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

Thanks in advance for your help!

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- May 13th 2010, 12:39 PMcombover1234Need 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

Thanks in advance for your help! - May 13th 2010, 01:38 PMojones
Wow, are you sure you have this right? Looks like something your professor made up while drunk. Assuming it is correct, rewrite it as:

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

$

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

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

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