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Homework Help: 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


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    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


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    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
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