StrictionLine::usage="Given the surface x = + u v, y = u + v, z = u + v, find the equation of the striction line. References: Usunariz B., Ubaldo. Problemas de Geometría analítica y Diferencial;Personal ";
2
u
2
u
Development:
In[]:=
s[u_,v_]:={+uv,u+v,u+v};c[u_]:={,u,u};w[u_]:={u,,1};
2
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2
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2
u
Demonstration that it is not a developable surface:
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ResourceFunction["PrincipalCurvatures"][s[u,v],{u,v}]
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+((1++)(1++6uv-14v++-4v(11+)+17(1+3)-2v(-50+7)+(12-30+)+(-7-3+))),-((1++)(1++6uv-14v++-4v(11+)+17(1+3)-2v(-50+7)+(12-30+)+(-7-3+)))/.{u1,v1}
-+2v+7v---(-9+)
6
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3
u
5
u
2
v
2
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2
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4
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2
v
3/2
(1+4+6uv-2v-4v++4+(-3+))
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3
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5
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2
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2
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2
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4
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2
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2
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4
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8
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7
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2
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4
v
3
u
2
v
6
u
2
v
5
u
2
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4
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2
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4
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2
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2
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4
v
3
(1+4+6uv-2v-4v++4+(-3+))
6
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3
u
5
u
2
v
2
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2
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4
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2
v
-+2v+7v---(-9+)
6
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3
u
5
u
2
v
2
u
2
v
4
u
2
v
3/2
(1+4+6uv-2v-4v++4+(-3+))
6
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3
u
5
u
2
v
2
u
2
v
4
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2
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2
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4
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8
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7
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2
v
4
v
3
u
2
v
6
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2
v
5
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2
v
4
u
2
v
4
v
2
u
2
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4
v
3
(1+4+6uv-2v-4v++4+(-3+))
6
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3
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5
u
2
v
2
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2
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4
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2
v
In[]:=
N+,-
7
8
2
57
2
8
7
8
2
57
2
8
Out[]=
{1.28604,-0.048599}
The ≠0,p[u] ≠ 0, it is not a developable surface, the point P = {u = 1, v = 1} or P = {2, 2, 2} it’s a hyperbolic point.
K
G
Striction line:
s[u,v]=c[u]+vw[u];
In[]:=
β[u]=c[u]-w[u]
(c'[u]w[u]).(w'[u]w[u])
2
(Norm[w'[u]w[u]])
β[u_]:=-,u-(u+2u(1-)-(-u+2)),u-;
2
u
u(u+2u(1-)-(-u+2))
2
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2
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3
u
1+4+
2
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4
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2
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2
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2
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3
u
1+4+
2
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4
u
u+2u(1-)-(-u+2)
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2
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3
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1+4+
2
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4
u
In[]:=
ManipulateClearAll;s[u_,v_]:={+uv,u+v,u+v};β[u_]:=-,u-(u+2u(1-)-(-u+2)),u-;Show[{ParametricPlot3D[s[u,v],{u,-3,3t1},{v,-3,3t1},PlotStyle->Opacity[0.3],ColorFunction"BlueGreenYellow",MeshFalse,PerformanceGoal"Quality",BoundaryStyleDirective[Black,Thick]],Graphics3D[{Black,Ball[{2,2,2},0.25]}],ParametricPlot3D[β[u],{u,-2,2t1},PlotStyle{Red,Thick}]},AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll,ViewPoint{-3,-10,20},ImageSize500],Style["Striction line (red colored):",Bold,Large],{{t1,0,"Value (t1)"},-1+0.01,1,0.01},ControlPlacementTop
2
u
2
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2
u
u(u+2u(1-)-(-u+2))
2
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2
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3
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1+4+
2
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4
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2
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2
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2
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3
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1+4+
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4
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u+2u(1-)-(-u+2)
2
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2
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3
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1+4+
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4
u
Out[]=

