Mathematica Asked on July 25, 2020
c = 5.; b = 3.; a = 2.; len = 7.5;
u[t_] := Reduce[
NSolve[(2 c + a Cos[t] - b Cos[u[t]])^2 + (a Sin[t] - b Sin[u[t]])^2 -
len^2 == 0], u, Reals]
Plot[u[t], {t, 0, 2 Pi}]
Please help to obtain a simplified output using Reduce. (4-Bar linkage).
You don't need Reduce
, try Weirstrass substitution u[t] -> 2 ArcTan[uu]
:
Leaving the parameters undefined try
eq = TrigExpand[(2 c + a Cos[t] - b Cos[u[t]])^2 + (a Sin[t] -
b Sin[u[t]])^2 - len^2 == 0]
sol = 2 ArcTan[uu] /.Solve[eq /. u[t] -> 2 ArcTan[uu] // TrigExpand, uu]
which gives the two solution branches u[t] in analytical form.
{2 ArcTan[(4 a b Sin[ t] - [Sqrt](-4 (a^2 + b^2 - 4 b c + 4 c^2 - len^2 - 2 a b Cos[t] + 4 a c Cos[t]) (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]) + 16 a^2 b^2 Sin[t]^2))/(2 (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]))], 2 ArcTan[(4 a b Sin[ t] + [Sqrt](-4 (a^2 + b^2 - 4 b c + 4 c^2 - len^2 - 2 a b Cos[t] + 4 a c Cos[t]) (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]) + 16 a^2 b^2 Sin[t]^2))/(2 (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]))]}
Plot[sol /. {c -> 5, b -> 3, a -> 2, len -> 15/2} //Evaluate, {t, 0, 2 Pi}]
Correct answer by Ulrich Neumann on July 25, 2020
Clear["Global`*"]
c = 5; b = 3; a = 2; len = 15/2;
eqn = (2 c + a Cos[t] - b Cos[u])^2 + (a Sin[t] - b Sin[u])^2 -
len^2 == 0 // Simplify
(* 227/4 + 40 Cos[t] == 12 (Cos[t - u] + 5 Cos[u]) *)
sol = u /.
Assuming[0 <= t <= 2 Pi,
Solve[{eqn, 0 <= t < 2 Pi}, u, Reals, Method -> Reduce] //
Simplify] /. C[1] -> 0;
u1[t_] = sol[[1]]
u2[t_] = sol[[2]]
FunctionDomain[#[t], t] & /@ {u1, u2}
%[[1]] // N
3.14159 < t < 4.86928 || 1.4139 < t < 3.14159
Plot[{u1[t], u2[t]}, {t, 0, 2 Pi}, PlotPoints -> 100,
MaxRecursion -> 5, PlotLegends -> Placed[{u1, u2}, {0.85, 0.7}]]
For comparison,
ContourPlot[Evaluate@eqn,
{t, 0, 2 Pi}, {u, -1.35, 1.35},
AspectRatio -> 1/GoldenRatio]
Answered by Bob Hanlon on July 25, 2020
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