New answers tagged parametric-functions
2
To plot the solution for a range of delc, make these changes to your code:
(1) Remove delc = 1; Leave delc undefined. It will be your parameter.
(2) Keep the ParametricDSolveValue command and its first argument, but change the other arguments to get
s = ParametricNDSolveValue[ ... ,
{1/2*(V11[t] + V22[t] - 2*V12[t])^(-1)}, {t, 0, 100}, delc];
...
2
The output of Solve (stored in sol) gives more than one solution (actually, there are 9 possible solutions):
sol = Solve[((I*(del + gamma*Abs[a1]^2) - k/2) a1 - (ke/2) a2 +
Sqrt[ke] s ==
0) && ((I*(del + gamma*Abs[a2]^2) - k/2) a2 - (ke/2) a1 +
Sqrt[ke] s == 0), {a1, a2}];
Length@%
9
You can plot them individually:
n = 1;
...
1
Here is method for examining the relationship between P, T and G as functions of r. First we find the functions.
Clear[P, T, G]
{P[r_], T[r_], G[r_]} =
Solve[
{P == T/r - 1/(2 π r^2) + 2/(π r^4),
T == 1/(4 π r) + 2 P r - 1/(4 π r^3),
G == r/4 - (2 Pi P r^3)/3 + 3/(4 r)},
{P, T, G}][[1, All, 2]]
Then we make an interactive plot of the ...
0
ClearAll[T, P, G, t, p, g]
t = 1/(4 Pi r) + 2 P r - 1/(4 Pi r^3);
p = T/r - 1/(2 Pi r^2) + 2/(Pi r^4);
g = r/4 - (2 Pi P r^3)/3 + 3/(4 r);
func = {G, p, T} /. Solve[Eliminate[{T == t, P == p, G == g}, P], {G, T}, Reals][[1]];
{(23 + r^2)/(12 r),
2/(π r^4) - 1/(2 π r^2) + (-15 + 3 r^2)/(
4 π r^4),
(-15 + 3 r^2)/(4 π r^3)}
Quiet @ ...
1
ParametricPlot3D[{{s, 0, Sin[s]}, {0, s, Sin[s]}, {s, s, Sin[s]}}, {s, 0, 4 Pi},
PlotStyle -> Tube[.07], ImageSize -> Large]
0
axes[x_, y_, z_, f_, a_] :=
Graphics3D[
Join[{Red, Arrowheads[a]},
Arrow[Tube[{{0, 0, 0}, #}]] & /@ {{x, 0, 0}, {0, y, 0}, {0, 0,
z}}, {Text[
Style["x", FontSize -> Scaled[f]], {0.9*x, 0.1*y, 0.1*z}],
Text[Style["y", FontSize -> Scaled[f]], {0.1 x, 0.9*y, 0.1*z}],
Text[Style["z", FontSize -> Scaled[f]], {0.1*x, 0.1*...
4
With little modifications it seems to wotk:
s = ParametricNDSolveValue[{Derivative[2][y][x] + (a + b*(2 + (2/Pi)*ArcTan[x]))*y[x] == 0, y[-10] == Exp[I*10*Sqrt[a + b]],
Derivative[1][y][-10] == (-I)*Sqrt[a + b]*Exp[I*10*Sqrt[a + b]]}, y, {x, -10, 10}, {a, b}]
u = ParametricNDSolveValue[{Derivative[2][z][x] + (a + b*(2 + (2/Pi)*ArcTan[x]))*z[x] == 0, z[...
2
Let me give a little example.
Since i am working with MMa version 8.0, where ParametricNDSolve is not jet implemented, here a little workaround. Please adapt it to ParametricNDSolve.
I add intial conditions and look for minimum, not maximum, here.
eqs = {alpha* Derivative[2, 0, 0][x][t, alpha, beta] +
y[t, alpha, beta]*Derivative[1, 0, 0][x][t, alpha, ...
3
I think you're going about this in such a way as to make this harder than it is. Polygon represents a closed curve where the final point will connect to the first by default so we can just use that here. No need to add a line or anything.
The way this will work is as such, first get a tabular set of your noisy data. I'll do this in a slow way (with lots of ...
3
To expand a little on MarcoB's answer, if you want the plot to extend to the point {0., -1.99877}, you must tweak the plot options appropriately. Like so:
With[{γ = 0.172969},
ParametricPlot[{hR[Tan[α], γ], Areac[Tan[α], γ]}, {α, FBc[γ], FAc[γ] - 0.001},
PlotRange -> {{0, 1.2}, {-10, -1}},
PlotRangePadding -> {{.1, Automatic}, Automatic},
...
4
The point you want corresponds to the very first lowest value of $\alpha$; playing around with the values of your expression shows that very small changes in the value of $\alpha$ in this region translate to huge changes in the abscissa of the parametric points you want plotted. In other words, it is extremely easy for the adaptive internal plotting routines ...
2
This is just to fill out what I already said in a comment to the question.
You have made a simple typing mistake. You are omitting two commas from the last two segments that you added in 2nd example. You need commas before 0.41 < t <= 0.49 and 0.49 < t <= 0.53 Here is how plot looks when those commas are inserted:
ParametricPlot[
Piecewise[
...
7
Use Cases to extract the GraphicsComplex from the plot:
c1 = RevolutionPlot3D[{t, -1*2 t}, {t, 0, 1}];
gc = Cases[c1, _GraphicsComplex, Infinity];
Now use GeometricTransformation to apply different transformation functions onto the graphics:
Graphics3D[
{
gc,
GeometricTransformation[gc /. _RGBColor :> Red, ReflectionTransform @ {0, ...
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