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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.
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