RedesConjugadas::usage=" 797. The elliptical paraboloid + = 2z, is cut by the planes x + y = C, where C is an arbitrary constant. Find the family of lines that form a conjugated net with these sections. A net of surface lines is said to be conjugated if at each point the tangential vectors the lines of different families of this net are conjugated.Reference: Fedenko, A.S. Problemas de Geometría Diferencial, Fedenko, Moscú: Editorial Mir, Moscú, p.104,(1981).";
2
x
2
a
2
y
2
b
Development:
Let a = 3, b = 2 y C = 4.
In[]:=
Solve+==2z,x+y==4,{x,y}
2
x
3
2
y
2
Out[]=
x(6-(4+(6+(4-
2
5
3
-8+5z
),y2
5
3
-8+5z
),x2
5
3
-8+5z
),y2
5
3
-8+5z
)The first curve:
ζ[z_]:=6-4+[z_]:=6+4-'[z]
2
5
3
-8+5z
,2
5
3
-8+5z
,z;ζ
1
2
5
3
-8+5z
,2
5
3
-8+5z
,z;ζ
1
Out[]=
,-,1
3
-8+5z
3
-8+5z
The planes should not be orthogonal (in the second plane is the conjugated curve):
In[]:=
Solvex+y==4,-==4,{x,y}
x
3
y
2
Out[]=
x,y-
36
5
16
5
In[]:=
z=+
1
2
2
36
5
3
2
-
16
5
2
Out[]=
56
5
Let the point P = {, -, }
36
5
16
5
56
5
In[]:=
,-,1/.z
3
-8+5z
3
-8+5z
56
5
Out[]=
,-,1
1
4
1
4
In[]:=
Solve+==2z,-==4,{x,y}
2
x
3
2
y
2
x
3
y
2
Out[]=
x(8-(-12-(8+(-12+
3
5
2
-48+5z
),y2
5
2
-48+5z
),x3
5
2
-48+5z
),y2
5
2
-48+5z
)The second curve (the conjugate of the first):
ζ
3
3
5
2
-48+5z
,2
5
2
-48+5z
,z;ζ
4
3
5
2
-48+5z
,2
5
2
-48+5z
,z;ζ
4
In[]:=
,,1/.z
3
2
-48+5z
2
-48+5z
56
5
Out[]=
,,1
3
4
1
2
In[]:=
ManipulateClearAll;ShowContourPlot3D+==2z,{x,-15,15},{y,-15,15},{z,-6,15},ContourStyle->{Opacity[0.1],Yellow},MeshFalse,PerformanceGoal"Quality",ContourPlot3D[x+y==4,{x,-15,15},{y,-15,15},{z,-15,15},ContourStyle->{Opacity[0.2],Red},MeshFalse,PerformanceGoal"Quality"],ContourPlot3D-==4,{x,-15,15},{y,-15,15},{z,-15,15},ContourStyle->{Opacity[0.2],Blue},MeshFalse,PerformanceGoal"Quality",Graphics3DBlack,Ball,-,,0.45,ParametricPlot3D6-4+,t1,PlotStyleRed,ParametricPlot3D6+4-,t1,PlotStyleRed,ParametricPlot3D8--12-,t1,PlotStyleBlue,ParametricPlot3D8+-12+,t1,PlotStyleBlue,Graphics3DRed,Point,-,,Arrowheads[.05],ArrowTube,-,,,-,+,-,1,Graphics3DBlue,Point,-,,Arrowheads[.05],ArrowTube,-,,,-,+,,1,AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll,ViewPoint{-8,-2,6},ImageSize500,Style["Conjugate nets:",Bold,Medium],{{t1,-15+0.02,"Value (u)"},-15+0.01,15,0.01},ControlPlacementTop
2
x
3
2
y
2
x
3
y
2
36
5
16
5
56
5
2
5
3
-8+5z
,2
5
3
-8+5z
,z,z,8
5
2
5
3
-8+5z
,2
5
3
-8+5z
,z,z,8
5
3
5
2
-48+5z
,2
5
2
-48+5z
,z,z,48
5
3
5
2
-48+5z
,2
5
2
-48+5z
,z,z,48
5
36
5
16
5
56
5
36
5
16
5
56
5
36
5
16
5
56
5
(t1+15)
3
1
4
1
4
36
5
16
5
56
5
36
5
16
5
56
5
36
5
16
5
56
5
(t1+15)
3
3
4
1
2
Out[]=
In the graph the red tones correspond to the curve on the plane x + y = 4, the blue shades are of their cnjugate on the plane - = 4.
x
3
y
2
Verification that the tangent vectors to each curve are conjugated:
In[]:=
1 4 | - 1 4 | 1 |
1 6 | 0 | 0 |
0 | 1 4 | 0 |
0 | 0 | 0 |
3 4 |
1 2 |
1 |