Chapter 1 • Theory & Derivations

Theory of Space Curves: Arc-Length, Parametrization & Tangent Lines

Foundations of curve theory in 3D Euclidean space: vector-valued functions of a single real variable, regular parametrizations and velocity vectors, arc-length as an intrinsic geometric parameter, unit tangent vectors, equations of tangent lines, the osculating plane and order of contact, and canonical Taylor approximations near a regular point.

§1.1Vector Functions of a Real Variable, Smooth Curves & Regular Parametrizations

1. Vector Functions and Curves in Euclidean 3-Space

Let $\mathbb{R}^3$ denote standard three-dimensional Euclidean space endowed with the Cartesian coordinate system and standard inner product $\langle \mathbf{u}, \mathbf{v} \rangle = \mathbf{u} \cdot \mathbf{v}$.

Definition 1.1 (Parametrized Space Curve): A parametrized space curve (or vector function of a real variable) is a continuous mapping:

\mathbf{r}: I \to \mathbb{R}^3, \quad t \mapsto \mathbf{r}(t) = \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix}

where $I \subseteq \mathbb{R}$ is an interval of the real line. The set of image points $C = \mathbf{r}(I) \subset \mathbb{R}^3$ is the trace (or geometric trajectory) of the curve.

2. Differentiability and Regularity

Definition 1.2 (Smoothness and Regular Points):

A curve $\mathbf{r}(t)$ is of class $C^k$ ($k \ge 1$) if its component functions $x(t), y(t), z(t)$ possess continuous derivatives up to order $k$ on $I$.

The velocity vector (or derivative vector) at $t$ is:

\mathbf{r}'(t) = \frac{d\mathbf{r}}{dt} = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h} = \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix}

The scalar speed is $v(t) = \|\mathbf{r}'(t)\| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}$.

A point $t_0 \in I$ is called a regular point if:

\mathbf{r}'(t_0) \ne \mathbf{0} \iff \|\mathbf{r}'(t_0)\| > 0

If $\mathbf{r}'(t_0) = \mathbf{0}$, the point $t_0$ is called a singular point (or stationary point).

A curve $\mathbf{r}: I \to \mathbb{R}^3$ is regular if every point in $I$ is regular: $\mathbf{r}'(t) \ne \mathbf{0}$ for all $t \in I$.

Remark (Significance of Regularity): Regularity guarantees that the curve possesses a well-defined direction of motion at every instant, avoiding cusps, halts, or instantaneous sharp corners. For instance, the curve $\mathbf{r}(t) = (t^2, t^3, 0)$ is $C^\infty$ smooth as a vector function, but has a singular point at $t = 0$ ($\mathbf{r}'(0) = \mathbf{0}$), forming a geometric cusp in the trace.

3. Reparametrization and Equivalence of Curves

Definition 1.3 (Admissible Reparametrization): Let $\mathbf{r}: I \to \mathbb{R}^3$ be a $C^k$ curve ($k \ge 1$). Let $J \subseteq \mathbb{R}$ be another interval, and let $\phi: J \to I$ be a $C^k$ bijective scalar function such that:

\phi'(\tau) \ne 0 \quad \forall \tau \in J

The composition $\tilde{\mathbf{r}} = \mathbf{r} \circ \phi: J \to \mathbb{R}^3$ is called a reparametrization of $\mathbf{r}$.

If $\phi'(\tau) > 0$ for all $\tau \in J$, $\phi$ is an orientation-preserving reparametrization.

If $\phi'(\tau) < 0$ for all $\tau \in J$, $\phi$ is an orientation-reversing reparametrization.

By the Chain Rule:

\tilde{\mathbf{r}}'(\tau) = \frac{d}{d\tau}[\mathbf{r}(\phi(\tau))] = \mathbf{r}'(\phi(\tau)) \cdot \phi'(\tau)

Since $\phi'(\tau) \ne 0$, $\tilde{\mathbf{r}}'(\tau) \ne \mathbf{0} \iff \mathbf{r}'(\phi(\tau)) \ne \mathbf{0}$. Thus regularity is an intrinsic property invariant under admissible reparametrizations.

§1.2Arc-Length as an Intrinsic Invariant Parameter & Arc-Length Reparametrization

1. Arc-Length of a Curve Segment

Let $\mathbf{r}: [a, b] \to \mathbb{R}^3$ be a regular $C^1$ space curve.

Definition 1.4 (Arc-Length Function): The arc-length of the curve between parameter values $t_0$ and $t$ ($t \ge t_0$) is defined by the definite Riemann integral:

s(t) = \int_{t_0}^t \|\mathbf{r}'(u)\| \, du = \int_{t_0}^t \sqrt{x'(u)^2 + y'(u)^2 + z'(u)^2} \, du

2. Properties of the Arc-Length Parameter

Theorem 1.1 (Fundamental Properties of Arc-Length):

By the Fundamental Theorem of Calculus:

\frac{ds}{dt} = \|\mathbf{r}'(t)\| = v(t) > 0

Since $\mathbf{r}$ is regular, $\frac{ds}{dt} > 0$ everywhere on $I$.

Therefore, the arc-length mapping $s: I \to [0, L]$ is strictly increasing and hence invertible.

The differential of arc-length satisfies the Pythagorean metric identity:

ds^2 = dx^2 + dy^2 + dz^2 = \langle d\mathbf{r}, d\mathbf{r} \rangle

3. Reparametrization by Arc-Length (Natural / Unit-Speed Parametrization)

Because $s(t)$ is strictly increasing, its inverse function $t = t(s) = s^{-1}(s)$ exists and is $C^1$ smooth by the Inverse Function Theorem:

\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{\|\mathbf{r}'(t)\|}

Definition 1.5 (Unit-Speed / Natural Parametrization): Reparametrizing the curve with respect to arc-length $s$ yields the natural parametrization:

\mathbf{r}(s) = \mathbf{r}(t(s))

Theorem 1.2 (Unit-Speed Characterization): A regular curve $\mathbf{r}(s)$ is parametrized by arc-length if and only if its velocity vector has constant unit magnitude at every point:

\left\| \frac{d\mathbf{r}}{ds} \right\| \equiv 1 \quad \forall s

Complete Line-by-Line Proof:

Using the Chain Rule:

\frac{d\mathbf{r}}{ds} = \frac{d\mathbf{r}}{dt} \frac{dt}{ds} = \mathbf{r}'(t) \cdot \frac{1}{\|\mathbf{r}'(t)\|}

Taking the Euclidean norm of both sides:

\left\| \frac{d\mathbf{r}}{ds} \right\| = \left\| \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \right\| = \frac{\|\mathbf{r}'(t)\|}{\|\mathbf{r}'(t)\|} = 1 \quad \blacksquare

Parametrization by arc-length is the natural metric parametrization of differential geometry because it frees all geometric quantities (curvature, torsion, normal vectors) from artificial velocity variations.

§1.3The Unit Tangent Vector, Tangent Lines & Order of Contact

1. The Unit Tangent Vector

Definition 1.6 (Unit Tangent Vector $\mathbf{T}$): Let $\mathbf{r}: I \to \mathbb{R}^3$ be a regular curve.

For an arbitrary parameter $t$:

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}

For an arc-length natural parameter $s$:

\mathbf{T}(s) = \mathbf{r}'(s) = \frac{d\mathbf{r}}{ds}

Clearly, $\|\mathbf{T}(s)\| = 1$ for all $s$.

2. Equation of the Tangent Line

Definition 1.7 (Tangent Line to a Space Curve): The tangent line to the curve $\mathbf{r}(t)$ at the point $P_0 = \mathbf{r}(t_0)$ is the straight line passing through $P_0$ collinear with the velocity vector $\mathbf{r}'(t_0)$. Its vector parametric equation is:

\mathbf{X}(\lambda) = \mathbf{r}(t_0) + \lambda \mathbf{r}'(t_0), \quad \lambda \in \mathbb{R}

In Cartesian coordinates, if $\mathbf{r}(t_0) = (x_0, y_0, z_0)$ and $\mathbf{r}'(t_0) = (x'_0, y'_0, z'_0)$:

\frac{X - x_0}{x'_0} = \frac{Y - y_0}{y'_0} = \frac{Z - z_0}{z'_0}

3. Order of Contact

How tightly does a straight line, plane, or surface hug a space curve?

Definition 1.8 (Order of Contact): Let $S$ be a smooth surface defined implicitly by $F(x, y, z) = 0$. Let $\mathbf{r}(t)$ be a curve meeting $S$ at $t = t_0$, so $F(\mathbf{r}(t_0)) = 0$. Define the composite scalar function $g(t) = F(\mathbf{r}(t))$. The curve and surface have contact of order $n$ at $t = t_0$ if:

g(t_0) = 0, \quad g'(t_0) = 0, \quad g''(t_0) = 0, \quad \dots, \quad g^{(n)}(t_0) = 0, \quad \text{and } g^{(n+1)}(t_0) \ne 0

A generic secant line intersects a curve with contact of order 0.

The tangent line is the unique straight line having contact of order $\ge 1$ with the curve at $t_0$.

Space Curves, Tangent Vector & Osculating Plane Visualizer

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§1.4The Osculating Plane (Plane of Curvature) & Analytical Equation

1. Geometric Definition of the Osculating Plane

The word osculating originates from the Latin osculari ("to kiss"). The osculating plane is the unique plane that "kisses" the curve most closely at a given point.

Definition 1.9 (Osculating Plane): Let $\mathbf{r}(t)$ be a regular curve of class $C^2$ such that $\mathbf{r}'(t) \times \mathbf{r}''(t) \ne \mathbf{0}$. The osculating plane at point $P_0 = \mathbf{r}(t_0)$ is defined equivalently as:

The limiting position of the plane passing through three distinct points $P_0, P_1, P_2$ on the curve as $P_1, P_2 \to P_0$.

The limiting position of the plane containing the tangent line at $P_0$ and a neighboring point $P_1$ as $P_1 \to P_0$.

The unique plane having contact of order $\ge 2$ with the curve at $P_0$.

2. Analytical Equation of the Osculating Plane

Theorem 1.3 (Osculating Plane Equation): Let $\mathbf{X} = (X, Y, Z)$ be an arbitrary point in the osculating plane to $\mathbf{r}(t)$ at $t = t_0$.

In vector scalar triple product form:

[\mathbf{X} - \mathbf{r}(t_0), \; \mathbf{r}'(t_0), \; \mathbf{r}''(t_0)] = 0 \iff (\mathbf{X} - \mathbf{r}(t_0)) \cdot (\mathbf{r}'(t_0) \times \mathbf{r}''(t_0)) = 0

In determinant form:

\begin{vmatrix} > X - x(t_0) & Y - y(t_0) & Z - z(t_0) \\ > x'(t_0) & y'(t_0) & z'(t_0) \\ > x''(t_0) & y''(t_0) & z''(t_0) > \end{vmatrix} = 0

For an arc-length parametrized curve $\mathbf{r}(s)$:

[\mathbf{X} - \mathbf{r}(s_0), \; \mathbf{r}'(s_0), \; \mathbf{r}''(s_0)] = 0 \iff (\mathbf{X} - \mathbf{r}(s_0)) \cdot (\mathbf{T}(s_0) \times \mathbf{T}'(s_0)) = 0

Line-by-Line Proof:

Let $\Pi$ be a plane passing through $\mathbf{r}(t_0)$ with unit normal $\mathbf{n}_\Pi$:

F(\mathbf{X}) = (\mathbf{X} - \mathbf{r}(t_0)) \cdot \mathbf{n}_\Pi = 0

Define the distance function along the curve:

g(t) = F(\mathbf{r}(t)) = (\mathbf{r}(t) - \mathbf{r}(t_0)) \cdot \mathbf{n}_\Pi

For $\Pi$ to have contact of order $\ge 2$ with the curve at $t_0$, we require:

$g(t_0) = (\mathbf{r}(t_0) - \mathbf{r}(t_0)) \cdot \mathbf{n}_\Pi = 0$ (satisfied automatically).

$g'(t_0) = \mathbf{r}'(t_0) \cdot \mathbf{n}_\Pi = 0$.

This implies that $\mathbf{n}_\Pi$ is perpendicular to the velocity vector $\mathbf{r}'(t_0)$.

$g''(t_0) = \mathbf{r}''(t_0) \cdot \mathbf{n}_\Pi = 0$.

This implies that $\mathbf{n}_\Pi$ is also perpendicular to the acceleration vector $\mathbf{r}''(t_0)$.

Since the normal vector $\mathbf{n}_\Pi$ is perpendicular to both $\mathbf{r}'(t_0)$ and $\mathbf{r}''(t_0)$, it must be collinear with their vector cross product:

\mathbf{n}_\Pi \parallel \mathbf{r}'(t_0) \times \mathbf{r}''(t_0)

Therefore, any displacement vector $\mathbf{X} - \mathbf{r}(t_0)$ in the plane $\Pi$ must satisfy:

(\mathbf{X} - \mathbf{r}(t_0)) \cdot (\mathbf{r}'(t_0) \times \mathbf{r}''(t_0)) = 0 \quad \blacksquare

3. Tangent and Normal Planes for Implicit Surfaces

Definition 1.10 (Tangent and Normal Planes of Surfaces): Let a surface be given implicitly by $F(x, y, z) = 0$, and let $P_0 = (x_0, y_0, z_0)$ be a regular point ($\nabla F(P_0) \ne \mathbf{0}$).

The normal vector to the surface at $P_0$ is the gradient vector:

\mathbf{n} = \nabla F(P_0) = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)_{P_0}

The tangent plane to the surface at $P_0$ has equation:

\nabla F(P_0) \cdot (\mathbf{X} - \mathbf{P}_0) = 0 \iff F_x(X - x_0) + F_y(Y - y_0) + F_z(Z - z_0) = 0

The normal line to the surface at $P_0$ is:

\frac{X - x_0}{F_x} = \frac{Y - y_0}{F_y} = \frac{Z - z_0}{F_z}

§1.5Canonical Local Form & Taylor Expansions Near a Regular Point

1. Taylor Expansion of a Space Curve

Let $\mathbf{r}(s)$ be an arc-length parametrized $C^3$ curve with $\mathbf{r}(0) = \mathbf{0}$. Expanding $\mathbf{r}(s)$ in a Taylor series about $s = 0$:

\mathbf{r}(s) = \mathbf{r}(0) + s \mathbf{r}'(0) + \frac{s^2}{2} \mathbf{r}''(0) + \frac{s^3}{6} \mathbf{r}'''(0) + O(s^4)

Recall:

$\mathbf{r}'(0) = \mathbf{T}$ (unit tangent vector).

$\mathbf{r}''(0) = \mathbf{T}' = \kappa \mathbf{N}$ (curvature $\times$ principal normal).

$\mathbf{r}'''(0) = (\kappa \mathbf{N})' = \kappa' \mathbf{N} + \kappa \mathbf{N}' = \kappa' \mathbf{N} + \kappa (-\kappa \mathbf{T} + \tau \mathbf{B}) = -\kappa^2 \mathbf{T} + \kappa' \mathbf{N} + \kappa \tau \mathbf{B}$.

2. The Canonical Coordinate Form

Adopting the Frenet trihedron $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ at $s = 0$ as a local Cartesian basis, any point on the curve has coordinates $\mathbf{r}(s) = x(s) \mathbf{T} + y(s) \mathbf{N} + z(s) \mathbf{B}$:

\begin{aligned} x(s) &= s - \frac{\kappa^2}{6} s^3 + O(s^4) \\ y(s) &= \frac{\kappa}{2} s^2 + \frac{\kappa'}{6} s^3 + O(s^4) \\ z(s) &= \frac{\kappa \tau}{6} s^3 + O(s^4) \end{aligned}

3. Geometric Projections onto the Fundamental Coordinate Planes

By eliminating $s$ in the lowest-order leading terms, we discover the local geometric silhouette of any space curve:

1. Projection onto the Osculating Plane (Span{$\mathbf{T}, \mathbf{N}$}, $xy$-plane):

y \approx \frac{\kappa}{2} x^2

To leading order, the curve resembles a parabola opening along the principal normal $\mathbf{N}$.

2. Projection onto the Rectifying Plane (Span{$\mathbf{T}, \mathbf{B}$}, $xz$-plane):

z \approx \frac{\kappa \tau}{6} x^3

To leading order, the curve resembles a cubic inflection crossing its tangent line.

3. Projection onto the Normal Plane (Span{$\mathbf{N}, \mathbf{B}$}, $yz$-plane):

z^2 \approx \frac{2 \tau^2}{9 \kappa} y^3 \iff y^3 \sim z^2

To leading order, the curve projects as a semicubical Neil's cusp!

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.

Tier 1 • Foundational Concept Example 1.1: Arc-Length Reparametrization & Osculating Plane of the Twisted Cubic

Consider the twisted cubic space curve given by:

\mathbf{r}(t) = \begin{pmatrix} t \\ t^2 \\ t^3 \end{pmatrix}, \quad t \in \mathbb{R}

Find the velocity vector $\mathbf{r}'(t)$, acceleration vector $\mathbf{r}''(t)$, and the cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$.

Determine the unit tangent vector $\mathbf{T}(t)$ and the Cartesian equation of the tangent line to the curve at $t = 1$.

Compute the exact Cartesian equation of the osculating plane to the curve at the point $t = 1$.

1. Velocity, Acceleration & Cross Product Computation

Differentiating $\mathbf{r}(t) = (t, t^2, t^3)^T$:

\mathbf{r}'(t) = \begin{pmatrix} 1 \\ 2t \\ 3t^2 \end{pmatrix}, \quad \mathbf{r}''(t) = \begin{pmatrix} 0 \\ 2 \\ 6t \end{pmatrix}

Now compute the cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$:

\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2t & 3t^2 \\ 0 & 2 & 6t \end{vmatrix} = \mathbf{i}(12t^2 - 6t^2) - \mathbf{j}(6t - 0) + \mathbf{k}(2 - 0) = \begin{pmatrix} 6t^2 \\ -6t \\ 2 \end{pmatrix}

2. Unit Tangent Vector and Tangent Line at $t = 1$

At $t = 1$:

\mathbf{r}(1) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{r}'(1) = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

The speed at $t = 1$ is:

\|\mathbf{r}'(1)\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

Thus, the unit tangent vector is:

\mathbf{T}(1) = \frac{\mathbf{r}'(1)}{\|\mathbf{r}'(1)\|} = \frac{1}{\sqrt{14}} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

The vector equation of the tangent line at $t = 1$ is:

\mathbf{X}(\lambda) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \lambda \in \mathbb{R}

In symmetric Cartesian form:

\frac{X - 1}{1} = \frac{Y - 1}{2} = \frac{Z - 1}{3} \quad \blacksquare

3. Cartesian Equation of the Osculating Plane at $t = 1$

At $t = 1$, the normal vector to the osculating plane is:

\mathbf{n}_{\text{osc}} = \mathbf{r}'(1) \times \mathbf{r}''(1) = \begin{pmatrix} 6(1)^2 \\ -6(1) \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \\ 2 \end{pmatrix} = 2 \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}

We can take the simplified normal vector $\mathbf{N}_0 = (3, -3, 1)^T$. The equation of the osculating plane through $\mathbf{r}(1) = (1, 1, 1)$ is:

\mathbf{N}_0 \cdot (\mathbf{X} - \mathbf{r}(1)) = 0

$$3(X - 1) - 3(Y - 1) + 1(Z - 1) = 0$$

3X - 3 - 3Y + 3 + Z - 1 = 0 \iff 3X - 3Y + Z - 1 = 0 \quad \blacksquare

Tier 2 • Advanced Structural Analysis Example 1.2: Rigorous Derivation of the Osculating Plane via Limiting Tangent Secants

Let $\mathbf{r}(t)$ be a regular $C^3$ curve with $\mathbf{r}'(t) \times \mathbf{r}''(t) \ne \mathbf{0}$.

Consider the plane $\Pi(h)$ containing the tangent line at $\mathbf{r}(t)$ and passing through a neighboring point $\mathbf{r}(t + h)$ with $h \ne 0$. Find the normal vector $\mathbf{n}(h)$ to this plane.

Compute $\lim_{h \to 0} \frac{\mathbf{n}(h)}{h^2/2}$ and prove that the limiting plane coincides with the osculating plane $[\mathbf{X} - \mathbf{r}(t), \mathbf{r}'(t), \mathbf{r}''(t)] = 0$.

Prove that the distance $d(h)$ from the neighboring point $\mathbf{r}(t + h)$ to the osculating plane satisfies $d(h) = O(h^3)$, confirming contact of order $\ge 2$.

1. Normal Vector to the Secant Plane $\Pi(h)$

The plane $\Pi(h)$ contains the point $\mathbf{r}(t)$, the direction vector of the tangent line $\mathbf{r}'(t)$, and the neighboring point $\mathbf{r}(t + h)$. Hence, two vectors lying parallel to $\Pi(h)$ are:

\mathbf{v}_1 = \mathbf{r}'(t) \quad \text{and} \quad \mathbf{v}_2(h) = \mathbf{r}(t + h) - \mathbf{r}(t)

A normal vector to $\Pi(h)$ is given by their cross product:

\mathbf{n}(h) = \mathbf{r}'(t) \times (\mathbf{r}(t + h) - \mathbf{r}(t))

2. Limiting Position as $h \to 0$

Expanding $\mathbf{r}(t + h)$ in a Taylor series about $t$:

\mathbf{r}(t + h) - \mathbf{r}(t) = h \mathbf{r}'(t) + \frac{h^2}{2} \mathbf{r}''(t) + \frac{h^3}{6} \mathbf{r}'''(t) + O(h^4)

Substituting this into $\mathbf{n}(h)$:

\begin{aligned} \mathbf{n}(h) &= \mathbf{r}'(t) \times \left( h \mathbf{r}'(t) + \frac{h^2}{2} \mathbf{r}''(t) + \frac{h^3}{6} \mathbf{r}'''(t) + O(h^4) \right) \\ &= h (\mathbf{r}'(t) \times \mathbf{r}'(t)) + \frac{h^2}{2} (\mathbf{r}'(t) \times \mathbf{r}''(t)) + \frac{h^3}{6} (\mathbf{r}'(t) \times \mathbf{r}'''(t)) + O(h^4) \end{aligned}

Since $\mathbf{r}'(t) \times \mathbf{r}'(t) = \mathbf{0}$:

\mathbf{n}(h) = \frac{h^2}{2} (\mathbf{r}'(t) \times \mathbf{r}''(t)) + \frac{h^3}{6} (\mathbf{r}'(t) \times \mathbf{r}'''(t)) + O(h^4)

Dividing by $\frac{h^2}{2}$:

\lim_{h \to 0} \frac{\mathbf{n}(h)}{h^2/2} = \mathbf{r}'(t) \times \mathbf{r}''(t)

Since the orientation of a plane is determined by the direction of its normal, the limiting normal to $\Pi(h)$ as $h \to 0$ is precisely:

\mathbf{n}_0 = \mathbf{r}'(t) \times \mathbf{r}''(t)

The limiting plane equation through $\mathbf{r}(t)$ is therefore:

(\mathbf{X} - \mathbf{r}(t)) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) = 0 \iff [\mathbf{X} - \mathbf{r}(t), \mathbf{r}'(t), \mathbf{r}''(t)] = 0 \quad \blacksquare

3. Contact Order and Distance Estimate

The distance from $\mathbf{r}(t + h)$ to the osculating plane is:

d(h) = \frac{|(\mathbf{r}(t + h) - \mathbf{r}(t)) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t))|}{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}

Substituting the Taylor expansion:

\mathbf{r}(t + h) - \mathbf{r}(t) = h \mathbf{r}'(t) + \frac{h^2}{2} \mathbf{r}''(t) + \frac{h^3}{6} \mathbf{r}'''(t) + O(h^4)

Taking the dot product with $\mathbf{r}'(t) \times \mathbf{r}''(t)$:

$h \mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) = 0$.

$\frac{h^2}{2} \mathbf{r}''(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) = 0$.

The first non-zero term is the cubic term:

(\mathbf{r}(t + h) - \mathbf{r}(t)) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) = \frac{h^3}{6} [\mathbf{r}'''(t), \mathbf{r}'(t), \mathbf{r}''(t)] + O(h^4)

Therefore:

d(h) = \frac{|[\mathbf{r}'(t), \mathbf{r}''(t), \mathbf{r}'''(t)]|}{6 \|\mathbf{r}'(t) \times \mathbf{r}''(t)\|} h^3 + O(h^4) = O(h^3)

Since $d(h) / h^2 \to 0$ as $h \to 0$, the osculating plane has contact of order at least 2 with the curve. $\blacksquare$

Tier 3 • Honors / Proof Challenge Example 1.3: Straight Line Characterization Theorem via Collinear Velocity & Acceleration

Prove the fundamental characterization theorem of straight lines in Euclidean 3-space: Let $\mathbf{r}: I \to \mathbb{R}^3$ be a regular $C^2$ curve on an open interval $I$.

Prove that $\mathbf{r}(I)$ is a segment of a straight line if and only if:

\mathbf{r}''(t) \times \mathbf{r}'(t) = \mathbf{0} \quad \forall t \in I

Show that if the condition holds, there exists a scalar function $\lambda(t)$ such that $\mathbf{r}''(t) = \lambda(t) \mathbf{r}'(t)$, and solve this differential equation explicitly to prove $\mathbf{r}(t) = \mathbf{r}_0 + \mu(t) \mathbf{v}_0$ for constant vectors $\mathbf{r}_0, \mathbf{v}_0 \in \mathbb{R}^3$.

1. Forward Direction: Straight Line $\implies \mathbf{r}''(t) \times \mathbf{r}'(t) = \mathbf{0}$

Suppose $\mathbf{r}(I)$ is a straight line. Then $\mathbf{r}(t)$ can be represented as:

\mathbf{r}(t) = \mathbf{r}_0 + f(t) \mathbf{v}_0

where $\mathbf{r}_0 \in \mathbb{R}^3$ is a fixed point, $\mathbf{v}_0 \in \mathbb{R}^3 \setminus \{\mathbf{0}\}$ is a constant direction vector, and $f: I \to \mathbb{R}$ is a $C^2$ function. Differentiating with respect to $t$:

\mathbf{r}'(t) = f'(t) \mathbf{v}_0

Since $\mathbf{r}$ is regular, $f'(t) \ne 0$ for all $t \in I$. Differentiating a second time:

\mathbf{r}''(t) = f''(t) \mathbf{v}_0

Computing the cross product:

\mathbf{r}''(t) \times \mathbf{r}'(t) = (f''(t) \mathbf{v}_0) \times (f'(t) \mathbf{v}_0) = f''(t) f'(t) (\mathbf{v}_0 \times \mathbf{v}_0) = f''(t) f'(t) \mathbf{0} = \mathbf{0} \quad \blacksquare

2. Reverse Direction: $\mathbf{r}''(t) \times \mathbf{r}'(t) = \mathbf{0} \implies$ Straight Line

Assume $\mathbf{r}''(t) \times \mathbf{r}'(t) = \mathbf{0}$ for all $t \in I$. Since $\mathbf{r}$ is regular, $\mathbf{r}'(t) \ne \mathbf{0}$ everywhere. Recall that the cross product of two vectors in $\mathbb{R}^3$ vanishes if and only if they are linearly dependent (collinear). Thus, for each $t \in I$, there exists a scalar $\lambda(t) \in \mathbb{R}$ such that:

\mathbf{r}''(t) = \lambda(t) \mathbf{r}'(t)

Now consider the unit tangent vector $\mathbf{T}(t)$:

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}

We compute the derivative of $\mathbf{T}(t)$ with respect to $t$ using the quotient rule:

\frac{d\mathbf{T}}{dt} = \frac{\mathbf{r}''(t) \|\mathbf{r}'(t)\| - \mathbf{r}'(t) \frac{d}{dt}\|\mathbf{r}'(t)\|}{\|\mathbf{r}'(t)\|^2}

Recall that $\frac{d}{dt}\|\mathbf{r}'(t)\| = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{\|\mathbf{r}'(t)\|}$. Substituting $\mathbf{r}''(t) = \lambda(t) \mathbf{r}'(t)$:

\mathbf{r}'(t) \cdot \mathbf{r}''(t) = \mathbf{r}'(t) \cdot (\lambda(t) \mathbf{r}'(t)) = \lambda(t) \|\mathbf{r}'(t)\|^2

So:

\frac{d}{dt}\|\mathbf{r}'(t)\| = \frac{\lambda(t) \|\mathbf{r}'(t)\|^2}{\|\mathbf{r}'(t)\|} = \lambda(t) \|\mathbf{r}'(t)\|

Now substitute this back into the derivative of $\mathbf{T}$:

\frac{d\mathbf{T}}{dt} = \frac{(\lambda(t) \mathbf{r}'(t)) \|\mathbf{r}'(t)\| - \mathbf{r}'(t) (\lambda(t) \|\mathbf{r}'(t)\|)}{\|\mathbf{r}'(t)\|^2} = \frac{\mathbf{0}}{\|\mathbf{r}'(t)\|^2} = \mathbf{0}

Since $\frac{d\mathbf{T}}{dt} = \mathbf{0}$ on the connected interval $I$, the unit tangent vector is constant:

\mathbf{T}(t) = \mathbf{T}_0 \quad \text{for some constant vector } \mathbf{T}_0 \in \mathbb{R}^3, \; \|\mathbf{T}_0\| = 1

Since $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$, we have:

\mathbf{r}'(t) = \|\mathbf{r}'(t)\| \mathbf{T}_0

Let $v(t) = \|\mathbf{r}'(t)\| > 0$. Integrating from an initial parameter $t_0$:

\mathbf{r}(t) - \mathbf{r}(t_0) = \int_{t_0}^t \mathbf{r}'(u) \, du = \left( \int_{t_0}^t v(u) \, du \right) \mathbf{T}_0

Defining $\mathbf{r}_0 = \mathbf{r}(t_0)$ and the scalar function $\mu(t) = \int_{t_0}^t v(u) \, du$:

\mathbf{r}(t) = \mathbf{r}_0 + \mu(t) \mathbf{T}_0

This is precisely the parametric equation of a straight line in $\mathbb{R}^3$! $\blacksquare$