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MathGroup Archive 2005

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Re: Re: Gray's Differential Geometry error?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63246] Re: [mg63215] Re: Gray's Differential Geometry error?
  • From: "Steven T. Hatton" <hattons at globalsymmetry.com>
  • Date: Tue, 20 Dec 2005 04:44:49 -0500 (EST)
  • References: <dnv48c$och$1@smc.vnet.net> <200512191201.HAA10976@smc.vnet.net> <7F9B8DFC-F317-4A5E-8341-457A8F7BBCC0@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

On Monday 19 December 2005 10:11, Andrzej Kozlowski wrote:
> *This message was transferred with a trial version of CommuniGate(tm) Pro*
>
> On 19 Dec 2005, at 21:01, Steven T. Hatton wrote:
> > The parametric form of the ellipse is given by:
> >
> > z[t_] = (a - b)/(2Exp[I t]) + ((a + b)Exp[I t])/2
> >
> > Gray asserts that the derivative of z wrt t can be expressed as:
> >
> > I/2 (s e + s/e)(s e - s/e) /. {s -> Sqrt[a + b], e -> Exp[t/2I]}
> >
> > I am inclined to believe this is correct (not a typo), but have not
> > yet show
> > it to be.  My suspicion is that it follows from some kind of
> > "completing
> > the square" manipulation.  Do you believe the second expression
> > correctly
> > expresses dz/dt?
>
> I do not have Gray's book here (I do own it but tend to  keep it on a
> different continent than the one I am on now ;-)) but it is trivial
> to show that the above can't possibly be correct.  Just put b = -a.
> Then obviously s is 0 so the expression for the derivative is 0.
> However, the function z[t] has a non zero derivative, unless a = b = 0.
>
> It is also quite easy to guess what Gray's expression should have
> been. Let's do it with the help of Mathematica (although I first
> worked it out by hand):
>
>
> FullSimplify[D[(a - b)/(2*Exp[I*t]) + ((a + b)*Exp[I*t])/2, t]]
>
>
> I*b*Cos[t] - a*Sin[t]
>
>
> FullSimplify[(I/2)*(s*e + c/e)*(s*e - c/e) /. {s -> Sqrt[a + b],c ->
> Sqrt[a - b], e -> Exp[(t/2)*I]}]
>
>
> I*b*Cos[t] - a*Sin[t]

So, I start again.  Define the ellipse in parametric form:
z[t_]:=(a - b)/(2*E^(I*t)) + ((a + b)*E^(I*t))/2

Look at the derivative in the form Mathematica gives it:
z'[t]
==>((-I/2)*(a - b))/E^(I*t) + (I/2)*(a + b)*E^(I*t)
F1=-Sqrt[a^2 - b^2]+0I
F1=Sqrt[a^2 - b^2]+0I

Express the displacement from F1 to z[t] as a vector:
z1[t_]:=-Sqrt[a^2 - b^2] + (a - b)/(2*E^(I*t)) + ((a + b)*E^(I*t))/2

Express the displacement from F2 to z[t] as a vector:
z2[t_]:=Sqrt[a^2 - b^2] + (a - b)/(2*E^(I*t)) + ((a + b)*E^(I*t))/2

Rewrite z1 as a square:
z11[t_]:=(Sqrt[a - b]/E^((I/2)*t) - Sqrt[a + b]*E^((I/2)*t))^2/2

Rewrite z2 as a square:
z21[t_]:=(Sqrt[a - b]/E^((I/2)*t) + Sqrt[a + b]*E^((I/2)*t))^2/2

Express z'[t] following your suggestion:
zp[t_]:=(I/2)*(-(Sqrt[a - b]/E^((I/2)*t)) + Sqrt[a + b]*E^((I/2)*t))*
 (Sqrt[a - b]/E^((I/2)*t) + Sqrt[a + b]*E^((I/2)*t))

zp[t]/z11[t] // Cancel

z21[t]/zp[t] // Cancel

Gives the result Gray intended.

Thank you for sharing your insight.  It was very helpful.

Steven


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