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Polar Coordinates

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx
    Problem in picture below

    2. Relevant equations



    3. The attempt at a solution
    [PLAIN]http://img28.imageshack.us/img28/7162/76013837.png [Broken]

    The answer for this is
    dy/dx = [tex]-cos\theta sin\theta + (1-sin\theta)cos\theta[/tex]/[tex]-cos^2\theta - (1-sin/theta)sin\theta [/tex] I cannot figure out why I'm not getting the same or equivalent for dy/dtheta.... and trig makes me woozy.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 4, 2011 #2

    Pengwuino

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    Your [itex]dy/d\theta[/itex] is equivalent. Take a look at the second to last line of your calculation before you used the double angle equation.
     
  4. Sep 5, 2011 #3
    I did Peng Peng, but I just can't see it -_- Which type of identity did they use?
     
  5. Sep 5, 2011 #4

    Pengwuino

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    They didn't! [itex]\cos(\theta) - 2\sin(\theta)\cos(\theta) = -\cos(\theta)\sin(\theta) +(1-\sin(\theta))\cos(\theta)[/itex]

    The tex translator seems to be pooping itself.
     
  6. Sep 5, 2011 #5
    @_@ I really need to take another trig class...let me try to work it out.
     
  7. Sep 5, 2011 #6

    Pengwuino

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    Just write out the right hand side. There is no trig involved. They just wrote down the answer in a way that the 'r' shows up again. They could have written it

    [itex]-\cos(\theta)\sin(\theta) +r\cos(\theta)[/itex]

    if they wanted to be more explicit.
     
  8. Sep 5, 2011 #7
    omg I see it now...maybe I need to take another algebra class -_-

    Thx peng!
     
  9. Sep 5, 2011 #8
    yes thx peng I saw it after I started to think "algebra" and not trig -_-
     
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