       Re: implicit function

• To: mathgroup at smc.vnet.net
• Subject: [mg109687] Re: implicit function
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 12 May 2010 07:32:23 -0400 (EDT)

```Karpenko Alexey wrote:
> 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.

Your function also depends on the value of b, so you will need to take
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-Sqrt[3-2*b^2*Sin[t]^2])/
(Sqrt+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:= Table[impfunc[b], {b,0.,1.,1/16}]

Out= {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

```

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