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MathGroup Archive 2005

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Re: Crossing of 3D functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56987] Re: Crossing of 3D functions
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Thu, 12 May 2005 02:32:20 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <d5sks6$o2t$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

Needs["Graphics`ImplicitPlot`"]

Fa = Exp[-(x^2 + y^2)];
Fb = 0.5;

ImplicitPlot[Fa == Fb, {x, -2, 2}, {y, -2, 2}]

???

Regards

  Jens

"Paawel" <fonfastik at interia.pl> schrieb im 
Newsbeitrag news:d5sks6$o2t$1 at smc.vnet.net...
> Hello
> I have two functions of surfaces
>
> F1=\!\(\(-7.516\)*\
>      Exp[\(-3.076\)*\((r - 1.184)\)]*\((1 -
>          8.28*10\^\(-5\)*\((180\  - \ 
> f)\)\^2)\)*0.7\^0.5 +
>    3.758*Exp[\(-6.152\)*\((r - 1.184)\)] + 
> 3.92\)
>
> and
>
> F2=\!\(\(-0.958\)*\ Exp[\(-4.993\)*\((r - 
> 1.375)\)]*\((1 - \
> 1.07*10\^\(-3\)*\((133\  - \ f)\)\^2)\)*1\^0.5 + 
> 0.479*
>          Exp[\(-9.986\)*\((r - 1.375)\)] + 
> 0.479\)
>
> variables are r and f
>
> I want to find a plot which describes crossing 
> of these functions.
>
> I tried Solve command but without any success
>
> I also tried a simpler example
> Fa=Exp[-(x^2 + y^2)]
> and
> Fb=0.5
> and when I used
> Solve[Fa == Fb, {y}]
> I received y1 and y2 - two parts of wanted 
> function
> How to obtain one function (circle) from the 
> above Fa nad Fb?
>
> Please Help
> Pawel
> 



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