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Rectangular Equation to Polar

  1. Mar 5, 2006 #1
    I need to convert [tex]x^2+y^2-3cos\Theta+4sin\Theta=0[/tex] to polar.

    Obviously the [tex]x^2+y^2[/tex] part would = [tex]r^2[/tex], but how can I get the cos and sin part to simplify?
     
    Last edited: Mar 5, 2006
  2. jcsd
  3. Mar 5, 2006 #2

    quasar987

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    what does it mean to have an expression written out in polar coordinates?
     
  4. Mar 5, 2006 #3
    Who said anything about coordinates. The polar form of the equation is what I need to achieve.
     
  5. Mar 5, 2006 #4

    quasar987

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    Those are synonyms as far as i know. Why, do you make a distinction btw an equn being written in polar coordinates and its "polar form"?
     
  6. Mar 5, 2006 #5
    Well, actually, I guess they are the same thing, but either way, I don't know how to do it :(

    As for your question, I don't see a purpose, but my teacher sure does.
     
  7. Mar 5, 2006 #6

    quasar987

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    The purpose of my question is that if you answer it, you'll probably see the answer to your question.

    Hint: Fill the "?" spots. 'An equation is written in rectangular coordinates if it is of the form [itex]F(x,y)=0[/itex]. Similarily, an equation is written in polar coordinates if it is of the form [itex]F(?,?)=0[/itex].'
     
  8. Mar 5, 2006 #7
    I don't want coordinates for when the value equals 0. All I want, is to convert the rectangular equation into polar form. So instead of having x and y I need r and [tex]\Theta[/tex].

    Currently, I have [tex]r^2-3cos\Theta+4sin\Theta=0[/tex] but I don't know how to convert the sin and cos into polar form.
     
  9. Mar 5, 2006 #8

    quasar987

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    This is not so surprising, since you do not even know what an 'equation being in polar form' means in the first place.

    I tried to help you with the definition. All you had to do was to write r and [itex]\Theta[/itex] where the ? were. The definition now tells you that an equation is in polar form if it is of the form [itex]F(r, \Theta)=0[/itex].

    What you have achieved so far is to get it into the form

    [tex]r^2-3cos\Theta+4sin\Theta=0[/tex].

    Does that match the definition?
     
  10. Mar 5, 2006 #9
    Yes it does. So what do I do next?
     
  11. Mar 5, 2006 #10

    quasar987

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    What do you conclude about which form the equation [itex]r^2-3cos\Theta+4sin\Theta=0[/itex] is in?
     
  12. Mar 6, 2006 #11

    HallsofIvy

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    I'm going to jump in and ask: what does the [itex]\theta[/itex] mean in the original question? Normally, one does not have the polar angle (that is, the [itex]\theta[/itex] of polar coordinates) in a cartesian coordinate equation so I would not assume that that is what is intended.
     
  13. Mar 6, 2006 #12
    [tex]r-3x+4y[/tex]

    Remember x=cos(r) and y = Sin(r).
     
  14. Mar 7, 2006 #13

    HallsofIvy

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    Now that's just nonsense!
     
  15. Mar 7, 2006 #14
    trying to go from [tex](x,y) = (r,theta)[\tex] right?

    x^2+y^2-3cosX+4sinX

    r^2-3cosx+4sinX

    cos(theta) = x/r
    Sin(theta) = y/r
    r^2-3(x/r)+4(y/r)
    just work from there.
     
  16. Mar 7, 2006 #15
    konartist,
    multiplying both sides by r will yield 3x+4x=r times r^1/2
     
  17. Mar 8, 2006 #16
    the equation is correct once it is in terms of R and [tex]\theta[/tex] there is nothing more you can do with it
     
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