Conversion of polar equation to rectangular equation

  • Thread starter Burger2010
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  • #1

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r= (15)/(3-2cos(theta))


I'm lost!!
Please Help!!
:confused:
 

Answers and Replies

  • #2
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in polar coordinates r=(x^2+y^2)^1/2 and also x=rcosQ ,y=rsinQ so putting these in above equation and cosQ=x/r we get( x^2+y^2)^1/2=15/(3-2*x/( x^2+y^2)^1/2) now this can be solved
 
  • #3
In case Vandanak's brilliant solution is hard to read, I've formatted his statements in TeX

Given:
[tex] r = \frac{15}{3-2cos(\theta)} [/tex]

[tex] r = \sqrt{x^2+y^2} [/tex]

[tex] x = rcos(\theta) [/tex]

[tex] y = rsin(\theta) [/tex]

substituting
[tex]\sqrt{x^2+y^2} = \frac{15}{3-2(\frac{x}{\sqrt{x^2+y^2}})} [/tex]
 
  • #4
arildno
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If I did this correctly, you should end up with:
[tex]y=\pm\frac{\sqrt{5(15-x)(x+3)}}{3}[/tex]

Thus, the solution is a closed curve.
 

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