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 Author Comment/Response William Duhe 07/26/12 09:37am So here I have several functions plotted, a[t], B[t],Rm[t]. What I need is syntax that will allow me to see the value of a[t] when T[t] is 1. This should be easy to see that at first when T[t] is 1 a[t] is also one, but as the function turns around T[t] becomes 1 again, at about t = 575, and a[t] has risen by a certain amount. I need to quantify this amount. The syntax is provided bellow. Thanks ahead of time and best wishes! M = 100; g = 1; \[Alpha] = 1; ti = 0;(*initial time*) tf = 700; (*final plot time*) m = 1; RB[t] = (m*T[t]^3*g)/(2*\[Pi])^(3/2)*(m/T[t])^(3/2)*E^(-m/T[t]); R = 0.0005; V = 5*10^(-12)*M^4; i = Sqrt[(g*m^4 - V)/(3*M^2)]*m; rR = g*T[t]^4;(*Energy Density-Radiation*) pR = g*T[t]^4;(*Pressure-Radiation*) pM = 0;(*Pressure-Phi Particle*) r = rR + Rm[t] - V;(*Overall Energy Density*) p = pR + V;(*Overall Pressure*) s1 = NDSolve[{a''[t] == -a[t]*1/(6*M^2) (2*g*T[t]^4 + 2 V + Rm[t]), T'[t] == (R*Rm[t])/ T[t]^3 - (Sqrt[3/2]*\[Alpha]^2)/( Sqrt[T[t]]*m^( 11/2))*(((g*T[t]^3*m)/(2*\[Pi])^(3/2)*(m/T[t])^(3/2)* E^(-m/T[t]))^2 - Rm[t]^2) - (a'[t]*T[t])/a[t], Rm'[t] == -R*Rm[t] + (Sqrt[3/2]*\[Alpha]^2*T[t]^(5/2))/m^( 11/2)*((g/(2*\[Pi])^(3/2)*(m/T[t])^(3/2)*E^(-m/T[t]))^2 - Rm[t]^2) - 3 a'[t]/a[t]*Rm[t], T[0] == m, Rm[0] == g/(2*\[Pi])^(3/2)*E^-1, a'[0] == i, a[0] == 1}, {a[t], T[t], Rm[t]}, {t, ti, tf}, Method -> "BDF"] Plot[Evaluate[Rm[t]/RB[t] /. s1], {t, ti, tf}, PlotRange -> Automatic, AxesLabel -> {Time, Rm[t]}] Plot[Evaluate[T[t] /. s1], {t, ti, tf}, PlotRange -> Automatic, AxesLabel -> {Time, T[t]}] Plot[Evaluate[a[t] /. s1], {t, ti, tf}, PlotRange -> Automatic, AxesLabel -> {Time, a[t]}] URL: ,

 Subject (listing for 'Relating Two Functions') Author Date Posted Relating Two Functions William Duhe 07/26/12 09:37am Re: Relating Two Functions Bill Simpson 07/26/12 11:59pm
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