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