I have two functions with two parameters (p & x). I am looking for the values of "x" that makes the two functions to be equal, varying "p" from 0 to 1. I have two solutions for this implicit solution, but I am only interested on drawing the first solution where there are equal. The problem looks something like this:

VF[x_, p_] = 
  50 (1 - p) x + 25 p x + 
   59 (0.044/(1/p)^3.7 + (0.52 (1 - p))/(1/p)^2.7) x^2.7;

VL[x_, p_] = -100 + 
   180 p x - (14 x^2.7)/(1/p)^3.7 + (-1 + 
      p) (-50 x + (17 x^2.7)/(1/p)^2.7);

 Plot[Evaluate@{VF[x, p], VL[x, p]}, {x, 0, 5}], {{p, 1, "p"}, 0.00001, 1}]

Two functions, varying "p", with x in the horizontal axis

Using ContourPlot I get the two curves for "p" within [0,1] and I haven't been able to get rid of the second curve (second solution for each "p") in the plot.

CountourPlot with two solutions for x, varying p from 0 to 1

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    – bbgodfrey
    Commented Apr 22, 2015 at 12:20
  • $\begingroup$ you should show the ContourPlot command you used. $\endgroup$
    – george2079
    Commented Apr 22, 2015 at 13:39

1 Answer 1


One approach:

 p[ x_ ] := p /. 
      First@FindRoot[ VF[x, p] - VL[x, p] == 0 , { p, 0.001 }]
 ParametricPlot[{p[x], x}, {x, .5, 5}, AspectRatio -> 1/GoldenRatio]

enter image description here

Note this is somewhat fortuitous that we can pick an initial guess at p that consistently yields the first solution..

Another approach is to operate on the graphics generated by contour plot to pull out the lines:

 lines = Cases[Normal@First@
       VF[x, p] - VL[x, p] == 0  , {p, 0, 1}, {x, 0, 
           5}], _GraphicsComplex, Infinity], _Line , Infinity];

      Graphics[ #  , Axes -> True, AspectRatio -> 1/GoldenRatio] & /@ lines]

enter image description here

enter image description here


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