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December 18, 2014

December 18, 2014

Posted by **Zac** on Sunday, January 6, 2013 at 6:19pm.

- Pre-Calc -
**Damon**, Sunday, January 6, 2013 at 6:33pmI 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

so

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

r= (x^2+y^2)^.5

so

(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

- Pre-Calc -
**Zac**, Sunday, January 6, 2013 at 6:55pmwouldnt the "1"in the 1+.... in the 4th step become r^2 when you multiply both sides by r^2.

- Pre-Calc -
**Damon**, Sunday, January 6, 2013 at 6:59pmYou are right

r^3 = r^2 + 6 x y

(x^2+y^2)^3 = (x^2+y^2)^2 + 6 x y

- Pre-Calc -
**Damon**, Sunday, January 6, 2013 at 7:07pmx^6 + 3 x^4y^2 + 3 x^2 y^4 + y^6

= x^4 + 2 x^2y^2 + y^4 + 6 x y

- Pre-Calc -
**Zac**, Sunday, January 6, 2013 at 8:11pmThanks a lot for your help

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