������������ ���� �� �������

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����������������
���������
Input the curve:
���������
α[�_] �= {� + ���[�]� ���[�]� � ���[� / �]}�
We want to use the formula κ(t) =
���������
���������
α' (t)×α'' (t)
α' (t) 3
for the curvature, so we compute the cross product:
�����[α �[�]� α ��[�]] // ������������
�
��
� �
� ���  + ���
� - � ���  � �
�
�
�
And take its (squared) norm:
���������
���������
����������������������� = �����[# � � � /@ %] // ������ // ������������
�� + � ���[�]
On the other hand, the denominator is the cube of the norm α’(t):
���������
���������
�������������� = ���������# � � � /@ α �[�] // ������ // ������������ � �
(� + � ���[�])�/�
And then the curvature is just:
���������
����[�����������������������] / ��������������
�� + � ���[�]
���������
(� + � ���[�])�/�
For the torsion, we want to use the formula
α' (t)·(α'' (t)×α''' (t))
. We already have the denominator, so we
α' (t)×α'' (t) 2
just need to compute the numerator:
���[α �[�]� �����[α ��[�]� α ���[�]]] // ������ // ������������
�
� ��� 
�
So the formula tells us that the torsion is:
% / �����������������������
� ��� �� 
�� + � ���[�]
2 ���
Sols.nb
���������
���������
�����[α]
Input the curve:
���������
α[�_� �_� �_] �= {(� � ���[�] + �) ���[�]� (� � ���[�] + �) ���[�]}�
Find the tangent vector:
���������
���������
������� = �[α[�� �� �]� �] // ��������
{- (� + � � ���[�]) ���[�]� � ���[�] + � � ���[� �]}
Squared norm of the tangent vector:
���������
���������
����������� = �����[# � � � /@ %] // ������ // ������������
� �� + �� + � � � ���[�]
So the unit tangent vector is the tangent vector divided by the norm; and the corresponding angle is
arctan of y/x:
������ @@ (������� / ����[�����������]) // ������������
������-
(� + � � ���[�]) ���[�]
�
� ���[�] + � � ���[� �]
� �� + �� + � � � ���[�]

� �� + �� + � � � ���[�]
Recall that the angle θ isn’t really well-defined, but its derivative θ’ is, so we differentiate to get θ’:
�[%� �] // ������������
� �� + �� + � � � ���[�]
� �� + �� + � � � ���[�]
And now the idea is to integrate θ’ from 0 to 2π. Unfortunately, this is an improper integral, so we have
to be a little bit careful. First of all, Mathematica will happily give us an indefinite integral which
increases smoothly:
θ = ���������
��
�
-
�
�
� �� + �� + � � � ���[�]
� �� + �� + � � � ���[�]
������
� �
(� � + �) ��� �� 
� � ��� ��  - � ��� �� 
+
�
�
������
(� � + �) ��� �� 
- � � ��� ��  + � ��� �� 

Sols.nb
����[θ /� {� → �� � → �}� {�� �� � π}]
10
8
6
4
2
1
2
3
4
5
6
-2
Now, we take the limit as θ goes to 2π from the left and subtract the limit as θ approaches 0 from the
right:
�����θ = �����[θ� � → � π� ��������� → �]
�
�
(� � - �)
�+
(� �+�)�
(-� �+�)�
π
��+�
�����θ = �����[θ� � → �� ��������� → - �]
(� � - �)
(� �+�)�
(-� �+�)�
π
� (� � + �)
(�����θ - �����θ) / (� π) // ������������
�
�
�+
(� � + �)�
(- � � + �)�
��
(� �+�)�
(-� �+�)�
��+�
Now get rid of annoying square roots:
% /� �����_ � �  �_ � � ⧴ � / ���[�]
�
�
�-
��
���[- � � + �]
+
��+�
���[- � � + �]
Then simplifying gives the rotation number:
������������[%]
�
�
� + ����[� � - �]
This is 2 when 2a > b, it’s 1 when 2a < b, and (interestingly) it is 3/2 when 2a = b, which is this case:
���
3
4 ���
Sols.nb
��������������[α[�� �� �]� {�� �� � π}]
2
1
1
2
3
4
-1
-2
You can see that there’s a cusp, where the tangent indicatrix instantly flips from pointing right to pointing
left, accounting for the fact that the rotation index is only a half-integer.
���������
Consider the following picture:
p
���������
θ
Then certainly
α(s)
π/2+θ
Sols.nb
π
2
���
5
+ θ(s) = arctan α2 ' (s) ,
α1 ' (s)
meaning that
θ(s) = arctan α2 ' (s)  - π2 .
α1 ' (s)
In turn, this means that
α ' (s)
sin(θ(s)) = sinarctan α2 ' (s)  - π  = -cosarctan α2 ' (s)  = -α1 ' (s)
α1 ' (s)
2
1
and
α ' (s)
cos(θ(s)) = cosarctan α2 ' (s)  - π  = sinarctan α2 ' (s)  = α2 ' (s)
α1 ' (s)
2
1
Moreover, since α(s) must lie on the line ℓ(p, θ) which is given by x cosθ + y sinθ = p, we know that
α1 (s) cos(θ(s)) + α2 (s) sin(θ(s)) = p(s)
and so, plugging in the expressions above, we get
p(s) = α1 (s) α2 ' (s) - α2 (s) α1 ' (s)
so we’ve succeeded in writing θ and p as functions of s.
���������
��������������
���������
�����[α]� �����[β]�
���������
α[�_] �= 
β[�_] �= 
�
�
�
�
� - � +
� - � +
�+�� 
�+�� 
�/�
�/�
�
�
�
�
�
�
- � +
� + � � �
- � +
� + � � �
�
�
�
�
�
- � +
� + � �  �
- � +
� + � �  + ��
�
(There’s no real mystery where these came from: α is just the arclength parametrization of the curve
4/3 t3/2 , t, t2 , which happens to be fairly easy to find an arclength parametrization of, and β is the same
thing shifted up by 1).
Plotting:
6 ���
Sols.nb
������������[{����������������[{α[�]� β[�]}� {�� �� �}� ��������� → #] � /@
{������ ����� {����� ���}}}� ��������� → ���]
2.0
2.0
0.0
0.5
1.0
1.5
1.5
1.0
1.0
1.0
0.5
0.0
1.0
0.5
0.0
2.0
1.5
0.5
0.5
0.0
0.0
0.5
1.0
0.5
0.0
1.0
1.0
0.5
1.0
0.5
0.0
0.0
0.0
Hopefully the three different perspectives are convincing that these are not planar curves, nor are they
helices, but of course we really need to prove this, which we can do by noting that plane curves have
vanishing torsion and helices have constant curvature, so this boils down to computing curvature and
torsion. First, curvature:
���������
�����������[�_] �= ��������[� ∈ ������ ������������[����[α ��[�]]]]
���������
�����������[�]
�+
���������
� ���-�+ �+� � 
�
���-�+ �+� � 
���[� + � �]�/�
which is obviously the same for both α and β, since they have the same derivative. We can see that the
curvature is nonconstant by evaluating at s=1 and s=2, showing that α and β aren’t helices:
���������
���������
{�����������[���]� �����������[���]}
{��������� ��������}
As for torsion:
���������
���������[�_] �=
α �[�]������[α ��[�]� α ���[�]]  �����[α �[�]� α ��[�]]������[α �[�]� α ��[�]]
Now, the torsion is not zero, as we can see by evaluating at one point, so α and β can’t be plane
curves, either:
���������
���������
���������[���]
- ��������
So we’ve seen that α and β aren’t helices and aren’t plane curves, so we’ll have proved they give a
counterexample if we can show they’re parametrized by arclength and lie at distance 1 from each other.
First, arclength parametrization. We compute |α’(s)|:
Sols.nb
���
����[�[α[�]� �]]
�

���[� + � �]
+ � ���
-� +
�+��
�
 + ���
�+��
-� +
�+��

�
�+��
And then some simplification shows that this is 1:
��������[� > �� ������������[%]]
�
Since β’(s)=α’(s), β is also parametrized by arclength.
And, finally, the fact that α(s) - β(s) = 1 is kind of obvious, since we can just subtract β(s) from α(s)
to get a vector which obviously has length 1:
α[�] - β[�] // ��������
{�� �� - �}
7
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