Re: ContourPlot non rectangular evaluation?
- To: mathgroup at smc.vnet.net
- Subject: [mg127840] Re: ContourPlot non rectangular evaluation?
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Sun, 26 Aug 2012 02:50:43 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
- Delivered-to: mathgroup-newsend@smc.vnet.net
- References: <20120825082641.60C04689F@smc.vnet.net>
Since you say you can find where the denominator g[x, y] == 0 and imply that tis is a "curve", I presume you already have, at least numerically, a function k[x] given implicitly by g[x, y] == 0, that is, satisfying g[x, k[x]] == 0.
The condition for a point {x, y} to be above the curve g[x, y] == 0 is therefore that y > k[x]. Then use that condition in the RegionFunction option.
For example:
f[x_, y_] := x^3y - 4 Exp[y -x]
g[x_, y_] := x^2 - y^3
h[x_, y_] := f[x, y]/g[x ,y]
k[x_] := Abs[x]^(2/3)
ContourPlot[h[x, y], {x, -2, 2}, {y, 0, 3},
Exclusions -> {g[x, y] == 0},
RegionFunction -> Function[{x, y}, y > k[x]],
Contours -> 10]
On Aug 25, 2012, at 4:26 AM, Sam McDermott <samwell187 at gmail.com> wrote:
>
> I'd like to make a plot of level curves of a function, but this =
function becomes singular and passes from infinity to negative infinity, =
which really uglifies the plot. It is easy for me to identify where the =
function becomes singular (i.e., when the denominator becomes 0!), but =
I'm having trouble telling Mathematica to stop evaluating before then, =
because this singularity is sensitive to both of the variables on the =
axes of my ContourPlot. Is there a simple way of telling Mathematica =
where to stop?
>
> In other words, I have some function
>
> h[x_,y_]:=f[x,y]/g[x,y]
>
> and I can numerically find the zeroes of g[x,y]. Can I make a =
ContourPlot such that
>
> ContourPlot[h[x,y],{x,xmin,xmax},{y,ymin,ymax}]
>
> does not evaluate below the curve
>
> g[x,y]=0
>
> ?
>
> I think I'm basically looking for some way of setting assumptions of =
evaluating an "If" conditional inside of ContourPlot but can't find a =
good way of doing it. Any help would be much appreciated!
---
Murray Eisenberg =
murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
- References:
- ContourPlot non rectangular evaluation?
- From: Sam McDermott <samwell187@gmail.com>
- ContourPlot non rectangular evaluation?