Talk:Brachistochrone curve

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The "History" section states "Five mathematicians responded with solutions:" but only lists four. Either the numbers five and four should be four and three, respectively, or the fifth mathematician (fourth published) should be identified.


Perhaps Snell's Law, while it is a clever technique to use in this situation, is not as instructive as an example should be for those looking at the entry.

I am also intrigued and a bit dubious about the use of Snell's Law as proof. Snell's Law is based on Fermat's principle of least time, but that itself could be said to be based on Quantum Electrodynamics, or alternatively in classical mechanics based on Huygens' wave construction. It seems to me we shouldn't get involved in that stuff here - it is really a problem in mathematics, the Physics part just gives a framing to it. Again, Fermat's principle of least time is just a variational principle so it comes back to the calculus of variations at bottom. So I would think the "Alternate Proof" given here is the correct one for this page, although I am no kind of expert on this subject. AdamWGibson (talk) 16:09, 8 May 2008 (UTC)

[edit] Alternate Proof

The aim is to minimise the integral \int{dt} = \int{\frac{ds}{v}}.

Starting with v=0 at y=0 (starting from rest): v=\sqrt{-2gy} (negative sign because y is down)

ds=\sqrt{1+y' ^2} dx =\sqrt{1+x' ^2} dy (we can use either one), therefore we minimise \int{\sqrt{\frac{1+y' ^2}{-2gy}}}dx = \int{\sqrt{\frac{x' ^2 +1}{-2gy}}}dy

Substtuting into the Euler-Lagrange equation (this is the most common tool in variational analysis; for this simple problem however, we can derive it here itself)

we get \frac{d}{dy}\frac{\partial}{\partial x'}\sqrt{\frac{x' ^2 +1}{-2gy}} - \frac{\partial}{\partial x}\sqrt{\frac{x' ^2 +1}{-2gy}} = 0 (second term does not contribute as it does not have x explicitly;

\frac{d}{dy} \frac{x'}{\sqrt{-2gy}\sqrt{x' ^2 +1}} = 0

Ck.mitra (talk) 05:33, 29 May 2008 (UTC) \frac{x'}{\sqrt{-2gy}\sqrt{x' ^2 +1}} = c

Rearranging gives x' =\sqrt{\frac{-2gcy}{1+2gcy}}

Letting k=2gc: y' ^2 = -\frac{1+ky}{ky} = -\frac{k^{-1}+y}{y} = -\frac{D+y}{y}