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How to continuously control line thickness in plotting functions?

Vitaliy Kaurov, WOLFRAM Research
Posted 12 years ago
I know that I can control color and opacity of lines in plotting functions – like in the example below. I also would like the line to change its thickness smoothly as we move in the positive x-direction. Is it possible to do via options to Plot? If not, - what could be the easiest approaches? 

Plot[Sin[x^2], {x, -3, 3}, PlotStyle -> Thickness[.04], PlotPoints -> 50,
  ColorFunction -> Function[{x, y}, Directive[ColorData["Rainbow"][x], Opacity[Sin[10 y + 15 x]]]]]
POSTED BY: Vitaliy Kaurov
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Hi Ahmed. Of course there was the objective of the code. Thanks.
POSTED BY: Mohammed Bennani

Very interesting techniques to control the thickness of the curves. It well suited in animating the orbit of celestial objects. Thanks for this sharing.

x[t_] = 50 Cos[t]; y[t_] = 50 Sin[t]; z[t_] = 0 t;
H[t_, a_] = 
  ParametricPlot3D[{x[t], y[t], z[t]}, {t, a, a - 1.8 Pi}, 
   PlotRange -> {{-60, 60}, {-60, 60}, {-5, 5}},
   PlotStyle -> {Red, Opacity[0.6], Tube[t/200]}];
S[t_] = Graphics3D[Sphere[{x[t], y[t], z[t]}, 2]];
Animate[Show [H[t, a], S[a]], {a, 0, 2 Pi}]
POSTED BY: Mohammed Bennani

Hi Mohammed, Your code didn't work at first trial, but with little modifications, it worked, hope this is what you intended to show
POSTED BY: Ahmed Elbanna
It crossed my mind that in some situations we can take advantage of Filling:

Table[Plot[Evaluate[
  Flatten[{#, # + amp (x + 3.5), # - amp (x + 3.5)} & /@ {Cos[x^2], .3 Sin[10 x]}]], {x, -3, 3},
  PlotPoints -> 50, Filling -> {3 -> {2}, 5 -> {6}}, PlotStyle -> {None, Gray, Gray, None, Gray, Gray},
  ColorFunction -> Function[{x, y}, Directive[ColorData["Rainbow"][x], Opacity[x]]],
  AspectRatio -> 1/4, ImageSize -> 500], {amp, .005, .15, .05}]




The controlled thickness is only vertical in this case. It will not work in some other cases, for example ParametricPlot does not have Filling option at all. But, in ParametricPlot case, for example, we can use the function itself, because it can show bounded areas:

ParametricPlot[{(v + 1) u Cos[u], (v + 1) u Sin[u]}, {u, 0, 6 Pi}, {v, 0.5, 1},
Mesh -> None, MaxRecursion -> 4, Axes -> False, BoundaryStyle -> GrayLevel[.9], 
 ColorFunction -> Function[{x, y, u, v}, Directive[ColorData["Rainbow"][u v], Opacity[v]]]]
POSTED BY: Vitaliy Kaurov
Using MeshShading will also work but not as good as ParametricPlot and using Filling but it doesn't change the function to use and original input.
Plot[Sin[x^2], {x, -3, 3}, Mesh -> {Range[-3, 3, 6/100]}, MeshStyle -> None,
MeshShading -> (Directive[Hue[Rescale[#, {-3, 3}, {0, 1}]],
AbsoluteThickness[Rescale[#, {-3, 3}, {1, 20}]]] & /@Range[-3, 3, 6/100]),
PlotRangePadding -> Scaled[0.05]]
POSTED BY: MinHsuan Peng
Min, that is an awful abuse of Mesh.  I like it. emoticon
POSTED BY: Brett Champion
This whole thread demonstrates mostly for better the stupifying generality of Mathematica. Awesome.
POSTED BY: Seth Chandler
Not a general solution, but this works some times:

ParametricPlot3D[{{x, Sin[x^2], x}, {x, Cos[x^2], x}}, {x, 0 - 3, 3},
PlotStyle -> {{Opacity[0.5], Tube[x/2000]}, {Opacity[0.3], Tube[x/2000]}},
ViewPoint -> {0, 0, 1000}, PlotRange -> All, ColorFunction -> "Rainbow"]



npts = 12;
ParametricPlot3D[{{Cos[2 t], Sin[2 t], Cos}, {Sin[2 t], Cos[2 t], Sin}}, {t, 0, 2 Pi},
  PlotStyle -> {{Red, Tube[{0.1 Sin[t/(2 Pi)], 0.1}, PlotPoints -> npts, Frame -> "Frenet"]},
                {Yellow, Tube[{0.1 Sin[t/(2 Pi)], 0.1}, Frame -> "Frenet", PlotPoints -> npts]}},
  PlotRange -> All, MaxRecursion -> 0, PlotPoints -> 200]
POSTED BY: Ulises Cervantes Pimentel
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