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Re: implicit function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg109693] Re: implicit function
  • From: cinnabar <kolbasa.sapiens at gmail.com>
  • Date: Wed, 12 May 2010 07:33:30 -0400 (EDT)
  • References: <hsbbg7$jmo$1@smc.vnet.net>

Hello, Alexey!

The first and most simple way to do this that came to my mind is:

x[b_, t_] = (Sqrt[3] - Sqrt[(3 - 2 b^2*Sin[t]^2)])/(Sqrt[3] +
     Sqrt[(3 - 2 b^2*Sin[t]^2)]);
Phi[x1_] = (1 + x1)^3 (1 - x1) Exp[-x1];

And then just plot the equation line with ContourPlot. Note that
NIntegrate is used to avoid computing integral in analytic form:

ContourPlot[
      NIntegrate[(Phi[x[b, tt]] Sin[tt]), {tt, t1, Pi/2}] ==
  1/2 NIntegrate[(Phi[x[b, tt]] Sin[tt]), {tt, 0, Pi/2}],
{b, 0, 1}, {t1, 0, Pi/2}, MaxRecursion -> 10]

ContourPlot can be used intrinsically to plot equations, as you can
see in help section on this function. Mathematica 7 has nice
Documentation Center, which is strongly suggested to read when a
question arises on a function definition, arguments, etc. It has a lot
of examples too and many-many guidelines, tutorials, demos etc.

=D0=A3=D1=81=D0=BF=D0=B5=D1=85=D0=BE=D0=B2 =D1=81 =D0=9C=D0=B0=D1=82=D0=B5=
=D0=BC=D0=B0=D1=82=D0=B8=D0=BA=D0=BE=D0=B9!
=D0 =D0=BE=D0=BC=D0=B0=D0=BD


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