# Polar rose

by JanClaesen
Tags: polar, rose
 P: 56 I'm trying to express the polar rose as an implicit function: r(t)=sin t x = sin t * cos t y = sin^2 t Since sin t * cos t = (1/2) * sin 2t and sin^2 t = (1/2) * (1-cos 2t) (2x)^2 + (1-2y)^2 = 1 4x^2 -4y + 4y^2 = 0 When I plot this, Maple plots a circle, where have I gone wrong?
 Sci Advisor HW Helper PF Gold P: 12,016 "(2x)^2 + (1-2y)^2 = 1" This is an equation for a circle. Your parametrization is not.
 Sci Advisor HW Helper Thanks P: 26,148 Hi JanClaesen! (have a theta: θ ) A rose is usually r = ksinθ or r = kcosθ … see http://en.wikipedia.org/wiki/Rose_(mathematics). For k = 1, it is a circle. (But you could have got the same equation if you'd just made it r2 = y )
 P: 56 Polar rose Do you have any hints on how to find the Cartesian equation for r(θ)=sin(2θ), I really can't seem to find it. :)
 Sci Advisor HW Helper Thanks P: 26,148 Hint: multiply both sides by r2.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,693 And use the identity $sin(\theta)= 2sin(\theta)cos(\theta)$.
HW Helper
Thanks
P: 26,148
 Quote by HallsofIvy And use the identity $sin(\theta)= 2sin(\theta)cos(\theta)$.
He knows that
 Quote by JanClaesen Since sin t * cos t = (1/2) * sin 2t …
(and have a theta: θ )
 P: 56 Wow, that was clever, thank you For those interested: xy = 0.5(x^2+y^2)(x^2+y^2)^(1/2) (where x^2+y^2 = sin^2 (2θ) )
 Sci Advisor HW Helper Thanks P: 26,148 (try using the X2 tag just above the Reply box ) That's it! And then expand it , and put it all on the left: (x2 + y2)3 - 2xy = 0.
P: 56
 Quote by tiny-tim (try using the X2 tag just above the Reply box ) That's it! And then expand it , and put it all on the left: (x2 + y2)3 - 2xy = 0.
Yep, thanks again

Is there a human way to do this also for sin(3θ)? Or would that be a computer job?
I'm trying to do this now, but I have a feeling it's quite tough.
HW Helper
Thanks
P: 26,148
 Quote by JanClaesen Is there a human way to do this also for sin(3θ)?
Hint: try it for cos(3θ) + isin(3θ)

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