       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.

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.
>