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Polar coordinates

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    change the following equation into polar form:


    3. The attempt at a solution
    r*sin(t) = r^2 * cos(t)^2

    stuck after this... my friend suggested that I cancel an r, but won't that get rid of one of the solutions?

    I'm not really sure how to proceed
  2. jcsd
  3. Jan 26, 2008 #2
    Well, from what I remember, I would say that your friend is right. Cancelling an r is the way to go, mostly because it's nice to have the answer in the "r=" form. It's really the same as if you had x^2=xy, to solve for y you'd simple cancel an x to get y=x.
  4. Jan 26, 2008 #3
    I thought that since "r" was a variable we couldn't cancel it out because we would be losing one of the solutions...

    but assuming we can, how should I proceed?
  5. Jan 27, 2008 #4
    I believe dividing both sides by r* cos(t)^2 gives you the equation in polar coordinates solved for r. Does the question ask you to do anything else?
  6. Jan 27, 2008 #5
    take care this part is not right, maybe it is just a typo, but it shoul read like this
    [tex] (cos(t))^{2}[/tex]
  7. Jan 27, 2008 #6
    By just dividing both sides by r you will defenitely loose one solution, the one when x takes negative values.
  8. Jan 27, 2008 #7
    yes, that is how it should read
  9. Jan 27, 2008 #8
    any tips on how to get started? :)
  10. Jan 27, 2008 #9
    well one solution is going to be zero. I think there are two ways of going about this
    r*sin(t) = r^2 * (cos(t))^2
    r^2*(cos(t))^2 -r*sin(t)=0 now factor a r out
    r( r*(cos(t))^2 -sin(t))=0, so which are the two solutions here?
    or you might want to solve this r^2*(cos(t))^2 -r*sin(t)=0 as a quadratic equation.
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