Related
Suppose I have a function like this.
u = (1 / 4 Sin[t] (1 - r^2)) ;
Plot[u,{r,0,1}]
The above command will plot "U" on Y-axis and "r" on X-axis, But I want it in the reverse direction. "U" on X-axis and "r" on Y-axis.
How to do this, I'm new to Mathematica.
Many thanks for considering my request.
You can tabulate the results using Table and reverse each data point
Clear[t, u]
u[r_] := (1/4 Sin[t] (1 - r^2));
t = 1.57;
ru = ListLinePlot[table = Table[{r, u[r]}, {r, 0, 1, 0.01}],
AspectRatio -> 1, ImageSize -> 200, AxesLabel -> {r, u}];
ur = ListLinePlot[Reverse /# table,
AspectRatio -> 1, ImageSize -> 200, AxesLabel -> {u, r}];
GraphicsRow[{ru, ur}]
Or you can generate an inverse function to use in Plot
Clear[t, u]
u[r_] := (1/4 Sin[t] (1 - r^2));
t = 1.57;
plotru = Plot[u[r], {r, 0, 1},
AspectRatio -> 1, ImageSize -> 200, AxesLabel -> {r, u}];
Clear[t, u]
v = Quiet[InverseFunction[(1/4 Sin[t] (1 - #^2)) &]];
t = 1.57;
plotur = Plot[-v[x], {x, 0, 0.25},
AspectRatio -> 1, ImageSize -> 200, AxesLabel -> {u, r}];
GraphicsRow[{plotru, plotur}]
I found the answer to this Question using the ParametricPlot command
ParametricPlot[{(1 - r^2) /. r -> Abs[r], r}, {r, -Pi, Pi}]
I have a simple Mathematica code below where I first introduce a scalar function ϕ = ϕ[x,y,z] and then calculate the gradient of ϕ. Now, I would like to evaluate the Gradient at point P by substituting in proper values for x, y, z. Please assist me with last step with plugging values into x and y into gradient. See code Below:
ϕ = y^2 + z^2 - 4;
varlist = {x, y, z}
Delϕ = Table[D[ϕ, varlist[[i]]], {j, 1, 1}, {i, 1, 3}]
Delϕ // MatrixForm
P = {2, 1, Sqrt (3)}
Thanks
Assuming you meant y^2 + z^2 - 4 x
φ = y^2 + z^2 - 4 x;
varlist = {x, y, z};
g = D[φ, #] & /# varlist
{-4, 2 y, 2 z}
p = {2, 1, Sqrt[3]};
grad = g /. Thread[varlist -> p]
{-4, 2, 2 Sqrt[3]}
another approach is to make your derivative a function:
\[Phi] = y^2 + z^2 - 4 x;
varlist = {x, y, z};
Del\[Phi][{x_, y_, z_}] = Table[D[\[Phi], varlist[[i]]], {i, 1, 3}];
then you can simply do this:
P = {2, 1, Sqrt[3]};
Del\[Phi][P]
{-4, 2, 2 Sqrt[3]}
I have two functions:
y = x - x^3and
x = y^3 - y
I need to plot them both in one plane, I wonder how to achieve this with Mathematica?
Thanks
One way is to use ContourPlot
let's Wrap it in Manipulate to make it more easy to adjust
Manipulate[
With[{f1 = y == x - x^3, f2 = x == y^3 - y},
ContourPlot[{f1, f2}, {x, -lim, lim}, {y, -lim, lim},
Frame -> True,
FrameLabel -> {{y, None}, {x, {f1, f2}}},
ImagePadding -> 30,
GridLines -> Automatic,
GridLinesStyle -> Directive[Thickness[.001], LightGray]]
],
{{lim, 3, "limit"}, .1, 10, .1, Appearance -> "Labeled"}
]
I can plot the curve corresponding to an implicit equation:
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
But I cannot find a way to color the contour line depending on the location of the point. More precisely, I want to color the curve in 2 colors, depending on whether x² + y² < k or not.
I looked into ColorFunction but this is only for coloring the region between the contour lines.
And I was not able to get ContourStyle to accept a location-dependent expression.
you could use RegionFunction to split the plot in two:
Show[{
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 < .5],
ContourStyle -> Red],
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 >= .5],
ContourStyle -> Green]
}]
Maybe something like this
pl = ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
points = pl[[1, 1]];
colorf[{x_, y_}] := ColorData["Rainbow"][Rescale[x, {-1, 1}]]
pl /. {Line[a_] :> {Line[a, VertexColors -> colorf /# points[[a]]]}}
which produces
This does not provide a direct solution to your question but I believe it is of interest.
It is possible to color a line progressively from within ContourPlot using what I think is an undocumented format, namely a Function that surrounds the Line object. Internally this is similar to what Heike did, but her solution uses the vertex numbers to then find the matching coordinates allowing styling by spacial position, rather than position along the line.
ContourPlot[
x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
BaseStyle -> {12, Thickness[0.01]},
ContourStyle ->
(Line[#, VertexColors -> ColorData["DeepSeaColors"] /# Rescale##] & ## # &)
]
For some of the less adept, less information is more. Time was wasted browsing for a way to set the color of contour lines until I chanced onto Roelig's edited answer. I just needed ContourStyle[].
Show[{ContourPlot[
x^2 + 2 x y Tan[2 # ] - y^2 == 1, {x, -3, 3}, {y, -3.2, 3.2},
ContourStyle -> Green] & /# Range[-Pi/4, Pi/4, .1]},
Background -> Black]
Why doesn't this work?
I have bypassed this before but i can't remember how i did it, and I never went on to figure out why this type of inputs didn't work. About time to get to know it!
For those who cant see the pic:
RegionPlot3D[
x^2 + 2 y^2 - 2 z^2 = 1 && -1 <= z <= 1, {x, -5, 5}, {y, -5,
5}, {z, -1, 1}]
Set::write: "Tag Plus in -2.+25.+50. is Protected"
And then there is just an empty cube without my surface.
If z is limited by other surfaces you could go like this:
RegionPlot3D[
x^2 + 2 y^2 - 2 z^2 < 1 && z < x + 2 y && z^2 < .5,
{x, -2, 2}, {y, -2, 2}, {z, -1, 1},
PlotPoints -> 50, MeshFunctions -> {Function[{x, y, z}, z]},
PlotStyle -> Directive[Red, Opacity[0.8]]]
Or with ContourPlot:
ContourPlot3D[
x^2 + 2 y^2 - 2 z^2 == 1,
{x, -2, 2}, {y, -2, 2}, {z, -1, 1},
RegionFunction -> Function[{x, y, z}, z < x + 2 y && z^2 < .5],
PlotPoints -> 50, MeshFunctions -> {Function[{x, y, z}, z]},
ContourStyle -> Directive[Red, Opacity[0.8]]]]
Try this
RegionPlot3D[x^2 + 2 y^2 - 2 z^2 < 1,
{x, -5, 5}, {y, -5, 5}, {z, -1, 1}]
Or, if you just want the surface
ContourPlot3D[x^2 + 2 y^2 - 2 z^2 == 1,
{x, -5, 5}, {y, -5, 5}, {z, -1, 1}]
Note the double equals sign, rather than the single equals sign.