New answers tagged

1

You can also use a single ContourPlot using Evaluate[1 - {{a11, a12, a13}, {a21, a22, a23}}.{P, B, i}] as the first argument: Manipulate[ContourPlot3D[Evaluate[1 - {{a11, a12, a13}, {a21, a22, a23}}.{P, B, i}], {P, 0, 1}, {B, 0, 1}, {i, 0, 1}, ContourStyle -> (Opacity[0.5, #] & /@ {Brown, Blue}), Mesh -> None, ViewPoint -> {2.5, 1, ...


3

Use one Manipulate Manipulate[ Module[{p1, p2, i, B, p}, p1 = ContourPlot3D[ 1 - a11 P - a12 B - a13 i == 0, {P, 0, 1}, {B, 0, 1}, {i, 0, 1}, ContourStyle -> Directive[Opacity[0.5], Brown], Mesh -> None ]; p2 = ContourPlot3D[ 1 - a21 P - a22 B - a23 i == 0, {P, 0, 1}, {B, 0, 1}, {i, 0, 1}, ContourStyle -> Directive[...


0

It appears that you didn't even look at what you built with Table when you defined xy. If you had you would have seen your Table expression doesn't build a list of coordinate pairs, which is what I think you are trying to get. Mathematica is highly interactive. That lets you check your code in each definition you make before you proceed to use it. You ...


0

S1i[t_, V1i_] := t*V1i; Manipulate[Plot[S1i[t, V1i], {t, 0, 10}, PlotRange -> {{0, 10}, {0, 100}}], {V1i, 0, 10}]


7

A workaround is to ReleaseHold the plot. Plot has gotten very complicated. I don't really know what they are trying to address exactly. Some of it has to do with data, Quantity[] etc.; some has to do with plot themes; and then there some other stuff I don't get. Workaround: Manipulate[ ReleaseHold@ Plot[0.4 + 0.001 x m, {x, 0, 1400}, PlotRange -&...


1

The red mark means that the value of the variable associated with the slider is outside of the range of the slider. You set the value to 0.1 but you set the range to start at 0.2.


1

Code ClearAll[f1, f2, f3] f1[x_] := 4.8 + Sin[x] f2[x_] := 3 + (-8 + x) (1/13 + (0.01 + 0.0022*(-4 + x))*(5 + x)) f3[x_] := 0.01*(x + 5)^2 viewpoint = {-2.7, 1.6, 1.3}; plt1 = Plot3D[f3[x], {x, -5, 5}, {y, 2, 6}, Mesh -> 20, PlotStyle -> Opacity[0], MeshStyle -> Opacity[.8], BoundaryStyle -> Directive[Thick, Gray], Lighting -> "...


1

Is this what you want ? Manipulate[ Grid[{{Show[B1, B2, B3, B4[a], AxesStyle -> Thick, Boxed -> False, AxesOrigin -> {0, 0, 0}, AxesLabel -> {x, y, z}, BoxRatios -> {1, 1, 1.3}], Show[t2, t3[a]]}}, ItemSize -> {{30, 30}}], {{a, 1}, -5, 5}] (Please format the item size accordingly)


1

As your code for some reason crashes my mathematica kernel, I'll give you an example with a shorter, lighter code that can be easily adapted to your case: plt1[a_] := Plot[Sin[a*x], {x, 0, 10}]; plt2[a_] := Plot[Cos[a*x], {x, 0, 10}]; Manipulate[ Grid[{{plt1[a], plt2[a]}}, ItemSize -> {{30, 30}}, Frame -> All, FrameStyle -> Red], {{a, 1}, -5, ...


4

Silvia already gave a pretty rigorous derivation (compare this with the treatment by Hall and Wagon), so let me show how to plot the desired trajectory of the rolling polygon's corner. One could certainly modify the code I gave here for this, but I will instead adapt this solution I previously wrote in OpenGL to Mathematica: With[{n = 4}, (* number of sides ...


21

This question is too interesting to resist, so I'll talk about how to analyze the problem. Take a look at sketch above. It describes an arbitrary moment during the rolling. From the kinematics view, $P$ is the "instant center of rotation". From the energy view, the square's center of mass $O$ keeps its height, thus the potential of the square doesn't change,...


0

Michael E2 inspired me to bite the bullet and move stuff out of the tracking functions into manipulate's first argument and instead use additional variables to allow testing whether the state of each has changed. I also added a new variable ("monkeyLives" in the code below) that can only be changed by Autorun and test that to see where the change is coming ...


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