EquationTangentPlane::usage="Find the equation of the plane tangent to the surface S at point P {1, 0, -1}, where S is the surface generated by turning the curve C: {z = - 2 y, x = 0} around the OZ axis. Obtain the geodesic curvature of C in S. Reason if the intersection curve of S with the plane z = 1 is a geodesic. Reason if there is some asymptotic line of S that passes through the point P. Reference: www2.caminos.upm.es; casanchi.org.mat>cgeodesica01; Personal.";
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Development:
How the curve is going to rotate around Z axis, we substitute ±+ in y:
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z=+±2+;
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Expand==
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Equation of the plane tangent to the surface at the point P{1, 0, -1}:
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F[x_,y_,z_]:=+2+-2z-2z+-4-4;[x_,y_,z_]=D[F[x,y,z],x];[x_,y_,z_]=D[F[x,y,z],y];[x_,y_,z_]=D[F[x,y,z],z];(x-1)[1,0,-1]+y[1,0,-1]+(z+1)[1,0,-1]==0
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-4(1+z)0;z==-1;
Geodesic curvature of the C in S:Geodesic curvature ( measures how much a curve, traced over a surface, deviates from being geodesic (the strightest or shortest possible path on that surface). It can be calculated like this =[C'[t],C''[t],N], where C’[t] is the tangent vector to C and N the normal vector to surface.
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Normal with respect to the surface in P {1, 0, -1}:
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N1={[1,0,-1],[1,0,-1],[1,0,-1]}
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{0,0,-4};
Tangent vector of the curve C at t = 0 ( C’[t] ):
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C1[t_]:={0,t,-2t};=Det[{C1'[0],C1''[0],{0,0,-4}}]
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At that point t = 0 the C curve is a geodesic.
Intersection curve of S with plane z = 1 is a geodesic?
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These two circles are not geodesic, because their geodesic curvature is =, where r is its radius, that is, the geodesic curvature is not zero.
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Below are shown the geodesics of the surface S and it is seen that the curve C is included as we had observed, there is no circumference as geodesic:
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geo=ResourceFunction["Geodesic"]u,v,++2+,{u,v},t,{0,0},θ
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5+4+4+8+-5+4+4+8++5+4+4+8++[t]0,5+4+4+8+-5+4+4+8++5+4+4+8++[t]0,u[0]0,v[0]0,[0]Cos[θ],[0]Sin[θ]
4u[t]1++++1++[t]
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solg=Flatten(NDSolve[#1,{u,v},{t,1,4}]&)/@Tablegeo,θ,,2π,0.1,1;ParametricPlot3DEvaluateu[t],v[t],++2+/.solg,{t,1,4}
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Pass some asymptotic line through the point P?
The point P{1, 0, -1} is a parabolic point, therefore if there must be only one asymptotic line that passes through there.
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Solve[+2+-2z-2z+-4-4==0,z]
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z+-2+,z++2+
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σ[u_,v_]:=u,v,+-2+;ResourceFunction["SecondFundamentalFormCoefficients"][σ[u,v],{u,v}]/.{u1,v0}
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{2,0,0}
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DSolve[2==0,u[v],v]
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{{u[v]}}
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As expected from a parabolic point only a real asymptotic like passes.
Verification that P{1, 0, -1} is a parabolic point:
Verification that P{1, 0, -1} is a parabolic point:
LN-==0;(2)(0)-0==0;
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Animation:
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ManipulateShowRevolutionPlot3D[{0,y,-2y},{y,-5,5},{θ,0,θ1},RevolutionAxis{0,0,1},PlotStyle->Opacity[0.2],MeshFalse,BoundaryStyleDirective[Black,Thick],PerformanceGoal"Quality",ColorFunction"BlueGreenYellow"],ParametricPlot3D[t{0,0,1},{t,-5,10},ColorFunction->"BlueGreenYellow"],ContourPlot3D[z==-1,{x,-5,5},{y,-5,5},{z,-5,5},MeshFalse,ContourStyle->{Yellow,Opacity[0.6]}],Graphics3D[{Black,Ball[{1,0,-1},0.15]}],ParametricPlot3Dx,-
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,PlotStyle{Orange,Thick},ParametricPlot3Dx,-2
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,PlotStyle{Orange,Thick},AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRange8,ViewPoint{1.3,-1,1},ImageSize900,Style["Surface of revolution:",Bold,Medium],{{θ1,0.00002,"Value (θ)"},0.00001,2π,0.00001},ControlPlacementLeft2
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