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Re: Re: Re: graphic

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104618] Re: [mg104583] Re: [mg104518] Re: graphic
  • From: "David Park" <djmpark at comcast.net>
  • Date: Thu, 5 Nov 2009 03:48:27 -0500 (EST)
  • References: <hcl3qc$btt$1@smc.vnet.net> <200911030752.CAA01048@smc.vnet.net> <27376704.1257318514372.JavaMail.root@n11>

Now that I see a nice example:

Needs["Presentations`Master`"]

f1[x_, y_] = x^2 + y^2 - 2;
f2[x_, y_] = x^2;

Draw3DItems[
 {ContourDraw[f1[x, y] == 0, {x, -2, 2}, {y, -2, 2}] //
   RaiseTo3D[f2[#1, #2] &]},
 NiceRotation]


David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/ 



From: DrMajorBob [mailto:btreat1 at austin.rr.com]

Or, more generally,

f1[x_, y_] = x^2 + y^2 - 2;
f2[x_, y_] = x^2;
cp = ContourPlot[f1[x, y] == 0, {x, -2, 2}, {y, -2, 2}];
Graphics3D[
  Replace[First[cp],
   GraphicsComplex[points_, data_] :>
    GraphicsComplex[{#1, #2, f2[#1, #2]} & @@@ points, data]]]

Correct?

Bobby

On Tue, 03 Nov 2009 01:52:03 -0600, Szabolcs Horv=E1t <szhorvat at gmail.com>
wrote:

> On 2009.11.01. 23:58, Nodar wrote:
>> Dear All,
>>
>>    May I build 3D curve graphic z=F2(x,y), where x and y are the
>> solutions of equation F1(x,y)==0 ?
>>    May I receive list of points from a "ImplicitPlot" graphic ?
>
> The simplest way would be to solve the F1(x,y)==0 equation.  Here's a
> hackish solution for the case when this is not possible:
>
> Suppose F1(x,y) = x^2+y^2-2 and F2(x,y) = x^2
>
> cp = ContourPlot[x^2 + y^2 - 2 == 0, {x, -2, 2}, {y, -2, 2}]
>
> Graphics3D[
>   Replace[First[cp],
>    GraphicsComplex[points_, data_] :>
>     GraphicsComplex[{#1, #2, #1^2} & @@@ points, data]]]
>
>
> This will work in Mathematica 6 or later.
>


--
DrMajorBob at yahoo.com




  • References:
    • Re: graphic
      • From: Szabolcs Horvát <szhorvat@gmail.com>
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