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Mathematica Graphics and Visualization

Is it possible to plot a piece-wise function with variable bounds?

Alec Thomson

Posted 10 years ago

Specifically I am trying to plot the following: As v_r vs. l, sampling d in digit intervals from d=5 to d=15. Please forgive the specific nature of this question but I am having trouble finding a solution. The domain of l only need be 0 to pi/2. For the sake of context here is the rest of my visualisation: Plot[{Evaluate[ Table[Sin[l](2500/Sqrt[100 + d^2 - 20dCos[l]] - 250), {d, 0, 5}]], Evaluate[ Table[Sin[l](2500/Sqrt[100 + d^2 - 20dCos[l]] - 250), {d, 15, 20}]]}, {l, 0, \[Pi]/2}, AxesLabel -> {l, v}] I have tried using a combination of the `Table` and `Piecewise` functions, but to no avail. I would greatly appreciate any help. Cheers.

POSTED BY: Alec Thomson

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S M Blinder

S M Blinder, WRI and Univ of Michigan

Posted 10 years ago

Sometimes an alternative to Piecewise is a construction something like f[x_]:=f1[x] Boole[condition1]+f2[x] Boole[condition2]

POSTED BY: S M Blinder

Kuba Podkalicki

Posted 10 years ago

Quick answer: Evaluate @ Table[ Sin[l] Piecewise[{ {250, 0 < l < ArcCos[(d^2 + 75)/(20 d)]}, {2500/Sqrt[100 + d^2 - 20 d Cos[l]] - 250, l > ArcCos[(d^2 + 75)/(20 d)]}}], {d, 5, 15, 2}] Basic answer: Plot[ Evaluate@ Table[Sin[ l] Piecewise[{{250, 0 < l < ArcCos[(d^2 + 75)/(20 d)]}, {2500/ Sqrt[100 + d^2 - 20 d Cos[l]] - 250, l > ArcCos[(d^2 + 75)/(20 d)]}}], {d, 5, 15, 2}], {l, 0, Pi/2}, PlotRange -> All, ImageSize -> 800, Axes -> False, Frame -> True, FrameTicks -> {{Automatic, Automatic}, {Range[0, Pi/2, Pi/12], Automatic}}, GridLines -> {Table[ArcCos[(d^2 + 75.)/(20 d)], {d, 5, 15, 2}], {}}, PlotStyle -> Thick, BaseStyle -> 18] Extended answer: Plot[ Evaluate@Table[ Sin[l] Piecewise[{ {250, 0 < l < ArcCos[(d^2 + 75)/(20 d)]}, {2500/Sqrt[100 + d^2 - 20 d Cos[l]] - 250, l > ArcCos[(d^2 + 75)/(20 d)]}}], {d, 5, 15, 2}], {l, 0, 2 Pi}, PlotRange -> All, ImageSize -> 800, Axes -> False, Frame -> True, FrameTicks -> {{Automatic, Automatic}, {Range[0, 2 Pi, Pi/2], None}}, PlotStyle -> Thick, BaseStyle -> 18 ] // Show[#, Epilog -> { {EdgeForm@Dashed, FaceForm@None, Rectangle[{0, 0}, {Pi/4, 140}]}, Inset[ Show[#, PlotRange -> {{0, Pi/4}, {0, 140}}, ImageSize -> 250, FrameTicks -> {{Automatic, Automatic}, {Range[0, Pi/4, Pi/12], None}}, GridLines -> {Table[ ArcCos[(d^2 + 75)/(20 d)], {d, 5, 15, 2}], {}}], ImageScaled[{.65, .3}]] }] &

Quick answer:

Evaluate @ Table[
  Sin[l] Piecewise[{
     {250, 0 < l < ArcCos[(d^2 + 75)/(20 d)]},
     {2500/Sqrt[100 + d^2 - 20 d Cos[l]] - 250,   l > ArcCos[(d^2 + 75)/(20 d)]}}],
  {d, 5, 15, 2}]

Basic answer:

Plot[     Evaluate@ Table[Sin[  l] Piecewise[{{250,   0 < l < ArcCos[(d^2 + 75)/(20 d)]}, {2500/   Sqrt[100 + d^2 - 20 d Cos[l]] - 250, 
       l > ArcCos[(d^2 + 75)/(20 d)]}}], {d, 5, 15, 2}],
 {l, 0, Pi/2},
 PlotRange -> All, ImageSize -> 800, Axes -> False, Frame -> True,
 FrameTicks -> {{Automatic, Automatic}, {Range[0, Pi/2, Pi/12],     Automatic}},
 GridLines -> {Table[ArcCos[(d^2 + 75.)/(20 d)], {d, 5, 15, 2}], {}},   PlotStyle -> Thick, BaseStyle -> 18]

enter image description here Extended answer:

Plot[   Evaluate@Table[    Sin[l] Piecewise[{  {250, 0 < l < ArcCos[(d^2 + 75)/(20 d)]},
       {2500/Sqrt[100 + d^2 - 20 d Cos[l]] - 250,   l > ArcCos[(d^2 + 75)/(20 d)]}}],   {d, 5, 15, 2}],
  {l, 0, 2 Pi},
  PlotRange -> All, ImageSize -> 800, Axes -> False, Frame -> True,
  FrameTicks -> {{Automatic, Automatic}, {Range[0, 2 Pi, Pi/2], None}},
  PlotStyle -> Thick,    BaseStyle -> 18
  ] //  Show[#, Epilog -> {    {EdgeForm@Dashed, FaceForm@None, Rectangle[{0, 0}, {Pi/4, 140}]},
           Inset[
            Show[#, PlotRange -> {{0, Pi/4}, {0, 140}}, ImageSize -> 250, 
             FrameTicks -> {{Automatic, Automatic}, {Range[0, Pi/4, Pi/12], 
                None}},
             GridLines -> {Table[
                ArcCos[(d^2 + 75)/(20 d)], {d, 5, 15, 2}], {}}],
            ImageScaled[{.65, .3}]]
           }] &

enter image description here

POSTED BY: Kuba Podkalicki

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