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Polar rose 
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#1
Nov709, 03:19 PM

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) * (1cos 2t) (2x)^2 + (12y)^2 = 1 4x^2 4y + 4y^2 = 0 When I plot this, Maple plots a circle, where have I gone wrong? 


#2
Nov709, 04:17 PM

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P: 12,016

"(2x)^2 + (12y)^2 = 1" This is an equation for a circle.
Your parametrization is not. 


#3
Nov709, 04:42 PM

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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 r^{2} = y ) 


#4
Nov709, 06:58 PM

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. :)



#6
Nov809, 04:29 AM

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PF Gold
P: 39,693

And use the identity [itex]sin(\theta)= 2sin(\theta)cos(\theta)[/itex].



#8
Nov809, 05:29 AM

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) ) 


#9
Nov809, 05:37 AM

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(try using the X^{2} tag just above the Reply box )
That's it! And then expand it , and put it all on the left: (x^{2} + y^{2})^{3}  2xy = 0. 


#10
Nov809, 06:33 AM

P: 56

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. 


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