# Finding unit tangent, normal, and binormal vectors for interpolated function

As an extension of this question, is it possible to find the unit tangent, normal, and binormal vectors for an interpolated function? eg

pts = {{1, 1, -1}, {2, 2, 1}, {3, 3, -1}, {3, 4, 1}};
f = Interpolation[Transpose[{N@Range[0, 1, 1/(Length[pts] - 1)], pts}]];


Length@Last@FrenetSerretSystem[f[t], t] outputs 2.

Motivation is as this question.

• You can either use NDSolve[] on the Frenet-Serret equations (like what was done here), or construct a Bishop frame instead (like what was done here). Mar 1 at 9:38
• @J. M. thank you, I'll have a go Mar 1 at 9:41

First we get you interpolating function:

pts = {{1, 1, -1}, {2, 2, 1}, {3, 3, -1}, {3, 4, 1}};
f = Interpolation[Transpose[{N@Range[0, 1, 1/(Length[pts] - 1)], pts}]];


Then we calculate the tangent, nornal and binormal:

utan[t_] = f'[t]/Norm[f'[t]];
normal[t_] = Module[{fu, x}, fu[x_] = f''[x] - (f''[x].utan[x]) utan[x];
fu[t]/Norm[fu[t]]];
binormal[t_] = Cross[utan[t], normal[t]];


And finally we make an interactive plot:

Manipulate[
Show[{ParametricPlot3D[f[x], {x, 0, 1}],
Graphics3D[{Arrow[{f[t], f[t] + utan[t]}], Red,
Arrow[{f[t], f[t] + normal[t]}], Green,
Arrow[{f[t], f[t] + binormal[t]}]}]
}, PlotRange -> {{0, 4}, {0, 4}, {-2, 2}}], {t, 0, 1}] • great! thanks :) Mar 1 at 11:57