Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

implicit function

21 views
Skip to first unread message

Karpenko Alexey

unread,
May 11, 2010, 6:28:23 AM5/11/10
to
Hi. Im newbie in Mathematica Product. I need to draw graph of implicit function. Implicit function have integral,and argument in lower case of it.
How can i do this in Mathematica 7.0?

functions:
x(b,t)=(sqrt(3)-sqrt(3-2b^2*sin^2(t)))/(sqrt(3)+sqrt(3-2b^2*sin^2(t)))
Phi(x(b,t))=(1+x)^3*(1-x)*e^(-x)
and general equation is (implicit function for t1):
integral_t1^pi/2{Phi(x(b,t))*sin(t)dt}=1/2*integral_0^pi/2{Phi(x(b,t))*sin(t)dt}

b in [0,1], t in [0, pi/2]
sorry 4 my bad english.
thank you.
With best regards, Karpenko Alexey.

Daniel Lichtblau

unread,
May 12, 2010, 7:32:16 AM5/12/10
to

Your function also depends on the value of b, so you will need to take
that into account. Below is one way to go about this. Notice that I
restrict definitions to evaluate only when passed explicitly numeric
values, so that you avoid a slew of useless messages, and wasted cycles,
from attempts to evaluate numeric functions (such as NIntegrate and
FindRoot) on symbolic values.

x[b_?NumberQ,t_?NumberQ] :=
(Sqrt[3]-Sqrt[3-2*b^2*Sin[t]^2])/
(Sqrt[3]+Sqrt[3-2*b^2*Sin[t]^2])

phi[x_] := (1+x)^3*(1-x)*E^(-x)

myInt[b_?NumberQ, t1_?NumberQ] :=
NIntegrate[phi[x[b,t]]*Sin[t], {t,t1,Pi/2}]

impfunc[b_?NumberQ] := t1 /.
FindRoot[myInt[b,t1] == myInt[b,0]/2, {t1,Pi/4}]

In[15]:= Table[impfunc[b], {b,0.,1.,1/16}]

Out[15]= {1.0472, 1.04729, 1.04757, 1.04804, 1.0487, 1.04954,
1.05056, 1.05176, 1.05313, 1.05466, 1.05632, 1.05809,
1.05991, 1.06169, 1.06328, 1.06438, 1.06444}

Daniel Lichtblau
Wolfram Research

cinnabar

unread,
May 12, 2010, 7:33:24 AM5/12/10
to
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

0 new messages