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I am using the general form of a second-degree plane curve:

$$Ax^2+2Bxy + Cy^2+2Dx + 2Ey + F = 0$$

I want to randomly generate plane curves of this form, so I am using RandomReal[{-1,1},6] to
generate the coefficients. I made the above equation into a function:

SecondDegreeCurve[{a_, b_, c_, d_, e_, f_}, x_, y_] := 
   a x^2 + 2 b x y + c y^2 + 2 d x + 2 e y + f == 0

However, when I try to plug the composition

 SecondDegreeCurve[RandomReal[{5}, 6], x, y], 
 {x, -2, 2}, {y, -2, 2}]

into ContourPlot, the image comes back blank.

When I run SecondDegreeCurve[RandomReal[{5}, 6], x, y], I get something like the following output:

1.4557 + 5.20582 x + 1.29609 x^2 + 9.37565 y + 6.73248 x y + 1.84528 y^2 == 0

and when I plug this into ContourPlot, the curve is displayed.

My question is, what is it about the initial composed expression that doesn't display the curve?

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up vote 15 down vote accepted

This has to do with the HoldAll attribute of ContourPlot. Try it with Evaluate inserted, like this:

   SecondDegreeCurve[RandomReal[{5}, 6], x, y] // Evaluate,
   {x, -2, 2}, {y, -2, 2}

you get ouput like this:

Mathematica graphics


Mathematica graphics

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That did the trick! So, since I ran the composed function and got the output, that was the evaluation step that I was missing initially? – tlehman Feb 5 '12 at 22:12
Yeah, if you don't use Evaluate your function is evaluated completely anew for every point the ContourPlot tries. That means the RandomReal is executed every time, yielding a hopeless mess of values. With Evaluate the RandomReal disappears from the resulting function, as only the numerical results remain. – Sjoerd C. de Vries Feb 5 '12 at 22:16

Evaluate works. I prefer Function:

SecondDegreeCurve[{a_, b_, c_, d_, e_, f_}, x_, y_] := 
  a x^2 + 2 b x y + c y^2 + 2 d x + 2 e y + f == 0

ContourPlot[#, {x, -2, 2}, {y, -2, 2}] & @
  SecondDegreeCurve[RandomReal[{5}, 6], x, y]

Mathematica graphics

share|improve this answer

Use Evaluate, i.e.

ContourPlot[Evaluate @ SecondDegreeCurve[RandomReal[{5}, 6], x, y] == 0, 
   {x, -2, 2}, {y, -2, 2}]
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