OrthogonalNet::usage="Demonstrate that on a straight helicoid {u Cos[v]​, u Sin[v]​, a v}​, the diferential equation
2
du
- (​
2
u
+
2
a
​)
2​
dv
== 0 defines the orthogonal net. References: Fedenko, A.S. Problemas de Geometría Diferencial. Editorial Mir: p.91, Moscú, 1981; Personal.";
Development:
Let a = 3, P = {u = 1, v = π} o P = {-1, 0, 3π}:
In[]:=
DSolve[
2
u'[v]
-(
2
u[v]
+9)==0,u[v],v]
Out[]=
u[v]-
3Tanh[v-

1
]
1-
2
Tanh[v-

1
]
,u[v]
3Tanh[v-

1
]
1-
2
Tanh[v-

1
]
,u[v]-
3Tanh[v+

1
]
1-
2
Tanh[v+

1
]
,u[v]
3Tanh[v+

1
]
1-
2
Tanh[v+

1
]

In[]:=
Solve1==-
3Tanh[π-

1
]
1-
2
Tanh[π-

1
]
,

1

In[]:=
N

1

π+ArcTanh
1
10
-π

2
if

2
∈

Out[]=


1

3.46904-(0.+3.14159)

2
if

2
∈

In[]:=
Solve1==
3Tanh[π-

1
]
1-
2
Tanh[π-

1
]
,

1

In[]:=
N

1

π-ArcTanh
1
10
-π

2
if

2
∈

Out[]=


1

2.81414-(0.+3.14159)

2
if

2
∈

In[]:=
Solve1==-
3Tanh[π+

1
]
1-
2
Tanh[π+

1
]
,

1

In[]:=
N

1

-π-ArcTanh
1
10
+π

2
if

2
∈

Out[]=


1

-3.46904+(0.+3.14159)

2
if

2
∈

In[]:=
Solve1==
3Tanh[π+

1
]
1-
2
Tanh[π+

1
]

In[]:=
N

1

-π+ArcTanh
1
10
+π

2
if

2
∈

Out[]=


1

-2.81414+(0.+3.14159)

2
if

2
∈

In[]:=
S[u_,v_]:={uCos[v],uSin[v],3v};
Demonstration that they are orthogonal to each other:
In[]:=
S-
3Tanh[v-3.4690428038270515`]
1-
2
Tanh[v-3.4690428038270515`]
,v
Out[]=

3Cos[v]Tanh[3.46904-v]
1-
2
Tanh[3.46904-v]
,
3Sin[v]Tanh[3.46904-v]
1-
2
Tanh[3.46904-v]
,3v
In[]:=
ζ
1
[v_]:=
3Cos[v]Tanh[3.4690428038270515`-v]
1-
2
Tanh[3.4690428038270515`-v]
,
3Sin[v]Tanh[3.4690428038270515`-v]
1-
2
Tanh[3.4690428038270515`-v]
,3v;​​
In[]:=
S
3Tanh[v-2.8141425033525347`]
1-
2
Tanh[v-2.8141425033525347`]
,v
Out[]=
-
3Cos[v]Tanh[2.81414-v]
1-
2
Tanh[2.81414-v]
,-
3Sin[v]Tanh[2.81414-v]
1-
2
Tanh[2.81414-v]
,3v
In[]:=
ζ
2
[v_]:=-
3Cos[v]Tanh[2.8141425033525347`-v]
1-
2
Tanh[2.8141425033525347`-v]
,-
3Sin[v]Tanh[2.8141425033525347`-v]
1-
2
Tanh[2.8141425033525347`-v]
,3v;
In[]:=
ζ
1
'[π].
ζ
2
'[π]
Out[]=
1.77636×
-15
10
That is to say zero, so we can do it with the other curves.
We will graph only two:
In[]:=
​​ManipulateClearAll;S[u_,v_]:={uCos[v],uSin[v],3v};
ζ
1
[v_]:=
3Cos[v]Tanh[3.4690428038270515`-v]
1-
2
Tanh[3.4690428038270515`-v]
,
3Sin[v]Tanh[3.4690428038270515`-v]
1-
2
Tanh[3.4690428038270515`-v]
,3v;
ζ
2
[v_]:=-
3Cos[v]Tanh[2.8141425033525347`-v]
1-
2
Tanh[2.8141425033525347`-v]
,-
3Sin[v]Tanh[2.8141425033525347`-v]
1-
2
Tanh[2.8141425033525347`-v]
,3v;ShowParametricPlot3D[S[u,v],{u,-30,30},{v,0,2π},PlotStyle->Opacity[0.2],MeshFalse,BoundaryStyleDirective[Black,Thick],PerformanceGoal"Quality",ColorFunction"BlueGreenYellow"],ParametricPlot3D[
ζ
1
[v],{v,0,t1},PlotStyleRed],ParametricPlot3D[
ζ
2
[v],{v,0,t1},PlotStyleBlue],​​Graphics3D[{Red,Ball[{-1,0,3π},1]}],​​Graphics3DRed,Point[{-1,0,3π}],Arrowheads[.05],ArrowTube{-1,0,3π},{-1,0,3π}+
t1
2
ζ
1
'[π],Graphics3DBlue,Point[{-1,0,3π}],Arrowheads[.05],ArrowTube{-1,0,3π},{-1,0,3π}+
t1
2
ζ
2
'[π],AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll,ImageSize500,ViewPoint{-3,-3,2},ImageSize500,Style["Orthogonal net:",Bold,Large],{{t1,1,"Value (t)"},0.01,2π,0.1},ControlPlacementTop
Out[]=
​
Orthogonal net:
Value (t)