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Curve length and very hard integral

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the length of the curve:

    [itex]\phi(t)=\left\{(5+\cos(3t))\cos(t), (5+\cos(3t))\sin(t) \right\}\mbox{ with } t\in [0, 2\pi][/itex]

    2. Relevant equations

    [itex]L_{\phi}= \int_{a}^{b}\sqrt{[x'(t)]^2+ [y'(t)]^2}\qquad (1.1)[/itex]

    Where

    [itex]x(t)= (5+\cos(3t))\cos(t)[/itex]

    [itex]y(t)= (5+\cos(3t))\sin(t)[/itex]

    [itex]a= 0\qquad b= 2\pi[/itex]


    3. The attempt at a solution


    Ok, i noticed that [itex]\phi(t)[/itex] is in this form:

    [itex]\phi(t)=(r(t)\cos(t), r(t)\sin(t))[/itex]


    so it can be expressed in polar form:

    [itex]r= r(t)\iff r=5+\cos(3t) \quad t\in[0,2\pi][/itex]

    so:

    [itex]L_{\phi}=\int_{0}^{2\pi}\sqrt{[r(t)]^2+[r'(t)]^2}= [/itex]

    [itex]= \int_{0}^{2\pi}\sqrt{(5+\cos(3t))^2+(-3\sin(3t))^2}dt[/itex]


    but this integral is not so easy to solve :(

    What can i do to solve it? I try to use wolfram but it gives me an approximate result.

    [Sorry, my English is not so good, forgive me if there are mistakes]
     
  2. jcsd
  3. Feb 20, 2013 #2

    CompuChip

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    So far I don't see any errors.
    Are you sure you need the result in closed form, and you're not allowed to do numeric integration?

    Also, are you sure it's not
    [tex]\phi(t) = \{ 5 + (3 \cos(3t) \cos(t)), 5 + (3 \cos(3t) \sin(t)) \}[/tex]
     
  4. Feb 20, 2013 #3
    Thanks for your reply CompuChip :)

    I can't use numeric integration, but if you tell that this integral can't be expressed in closed form I'll go to my teacher to check it.
     
  5. Feb 20, 2013 #4

    CompuChip

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    I'm not sure there is no clever substitution that will do the trick. But there is no immediately obvious one; and if the best Wolfram Alpha does is also give you a numeric answer, that's a pretty strong hint that there is no "nice" form.
     
  6. Feb 20, 2013 #5
    Yes, you are right, i have the same feeling. Thank you for your time!
     
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