Rectangular Equation to Polar

  • #1
sportsguy3675
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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?
 
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  • #2
quasar987
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what does it mean to have an expression written out in polar coordinates?
 
  • #3
sportsguy3675
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Who said anything about coordinates. The polar form of the equation is what I need to achieve.
 
  • #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"?
 
  • #5
sportsguy3675
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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.
 
  • #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].'
 
  • #7
sportsguy3675
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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.
 
  • #8
quasar987
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sportsguy3675 said:
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.
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?
 
  • #9
sportsguy3675
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Yes it does. So what do I do next?
 
  • #10
quasar987
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quasar987 said:
The definition now tells you that an equation is in polar form if it is of the form [itex]F(r, \Theta)=0[/itex].

Does [itex]r^2-3cos\Theta+4sin\Theta=0[/itex] match the definition?

sportsguy3675 said:
Yes it does.

What do you conclude about which form the equation [itex]r^2-3cos\Theta+4sin\Theta=0[/itex] is in?
 
  • #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.
 
  • #12
konartist
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sportsguy3675 said:
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?

[tex]r-3x+4y[/tex]

Remember x=cos(r) and y = Sin(r).
 
  • #13
HallsofIvy
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konartist said:
[tex]r-3x+4y[/tex]

Remember x=cos(r) and y = Sin(r).
Now that's just nonsense!
 
  • #14
konartist
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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.
 
  • #15
Plastic Photon
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konartist,
multiplying both sides by r will yield 3x+4x=r times r^1/2
 
  • #16
BananaMan
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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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