Talk:Pedal curve

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Hm. Contrary to its pedal curve page, this mathworld page 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 P = 0 and z = kcis(t) + cis(kt) (epicycloids have k > 0, hypocycloids k < 0) we get
z'=ik\left({\rm cis}(t)+{\rm cis}(kt)\right)
|z'|^2=z'\bar z'=2k^2\left(1+\cos(kt-t)\right)
\langle z',P-z\rangle=-\Re(z'\bar z)=k(k-1)\sin(kt-t)
{\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)}
and, finally, the pedal curve is
w=z+z'{\langle z',P-z\rangle\over|z'|^2}={k+1\over2}\left({\rm cis}(t) + {\rm cis}(kt)\right)
Noting {\rm cis}A+{\rm cis}B=2\cos{A-B\over2}{\rm cis}{A+B\over2} and letting t = 2θ / (k + 1)
w=(1+k)\cos\left({1-k\over 1+k}\theta\right){\rm cis}(\theta)
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)

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