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: Arc length of a regular parametrized curve

  1. Aug 31, 2013 #1
    Given [tex]t\in I[/tex]the arc length of a regular parametrized curve [tex]\alpha : I \to \mathbb{R}^3[/tex] from the point [tex]t_0[/tex] is by definition [tex]s(t) = \int^t_{t_0}|\alpha'(t)|dt[/tex] where [tex]|\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}[/tex] is the length of the vector [tex]\alpha'(t).[/tex] Since [tex]\alpha'(t) \ne 0[/tex] the arc length [tex]s[/tex] is a differentiable function of and [tex]ds/dt = |\alpha'(t)|.[/tex]

    This is where I get confused.

    It can happen that the parameter [tex]t[/tex]is already the arc length measured from some point. In this case, [itex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if [tex]|\alpha'(t)| = 1[/tex] then [tex]s = \int_{t_0}^t dt = t - t_0.[/tex]

    How did they get that it equals 1? I am not sure what they are saying?
  2. jcsd
  3. Aug 31, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If t = s, where s is the arclength, then ds/dt = 1.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted