1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    [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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted