MHB How to Convert Polar to Rectangular Coordinates in Calculus?

[karush]

$r=3-\cos\left({\theta}\right)$
${r}^{2}=3r-r\cos\left({\theta}\right)$
${x}^{2}+{y}^{2}=3r+x$
How you deal with 3r ?

 

[topsquark]

$r=3-\cos\left({\theta}\right)$
${r}^{2}=3r-r\cos\left({\theta}\right)$
${x}^{2}+{y}^{2}=3r+x$
How you deal with 3r ?

You have a slight mistake: [math]x^2 + y^2 = 3r - x[/math]

As always [math]r = \sqrt{x^2 + y^2}[/math].

Continuing:
[math]x^2 + y^2 = 3 \sqrt{x^2 + y^2} - x[/math]

[math]x^2 + x + y^2 = 3 \sqrt{x^2 + y^2}[/math]

[math]\left ( x^2 + x + y^2 \right ) ^2 = 9(x^2 + y^2)[/math]

etc.

Yes, it's ugly.

-Dan

 

[karush]

Funny I had that answer and thot it was wrong, quess intuition is always right?