Re: ContourPlot, equation and R.H. side of equation_Plotting problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg99087] Re: [mg99051] ContourPlot, equation and R.H. side of equation_Plotting problem*From*: "David Park" <djmpark at comcast.net>*Date*: Sun, 26 Apr 2009 01:40:52 -0400 (EDT)*References*: <15500446.1240650564504.JavaMail.root@n11>

This works: eqn1 = 3 x^2 + 6 y^2 == 6 ContourPlot[Evaluate@eqn1, {x, -2, 2}, {y, -2, 2}, Axes -> True] I think this is what might be called a feature. ContourPlot has the attribute HoldAll. Without the Evaluate it see no x and or y to substitute values. So you obtain no numeric result and no image. When you enter the form with the explicit equal sign, Mathematica follows a path that does evaluate and you get the image of the curve. The lesson is: when you don't obtain an image when you expect one, and the plotting parameters are not manifest in the expression, use Evaluate. David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: Bill [mailto:WDWNORWALK at aol.com] Hi: 1a.) When I assign 3 x^2 + 6 y^2 == 6 to eqn1, in Mathematica like this: eqn1=3 x^2 + 6 y^2 == 6; I can't get ContourPlot to plot eqn1 using this code: ContourPlot[eqn1, {x, -2, 2}, {y, -2, 2}, Axes -> True] 1b.) If I assign the equation like this without the constant on the R.H. side in eqn2, ContourPlot will plot the equation as expected, using the following syntax: eqn2=3 x^2 + 6 y^2; ContourPlot[eqn2 == 6, {x, -2, 2}, {y, -2, 2}, Axes -> True] Question: How can I get method 1a to work? Could you please give me code for this? Thanks, Bill PS. I'm using Mathematica 6.0.1 w/ Win XP on a PC.