# How to plot implicit function with intermediate variable

Posted 7 months ago
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 Here, I intend to plot a 3D picture of with two implicit functionswhere x, y, t, \xi, \eta, are variables. Of course, these variables can be constrained to certain interval that will guarantee the function to be well-posed.How can I draw this one? Attachments:
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Posted 7 months ago
 When t=0 the function you want to plot is identically zero.
Posted 7 months ago
 Here is an attempt: func[\[Xi]_, \[Eta]_, t_] = -((1350. Sech[\[Eta]]^2 Tanh[\[Eta]] (1. Sech[ 0.1 t - \[Xi]]^2 Tanh[0.1 t - \[Xi]] + 0.6 Sech[0.3 t - \[Xi]]^2 Tanh[0.3 t - \[Xi]] + 1.6 Sech[0.1 t + \[Xi]]^2 Tanh[0.1 t + \[Xi]]))/((1 - 1.5 Sech[\[Eta]]^2) (1 - Sech[0.1 t - \[Xi]]^2 - 0.5 Sech[0.3 t - \[Xi]]^2 - 1.5 Sech[0.1 t + \[Xi]]^2) (0.3 Sech[\[Eta]]^2 + 0.5 Sech[0.1 t - \[Xi]]^2 + 0.3 Sech[0.3 t - \[Xi]]^2 - 0.8 Sech[0.1 t + \[Xi]]^2)^2)) // Rationalize // Simplify; xyToXiEta[x_?NumericQ, y_, t_] := NSolve[{x == \[Xi] - 1/2 Tanh[\[Xi] - 3/10 t] - Tanh[\[Xi] - 1/10 t] - 3/2 Tanh[\[Xi] + 1/10 t], y == \[Eta] - 3/2 Tanh[\[Eta]]}, {\[Xi], \[Eta]}, Reals]; With[{t = 1}, ListPointPlot3D[ Flatten[ Table[{x, y, func[\[Xi], \[Eta], t]} /. xyToXiEta[x, y, t], {x, -1, 1, 1/20}, {y, -1, 1, 1/20}], 2], BoxRatios -> {1, 1, 1}, Axes -> True, PlotRange -> {All, All, 2000 {-1, 1}}, AxesLabel -> {x, y, z}]] 
Posted 7 months ago
Posted 7 months ago
 Unfortunately, NSolve[{x == \[Xi] - 1/2 Tanh[\[Xi] - 3/10 t] - Tanh[\[Xi] - 1/10 t] - 3/2 Tanh[\[Xi] + 1/10 t], y == \[Eta] - 3/2 Tanh[\[Eta]]}, {\[Xi], \[Eta]}, Reals] does not function.
Posted 7 months ago
 Sorry for the mistake, it does function.
Posted 7 months ago
 With : t=1; With[{t = 1}, ListPlot3D[ Flatten[Table[{x, y, func[\[Xi], \[Eta], t]} /. xyToXiEta[x, y, t], {x, -1, 1, 1/20}, {y, -1, 1, 1/20}], 2], BoxRatios -> {1, 1, 1}, Axes -> True, PlotRange -> {All, All, 2000 {-1, 1}}, AxesLabel -> {x, y, z}]] `
Posted 7 months ago