Manipulating PlotRange in ParametricPlot

Question

why does the following code not work?

r = {2.30 Cos[5s], -2.30 Sin[5s]};

Manipulate[ParametricPlot[{Sin[3 t], Cos[2 t]}, {t, 0, 2 Pi}, 
PlotRange -> {{-r[[1]] , r[[1]]}, {-r[[2]] , r[[2]]}}], {s, 0.1, 2 Pi, 0.01}]

but if I put r[[1]] and r[[2]] by hand it works

Manipulate[ParametricPlot[{Sin[3 t], Cos[2 t]}, {t, 0, 2 Pi}, 
PlotRange -> {{-2.3 Cos[5 s],2.3 Cos[5 s]}, {2.3 Sin[5 s] , -2.3 Sin[5 s]}}]
,{s, 0.1, 2 Pi,0.01}]

It does not work because Manipulate does not "see" that $r$ depends on $s$, the manipulated variable. The quickest fix would be to move the definition of r to inside the Manipulate, rather than outside it: Manipulate[r = ..; ParametricPlot[..], {s, 0.1, 2 Pi, 0.01}]. — MarcoB, Dec 18 '19 at 18:33
Possible duplicate of How are parameters evaluated for a Plot in Manipulate. See also Manipulate Plot Sliders Not Working and Manipulate using a variable defined outside of the manipulate function. — MarcoB, Dec 18 '19 at 18:36
Finally, your PlotRange definition could be represented more briefly using PlotRange -> ({-#, #} & /@ r). — MarcoB, Dec 18 '19 at 18:41

score 0 · Answer 1 · answered Dec 18 '19 at 22:56

0

Try this:

f[s_] := {2.30 Cos[5 s], -2.30 Sin[5 s]};
Manipulate[
ParametricPlot[{Sin[3 t], Cos[2 t]}, {t, 0, 2 Pi}, 
PlotRange -> {{-f[s][[1]], f[s][[1]]}, {-f[s][[2]], f[s][[2]]}}],
{s, 0.1, 2 Pi, 0.01}
]

answered Dec 18 '19 at 22:56

Tugrul Temel

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Manipulating PlotRange in ParametricPlot

1 Answers1