# Contourplot with three variables

I have the following equation to generate a contour plot for it:

ContourPlot[{(6 x^2 (-0.00001 + (z)^2))/(6 y^2) == 0.2}, {x, -1, 1}, {y, 0.1, 20}]


It has three variables. The range of the two variables is given above but the variable z has range of {10, 100}. My question is how to introduce the z range here while I want the usual contour plot in two dimensions.

• Similar to 77039.
– Syed
Commented Dec 1, 2023 at 1:23

\$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global*"]


Either, plot a family of curves,

ContourPlot[
Evaluate@Table[
(6 x^2 (-0.00001 + (z)^2))/(6 y^2) == 0.2,
{z, 10, 100, 15}],
{x, -1, 1}, {y, 0.1, 20},
FrameLabel -> (Style[#, 14] & /@ {x, y}),
PlotLegends -> LineLegend[Range[10, 100, 15],
LegendLabel -> Style[z, 14]]]


Or Manipulate the z value

Manipulate[
ContourPlot[
(6 x^2 (-0.00001 + (z)^2))/(6 y^2) == 0.2,
{x, -1, 1}, {y, 0.1, 20},
FrameLabel -> (Style[#, 14] & /@ {x, y})],
{{z, 10}, 10, 100, 1, Appearance -> "Labeled"},
SynchronousUpdating -> False,
TrackedSymbols :> {z}]


Look at a ContourPlot3D from above:

ContourPlot3D[(6 x^2 (-0.00001 + (z)^2))/(6 y^2) == 0.2,
{x, -1, 1}, {y, 0.1, 20}, {z, 10, 100},
ViewPoint -> {0, 0, \[Infinity]}, MeshFunctions -> {#3 &}, Mesh -> 5,
ContourStyle -> None, Ticks -> {Automatic, Automatic, None}]
`

• Further, I want the y-axis to be a log scale. I tried the "Scaling function", but it does not work. Any suggestion this, please? Commented Dec 1, 2023 at 8:09