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

**References**:**Re: Gray's Differential Geometry error?***From:*"Steven T. Hatton" <hattons@globalsymmetry.com>