Results 1 to 4 of 4
Like Tree2Thanks
  • 2 Post By SlipEternal

Math Help - arc length problem

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Tucson, AZ
    Posts
    10

    arc length problem

    This is actually an arc length problem for the polar equation r=1+sin(x) (x being theta) After some work you get:
    L=integral from 0 to 2 times pi (2+2*sin(x))^(1/2) dx. It gets reduced to

    L=2^(1/2) times integral from 0 to 2 times pi (cos(x)/(1-sin(x))^1/2) dx

    But if u=1-sin(x) don't the limits of integration both equal 1?

    The book says the answer is L=8 The figure is a cardioid.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: arc length problem

    Instead of simplifying \sqrt{\cos^2{x}}=\cos{x}, you should have used \sqrt{\cos^2{x}}=|\cos{x}|, and then the integral splits into three parts: integrating from 0 to \frac{\pi}{2}, from \frac{\pi}{2} to \frac{3\pi}{2}, and from \frac{3\pi}{2} to 2\pi. The answer 8 is correct.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,858
    Thanks
    721

    Re: arc length problem

    To make it a little more clear what you are saying:

    L = \int_0^{2\pi} \sqrt{2+2\sin \theta}d\theta = \sqrt{2} \int_0^{2\pi} \left| \cos \theta \right| \dfrac{1}{\sqrt{1-\sin \theta}} d\theta

    Note the absolute value around \cos \theta since \sqrt{\cos^2 \theta} \ge 0 for all \theta.

    Since \cos \theta \ge 0 on \left[0,\dfrac{\pi}{2} \right]\cup \left[\dfrac{3\pi}{2}, 2\pi \right] and \cos \theta \le 0 on \left[ \dfrac{\pi}{2},\dfrac{3\pi}{2} \right], you need to break up the integral:

    L & = \sqrt{2}\left[ \int_0^{\tfrac{\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta - \int_{\tfrac{\pi}{2}}^{\tfrac{3\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta + \int_{\tfrac{3\pi}{2}}^{2\pi} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta \right]

    Using the substitution u = 1-\sin \theta, du = -\cos \theta d\theta, you get:

    \begin{align*}L & = \sqrt{2}\left[ -\int_1^0 u^{-1/2}du + \int_0^2 u^{-1/2}du - \int_2^1 u^{-1/2}du \right] \\ & = \sqrt{2}\left[ \int_0^1 u^{-1/2}du + \int_0^2 u^{-1/2}du + \int_1^2 u^{-1/2}du \right] \\ & = 2\sqrt{2}\int_0^2 u^{-1/2}du \\ & = 4\sqrt{2}(\sqrt{2}-0) = 8\end{align*}

    Edit: Someone else beat me to it.
    Thanks from hollywood and ruggutah
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2012
    From
    Tucson, AZ
    Posts
    10

    Re: arc length problem

    Wow ! thanks to both of you. I never considered that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arc Length Problem Issue
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 26th 2012, 10:33 PM
  2. Arc Length Problem with e
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 13th 2011, 05:30 AM
  3. Arc length problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 13th 2009, 08:30 PM
  4. Another Arc Length Problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 9th 2009, 12:20 AM
  5. Length of triangle problem?
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 19th 2007, 02:59 PM

Search Tags


/mathhelpforum @mathhelpforum