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MathGroup Archive 2005

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Re: Crossing of 3D functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57005] Re: Crossing of 3D functions
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 12 May 2005 02:33:05 -0400 (EDT)
  • Organization: Cablecom Newsserver
  • References: <d5sks6$o2t$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Pawel,
see below.
Sincerely, Daniel

Paawel wrote:
> 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. 

First load the package: Graphics`ImplicitPlot`
Then the following will plot the crossing:

ImplicitPlot[F1[r, f] == F2[r, f], {r, -1, 2}, AspectRatio -> 1/GoldenRatio]

> 
> 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?

The circle has two branches,one above the x-axis, the other below. 
Therefore it is not possible to specify y[x] as a single valued function 
of x, you have to live with two functions.
> 
> Please Help
> Pawel
> 


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