Posted by Zac on Sunday, January 6, 2013 at 6:19pm.
I assume you mean
r = 1 + 3 sin 2t
where t = theta
x = r cos t
y = r sin t
sin 2 t = 2 sin t cos t
r = 1 + 6 sin t cos t
r = 1 + 6 (y/r)(x/r)
r^3 = 1 + 6 x y
but r^2 = x^2+y^2
(x^2+y^2)^(3/2) = 1 + 6 x y
(x^2+y^2)^3 = 1 + 12 x y + 36 x^2 y^2
you can multiply the left out
wouldnt the "1"in the 1+.... in the 4th step become r^2 when you multiply both sides by r^2.
You are right
r^3 = r^2 + 6 x y
(x^2+y^2)^3 = (x^2+y^2)^2 + 6 x y
x^6 + 3 x^4y^2 + 3 x^2 y^4 + y^6
= x^4 + 2 x^2y^2 + y^4 + 6 x y
Thanks a lot for your help
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