have resulted in a product of two derivatives.
     
  10. Mar 28, 2013 #9
    So

    [itex]\eta (t) = (x(t), y(t))[/itex] is any parametrization. Then let

    [itex]h(s) = \int^{s}_{a} | \eta ' (t) | dt [/itex]

    Now consider
    [itex]\gamma (t) = \eta (h^{-1} (s)) = (x(h^{-1} (s)), y(h^{-1} (s))).[/itex]

    We have

    [itex]| \gamma ' (t) | = | \gamma ' (h^{-1} (s)) | = \sqrt{(x'(h^{-1} (s))([h^{-1}]' (s))^{2} + (y'(h^{-1} (s))([h^{-1}]' (s))^{2}}[/itex]

    [itex] = ... = |\eta ' (s) | \sqrt{ x'(h^{-1} (s))^{2} + y'(h^{-1} (s))^{2}}[/itex]

    but now what?
     
  11. Mar 28, 2013 #10
    That is not correct. ## (h^{-1})' \ne |\eta'| ##.
     
  12. Mar 28, 2013 #11
    Well we defined [itex]h(s) = \int^{s}_{a} | \eta ' (t) |dt[/itex] didn't we? So by the FTC we end up with [itex] | \eta ' (t) | or | \eta ' (s) | [/itex]?

    Chain rule is really irritating me right now....
     
  13. Mar 28, 2013 #12
    Yes, ## h' ## is indeed equal to ## |\eta'| ##. But you have ## h^{-1} ## to differentiate.
     
  14. Mar 28, 2013 #13
    Right. So it'd be |n'|', but that still doesn't seem useful.
     
  15. Mar 28, 2013 #14
    Furthermore, how am I going to deal with the square root stuff?
     
  16. Mar 28, 2013 #15
    Why would that be so? Say you have y = f(x), and you also have its inverse x = g(y): g(f(x)) = x. Is there an equation linking f' and g'?
     
  17. Mar 28, 2013 #16
    I think thats enough for me today. I'll hand in wha I've got. Thanks for your help!
     
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