# Plot a space curve and its curvature

The vector is < te^t, e^-t, Sqrt [2]t >, where -5 <= t <= 5 I have to plot the space curve and its curvature function k(t).

I already found the curvature, but do not know how to plot the space curve and its curvature function.

Curvature is:

Sqrt[2 E^(-2 t) + (3 + 2 t)^2 + (2 Sqrt[2] E^t + Sqrt[2] E^t t)^2]/[(2 + E^(-2 t) + (E^t + E^t t)^2)^(3/2)]

• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – user9660 Nov 14 '14 at 17:30

ArcCurvature is already built-in to Mathematica, so there is no need to compute curvature manually:

f[t_] := {t Exp[t], Exp[-t], Sqrt[2] t}
Simplify@ArcCurvature[f[t], t]
Plot[%, {t, -3, 3}]
ParametricPlot3D[f[t], {t, -1, 1}]


which produces

$$\kappa(t)=\frac{e^{2 t} \sqrt{2 e^{4 t} (t+2)^2+e^{2 t} (2 t+3)^2+2}}{\left(e^{4 t} (t+1)^2+2 e^{2 t}+1\right)^{3/2}}$$

fun = {u E^u, E^-u, Sqrt[2] u};

cur = Sqrt[2/E^(2*u) + (3 + 2*u)^2 + (2*Sqrt[2]*E^u + Sqrt[2]*E^u*u)^2]/
(2 + E^(-2*u) + (E^u + E^u*u)^2)^(3/2);

plo = Plot[cur, {u, -5, 5}, PlotRange -> All]


range = Last[PlotRange /. AbsoluteOptions[plo, PlotRange]];

ParametricPlot3D[fun, {u, -5, 5},
BoxRatios -> 1,
ColorFunction -> Function[{x, y, z, u, v},
ColorData["Rainbow"][Rescale[cur, range]]],
ColorFunctionScaling -> False,
PlotStyle -> Thickness[0.015]]