Results 1 to 5 of 5

Math Help - Arc length from parametric

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    11

    Arc length from parametric

    I have put in the formula for arc length and am fairly sure I have the derivatives figured correctly. The problem is, I'm unable to put the equation into a form that is integratable.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    You forgot to square dx/dt and dy/dt
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    11
    Thanks for pointing that out. Anyway, I seem to have made another mess of it. I've been doing this calculus for hours now and can't seem to see straight anymore so I'm going to take a break. This is what I have so far but it is not pretty. Maybe trig substitution isn't the way to go but I can't see how else to do it.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2008
    Posts
    11
    It looks like I may have gotten a sign wrong. Now I'm left with something that looks quite simple, I might be on the right track.

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by CrazyLond View Post
    I have put in the formula for arc length and am fairly sure I have the derivatives figured correctly. The problem is, I'm unable to put the equation into a form that is integratable.

    dx=\frac{1}{\sqrt{1-t^2}}

    dy=\frac{-t}{1-t^2}

    Arc length then is

    \int_0^{\frac{1}{2}}\sqrt{\frac{1}{1-t^2}+\frac{t^2}{(1-t^2)^2}}dt

    Seeing that our quantity inside our squareroot is equal to

    \frac{(1-t^2)}{(1-t^2)^2}+\frac{t^2}{(1-t^2)^2}=\frac{1}{(1-t^2)^2}

    So we have

    \int\frac{dt}{1-t^2}

    if you look all the values within our limits of integration give a positive result, so we can forgoe our absolute value. So we have

    \int_1^{\frac{1}{2}}\frac{dt}{1-t^2}={\rm{arctanh}}(t)\bigg|_{0}^{\frac{1}{2}}=\fr  ac{\ln(3)}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric Arc Length
    Posted in the Calculus Forum
    Replies: 20
    Last Post: August 11th 2010, 01:16 PM
  2. Parametric Arc Length
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 2nd 2009, 05:27 PM
  3. parametric arc length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 10th 2009, 01:12 PM
  4. Length of Curve (Parametric)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 26th 2009, 05:09 AM
  5. arc length parametric
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 10th 2008, 10:54 PM

Search Tags


/mathhelpforum @mathhelpforum