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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)]
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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

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

enter image description here

enter image description here

| improve this answer | |
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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]

enter image description here

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]]

enter image description here

| improve this answer | |
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