7
To get an overview:
ContourPlot[{y == 6/x, y == x + 4, y == x - 4}, {x, -10, 10}, {y, -10, 10}]
Define it as a region through logical combinations, the signs of which you have to guess a bit, based on the above plot:
ImplicitRegion[x < 0 && y > 6/x && x - 4 < y < x + 4 ||
x >= 0 && y < 6/x && ...
6
f = x1 + x2*Sin[x1];
g = x1*x2 + Cos[x2]*x1;
ContourPlot[f - g, {x1, -10, 10}, {x2, -10, 10}, Contours -> {0},
ContourShading -> {None, Red}, PlotPoints -> 80]
5
It is quite difficult to understand what problems you faced, since you have not provided explicit code, so I am giving a very simple command just to get you started.
Assume the following functions
f[x1_][x2_] := x1^2 + 2*x2
g[x1_][x2_] := x1 + 2/x2
and then we can use
RegionPlot[f[x1][x2] >= g[x1][x2], {x1, 0, 4}, {x2, 0, 11}]
which returns a nice plot.
...
4
This problem is easy to solve by changing to a different coordinate system. Let
$$p=\tfrac12(x+y),\\q=y-x.$$
The Jacobian is easy to show to be equal to 1.
Next, we plot the region to determine the integration domain.
fig = ContourPlot[{y == 6/x, y == x + 4, y == x - 4, y == x, y == -x},
{x, -10, 10}, {y, -10, 10},
PlotLegends -> {p^2 == 6 + ...
3
Clear["Global`*"]
R = 25 (69634*^3);
ri = 6 R;
r0 = 7*^4 R;
T = 4000;
h = QuantityMagnitude[Quantity[1, "PlanckConstant"] // UnitConvert];
c = QuantityMagnitude[Quantity[1, "SpeedOfLight"] // UnitConvert];
k = QuantityMagnitude[Quantity[1,
"BoltzmannConstant"] // UnitConvert];
F[v_?NumericQ] := 4 Pi h v^4/...
3
The damped pendulum will always, after some initial time, obtain the frequency of the driver force, although with different phase shifts.
The following can serve as a start for further experimentation (Note, I reduced the number of parameters by choosing units to set the force constant to 1):
tmax = 500;
omegamax = .2;
dampmax = 0.1;
Manipulate[
force[t_] = ...
3
Edit
We can directly use RegionFunction
cond = (1 - 4 q^2)^2 (-19 - 88 q^2 + 144 q^4 +
12 (1 - 4 q^2)^2 Cos[a] + (-1 - 8 q^2 + 48 q^4) Cos[
2 a])^2 (-1803 + 7732 q^2 + 5744 q^4 + 5568 q^6 +
12 (33 + 644 q^2 + 816 q^4 + 704 q^6) Cos[a] +
4 (35 + 16 q^2 (4 + 79 q^2 + 56 q^4)) Cos[2 a] +
12 (-1 - 20 q^2 + 80 q^4 + 64 q^6) Cos[3 ...
2
I have plotted the results using the streamplot command but that
wasn't right
It will help if you show what you tried, so one can see what the problem is and explain better why what you saw "wasn't right"
m = 1/2;
G = 1;
(*state space representation *)
x1d = x2
x2d = -3 Sqrt[8 Pi G/3] Sqrt[1/2 x2^2 + 1/2 m^2 x1^2] - m^2 x1;
ic = {1, 0}; (*maps ...
2
To see Tooltips it is easier to use ContourPlot
Clear["Global`*"]
parabolic[r_, phi_] := {1/2 (r^2 - phi^2), phi r};
sol[1] = Solve[{x, y} == parabolic[r, phi], {r, phi}, Reals];
sol[2] = Solve[{x, y} == -parabolic[r, phi], {r, phi}, Reals];
{cpr1, cpr2} = ContourPlot[r /. sol[#],
{x, -0.5, 0.5}, {y, -1, 1},
Contours -> Range[0, ...
2
Workaround using RegionPlot3D
F[x_, y_, z_] := 2*(x^2 - 2 x*y + y^2 - y*z)^2 - y^4 - z^4
grad = Grad[F[x, y, z], {x, y, z}]
RegionPlot3D[-.02 Sqrt[grad . grad ] <=F[x, y, z] <= .02 Sqrt[grad . grad ]
, {x, -2, 2}, {y, -2,2}, {z, -2, 2} , PlotPoints -> 100, MaxRecursion -> 4]
2
Clear["Global`*"]
u0[r_, phi_] := Sum[I^(-n) BesselJ[n, r] Exp[I n phi], {n, -5, 5}]
Map[Plot3D[#[u0[r, phi]], {r, 0, 4}, {phi, 0, 2 Pi},
PlotRange -> All,
AxesLabel -> Automatic,
AxesStyle -> 12,
ColorFunction -> Hue,
PlotLabel -> Style[StringForm["``@u0", #], 14, Bold],
PlotPoints -> 50,
...
2
Clear["Global`*"]
$Version
(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)
inttau[r_] = -(1/
Sqrt[0.0345106153943703 - ((37.3042 - r) (-25.578 + r) (62.8822 +
r))/(3000 r)]);
tauanalytical[r_] = Integrate[inttau[r], r];
tauanalytical /@ Range[5, 30, 5]
(* {-1.80511 + 99.3459 I, -5.63104 + 99.3459 I, -...
1
I'm not 100% clear on what you want, but I assume that you want to actually plot grad on the y-axis, not lis. It also looks like you're trying to get "AvocadoColors" to scale from 0 to 9 instead of just the range of the data (1 to 9). If that's the case, my best guess at what you want is this:
n = 9;
lis = Subdivide[n];
grad = {1, 9, 9, 9, 5, 5, 9, ...
1
How about
ContourPlot[Boole[x^2 + y^2 == 1 || x^2 + y^2 == 2], {x, -2, 2}, {y, -2, 2}]
? Please, play with color on your own.
1
This is meant as a comment but I do not know how to add commands to comments.
My idea was to use the ImplicitRegion command.
test = ImplicitRegion[x^2 + y^2 == 1 || x^2 + y^2 == 2, {x, y}]
RegionPlot[test]
The result of the final command is shown below.
Is this what you had in mind?
1
xy = Transpose[{xcor, ycor}];
positions = Flatten[Position[#, 1]] & /@ Mmat;
lp = ListPlot[Extract[xy, List /@ positions],
PlotTheme -> "OpenMarkersThick",
PlotLegends -> Range[Length @ positions]]
lp2 = ListPlot[Extract[xy, List /@ positions]] /.
p_Point :> {Dynamic@EdgeForm[{Thin, CurrentValue["Color"]}],
...
1
Adding the option RegionFunction -> (-1 <= # <= 0 &) gives the desired result:
Plot[{x, -x}, {x, -5, 5},
PlotRange -> {{-1, 0}, {-3, 3}},
Frame -> True,
PlotLegends ->
Placed[LineLegend[{"Y1", "Y2"}, LegendLayout -> {"Row", 1},
LegendMarkerSize -> 20], {{0.5, 0.5}, {0, -1.8}}],
...
1
At first we use InterpolationOrder -> 0 to draw the piecewise function.
data = Abs[amn];
fig = ListPlot3D[data, PlotRange -> All, InterpolationOrder -> 0,
Mesh -> None, BoundaryStyle -> None, AspectRatio -> Automatic,
Boxed -> False, Axes -> None];
Then we replace all the space rectangles to pyramids by
pts = {{0, 0, 2}, {1, ...
1
To do this you might have to add extra values to your matrix, essentially to "pin down" the values of ListPlot3D to 0 between your "actual" matrix values. We can do that as follows:
mRiffle0[m_?MatrixQ] :=
Riffle[Riffle[#, 0, {1, -1, 2}] & /@
m, {ConstantArray[0, 1 + 2*Dimensions[m][[2]]]}, {1, -1, 2}]
ListPlot3D[mRiffle0[...
1
Clear["Global`*"]
G = 6.67259*10^(-11);
Manipulate[
F = G*(M*m)/r^2;
Show[Plot[G*(M*m)/R^2, {R, 0, 1},
PlotRange -> {{0, 1.2}, {0, 2*10^(-8)}}],
Graphics[{Red, PointSize[Large],
Tooltip[Point[#], #] &[{r, G*(M*m)/r^2}]}]],
{{M, 5}, 1, 10, 0.1, Appearance -> "Labeled"},
{{m, 1}, 1, 10, 0.01, Appearance -> "...
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