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Arclength math question

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Given [tex]r(t) = x(t)\textbf{i} + y(t)\textbf{j}[/tex] and

    [tex]\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = t.[/tex]

    find, if possible, a closed form expression for x(t) and y(t).

    2. Relevant equations

    3. The attempt at a solution
    I started by applying the fundamental theorem of calculus


    then evaluated and simplified and got

    [tex]x(t)^2 + y(t)^2 = 1.[/tex]

    Is it possible to continue from here?
  2. jcsd
  3. Feb 28, 2008 #2
    How are you getting [itex]x(t)^2 + y(t)^2 = 1[/itex]?
  4. Feb 28, 2008 #3


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    Staff: Mentor

    Not sure how you determined this from the fundamental theorem.

    [tex]x(t)^2 + y(t)^2 = 1.[/tex] This applies to a unit circle.

    Why t in this equation?
    [tex]\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = t.[/tex]

    The integral should provide length, not time.
  5. Feb 28, 2008 #4
    Oops. I meant

    [tex]x'(t)^2 + y'(t)^2 = 1.[/tex]

    As to how I got there:
    [tex]\frac{d}{dt}\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = \frac{dt}{dt}[/tex]

    [tex]\left\|\frac{dr}{dt}\right\| = 1[/tex]

    [tex]\sqrt{x'(t)^2 + y'(t)^2} = 1[/tex]

    [tex]x'(t)^2 + y'(t)^2 = 1.[/tex]

    I'm basically trying to find some vector-valued function r(t) in [tex]\texttt{r}^2[/tex] such that [tex]\forall t, s(t) = t[/tex], where [tex]s(t)[/tex] is arclength.
  6. Feb 28, 2008 #5
    That's better. You can play around with this by integrating with respect to t. However, you have two unknowns and one equation. I suspect that, without some additional assumptions, you won't be able to determine x(t) and y(t).

    Err, why would you do that? Since the integrator is [itex]d\tau[/itex], you can take |dr/dt| out because it doesn't involve [itex]\tau[/itex].
  7. Feb 28, 2008 #6
    This makes no sense to me. Where did s(t) come from? How are you going from a vector-valued function r(t) to a scalar-valued function s(t)?
  8. Feb 28, 2008 #7
    Again, a mistake. >_<
    [tex]\frac{d}{dt}\int_0^t \left\|\frac{dr}{d\tau}\right\| d\tau = \frac{dt}{dt}[/tex]

    [tex]s(t) := \int_0^t \left\|\frac{dr}{d\tau}\right\| d\tau[/tex]
  9. Feb 28, 2008 #8
    I will assume that the derivative inside the integral is [itex]d/d\tau r(\tau)[/itex] for otherwise the integral would evaluate to 0.

    The integral pretty ugly. Taking the derivative of the integral with respect to t doesn't help.
  10. Feb 29, 2008 #9


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    Staff Emeritus
    Science Advisor

    Yes, it does. By the fundamental theorem, mentioned before,
    [tex]\left|\left|\frac{dr}{dt}\right|\right|= 1[tex]
    just as he said before. That gives, just as he said before,
    [tex]\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)= 1[/itex].

    In fact, that trajectory is the unit circle.
  11. Feb 29, 2008 #10
    If you bring the d/dt inside the integral and evaluate the inside of the integral before integrating, then since the integrand is with respect to tau, the derivative with respect to t will be 0.

    Am I missing something?
  12. Feb 29, 2008 #11
    Yes, the fundamental theorem of calculus which says

    [tex]\frac{d}{dx}\int_a^x f(t) dt = f(x)[/tex]

    where a is some constant.

    [tex]\int_a^x f(t) dt[/tex] is a function of x. In other words, you can't bring the [tex]\frac{d}{dx}[/tex] inside the integral because that would be like saying [tex]\frac{d}{dx}\left [x^2\right] = \left[\frac{d}{dx}(x)\right]^2,[/tex] which is obviously untrue.
  13. Feb 29, 2008 #12
    Right. So it does simplify to [itex]x(t)^2 + y(t)^2 = 1[/itex]. Thus, x(t) = cos t and y(t) = sin t is a possible solution right?
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