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The Archimedes spiral equation in parametric form:

fx[t_] := Cos[t]*t;
fy[t_] := Sin[t]*t;

How can it be converted to implicit form?

Is there a general method to convert any parametric equation to implicit form?

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  • 1
    $\begingroup$ The spiral of Archimedes can indeed be expressed as an implicit Cartesian equation, but it isn't pretty or more useful than the parametric or polar one: x Tan[Sqrt[x^2 + y^2]] == y $\endgroup$ – J. M. is in limbo Nov 15 '19 at 16:39
  • $\begingroup$ @J.M.willbebacksoon I included the spiral of Archimedes for possible input. The solutions to this issue that I found here could not cope with this curve. $\endgroup$ – PavelDev Nov 15 '19 at 16:50
  • $\begingroup$ In principle, if the components of the parametric equations only have trigonometric functions of the parameter, one can derive an implicit Cartesian equation (e.g. this). Otherwise, there isn't any general method that would work on e.g. {x == Exp[t] + t^2, y == t - Cos[t]}. $\endgroup$ – J. M. is in limbo Nov 15 '19 at 16:56
  • $\begingroup$ @J.M.willbebacksoon "Otherwise, there isn't any general method that would work" How is it prove in mathematics? $\endgroup$ – PavelDev Nov 15 '19 at 17:13
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fx[t_] := Cos[t]*t;
fy[t_] := Sin[t]*t;

ParametricPlot[{fx[t], fy[t]}, {t, 0, 10}]

enter image description here

sol = Assuming[Element[{x, y}, Reals],
  Eliminate[{x == fx[t], y == fy[t]}, t,
    InverseFunctions -> True] //
   FullSimplify]

t^2 == x^2 + y^2 && x == t Cos[t]

Since t was not eliminated

solxy = sol[[-1]] /. Solve[sol[[1]], t, Reals] // FullSimplify

(* {x + Sqrt[x^2 + y^2] Cos[Sqrt[x^2 + y^2]] == 0, 
 x == Sqrt[x^2 + y^2] Cos[Sqrt[x^2 + y^2]]} *)

The result is the original spiral and its mirror

ContourPlot[Evaluate@solxy,
 {x, -10, 10}, {y, -10, 10},
 AspectRatio -> 1,
 PlotPoints -> 100,
 MaxRecursion -> 5]

enter image description here

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