Chapter 3 • Theory & Derivations

Unit 3: Helices, Involutes, Evolutes & Bertrand Curves

Advanced geometric analysis of special space curve classes: general and cylindrical helices, Lancret's theorem, circular helices, spherical indicatrices of the Frenet frame on the unit sphere S^2, string unwinding construction of involutes and evolutes, and the complete characterization of Bertrand curve pairs.

§3.1 Cylindrical and General Helices: Lancret's Theorem

1. General and Cylindrical Helices

A helix is one of the most fundamental curved structures in mathematics, nature (DNA double helix), and mechanical engineering.

Definition 3.1 (General Helix): A regular space curve $\mathbf{r}(s)$ of class $C^3$ with $\kappa(s) > 0$ is called a general helix (or cylindrical helix) if its tangent lines make a constant angle $\alpha$ with a fixed non-zero direction vector $\mathbf{u} \in \mathbb{R}^3$:

\mathbf{T}(s) \cdot \mathbf{u} = \cos \alpha = \text{constant} \quad \forall s \in I

where $\mathbf{u}$ is a unit vector ($\|\mathbf{u}\| = 1$) called the axis of the helix, and $0 < \alpha < \pi/2$.

2. Lancret's Theorem

In 1802, Michel Ange Lancret formulated the definitive criterion characterizing all general helices through their curvature and torsion.

Theorem 3.1 (Lancret's Theorem): A regular space curve $\mathbf{r}(s)$ with $\kappa(s) > 0$ is a general helix if and only if the ratio of its torsion to its curvature is constant:

\frac{\tau(s)}{\kappa(s)} = c = \cot \alpha = \text{constant} \quad \forall s \in I

Complete Line-by-Line Proof:

$(\implies)$ Assume $\mathbf{r}(s)$ is a general helix. Then there exists a constant unit vector $\mathbf{u}$ and an angle $\alpha$ such that:

\mathbf{T}(s) \cdot \mathbf{u} = \cos \alpha = \text{const}

Differentiating with respect to arc-length $s$:

\mathbf{T}'(s) \cdot \mathbf{u} = 0 \iff (\kappa(s) \mathbf{N}(s)) \cdot \mathbf{u} = 0

Since $\kappa(s) > 0$, this implies:

\mathbf{N}(s) \cdot \mathbf{u} = 0 \quad \forall s \in I

Thus, the fixed axis vector $\mathbf{u}$ is strictly orthogonal to the principal normal $\mathbf{N}(s)$ at every point! Because $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ forms an orthonormal basis, $\mathbf{u}$ must lie entirely in the rectifying plane (the plane spanned by $\mathbf{T}$ and $\mathbf{B}$):

\mathbf{u} = (\mathbf{u} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{u} \cdot \mathbf{B}) \mathbf{B} = (\cos \alpha) \mathbf{T}(s) + (\sin \alpha) \mathbf{B}(s)

Since $\mathbf{u}$ is a constant vector, its derivative with respect to $s$ must be zero:

\frac{d\mathbf{u}}{ds} = (\cos \alpha) \frac{d\mathbf{T}}{ds} + (\sin \alpha) \frac{d\mathbf{B}}{ds} = \mathbf{0}

Applying the Frenet-Serret formulas $\mathbf{T}' = \kappa \mathbf{N}$ and $\mathbf{B}' = -\tau \mathbf{N}$:

\cos \alpha (\kappa(s) \mathbf{N}(s)) + \sin \alpha (-\tau(s) \mathbf{N}(s)) = \mathbf{0}

[\kappa(s) \cos \alpha - \tau(s) \sin \alpha] \mathbf{N}(s) = \mathbf{0}

Since $\|\mathbf{N}(s)\| = 1 \ne 0$:

\kappa(s) \cos \alpha - \tau(s) \sin \alpha = 0 \iff \frac{\tau(s)}{\kappa(s)} = \frac{\cos \alpha}{\sin \alpha} = \cot \alpha = \text{constant} \quad \blacksquare

$(\impliedby)$ Assume $\frac{\tau(s)}{\kappa(s)} = c = \text{constant}$. Choose an angle $\alpha \in (0, \pi/2)$ such that $\cot \alpha = c$, so $\cos \alpha = \frac{c}{\sqrt{1 + c^2}}$ and $\sin \alpha = \frac{1}{\sqrt{1 + c^2}}$. Define the vector field:

\mathbf{u}(s) = (\cos \alpha) \mathbf{T}(s) + (\sin \alpha) \mathbf{B}(s)

Differentiating $\mathbf{u}(s)$ with respect to $s$:

\mathbf{u}'(s) = (\cos \alpha) \mathbf{T}'(s) + (\sin \alpha) \mathbf{B}'(s) = (\cos \alpha) \kappa(s) \mathbf{N}(s) - (\sin \alpha) \tau(s) \mathbf{N}(s)

= [\kappa(s) \cos \alpha - \tau(s) \sin \alpha] \mathbf{N}(s) = \kappa(s) \left[ \cos \alpha - \frac{\tau(s)}{\kappa(s)} \sin \alpha \right] \mathbf{N}(s) = \kappa(s) [\cos \alpha - (\cot \alpha) \sin \alpha] \mathbf{N}(s) = \mathbf{0}

Since $\mathbf{u}'(s) = \mathbf{0}$, $\mathbf{u}$ is a constant unit vector! Finally:

\mathbf{T}(s) \cdot \mathbf{u} = \mathbf{T}(s) \cdot [(\cos \alpha) \mathbf{T}(s) + (\sin \alpha) \mathbf{B}(s)] = \cos \alpha \|\mathbf{T}\|^2 + 0 = \cos \alpha = \text{constant}

Thus, $\mathbf{r}(s)$ is a general helix with axis $\mathbf{u}$. $\blacksquare$

§3.2 The Circular Helix: Metrics, Intrinsic Equations & Geodesic Property

1. Parametrization and Metric Relations

A circular helix is a general helix drawn on the surface of a right circular cylinder of radius $a$:

\mathbf{r}(t) = \begin{pmatrix} a \cos t \\ a \sin t \\ b t \end{pmatrix}, \quad a > 0, \; b > 0

The cylinder radius is $a$.
The pitch (vertical ascent per full revolution $t \in [0, 2\pi]$) is:

h = 2\pi b

The speed is constant $c = \sqrt{a^2 + b^2}$, so arc-length is $s = c t$.

2. Intrinsic Curvatures of the Circular Helix

As derived in Unit 2:

\kappa = \frac{a}{a^2 + b^2} = \text{constant}, \quad \tau = \frac{b}{a^2 + b^2} = \text{constant}

Notice that:

As $b \to 0$, the helix flattens into a circle of radius $a$: $\kappa \to 1/a$, $\tau \to 0$.
As $a \to 0$, the helix straightens into a vertical line: $\kappa \to 0$, $\tau \to 0$.

Theorem 3.2 (Characterization of Circular Helices): A space curve is a circular helix if and only if both its curvature $\kappa$ and its torsion $\tau$ are non-zero constants.

Proof: By Lancret's theorem, $\tau/\kappa = \text{const}$ implies it is a general helix. If $\kappa = \text{const}$, then the radius of curvature $\rho = 1/\kappa = \text{const}$. The projection of the curve onto a plane perpendicular to the axis is a planar curve of constant curvature, which is a circle of radius $a = \frac{\kappa}{\kappa^2 + \tau^2}$. Thus the curve is a circular helix. $\blacksquare$

3. Geodesic Property on the Cylinder

If we slit the circular cylinder along a vertical generator and flatten it onto the Euclidean plane, the circular helix unrolls into a straight line! Because straight lines minimize distance on the Euclidean plane, the circular helix is a geodesic (shortest path) on the cylindrical surface.

§3.3 Spherical Indicatrices of the Frenet Frame

1. The Concept of a Spherical Indicatrix

Given a space curve $\mathbf{r}(s)$, we can map the vectors of its moving Frenet frame $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ to points on the unit sphere $S^2 = \{ \mathbf{x} \in \mathbb{R}^3 : \|\mathbf{x}\| = 1 \}$.

Definition 3.2 (The Three Spherical Indicatrices):

Tangent Spherical Indicatrix:

The curve traced on $S^2$ by the unit tangent vector:

\mathbf{r}_T(s) = \mathbf{T}(s)

Principal Normal Spherical Indicatrix:

The curve traced on $S^2$ by the principal normal vector:

\mathbf{r}_N(s) = \mathbf{N}(s)

Binormal Spherical Indicatrix:

The curve traced on $S^2$ by the binormal vector:

\mathbf{r}_B(s) = \mathbf{B}(s)

2. Metrics and Curvatures of the Indicatrices

Theorem 3.3 (Arc-Length Differentials of Spherical Indicatrices): Let $s_T, s_N, s_B$ denote the arc-lengths of the tangent, normal, and binormal indicatrices, respectively. Then:

Tangent Indicatrix:

\frac{ds_T}{ds} = \left\| \frac{d\mathbf{T}}{ds} \right\| = \|\kappa \mathbf{N}\| = \kappa(s) \implies ds_T = \kappa(s) \, ds

Binormal Indicatrix:

\frac{ds_B}{ds} = \left\| \frac{d\mathbf{B}}{ds} \right\| = \|-\tau \mathbf{N}\| = |\tau(s)| \implies ds_B = |\tau(s)| \, ds

Principal Normal Indicatrix:

\frac{ds_N}{ds} = \left\| \frac{d\mathbf{N}}{ds} \right\| = \|-\kappa \mathbf{T} + \tau \mathbf{B}\| = \sqrt{\kappa(s)^2 + \tau(s)^2} \implies ds_N = \sqrt{\kappa^2 + \tau^2} \, ds

Remarkable Geometric Interpretation:

The total length of the tangent indicatrix is $\int \kappa \, ds$, which is the total curvature of the curve.
The total length of the binormal indicatrix is $\int |\tau| \, ds$, which is the total torsion of the curve.
The principal normal indicatrix advances at the speed of the Darboux vector: $\|\boldsymbol{\omega}\| = \sqrt{\kappa^2 + \tau^2}$!

§3.4 Involutes and Evolutes of Space Curves

1. Involutes of a Space Curve

An involute of a curve $\mathbf{r}(s)$ is the trajectory traced by the end of a taut string being unwound from the curve.

Definition 3.3 (Involute): Let $\mathbf{r}(s)$ be an arc-length parametrized $C^2$ curve. An involute $\mathbf{r}^*(s)$ is defined by:

\mathbf{r}^*(s) = \mathbf{r}(s) + (c - s) \mathbf{T}(s)

where $c$ is an arbitrary constant (the total length of the unwinding string).

Theorem 3.4 (Orthogonality Property of Involutes): The tangent line to the original curve $\mathbf{r}(s)$ is orthogonal to the velocity vector of the involute $\mathbf{r}^*(s)$.

Proof: Differentiating $\mathbf{r}^*(s)$ with respect to $s$:

\frac{d\mathbf{r}^*}{ds} = \mathbf{r}'(s) - \mathbf{T}(s) + (c - s) \mathbf{T}'(s) = \mathbf{T}(s) - \mathbf{T}(s) + (c - s) \kappa(s) \mathbf{N}(s) = (c - s) \kappa(s) \mathbf{N}(s)

Taking the dot product with $\mathbf{T}(s)$:

\frac{d\mathbf{r}^*}{ds} \cdot \mathbf{T}(s) = (c - s) \kappa(s) (\mathbf{N}(s) \cdot \mathbf{T}(s)) = 0 \quad \blacksquare

2. Evolutes of a Space Curve

Definition 3.4 (Evolute): A curve $E$ is an evolute of a curve $C$ if $C$ is an involute of $E$. In space, the tangents to an evolute are normal lines to the original curve. An evolute lies on the envelope of normal planes of the original curve and has equation:

\mathbf{r}_E(s) = \mathbf{r}(s) + \rho(s) \mathbf{N}(s) + \rho(s) \cot\left( \int \tau \, ds + c \right) \mathbf{B}(s)

where $\rho = 1/\kappa$ is the radius of curvature.

§3.5 Bertrand Curves and Bertrand Mates: Linear Identity akappa + btau = 1

1. Definition of Bertrand Curves

In 1850, Joseph Bertrand investigated curve pairs whose principal normal lines coincide everywhere in space.

Definition 3.5 (Bertrand Curve and Bertrand Mate): A regular space curve $\mathbf{r}(s)$ is a Bertrand curve if there exists another distinct curve $\mathbf{r}^(s)$ such that the principal normal line to $\mathbf{r}$ at $s$ is identical to the principal normal line to $\mathbf{r}^$ at the corresponding point $s^$. The curve $\mathbf{r}^$ is called a Bertrand mate (or conjugate curve) of $\mathbf{r}$.

2. The Bertrand Characterization Theorem

Theorem 3.5 (Bertrand Curve Characterization): A regular space curve $\mathbf{r}(s)$ with $\kappa(s) > 0$ and $\tau(s) \ne 0$ is a Bertrand curve if and only if there exist non-zero real constants $a$ and $b$ such that:

a \kappa(s) + b \tau(s) = 1 \quad \forall s \in I

Complete Line-by-Line Proof:

$(\implies)$ Let $\mathbf{r}^*(s)$ be a Bertrand mate of $\mathbf{r}(s)$. Since their principal normals coincide, $\mathbf{r}^*(s)$ must lie along the principal normal line of $\mathbf{r}(s)$:

\mathbf{r}^*(s) = \mathbf{r}(s) + a(s) \mathbf{N}(s)

for some scalar function $a(s)$. Differentiating with respect to $s$:

\frac{d\mathbf{r}^*}{ds} = \mathbf{T}(s) + a'(s) \mathbf{N}(s) + a(s) \mathbf{N}'(s) = \mathbf{T} + a' \mathbf{N} + a(-\kappa \mathbf{T} + \tau \mathbf{B}) = (1 - a\kappa) \mathbf{T} + a' \mathbf{N} + a\tau \mathbf{B}

By definition of Bertrand mates, the principal normal $\mathbf{N}^(s)$ must be collinear with $\mathbf{N}(s)$: $\mathbf{N}^ = \pm \mathbf{N}$. Since $\mathbf{N}^$ is orthogonal to the tangent vector $\mathbf{T}^ = \frac{d\mathbf{r}^/ds}{\|d\mathbf{r}^/ds\|}$, we have:

\frac{d\mathbf{r}^*}{ds} \cdot \mathbf{N}(s) = 0

Evaluating the dot product:

\left[ (1 - a\kappa) \mathbf{T} + a' \mathbf{N} + a\tau \mathbf{B} \right] \cdot \mathbf{N} = a'(s) = 0 \implies a(s) = a = \text{constant}

Thus the distance $a$ between corresponding points of Bertrand mates is strictly constant! Now:

\frac{d\mathbf{r}^*}{ds} = (1 - a\kappa) \mathbf{T} + a\tau \mathbf{B}

Since $\mathbf{T}^$ is orthogonal to $\mathbf{N}^ = \pm \mathbf{N}$, $\mathbf{T}^*$ lies in the plane spanned by $\mathbf{T}$ and $\mathbf{B}$. Let $\alpha$ be the angle between $\mathbf{T}^*$ and $\mathbf{T}$:

\mathbf{T}^* = (\cos \alpha) \mathbf{T} + (\sin \alpha) \mathbf{B}

One can show (by differentiating $\mathbf{T} \cdot \mathbf{T}^*$) that the angle $\alpha$ is constant. Therefore:

\frac{d\mathbf{r}^*}{ds} = \frac{ds^*}{ds} \mathbf{T}^* = \frac{ds^*}{ds} [(\cos \alpha) \mathbf{T} + (\sin \alpha) \mathbf{B}]

Equating components:

1 - a\kappa(s) = \frac{ds^*}{ds} \cos \alpha \quad \text{and} \quad a\tau(s) = \frac{ds^*}{ds} \sin \alpha

Eliminating the term $\frac{ds^*}{ds}$:

(1 - a\kappa(s)) \sin \alpha = a\tau(s) \cos \alpha \iff \sin \alpha - a\sin\alpha \kappa(s) - a\cos\alpha \tau(s) = 0

Dividing by $\sin \alpha \ne 0$:

a \kappa(s) + (a \cot \alpha) \tau(s) = 1

Letting $b = a \cot \alpha$, we obtain the universal linear relation:

a \kappa(s) + b \tau(s) = 1 \quad \blacksquare

$(\impliedby)$ If $a\kappa(s) + b\tau(s) = 1$ with $a \ne 0$, defining $\mathbf{r}^(s) = \mathbf{r}(s) + a\mathbf{N}(s)$ and differentiating shows that its principal normal $\mathbf{N}^(s)$ is parallel to $\mathbf{N}(s)$. Thus $\mathbf{r}^*(s)$ is a Bertrand mate. $\blacksquare$

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational / Problem 3.1 Example 3.1: Tangent and Binormal Indicatrices of a Circular Helix

Consider the circular helix:

\mathbf{r}(t) = \begin{pmatrix} a \cos t \\ a \sin t \\ b t \end{pmatrix}, \quad a > 0, \; b > 0

where $c = \sqrt{a^2 + b^2}$.

Find the explicit parametric vector equations for the tangent spherical indicatrix $\mathbf{r}_T(t)$ and binormal spherical indicatrix $\mathbf{r}_B(t)$ on the unit sphere $S^2$.
Prove that both the tangent and binormal indicatrices are circles on $S^2$.
Compute the radius and the total perimeter of each indicatrix over one full turn $t \in [0, 2\pi]$.

1. Parametric Equations of the Indicatrices

From Unit 2, the unit tangent vector is:

\mathbf{r}_T(t) = \mathbf{T}(t) = \begin{pmatrix} -\frac{a}{c} \sin t \\ \frac{a}{c} \cos t \\ \frac{b}{c} \end{pmatrix}, \quad \text{where } c = \sqrt{a^2 + b^2}

The binormal vector is:

\mathbf{r}_B(t) = \mathbf{B}(t) = \begin{pmatrix} \frac{b}{c} \sin t \\ -\frac{b}{c} \cos t \\ \frac{a}{c} \end{pmatrix}

2. Proof that Both Indicatrices are Circles on $S^2$

Tangent Indicatrix $\mathbf{r}_T(t)$:

The third component is constant: $Z_T = \frac{b}{c}$. The first two components satisfy:

X_T^2 + Y_T^2 = \left( -\frac{a}{c} \sin t \right)^2 + \left( \frac{a}{c} \cos t \right)^2 = \frac{a^2}{c^2} (\sin^2 t + \cos^2 t) = \frac{a^2}{c^2}

Thus, $\mathbf{r}_T(t)$ is the intersection of the unit sphere $S^2$ ($X^2 + Y^2 + Z^2 = 1$) with the horizontal plane $Z = \frac{b}{c}$. This intersection is a circle of radius $R_T = \frac{a}{c} = \frac{a}{\sqrt{a^2 + b^2}}$.

Binormal Indicatrix $\mathbf{r}_B(t)$:

The third component is constant: $Z_B = \frac{a}{c}$. The first two components satisfy:

X_B^2 + Y_B^2 = \left( \frac{b}{c} \sin t \right)^2 + \left( -\frac{b}{c} \cos t \right)^2 = \frac{b^2}{c^2} = \frac{b^2}{a^2 + b^2}

Thus, $\mathbf{r}_B(t)$ is a circle of radius $R_B = \frac{b}{c} = \frac{b}{\sqrt{a^2 + b^2}}$ in the plane $Z = \frac{a}{c}$. $\blacksquare$

3. Radius and Perimeter over One Full Turn

Over $t \in [0, 2\pi]$:

For the tangent indicatrix:

\text{Radius } R_T = \frac{a}{\sqrt{a^2 + b^2}}

\text{Perimeter } L_T = 2\pi R_T = \frac{2\pi a}{\sqrt{a^2 + b^2}} = 2\pi a \kappa

For the binormal indicatrix:

\text{Radius } R_B = \frac{b}{\sqrt{a^2 + b^2}}

\text{Perimeter } L_B = 2\pi R_B = \frac{2\pi b}{\sqrt{a^2 + b^2}} = 2\pi b \tau \quad \blacksquare

Advanced / Problem 3.2 Example 3.2: Complete Deductive Chain of Lancret's Theorem for General Helices

Provide a complete, self-contained proof of Lancret's Theorem: A space curve $\mathbf{r}(s)$ of class $C^3$ with $\kappa(s) > 0$ has the property that its tangent vector makes a constant angle with a fixed non-zero direction $\mathbf{u}$ if and only if $\tau(s) / \kappa(s) = \text{constant}$. Also demonstrate that if $\tau / \kappa \equiv 0$, the curve is planar, and if both $\kappa$ and $\tau$ are non-zero constants, the curve is a circular helix.

1. Forward Direction

Let $\mathbf{u}$ be a constant unit vector such that $\mathbf{T}(s) \cdot \mathbf{u} = \cos \alpha$ for constant $\alpha \in (0, \pi/2)$. Differentiating with respect to arc-length $s$:

\mathbf{T}'(s) \cdot \mathbf{u} = 0 \iff (\kappa(s) \mathbf{N}(s)) \cdot \mathbf{u} = 0

Since $\kappa(s) > 0$, we have:

\mathbf{N}(s) \cdot \mathbf{u} = 0 \quad \forall s

Since $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ is an orthonormal basis, $\mathbf{u}$ has zero component along $\mathbf{N}$. Therefore, $\mathbf{u}$ lies in the plane of $\mathbf{T}$ and $\mathbf{B}$:

\mathbf{u} = (\mathbf{u} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{u} \cdot \mathbf{B}) \mathbf{B}

Since $\|\mathbf{u}\| = 1$ and $\mathbf{u} \cdot \mathbf{T} = \cos \alpha$, we have $\mathbf{u} \cdot \mathbf{B} = \pm \sin \alpha$. Choosing orientation so that $\mathbf{u} \cdot \mathbf{B} = \sin \alpha$:

\mathbf{u} = (\cos \alpha) \mathbf{T}(s) + (\sin \alpha) \mathbf{B}(s)

Since $\mathbf{u}$ is a constant vector:

\mathbf{0} = \frac{d\mathbf{u}}{ds} = (\cos \alpha) \mathbf{T}'(s) + (\sin \alpha) \mathbf{B}'(s)

Applying Frenet-Serret formulas $\mathbf{T}' = \kappa \mathbf{N}$ and $\mathbf{B}' = -\tau \mathbf{N}$:

\mathbf{0} = (\cos \alpha) \kappa(s) \mathbf{N}(s) - (\sin \alpha) \tau(s) \mathbf{N}(s) = [\kappa(s) \cos \alpha - \tau(s) \sin \alpha] \mathbf{N}(s)

Since $\|\mathbf{N}(s)\| = 1$:

\kappa(s) \cos \alpha - \tau(s) \sin \alpha = 0 \implies \frac{\tau(s)}{\kappa(s)} = \frac{\cos \alpha}{\sin \alpha} = \cot \alpha = \text{constant} \quad \blacksquare

2. Reverse Direction

Suppose $\frac{\tau(s)}{\kappa(s)} = c = \text{constant}$. Define $\alpha \in (0, \pi/2)$ such that $\cot \alpha = c$. Define the vector field:

\mathbf{u}(s) = (\cos \alpha) \mathbf{T}(s) + (\sin \alpha) \mathbf{B}(s)

Differentiating with respect to $s$:

\mathbf{u}'(s) = (\cos \alpha) \mathbf{T}'(s) + (\sin \alpha) \mathbf{B}'(s) = [(\cos \alpha) \kappa(s) - (\sin \alpha) \tau(s)] \mathbf{N}(s)

Since $\tau(s) = c \kappa(s) = (\cot \alpha) \kappa(s)$:

(\cos \alpha) \kappa(s) - (\sin \alpha) (\cot \alpha) \kappa(s) = \kappa(s) [\cos \alpha - \cos \alpha] = 0

Hence $\mathbf{u}'(s) = \mathbf{0}$, meaning $\mathbf{u}$ is a constant vector. Its magnitude is $\|\mathbf{u}\|^2 = \cos^2 \alpha + \sin^2 \alpha = 1$. Finally, $\mathbf{T}(s) \cdot \mathbf{u} = \cos \alpha = \text{constant}$. Thus the curve is a general helix. $\blacksquare$

3. Limiting Cases

If $\tau / \kappa \equiv 0$: Since $\kappa > 0$, this requires $\tau(s) \equiv 0$.

By Problem 2.2, a curve with $\tau \equiv 0$ lies entirely in a plane. Thus planar curves are degenerate helices with $\alpha = \pi/2$.

If $\kappa = \text{const}$ and $\tau = \text{const} \ne 0$:

By Theorem 3.2, the curve is a circular helix. $\blacksquare$

Honors / Problem 3.3 Example 3.3: Comprehensive Proof of the Bertrand Curve Characterization Theorem

Prove the Bertrand Curve Characterization Theorem: A regular space curve $\mathbf{r}(s)$ of class $C^3$ with $\kappa(s) > 0$ and $\tau(s) \ne 0$ has a Bertrand mate $\mathbf{r}^*(s)$ if and only if there exist real constants $a \ne 0$ and $b$ such that:

a \kappa(s) + b \tau(s) = 1 \quad \forall s \in I

Furthermore, prove that:

The distance between corresponding points of a Bertrand curve and its mate is constant ($a = \text{const}$).
The angle $\alpha$ between their corresponding unit tangent vectors $\mathbf{T}$ and $\mathbf{T}^*$ is constant.
The product of the torsions of a Bertrand curve and its mate satisfies $\tau(s) \tau^(s^) = \frac{\sin^2 \alpha}{a^2} = \text{constant}$.

1. Proof that Distance $a$ is Constant

Let $\mathbf{r}^(s^)$ be a Bertrand mate of $\mathbf{r}(s)$. Since their principal normals coincide at corresponding points:

\mathbf{r}^*(s) = \mathbf{r}(s) + a(s) \mathbf{N}(s)

for some scalar function $a(s) \ne 0$. Differentiating with respect to $s$:

\frac{d\mathbf{r}^*}{ds} = \mathbf{T}(s) + a'(s) \mathbf{N}(s) + a(s) [-\kappa(s) \mathbf{T}(s) + \tau(s) \mathbf{B}(s)] = (1 - a\kappa) \mathbf{T} + a' \mathbf{N} + a\tau \mathbf{B}

Let $\mathbf{T}^$ and $\mathbf{N}^$ be the unit tangent and principal normal of $\mathbf{r}^*$. By definition of Bertrand mates, $\mathbf{N}^(s^) = \pm \mathbf{N}(s)$. Since $\mathbf{T}^ \perp \mathbf{N}^$, we must have $\frac{d\mathbf{r}^*}{ds} \perp \mathbf{N}(s)$:

\frac{d\mathbf{r}^*}{ds} \cdot \mathbf{N}(s) = 0 \iff a'(s) = 0

Therefore, $a(s) = a = \text{constant}$! $\blacksquare$

2. Proof of Constant Angle $\alpha$ and the Linear Relation

Since $a' = 0$:

\frac{d\mathbf{r}^*}{ds} = (1 - a\kappa) \mathbf{T} + a\tau \mathbf{B}

Since $\frac{d\mathbf{r}^}{ds} = \frac{ds^}{ds} \mathbf{T}^$, $\mathbf{T}^$ lies in the span of $\mathbf{T}$ and $\mathbf{B}$. Let $\alpha(s)$ be the angle between $\mathbf{T}^*$ and $\mathbf{T}$:

\mathbf{T}^* = (\cos \alpha) \mathbf{T} + (\sin \alpha) \mathbf{B}

Differentiating with respect to $s$:

\frac{d\mathbf{T}^*}{ds} = \frac{d\mathbf{T}^*}{ds^*} \frac{ds^*}{ds} = \kappa^* \mathbf{N}^* \frac{ds^*}{ds}

Since $\mathbf{N}^ = \pm \mathbf{N}$, $\frac{d\mathbf{T}^}{ds}$ is entirely parallel to $\mathbf{N}$. Computing $\frac{d\mathbf{T}^*}{ds}$ from $(\cos \alpha) \mathbf{T} + (\sin \alpha) \mathbf{B}$:

\frac{d\mathbf{T}^*}{ds} = (-\alpha' \sin \alpha) \mathbf{T} + (\cos \alpha) \kappa \mathbf{N} + (\alpha' \cos \alpha) \mathbf{B} - (\sin \alpha) \tau \mathbf{N}

= (-\alpha' \sin \alpha) \mathbf{T} + (\kappa \cos \alpha - \tau \sin \alpha) \mathbf{N} + (\alpha' \cos \alpha) \mathbf{B}

For this vector to be parallel to $\mathbf{N}$, the components along $\mathbf{T}$ and $\mathbf{B}$ must vanish:

-\alpha' \sin \alpha = 0 \quad \text{and} \quad \alpha' \cos \alpha = 0 \implies \alpha'(s) = 0

Thus the angle $\alpha$ is strictly constant! Now equating components in $\frac{ds^}{ds} \mathbf{T}^ = (1 - a\kappa) \mathbf{T} + a\tau \mathbf{B}$:

\frac{ds^*}{ds} \cos \alpha = 1 - a\kappa(s) \quad \text{and} \quad \frac{ds^*}{ds} \sin \alpha = a\tau(s)

Dividing the two equations:

\tan \alpha = \frac{a\tau(s)}{1 - a\kappa(s)} \iff 1 - a\kappa(s) = a\tau(s) \cot \alpha \iff a\kappa(s) + (a\cot\alpha)\tau(s) = 1

Setting $b = a\cot\alpha$ establishes the linear relation:

a\kappa(s) + b\tau(s) = 1 \quad \blacksquare

3. Torsion Product Relation

From the dual relation on the Bertrand mate $\mathbf{r}^(s^)$, $\mathbf{r}(s) = \mathbf{r}^(s^) - a \mathbf{N}^(s^)$. The same analysis applied in reverse yields:

-a \kappa^*(s^*) + (a \cot \alpha) \tau^*(s^*) = 1

and $\frac{ds}{ds^} \sin(-\alpha) = -a \tau^(s^) \implies \frac{ds}{ds^} \sin \alpha = a \tau^(s^)$. Multiplying the two differential relations:

\left( \frac{ds^*}{ds} \sin \alpha \right) \left( \frac{ds}{ds^*} \sin \alpha \right) = (a\tau(s)) (a\tau^*(s^*))

\sin^2 \alpha = a^2 \tau(s) \tau^*(s^*) \iff \tau(s) \tau^*(s^*) = \frac{\sin^2 \alpha}{a^2} = \text{constant} \quad \blacksquare