Conversion of polar equation to rectangular equation

  • Context: High School 
  • Thread starter Thread starter Burger2010
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    Polar Rectangular
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Discussion Overview

The discussion revolves around converting a polar equation, specifically r = (15)/(3-2cos(theta)), into its rectangular form. Participants explore the mathematical transformations involved in this conversion, including the relationships between polar and rectangular coordinates.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses confusion about the conversion process and seeks assistance.
  • Another participant provides a method for conversion using the relationships r = (x^2+y^2)^(1/2), x = rcos(theta), and y = rsin(theta), suggesting that substituting these into the original equation can lead to a solution.
  • A third participant reformats the previous contributions into TeX for clarity, reiterating the substitution steps involved in the conversion.
  • A later reply proposes a potential outcome of the conversion, suggesting that the result is y = ±(√(5(15-x)(x+3)))/3, indicating that the solution represents a closed curve.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the conversion process, as there is a mix of confusion, proposed methods, and a suggested outcome without agreement on the correctness of the steps taken.

Contextual Notes

The discussion includes various assumptions about the relationships between polar and rectangular coordinates, and the mathematical steps involved in the conversion remain unresolved. There may be dependencies on specific definitions and interpretations of the equations used.

Burger2010
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r= (15)/(3-2cos(theta))


I'm lost!
Please Help!
:confused:
 
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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
 
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]
 
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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