# Need some help with polar!

• May 13th 2010, 12:39 PM
combover1234
Need some help with polar!
I've been having alot of trouble with this equation. I need to convert it into standard y intercept form...

http://i963.photobucket.com/albums/a...34/Capture.png
• May 13th 2010, 01:38 PM
ojones
Wow, are you sure you have this right? Looks like something your professor made up while drunk. Assuming it is correct, rewrite it as:

$r=8\cos^4 \theta [1 - \sqrt{(1- \cos \theta )/2}]^3+3\sin^2 \theta
$

Multiplying both sides by $r^{11/2}$ gives:

$r^{13/2}=8r^4\cos^4 \theta [\sqrt r - \sqrt{(r- r\cos \theta )/2}]^3+3r^{7/2}r^2\sin^2 \theta
$

Now substitute $x=r\cos \theta, y=r\sin \theta,$ and $r=\sqrt{x^2+y^2}.$

Note: I took the posive square root in the half-angle identity $\sin \theta/2 = \pm \sqrt{(1-\cos \theta)/2}.$ Convince yourself that this is justified.