Talk:Pedal curve
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Hm. Contrary to its pedal curve page, this mathworld page (http://mathworld.wolfram.com/EpicycloidPedalCurve.html) says the epicycloid does not give a rose. But hypo and epicycloids are such close cousins they should both or neither. I'll check carefully. 142.177.20.80 01:08, 2 Aug 2004 (UTC)
- The page is mistaken. Working in the complex plane and starting with <math>P=0<math> and <math>z=k{\rm cis}(t)+{\rm cis}(kt)<math> (epicycloids have <math>k>0<math>, hypocycloids <math>k<0<math>) we get
- <math>z'=ik\left({\rm cis}(t)+{\rm cis}(kt)\right)<math>
- <math>|z'|^2=z'\bar z'=2k^2\left(1+\cos(kt-t)\right)<math>
- <math>\langle z',P-z\rangle=-\Re(z'\bar z)=k(k-1)\sin(kt-t)<math>
- <math>{\langle z',P-z\rangle\over|z'|^2}={k-1\over2k}\tan\left({k-1\over2}t\right)={k-1\over2ik}{{\rm cis}(kt)-{\rm cis}(t)\over{\rm cis}(kt)+{\rm cis}(t)}<math>
- and, finally, the pedal curve is
- <math>w=z+z'{\langle z',P-z\rangle\over|z'|^2}={k+1\over2}\left({\rm cis}(t) + {\rm cis}(kt)\right)<math>
- Noting <math>{\rm cis}A+{\rm cis}B=2\cos{A-B\over2}{\rm cis}{A+B\over2}<math> and letting <math>t=2\theta/(k+1)<math>
- <math>w=(1+k)\cos\left({1-k\over 1+k}\theta\right){\rm cis}(\theta)<math>
- which is obviously a rose. 142.177.126.230 16:25, 2 Aug 2004 (UTC)
- Mathworld's been corrected =) 142.177.126.230 23:07, 5 Aug 2004 (UTC)
More work needed on contrapedal curve; mainly, what's done in higher spaces? While one could sensibly use the curvature vector, one could also use the perpendicular subspace...and somehow the latter is more appealing. Kwantus 18:53, 2 Aug 2004 (UTC)