Need some help with polar!

May 2010
1
0
I've been having alot of trouble with this equation. I need to convert it into standard y intercept form...


Thanks in advance for your help!
 
May 2010
274
67
Los Angeles, California
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.
 
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