3
$\begingroup$

I am trying to show vector of fixed length which is the tangent to a circle which rotates. Any suggestions on how to simplify this ? My question is not about using Manipulate etc but more is there an easier way to compute and draw the tangent ?

x[t_] := Cos[t];
y[t_] := Sin[t];
d[x_, y_] := If[y == 0, 0, -x/y];
tangent[1, 0] = Arrow[{{1, 0}, {1, 1}}];
tangent[-1, 0] = Arrow[{{-1, 0}, {-1, -1}}];
tangent[0, -1] = Arrow[{{0, -1}, {1, -1}}];
tangent[0, 1] = Arrow[{{0, 1}, {1, 1}}];
tangent[x_, y_] = 
  If[y > 0, 
   Arrow[{{x, y}, {x - Cos[ArcTan[d[x, y]]], 
      y - Sin[ArcTan[ d[x, y]]]}}],
   Arrow[{{x, y}, {x + Cos[ArcTan[d[x, y]]], 
      y + Sin[ArcTan[ d[x, y]]]}}]];
Manipulate[
 p = ParametricPlot[{x[t], y[t]}, {t, 0, 2 π}, 
   PlotRange -> {-2, 2}];
 g = Graphics[{{Blue, Line[{{0, 0}, {x[a], y[a]}}]}, {Red, 
     tangent[x[a], y[a]]}}];
 Show[{p, g}],
 {a, 0, 2 π}]

Moving tangent

I extended my example above to include a 3D solution using suggested solutions -

Module[{x,y,z,tangent},
x[t_]:=Cos[t];
y[t_]:=Sin[t];
z[t_]:=0;
r ={{-2,2},{-2,2},{-2,2}};
p=ParametricPlot3D[{x[t],y[t],z[t]},{t,0,2π},PlotRange->r,BoxRatios->1];
tangent[x_,y_,z_]:={Red,Arrow[{{x,y,z},{x-y,x+y,z}}]};
Manipulate[
g :=Graphics3D[{Sphere[{x[t],y[t],z[t]},0.2],
{Green,Arrow[{{x[t],y[t],z[t]},{x[t],y[t],1}}]},
{Blue,Arrow[{{0,0,0},{x[t],y[t],z[t]}}]},
tangent[x[t],y[t],0]
},PlotRange->r,BoxRatios->1];
Show[p,g],
{t,0,4π}
]
]

3D

$\endgroup$
3
$\begingroup$

If all you need is a tangent to a circle, and not a tangent to an arbitrary curve, you can just rotate an arrow:

Manipulate[
 Graphics[{Circle[], Rotate[Arrow[{{1, 0}, {1, 1}}], φ, {0, 0}]}],
 {φ, 0, 2 Pi}
]

You may also be interested in the two-argument form of ArcTan[x,y] (look it up in the docs).

$\endgroup$
  • $\begingroup$ Thank you. Will checkout both suggestions. $\endgroup$ – David McHarg Mar 27 '13 at 17:26
2
$\begingroup$

Without Rotate:

ctangent[x_, y_] := Arrow[{{x, y}, {x - y, x + y}}]
Manipulate[
 Graphics[{Circle[], {Blue, Line[{{0, 0}, {x[a], y[a]}}]}, {Red, 
    ctangent[x[a], y[a]]}}, 
    PlotRange -> {{-2, 2}, {-2, 2}}, 
    Axes -> True], {a, 0, 2 π} 
]
$\endgroup$
  • $\begingroup$ Thanks. Never thought of that. Excellent. As i can use this solution in 3D as well. $\endgroup$ – David McHarg Mar 27 '13 at 17:43
1
$\begingroup$
a = {{0, 0}, {1, 0}, {1, 1}}; 
Manipulate[
 Graphics[{Circle[], 
           Rotate[{Line@a[[;;2]], Arrow[a[[2;;]]]}, x, {0, 0}]}, 
           PlotRange -> {{-2, 2}, {-2, 2}}, Axes -> True], 
{x, 0, 2 Pi}]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.