1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    [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]


    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook