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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg99071] Re: [mg99051] ContourPlot, equation and R.H. side of
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 26 Apr 2009 01:37:49 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Try these

ContourPlot[3 x^2 + 6 y^2 == 6,
 {x, -2, 2}, {y, -2, 2}, Axes -> True]

Attributes[ContourPlot]

{HoldAll,Protected}

eqn1 = 3 x^2 + 6 y^2 == 6;

ContourPlot[Evaluate[eqn1],
 {x, -2, 2}, {y, -2, 2}, Axes -> True]

eqn2 = 3 x^2 + 6 y^2;

ContourPlot[eqn2 == 6,
 {x, -2, 2}, {y, -2, 2}, Axes -> True]

ContourPlot[eqn2,
 {x, -2, 2}, {y, -2, 2}, Axes -> True,
 Contours -> {6}]


Bob Hanlon

---- Bill <WDWNORWALK at aol.com> wrote: 

=============
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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