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Re: How to set y always greater than or equal to x for a function?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg46630] Re: How to set y always greater than or equal to x for a function?
*From*: drbob at bigfoot.com (Bobby R. Treat)
*Date*: Thu, 26 Feb 2004 17:53:19 -0500 (EST)
*References*: <c1h0s4$9sc$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
That works great, this is just a quibble, but...
Turning a Rule into a function and using MapAt seemed distasteful to
me, so I came up with an equivalent solution, learning a thing or two
in the process:
f[x_, y_] := If[y > x, Sqrt[y - x], 0]
plot1 = Graphics3D@Plot3D[f[x, y], {x, -100, 100},
{y, -100, 100}, DisplayFunction -> Identity];
rule = {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y ? x -> Sequence[];
plot2 = Head[plot1][First@plot1 /. rule, Sequence @@ Rest@plot1];
Show[plot2, DisplayFunction -> $DisplayFunction];
Bobby
"David Park" <djmp at earthlink.net> wrote in message news:<c1h0s4$9sc$1 at smc.vnet.net>...
> Albert,
>
> I think this is an interesting plotting problem whose solution is not
> totally obvious. If you are just beginning to learn Mathematica, then it is
> probably not the first problem to tackle.
>
> Plot3D[Sqrt[y - x], {x, -100, 100}, {y, -100, 100}];
>
> gives warning messages and a jagged edge to the surface.
>
> We could define a function
>
> f[x_, y_] := If[y > x, Sqrt[y - x], 0]
>
> Then
>
> Plot3D[f[x, y], {x, -100, 100}, {y, -100, 100}];
>
> works more or less. But it has the extra xy-plane surface, which is not part
> of the surface you are specifying.
>
> A few years ago I learned the following trick from MathGroup. (I can't
> remember exactly who posted it.) Generate the plot above and then throw out
> all points that have y <= x.
>
> plot1 = Graphics3D@
> Plot3D[f[x, y], {x, -100, 100}, {y, -100, 100},
> DisplayFunction -> Identity];
>
> plot2 = MapAt[# /. {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y <= x ->
> Sequence[] &, plot1, 1];
>
> We had to convert the plot from SurfaceGraphics to Graphics3D. We had to map
> our function onto the first part of plot1. The first part contains all the
> primitive graphics for the surface. Any point in a Polygon for which y <=x
> gets replaced by Sequence[], which effectly gets rid of it. We then Show the
> new plot.
>
> Show[plot2, DisplayFunction -> $DisplayFunction];
>
> That gives a pretty good representation of the surface.
>
> The DrawGraphics package at my web site gives some other techniques for
> handling plotting problems of this type. One of its advantages is that it
> automatically extracts the primitive graphics from any plot and it is then
> easy to manipulate or combine the various elements.
>
> Needs["DrawGraphics`DrawingMaster`"]
>
> The following implements the same trick as above.
>
> Module[{g},
> g = Draw3D[f[x, y], {x, -100, 100}, {y, -100, 100}];
> Draw3DItems[
> {g /. {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y <= x -> Sequence[]},
> BoxRatios -> {1, 1, 1/2},
> Axes -> True]];
>
> Another method is to reparametrize the surface. If we were going to
> integrate a two variable function we could use
> Integrate[f[x,y],{x,-100,100},{y,x,100}]. The second iterator depends on x.
> But we can't directly do that when plotting surfaces. DrawGraphics has the
> command
>
> IteratorSubstitution[{y, Sqrt[y - x]}, {y, x, 100}]
> {{(-w)*(-100 + x) + x, Sqrt[(-w)*(-100 + x)]}, {w, 0, 1}}
>
> It reparametrizes y and Sqrt[y-x] for {y,x,100} in terms of x and w. w has
> the fixed iterator {w,0,1}. We can then plot this new function and then use
> DrawingTransform3D to get back from the xw-plane to the xy-plane.
>
> Draw3DItems[
> {Draw3D[Sqrt[(-w)*(-100 + x)], {x, -100, 100}, {w, 0, 1}] /.
> DrawingTransform3D[#1 & , (-#2)*(-100 + #1) + #1 & , #3 & ]},
>
> BoxRatios -> {1, 1, 1/2},
> Axes -> True,
> Background -> Linen,
> ImageSize -> 450];
>
> But this isn't too nice for this particular case because the polygons are
> crimed together at one corner.
>
> Another idea is to plot the function Sqrt[y], where we can use fixed ranges
> and then rotate it by 45 degrees. I plot it in two sections to get a better
> representation of the rising section.
>
> Draw3DItems[
> {{Draw3D[Sqrt[y], {x, -150, 150}, {y, 0, 2}, PlotPoints -> {25, 2}],
> Draw3D[Sqrt[y], {x, -150, 150}, {y, 2, 150}, PlotPoints -> 25]} //
> UseRotateShape[-45°, 0, 0]},
> BoxRatios -> {1, 1, 1/2},
> PlotRange -> {{-100, 100}1.01, {-100, 100}1.01, {-0.1, 15.1}},
> Axes -> True,
> Background -> Linen,
> ImageSize -> 450];
>
> This is probably a little nicer than the non DrawGraphics plot because the
> "mesh" lines are either level, or the sqrt curves.
>
> As an extra, here is the same surface combined with -Sqrt[y-x] and then spun
> around for better viewing. g is just the primitive graphics for the surface
> we plotted above. We draw it and then we draw it again with the z component
> reversed.
>
> plot3 =
> Module[{g},
> g = {Draw3D[Sqrt[y], {x, -150, 150}, {y, 0, 2}, PlotPoints -> {25,
> 2}],
> Draw3D[Sqrt[y], {x, -150, 150}, {y, 2, 150}, PlotPoints -> 25]}
> //
> UseRotateShape[-45°, 0, 0];
> Draw3DItems[
> {g,
> g /. DrawingTransform3D[#1 &, #2 &, -#3 &]},
> BoxRatios -> {1, 1, 1},
> Boxed -> False,
> PlotRange -> {{-100, 100}1.01, {-100, 100}1.01, {-12.1, 12.1}},
> Background -> Linen,
> ViewPoint -> {1.779, -2.844, 0.447},
> ImageSize -> 450]];
>
> SpinShow[plot3]
> SelectionMove[EvaluationNotebook[], All, GeneratedCell]
> FrontEndTokenExecute["OpenCloseGroup"]; Pause[0.5];
> FrontEndExecute[{FrontEnd`SelectionAnimate[200, AnimationDisplayTime -> 0.1,
> AnimationDirection -> Forward]}]
>
> David Park
> djmp at earthlink.net
> http://home.earthlink.net/~djmp/
>
>
>
>
>
>
> From: Albert Franco [mailto:francoatnsuok at earthlink.net]
To: mathgroup at smc.vnet.net
>
> I recently purchased Mathematica 5.0, and I am in the process of
> learning how to take advantage of it's many features.
>
> This is an example of one line I would like to calculate in Mathematica:
>
> Plot3D[Sqrt[y-x],{x,-100,100},{y,-100,100}];
>
>
> How can I best edit this line to make y always greater than or equal to x
> during
> calculation so the program doesn't try to take the square root of a
> negative number?
>
> Any help will be appreciated.
>
> Albert Franco
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