I have a function f[t,x,y]= x^2 y^2 t^(-2) Exp[(y/t)]^-1 and I want a data list of the variables x vs. y that satisfy f[t,x,y]>1, so that I can draw a boundary line that distinguishes between those two regions (f[t,x,y]>1 and f[t,x,y]<1). The first variable t can be chosen arbitrarily. How shall I proceed?
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3 Answers
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Clear[f, t];
f[t_, x_, y_] = x^2 y^2 t^(-2) Exp[(y/t)]^-1;
t = 1;
ContourPlot[f[t, x, y], {x, -5, 5}, {y, -5, 5}, Contours -> {1},
ContourStyle -> {AbsoluteThickness[2], Red},
ContourShading -> {LightGreen, LightBlue}, PlotPoints -> 50,
MaxRecursion -> 2, PlotRange -> All, PlotLegends -> Automatic]
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As t can be chosen arbitrarily, we set it simply to one. Then the function simplifies to:
f[x_, y_] = x^2 y^2 Exp[(y)]^-1
The boundary between f>1 and f<1 can be obtained by:
ContourPlot[f[x, y] == 1, {x, -1, 1}, {y, -5, 0}]
Or using RegionPlot:
RegionPlot[
{f[x, y] < 1, f[x, y] == 1, f[x, y] > 1},
{x, -1, 1}, {y, -5, 0}, MaxRecursion -> 7
]
answered Jul 12, 2022 at 14:51
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$\begingroup$ If I have to vary
tthen how should I do it? $\endgroup$ Jul 13, 2022 at 4:35 -
$\begingroup$ Simply introduce t again as a parameter. $\endgroup$ Jul 13, 2022 at 7:53
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$\begingroup$ But a function with three variables in either
ContourPlotorRegionPlotdo't work. $\endgroup$ Jul 13, 2022 at 8:33 -
$\begingroup$ I said "as parameter", not as argument. To plot t must have a numerical value that you can insert in the definition of f. $\endgroup$ Jul 13, 2022 at 8:35
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$\begingroup$ I tried to generate random numbers for each vairables and then solved the function. Then I used
ListContourPlotwithRegionFunction -> Function[{x, y, z}, z > 1]. I does look a bit somthing I wanted earlier. $\endgroup$ Jul 13, 2022 at 8:36
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With[
{
reg=ImplicitRegion[x^2 y^2 t^(-2) Exp[(y/t)]^-1<1,{x,y,t}]
}
,RegionPlot3D[reg]
]
