Mathematica Asked by Matandra on February 16, 2021
I’m trying to write a code that moves a circle on the graph of a specific function, but I can’t find the error in my code. I added values to both coordinates, but the Manipulate fails to count the value from the formula. I’m moving the circle with a vector that is perpendicular to the function, and the length of the vector is 1 because the radius of the circle is also 1.
How do I solve this problem? I’m a beginner.
f1[x_] := x
Manipulate[
Show[
Plot[f1, {x, -1, 8}, AspectRatio -> Automatic],
ParametricPlot[
{Cos[x] - Sqrt[1/((f1'[t])^2 + 1)]*f1'[t] + t,
Sin[x] + Sqrt[1/((f1'[t])^2 + 1)]}, {x, 0, 2 Pi},
PlotRange -> {{-4, 4}, {-4, 4}}
]
], {t, 0, 2 Pi}
]
Use your function to define a parametric curve (say, curve
) and use it with ParametricPlot
and add the moving circle asEpilog
. The normal to the curve at point t
is Cross[Normalize[curve'[t]]]
so we can easily find the center of moving circle:
ClearAll[curve]
curve[x_] := {x, x}
Manipulate[ParametricPlot[curve[x], {x, -Pi, 3 Pi},
Epilog -> {Red, Circle[curve[t] + Cross[Normalize[curve'[t]]]]},
PlotRange -> {{-Pi, 3 Pi}, {-Pi, 3 Pi}},
PlotRangePadding -> Scaled[.1]],
{{t, 0}, -Pi, 3 Pi, Appearance -> "Open"}]
ClearAll[curve2]
curve2[x_] := 5 {Cos @ x, Sin @ x};
Manipulate[ParametricPlot[curve2[x], {x, 0, 2 Pi},
Epilog -> {Red, Circle[curve2[t] + Cross[Normalize[curve2'[t]]]]},
PlotRange -> 5 {{-1, 1}, {-1, 1}},
PlotRangePadding -> Scaled[.1]],
{{t, 0.}, 0., 2. Pi, Appearance -> "Open"}]
ClearAll[curve3]
curve3[x_] := 7 {Sin[4 Sin @ x], Sin[4 Cos @ x]};
frames = Table[ParametricPlot[curve3[x], {x, 0, 2 Pi}, PlotStyle -> Thick,
Epilog -> {Red, Thick, Circle[curve3[t] + Cross[Normalize[curve3'[t]]]]},
PlotRange -> 7 {{-1, 1}, {-1, 1}},
PlotRangePadding -> Scaled[.15], ImageSize -> Large,
Axes -> False], {t, 0., 2 Pi, 2 Pi/100}];
Export["circleoncurve.gif", frames]
Use Circle[curve3[t] - Cross[Normalize[curve3'[t]]]]
to place the circle outside the curve:
Answered by kglr on February 16, 2021
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