Mathematica 9 is now available
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: [mg75087] Re: ImplicitPlot errors
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Mon, 16 Apr 2007 20:16:04 -0400 (EDT)
  • References: <evvbfs$9q5$1@smc.vnet.net>

Hi.

Quit;

<< "Graphics`ImplicitPlot`"

ImplicitPlot[{x^2 + x*y + y^2 == 7, y == 2*Sqrt[7/3]}, {x, -5, 5}]
(*works*)

ImplicitPlot[x+y==2,{x,-4,4}]
(*works*)

Dimitris


=CF/=C7 David Rees =DD=E3=F1=E1=F8=E5:
> Hi,
>
> I've been trying to get ImplicitPlot to plot an implicit function (natura=
lly
> ;) ), but it throws errors to cryptic for me, even when copying and pasti=
ng
> 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 n=
ot 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 sing=
le
> 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 fails.
>
> Thanks



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