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Re: simple plot
*To*: mathgroup at smc.vnet.net
*Subject*: [mg22393] Re: simple plot
*From*: Brian Higgins <bghiggins at ucdavis.edu>
*Date*: Sun, 27 Feb 2000 18:55:33 -0500 (EST)
*References*: <89a592$9gp@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <89a592$9gp at smc.vnet.net>,
"Steve" <sdunkel at ucsd.edu> wrote:
> How do I plot the following function?
>
> x^2+y^2-z^2=0
>
> I know this is a simple question but I'm new to Mathematica.
>
> Thanks
>
>
Steve,
For your particular problem you can always use Plot3D
Plot3D[Sqrt[x^2 + y^2] , {x, -3, 3}, {y, -3, 3}]
But for the more general case you may want to try ContourPlot3D. You
will need to load in the package Graphics`ContourPlot3D`. For example:
<< Graphics`ContourPlot3D`
ContourPlot3D[x^2 + y^2 - z^2, {x, -4, 4}, {y, -4, 4}, {z, 0, 10},
PlotPoints -> {4, 4}, Contours -> {0}, Axes -> True]
will show the contour for f(x,y,z)=x^2+y^2-z^2=0. Here is an example
that shows the contours for f(x,y,z)=0 and f(x,y,z)=6
ContourPlot3D[x^2 + y^2 - z^2, {x, -4, 4}, {y, -4, 4}, {z, 0, 10},
PlotPoints -> {4, 4}, Contours -> {0, 6}, Lighting -> False,
ContourStyle -> {{RGBColor[0, 0, 1]}, {RGBColor[1, 0, 0]}}, Axes ->
True]
Cheers,
Brian
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