Hey everyone, this is what I got. Email me all your criticism. Bryan

Note, I'm using t for t_0, and c for omega_0.

click here to get the Mathematica notebook file

In[1]:=

Out[1]=

$If[c/(-1 + c) 0, (2 (1 - c)^(1/2) + (-1 + 1/c^(1/2) c^(1/2)) c Log[4] - 2 c Log[1 + (1 ... - c)^(3/2)), Integrate[x/(c + x - c x)^(1/2), {x, 0, 1}, Assumptionsc/(-1 + c) ≠0]]$

Despite what the If statement says, I think the first expression is what we want when 0<c<1. I think we can make a few other simplifications, also. It comes out to the function below.

In[2]:=

$Plot[1/(1 - c) + (c Log[c^(1/2)/(1 + (1 - c)^(1/2))])/(1 - c)^(3/2), {c, 0, 1}, PlotRange-> {0, 1}, AxesLabel {"c", "t"}] ;$

[Graphics:HTMLFiles/index_4.gif]

The graph fits the two points we knew beforehand (c,t) = (1, 2/3) and (c,t) = (0, 1). Also, t does increase as c decreases.

In[3]:=

$FindRoot[1/(1 - c) + (c Log[c^(1/2)/(1 + (1 - c)^(1/2))])/(1 - c)^(3/2) == .87 , {c, .5}]$

Out[3]=

So when c > .1497, t > .87. That wasn't too bad.

Created by Mathematica (October 26, 2003)

final project