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!

Non-linear ODE, plane-polar coordinates.

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I have:


    And I need to find [itex]\dot{r}[/itex] and [itex]\dot{\theta}[/itex]

    2. The attempt at a solution

    I got as far as:


    How do I go from here to [itex]\dot{r}[/itex] and [itex]\dot{\theta}[/itex] ?
    Last edited: Mar 13, 2012
  2. jcsd
  3. Mar 13, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For the left sides of these last two equations you have ##x = r\cos\theta,\ y = r\sin\theta## to work with. Remembering that everything is a function of ##t##, you can caculate ##\dot x## and ##\dot y## by differentiating those, using the chain rule.
  4. Mar 13, 2012 #3
    Ok, so I have:

    [itex]\dot{r}\cos(\theta)-r\sin(\theta)\, \dot{\theta}=r(\sin(\theta)-\cos(\theta)(r^2-4))[/itex]


    Is there a "non-tedious" way of solving this?
  5. Mar 13, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    I can't speak for what you would consider tedious but the standard method would be to multiply the first equation by [itex]cos(\theta)[/itex], the second equation by [itex]sin(\theta)[/itex], and add the two equations. Then turn around and multiply the first equation by [itex]sin(\theta)[/itex], the second equation by [itex]cos(\theta)[/itex], then subtract.
  6. Mar 13, 2012 #5
    Thank you, sir.
  7. Mar 13, 2012 #6
    Have I got this second part right?

    You mean, I presume, [itex]\dot{r}[/itex], not r, on the left of both of those two equations. Now, how did you get "-1" on the right of the last equation? It looks to me like you have r on the right of both first two equations. Isn't r- r= 0?
    Last edited by a moderator: Mar 13, 2012
  8. Mar 13, 2012 #7


    [itex]-r(\sin^2 \theta+\cos^2 \theta)\dot{\theta}=r(\sin^2 \theta+\cos^2 \theta)[/itex]


    Last edited: Mar 13, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Non-linear ODE, plane-polar coordinates.
  1. Non Linear ODE (Replies: 0)

  2. Non-linear ODE (Replies: 1)