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Homework Help: Cardioid Length

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the length of the curve:
    x = 2a cos t - a cos 2t
    y = 2a sin t - a sin 2t


    2. Relevant equations

    [tex] L = \int_{C} dl [/tex]

    3. The attempt at a solution

    Well, after a lot of manipulating and using trig identities, I get this:

    [tex]L= 4a \int_{0}^{2\pi} \sin{t} dt [/tex]
    When I solve, I get 0.
    But - if I change the range to be not 0<t<2π but 0<t<π, and then multiply by 2 - I get 16a as the answer, which is correct.
    My question is why? I can see why it is like that "numerically" but not "philosophically".
    How do I identify a situation where I need to do it, without looking at the answers, next time?
    And was I doing everything correct? Or maybe it just turned out to be the correct answer by accident?
  2. jcsd
  3. Sep 15, 2009 #2


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    Science Advisor
    Homework Helper
    Gold Member

    Likely you have a mistake somewhere. I get:

    [tex]ds = \sqrt{x'(t)^2 +y'(t)^2}\ dt = 2\sqrt2 a \sqrt{1 - cos(t)}\ dt[/tex]

    which gives the correct answer.
  4. Sep 15, 2009 #3
    I had that too.
    [tex] 2\sqrt2 a \sqrt{1 - cos(t)}\ dt = 2\sqrt2 a \sqrt{2 sin^2(\frac{t}{2}} dt = 4a sin(\frac{t}{2})[/tex]

    Assuming that the trig identity is correct, what went wrong here?
    The latter form is much easier to integrate.
  5. Sep 15, 2009 #4
    oh. in the original integral I missed the [itex]\frac{t}{2}[/itex].
    nevermind, problem solved.
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