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
>