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I want to see the image of a tube made from a space curve. I define the surface using Frenet-Serret frame but the software does not show me the plot. What might be wrong with my code?

Thank you.

J. M.'s missing motivation
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Baran Cimen
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  • can you post the code for the plot please, thx – penguin77 Apr 12 '15 at 13:01
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  • Perhaps adding the underscores in the definition of T2 might help. – Sjoerd C. de Vries Apr 12 '15 at 14:40
  • possible duplicate of Extruding along a path – Michael E2 Apr 12 '15 at 16:06
  • It is not because I wasn't looking for the use of Tube command, I needed a mathematical expression. – Baran Cimen Apr 12 '15 at 16:09
  • Not all the answers use Tube but plot a shape (e.g. a circle) normal to a parametrized path. Also related: http://mathematica.stackexchange.com/questions/29444/plotting-a-2d-shape-along-a-3d-parametric-function-curve, http://mathematica.stackexchange.com/questions/69831/moving-polygon-on-path-curve-to-create-3d-shape – Michael E2 Apr 12 '15 at 17:29

1 Answers1

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You need to define c and to use the correct syntax in the definition of T2:

T2[s_, y_] := r[s] + 2*(c[s]*Sin[y] - j[s]*Cos[y])
c[s_] := s; (* for example *)
ParametricPlot3D[Evaluate@T2[s, y], {s, 0, 2*Pi}, {y, 0, 2*Pi}]

enter image description here

Update: For the updated version of the question:

ClearAll[r, s, j, v, T2, y]
r[s_] := {-(5 + 2*Cos[2*s])*Sin[3*s], (5 + 2*Cos[2*s])*Cos[3*s], 
  2*Sin[2*s]}
j[s_] := Normalize[Cross[r'[s], r''[s]]]
v[s_] := Normalize[Cross[j[s], Normalize[r'[s]]]]
T2[s_, y_] := r[s] + 1*(j[s]*Sin[y] - v[s]*Cos[y])

ParametricPlot3D[Evaluate@T2[s, y], {s, 0, 2*Pi}, {y, 0, 2*Pi}]

enter image description here

kglr
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