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?
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 )
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. :)
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) )
(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.
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.