Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: ImplicitPlot errors

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75146] Re: ImplicitPlot errors
  • From: "David Rees" <w3bdevilREMOVE at THISw3bdevil.com>
  • Date: Wed, 18 Apr 2007 05:07:30 -0400 (EDT)
  • References: <evvbfs$9q5$1@smc.vnet.net>

To all of those who replied, thanks, but I found the problem.

I entered the equation as "2xy" rather than "2x*y", when I do that 
Mathematica has no complaint.

I also had a problem with an equation involving Euler's constant, I thought 
the "e" character (or the stylised "e" symbol in the toolbox) automatically 
represented Euler's constant in this context. My bad, I didn't realise it 
was uppercase "E" (my calculator, a Ti-83 uses uppercase E to define Power 
10 exponents and lowercase e for Euler's).

Once again, thanks.

"David Rees" <w3bdevilREMOVE at THISw3bdevil.com> wrote in message 
news:evvbfs$9q5$1 at smc.vnet.net...
> Hi,
>
> I've been trying to get ImplicitPlot to plot an implicit function 
> (naturally
> ;) ), but it throws errors to cryptic for me, even when copying and 
> pasting
> from the Mathematica function reference.
>
> In[40]:= ImplicitPlot[{x^2 + x*y + y^2 == 7, y == 2*Sqrt[7/3]}, {x, -5, 
> 5}]
>
> ImplicitPlot::var :
>
> Equation x^2+x
> Function[x,x^2+2xy-3y^2-16]+Function[x,x^2+2xy-3<<1>>-16]^2==7 does not 
> have
> a single variable other than x
>
> ImplicitPlot::var :
>
> Equation Function[x,x^2+2xy-3y^2-16]==2Sqrt(7/3) does not have a single
> variable other than x
>
> Out[40]:=\!\(ImplicitPlot[{x\^2 + x\ Function[x, x\^2 + 2\
>          xy - 3\ y\^2 - 16] + Function[
>              x, x\^2 + 2\ xy - 3\ y\^2 - 16]\^2 == 7, Function[x, x\^2 + 
> 2\
>              xy - 3\ y\^2 - 16] == 2\ \@\(7\/3\), x == 2\ \@\(7\/3\)}, {x,
> \
> \(-5\), 5}, {Function[x, x\^2 + 2\ xy - 3\ y\^2 - 16], \(-5\), 5}]\)
>
>
>
> What am I doing wrong? Even ImplicitPlot[x+y==2,{x,-4,4}] fails.
>
> Thanks
>
>
> 




  • Prev by Date: Re: minmum of a function
  • Next by Date: Re: how make function of solution by NDSolve depending on parameter?
  • Previous by thread: Re: ImplicitPlot errors
  • Next by thread: Re: ImplicitPlot errors