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Here is the graph of 10x^3 + 10x^2 - 10x - 5y^3 + 100y = 5. How can I find its points of inflection or those parts that the trend of the graph bends or sways?

Reference:

f(x)

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  • $\begingroup$ Do you want to find minimum/maximum of the curve? $\endgroup$
    – zhk
    Feb 17, 2017 at 13:38
  • $\begingroup$ Maybe points where curvature vanishes? $\endgroup$
    – Daniel Lichtblau
    Feb 17, 2017 at 16:22

3 Answers 3

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f = 10 x^3 + 10 x^2 - 10 x - 5 y^3 + 100 y - 5;

EDIT: Simplified calculation of points based on comment by mikuszefski

(pts1 = {x, y} /. Solve[D[f, y] == 0 && f == 0, {x, y}, Reals] // 
    Simplify) // N

(*  {{-3.08289, 2.58199}, {2.42487, -2.58199}}  *)

(pts2 = {x, y} /. Solve[D[f, x] == 0 && f == 0, {x, y}, Reals] // 
    Simplify) // N

(*  {{-1., -4.44692}, {-1., -0.0500063}, {-1., 
  4.49693}, {0.333333, -4.50601}, {0.333333, 0.0685346}, {0.333333, 
  4.43747}}  *)

Show[
 ContourPlot[f, {x, -8, 8}, {y, -10, 10}, Contours -> {0}, 
  ContourShading -> None],
 Graphics[{AbsolutePointSize[6],
   Red, Point[pts1],
   Blue, Point[pts2]}]]

enter image description here

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  • $\begingroup$ @teokan-duran-demircan Hmmm, isn't this answer wrong? Isn't an inflection point a point of vanishing curvature? And why $D[f, x]/D[f, y] == 0$ that's the same as $D[f, x] == 0$ So basically this solution gives the points with horizontal and vertical slope, does it not? $\endgroup$
    – mikuszefski
    Mar 14, 2017 at 7:09
  • $\begingroup$ @mikuszefski - yes calculation of derivatives can be simplified. I leave it to you and OP as to what constitutes "those parts that the trend of the graph bends or sways." $\endgroup$
    – Bob Hanlon
    Mar 14, 2017 at 19:37
  • $\begingroup$ Sure, OP decides what OP wants. Cheers. $\endgroup$
    – mikuszefski
    Mar 15, 2017 at 6:53
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I assume an inflection point has zero curvature. Here is my attempt:

  eq1 = 10*x^3 + 10*x^2 - 10*x - 5*y^3 + 100*y == 5;
  soln1 = {ToRules[Reduce[eq1, y]]};
  eq2 = FullSimplify[ArcCurvature[{x, y /. soln1[[#]]}, x, "Cartesian"]] == 0 & /@ Range@Length@soln1; (* this may take a while *)
  soln2 = NSolve[#, x, Reals] & /@ eq2  // Flatten[#, 1] &

  pts = {x, y} /. soln1  /. soln2[[#]] & /@ Range@3 // Flatten[#, 1] &

{{-0.343223, -4.4761}, {-0.343223, 0.00794074}, {-0.343223, 4.46816}, {-0.333251, -4.47676}, {-0.333251, 0.00927029}, {-0.333251, 4.46749}, {-0.323341, -4.47742}, {-0.323341, 0.0105916}, {-0.323341, 4.46683}}

Inflection points seem to be clustered together.

 ContourPlot[Evaluate[eq1], {x, -8, 8}, {y, -10, 10}, Epilog -> {Red, PointSize[0.04], Point[pts]}, GridLines -> Automatic]

enter image description here

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  • $\begingroup$ I think you should have pts = ({x, y} /. soln1[[#]] /. soln2[[#]]) & /@ Range@3 (that's why the extra points) $\endgroup$
    – george2079
    Feb 17, 2017 at 22:06
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Finding points of vanishing curvature on implicit curve:

f[x_, y_] := -5 - 10 x + 10 x^2 + 10 x^3 + 100 y - 5 y^3
h = D[f[x, y], {{x, y}, 2}];
a = D[f[x, y], {{y, x}}] {1, -1};
k[u_, v_] := 
 Expand[a.h.a]/(D[f[x, y], {{x, y}}].D[f[x, y], {{x, y}}])^(3/
      2) /. {x -> u, y -> v}
p = {u, v} /. NSolve[k[u, v] == 0 && f[u, v] == 0, {u, v}, Reals]
ContourPlot[f[x, y], {x, -10, 10}, {y, -10, 10}, Contours -> {{0.}}, 
 ContourShading -> None, Epilog -> {Red, PointSize[0.02], Point[p]}]

enter image description here

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