manipulate - Tangent to a circle

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}
]

]

Comments

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...

Blog

Search This Blog