1. Not finding help here? Sign up for a free 30min 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!

Writing x^2 + y^2 = 1 + sin^2(xy) in polar form

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Write the equation

    [tex]x^2 + y^2 = 1 + sin^2(xy)[/tex]

    in polar form assuming

    [tex]x = rcos(\phi)[/tex]
    [tex]y = rsin(\phi)[/tex]

    [tex]0<r, 0<= \phi < 2pi[/tex]

    solve for r as a function of [itex]\phi[/itex]

    3. The attempt at a solution


    [tex](rcos(\phi))^2 + (rsin(\phi))^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

    [tex]r^2(cos^2(\phi) + sin^2(\phi)) = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

    [tex]r^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

    At this point I'm feeling pretty lost, since I have no idea how to get the all r:s alone on one side of the equation. More specifically I don't understand how to get them out of the trig function.
     
  2. jcsd
  3. Oct 10, 2012 #2
    Hello.
    I continued your calculation. it should be
    φ = 1/2 * arcsin [2 arcsin{sqrt(r^2 - 1)} / r^2 ]
    Regards.
     
  4. Oct 10, 2012 #3
    Thanks for the help, but I'm afraid I don't really understand exactly what you have done.

    It looks like you applied arcsin twice as well as some other operations to solve for [itex]\phi[/itex] rather than r.

    Wouldn't arcsin be a problem since that isn't defined for [0 to 2pi]?
     
  5. Oct 10, 2012 #4
    Hi.
    The function stands for both φ and -φ, i.e. even function or symmetric for x axis.
    The function stands for both φ and π-φ, i.e. even function or symmetric for y axis.
    The function stands for both φ and π/2-φ, i.e. symmetric for y=x
    So you can narrow the range of φ to [0,π/4] in your estimation.
    Regards.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook