# Using Plot3D for finding trend clearly

I have a function of two positive variables $t$ and $C$ with $t$ and $C$ in a specific range as below. $0 < C < 10^{-5}$ and $0< t <4.54 \times 10^{-6}$. I want to see which variable of $C$ where the value of the function doesn't depend much on the variable $t$. So I used Plot3D[] function. However, from the plot I can vaguely see the trend. Is there a way to see it more clearly in Mathematica?

Function:

(0.0002607680962081059*(-1 + E^((63.9137490706142*Sqrt[6.799999999999999*^-6 - 144*C]*t)/C)))/(E^((12254.901960784315*(6.799999999999999*^-6 + 0.002607680962081059*Sqrt[6.799999999999999*^-6 - 144*C])*t)/C)*Sqrt[6.799999999999999*^-6 - 144*C])


Plot:

Plot3D[(0.0002607680962081059 E^(-((12254.901960784315 (6.799999999999999*^-6+0.002607680962081059 Sqrt[6.799999999999999*^-6-144 C]) t)/C)) (-1+E^((63.9137490706142 Sqrt[6.799999999999999*^-6-144 C] t)/C)))/Sqrt[6.799999999999999*^-6-144 C],{C,0,10^-5},{t,0,4.54 10^-6},AxesLabel->{C,t,VL}, PlotRange-> All,Mesh-> None,Background->Black,AxesStyle->White,Boxed->False]

• I'd suggest reducing the C range to {C, 0, 10^-6} to zoom in on a more interesting region of your plot; I would also suggest using DensityPlot or ContourPlot instead of Plot3D to capture some features of your 3D graph as a 2D projection, which may be more readable. – MarcoB Mar 1 '16 at 8:00

I want to see which variable of C where the value of the function doesn't depend much on the variable t

For this, I would plot the derivative of the function with respect to t, in those regions where it is close to 0, then the dependence on t is weakest.

f[C_, t_] := (0.0002607680962081059 E^(-((12254.901960784315 \
(6.799999999999999*^-6 +
0.002607680962081059 Sqrt[
6.799999999999999*^-6 - 144 C]) t)/C)) (-1 +
E^((63.9137490706142 Sqrt[6.799999999999999*^-6 - 144 C] t)/
C)))/Sqrt[6.799999999999999*^-6 - 144 C];
With[
{dfdt = D[f[C, t], t]},
Plot3D[{dfdt, 0}, {C, 0, 10^-5}, {t, 0, 4.54 10^-6},
PlotRange -> {-10^4, 10^4}, AxesLabel -> {C, t, VL},
BoxRatios -> {1, 1, 1}]
] As you can see, nowhere in this range is the derivative with respect to t equal to zero, so the function always has some dependence on t. But that dependence is strongest between C=0 and C=5*10^(-6).

You can view the derivative using a regular Plot as well,

With[
{dfdt = D[f[C, t], t], cvalues = Range[2 10^-6, 10^-5, 2.0 10^-6]},
Plot[Evaluate[
dfdt /. C -> # & /@ cvalues
], {t, 0, 4.54 10^-6},
PlotLegends -> (Row[{"C = ", #}] & /@ cvalues)]] • Well, extremely impressive! Thank you very much. – anhnha Mar 1 '16 at 10:05