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Chapter 2 โ€ข Theory & Derivations

Unit 2: Curvature, Torsion & The Frenet-Serret Frame

Exhaustive treatment of differential geometry of space curves: the TNB moving trihedron, general parametric and Cartesian curvature formulas, osculating circles, evolutes, torsion, and the Frenet-Serret equations.

ยง2.1 The Moving Trihedron (TNB Frame) and Fundamental Planes

At each point along a smooth curve $C$ in $\mathbb{R}^3$, we can attach a moving, right-handed orthonormal coordinate system known as the Frenet-Serret Moving Trihedron (or TNB Frame).


1. Construction of the TNB Frame

1. Unit Tangent Vector $\vec{T}(t)$:

Points in the instantaneous direction of motion:

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

2. Principal Unit Normal Vector $\vec{N}(t)$:

Because $|\vec{T}(t)| = 1$ is constant, $\vec{T}'(t)$ is orthogonal to $\vec{T}(t)$ ($\vec{T} \cdot \vec{T}' = 0$). The unit vector in this orthogonal direction is:

$$\mathbf{\vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|}}$$

$\vec{N}(t)$ points directly in the direction that the curve is turning.

3. Binormal Unit Vector $\vec{B}(t)$:

Defined by the vector cross product to complete a right-handed orthonormal triad:

$$\mathbf{\vec{B}(t) = \vec{T}(t) \times \vec{N}(t)}$$

Because $\vec{T}$ and $\vec{N}$ are orthogonal unit vectors, $|\vec{B}| = |\vec{T}| |\vec{N}| \sin(\pi/2) = 1$, and $\vec{B}$ is perpendicular to both $\vec{T}$ and $\vec{N}$.


2. The Three Fundamental Osculating Planes

At any point $P$ on the curve, the three mutually orthogonal pairs of vectors define three fundamental planes:

1. The Osculating Plane (Spanned by $\vec{T}$ and $\vec{N}$):

  • Normal Vector: $\vec{B}(t)$
  • Equation: $(\vec{r} - \vec{r}_0) \cdot \vec{B} = 0$
  • Geometric Meaning: The plane that comes closest to containing the curve locally; it contains the instantaneous circle of curvature.

2. The Normal Plane (Spanned by $\vec{N}$ and $\vec{B}$):

  • Normal Vector: $\vec{T}(t)$
  • Equation: $(\vec{r} - \vec{r}_0) \cdot \vec{T} = 0$
  • Geometric Meaning: The plane orthogonal to the curve; all lines normal to the curve lie in this plane.

3. The Rectifying Plane (Spanned by $\vec{T}$ and $\vec{B}$):

  • Normal Vector: $\vec{N}(t)$
  • Equation: $(\vec{r} - \vec{r}_0) \cdot \vec{N} = 0$

ยง2.2 Curvature Formulas, Radius of Curvature and Osculating Circles

1. Geometric Definition of Curvature $\kappa$

The curvature $\kappa$ of a smooth curve measures how rapidly the curve changes its direction per unit change in arc length:

$$\mathbf{\kappa = \left| \frac{d\vec{T}}{ds} \right|}$$

Using the chain rule with parameter $t$:

$$\frac{d\vec{T}}{ds} = \frac{d\vec{T}/dt}{ds/dt} = \frac{\vec{T}'(t)}{|\vec{r}'(t)|} \implies \mathbf{\kappa = \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|}}$$

2. General Parametric Curvature Formula in $\mathbb{R}^3$

Theorem: For any smooth space curve parameterized by $\vec{r}(t)$:

$$\mathbf{\kappa(t) = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3}}$$
Proof:

Since $\vec{v}(t) = \vec{r}'(t) = v \vec{T}$ where $v = |\vec{r}'(t)| = \frac{ds}{dt}$: Differentiating velocity to get acceleration:

$$\vec{r}''(t) = \frac{d}{dt}[v \vec{T}] = v' \vec{T} + v \vec{T}'(t)$$

Since $\frac{d\vec{T}}{ds} = \kappa \vec{N} \implies \vec{T}'(t) = \frac{ds}{dt} \frac{d\vec{T}}{ds} = v \kappa \vec{N}$:

$$\vec{r}''(t) = v' \vec{T} + \kappa v^2 \vec{N}$$

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

$$\vec{r}' \times \vec{r}'' = (v \vec{T}) \times (v' \vec{T} + \kappa v^2 \vec{N}) = v v' (\vec{T} \times \vec{T}) + \kappa v^3 (\vec{T} \times \vec{N})$$

Since $\vec{T} \times \vec{T} = \vec{0}$ and $\vec{T} \times \vec{N} = \vec{B}$:

$$\vec{r}'(t) \times \vec{r}''(t) = \kappa v^3 \vec{B}$$

Taking the norm of both sides (since $|\vec{B}| = 1$):

$$|\vec{r}' \times \vec{r}''| = \kappa v^3 |\vec{B}| = \kappa |\vec{r}'|^3$$

Dividing by $|\vec{r}'|^3$:

$$\kappa = \frac{|\vec{r}' \times \vec{r}''|}{|\vec{r}'|^3} \quad \blacksquare$$

3. Special Curvature Formulas

1. Plane Curve in Cartesian Form $y = f(x)$:

Parameterize as $\vec{r}(x) = \langle x, \; f(x), \; 0 \rangle$. Then $\vec{r}'(x) = \langle 1, \; y', \; 0 \rangle$ and $\vec{r}''(x) = \langle 0, \; y'', \; 0 \rangle$. $\vec{r}' \times \vec{r}'' = \langle 0, \; 0, \; y'' \rangle \implies |\vec{r}' \times \vec{r}''| = |y''|$. $|\vec{r}'| = \sqrt{1 + y'^2}$.

$$\mathbf{\kappa(x) = \frac{|y''|}{(1 + y'^2)^{3/2}}}$$

2. Plane Curve in Polar Coordinates $r = f(\theta)$:

$$\mathbf{\kappa(\theta) = \frac{|r^2 + 2r'^2 - r r''|}{(r^2 + r'^2)^{3/2}}}$$

4. Radius of Curvature, Center of Curvature and Evolutes

  • Radius of Curvature $\rho$: The reciprocal of curvature:
$$\rho = \frac{1}{\kappa}$$
  • Osculating Circle (Circle of Curvature): The circle in the osculating plane that has the same tangent, normal, and curvature as the curve at that point. Its radius is $\rho$.
  • Center of Curvature $(\alpha, \beta)$ for plane curves:
$$\alpha = x - \frac{y'(1 + y'^2)}{y''}, \qquad \beta = y + \frac{1 + y'^2}{y''}$$
  • Evolute: The locus of the centers of curvature of a given curve is called its evolute.

ยง2.3 Torsion, the Frenet-Serret Formulas & Acceleration Components

1. Torsion $\tau$ of a Space Curve

While curvature $\kappa$ measures the rate at which $\vec{T}$ turns away from the tangent line, torsion $\tau$ measures how sharply the space curve twists out of its osculating plane.

Because $\vec{B}(s) \cdot \vec{B}(s) = 1$, $\frac{d\vec{B}}{ds}$ is perpendicular to $\vec{B}$. Furthermore, differentiating $\vec{B} \cdot \vec{T} = 0$:

$$\frac{d\vec{B}}{ds} \cdot \vec{T} + \vec{B} \cdot \frac{d\vec{T}}{ds} = 0 \implies \frac{d\vec{B}}{ds} \cdot \vec{T} + \vec{B} \cdot (\kappa\vec{N}) = 0 \implies \frac{d\vec{B}}{ds} \cdot \vec{T} = 0$$

Thus, $\frac{d\vec{B}}{ds}$ is perpendicular to both $\vec{B}$ and $\vec{T}$, which means it must be parallel to $\vec{N}$. We define the scalar torsion $\tau$ by:

$$\mathbf{\frac{d\vec{B}}{ds} = -\tau \vec{N}}$$

The negative sign is conventional so that a right-handed screw has positive torsion.

General Parametric Formula for Torsion:
$$\mathbf{\tau(t) = \frac{(\vec{r}'(t) \times \vec{r}''(t)) \cdot \vec{r}'''(t)}{|\vec{r}'(t) \times \vec{r}''(t)|^2}}$$

Criterion for Planar Curves: A space curve is a plane curve if and only if $\tau(t) \equiv 0$ for all $t$.


2. The Complete Frenet-Serret Formulas

The rate of change of the moving frame $\{\vec{T}, \vec{N}, \vec{B}\}$ with respect to arc length $s$ is governed by the celebrated Frenet-Serret Formulas:

$$\mathbf{\begin{cases} \dfrac{d\vec{T}}{ds} = \kappa \vec{N} \ \dfrac{d\vec{N}}{ds} = -\kappa \vec{T} + \tau \vec{B} \ \dfrac{d\vec{B}}{ds} = -\tau \vec{N} \end{cases}}$$

In matrix notation:

$$\begin{bmatrix} d\vec{T}/ds \\ d\vec{N}/ds \\ d\vec{B}/ds \end{bmatrix} = \begin{bmatrix} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix} \begin{bmatrix} \vec{T} \\ \vec{N} \\ \vec{B} \end{bmatrix}$$

Notice that the coefficient matrix (the Darboux matrix) is skew-symmetric, which is a mathematical guarantee that the orthonormal nature of the basis is preserved along the entire curve.


3. Tangential and Normal Components of Acceleration

In physical kinematics, the acceleration of a particle can be decomposed uniquely into orthogonal components along the tangent and principal normal:

$$\mathbf{\vec{a} = a_T \vec{T} + a_N \vec{N}}$$

where:

  • Tangential Acceleration: Rate of change of speed:
$$a_T = \frac{dv}{dt} = v' = \frac{\vec{r}'(t) \cdot \vec{r}''(t)}{|\vec{r}'(t)|}$$
  • Normal (Centripetal) Acceleration: Tendency to change direction:
$$a_N = \kappa v^2 = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|}$$

Notice that the binormal component of acceleration is always identically zero ($a_B = 0$). Acceleration always lies entirely in the osculating plane!

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Solved Problem Example 2.1: Curvature and Osculating Circle of a Parabola at its Vertex

Find the curvature $\kappa$, radius of curvature $\rho$, and center of curvature of the parabola $y = x^2$ at its vertex $(0, 0)$.

Step 1: Compute derivatives of $y = x^2$

$$y' = \frac{dy}{dx} = 2x$$
$$y'' = \frac{d^2y}{dx^2} = 2$$

At the vertex $x = 0$:

$$y'(0) = 0, \qquad y''(0) = 2$$

Step 2: Compute curvature $\kappa$ Using the Cartesian curvature formula:

$$\kappa(x) = \frac{|y''|}{(1 + y'^2)^{3/2}}$$

Substitute $x = 0$:

$$\kappa(0) = \frac{|2|}{(1 + 0^2)^{3/2}} = \frac{2}{1} = 2$$

Step 3: Compute radius of curvature $\rho$

$$\rho = \frac{1}{\kappa} = \frac{1}{2} = 0.5$$

Step 4: Compute center of curvature $(\alpha, \beta)$ Using the center of curvature formulas:

$$\alpha = x - \frac{y'(1 + y'^2)}{y''} = 0 - \frac{0(1 + 0)}{2} = 0$$
$$\beta = y + \frac{1 + y'^2}{y''} = 0 + \frac{1 + 0}{2} = \frac{1}{2}$$

Thus, the center of curvature is $(0, 1/2)$. The equation of the osculating circle at the vertex is:

$$(x - 0)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 \implies x^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4}$$
Solved Problem Example 2.2: Curvature and Evolute of the Canonical Ellipse

For the canonical ellipse parameterized by:

$$\vec{r}(t) = \langle a\cos t, \; b\sin t, \; 0 \rangle \quad (a > b > 0)$$

(a) Find the curvature $\kappa(t)$ as an explicit function of parameter $t$. (b) Find the maximum and minimum curvatures and the points where they occur. (c) Deduce the radii of curvature at the major and minor vertices.

Part (a): Compute Curvature Formula Compute first and second derivatives:

$$\vec{r}'(t) = \langle -a\sin t, \; b\cos t, \; 0 \rangle$$
$$\vec{r}''(t) = \langle -a\cos t, \; -b\sin t, \; 0 \rangle$$

Compute the cross product $\vec{r}' \times \vec{r}''$:

$$\vec{r}' \times \vec{r}'' = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -a\sin t & b\cos t & 0 \\ -a\cos t & -b\sin t & 0 \end{vmatrix}$$
$$= \hat{k} [(-a\sin t)(-b\sin t) - (b\cos t)(-a\cos t)] = \hat{k} [ab\sin^2 t + ab\cos^2 t] = ab\hat{k}$$

Magnitude:

$$|\vec{r}' \times \vec{r}''| = ab$$

Now compute $|\vec{r}'(t)|$:

$$|\vec{r}'(t)| = \sqrt{a^2\sin^2 t + b^2\cos^2 t}$$

Thus, the curvature is:

$$\mathbf{\kappa(t) = \frac{ab}{(a^2\sin^2 t + b^2\cos^2 t)^{3/2}}}$$

Part (b): Extreme Values of Curvature Rewrite the denominator:

$$a^2\sin^2 t + b^2\cos^2 t = b^2 + (a^2 - b^2)\sin^2 t$$

Since $a > b$, this denominator is minimized when $\sin t = 0$ ($t = 0, \pi$) and maximized when $\sin^2 t = 1$ ($t = \pi/2, 3\pi/2$).

  • Maximum Curvature: Occurs at $t = 0, \pi$ (vertices $(\pm a, 0)$):
$$\kappa_{max} = \frac{ab}{(b^2)^{3/2}} = \frac{ab}{b^3} = \mathbf{\frac{a}{b^2}}$$
  • Minimum Curvature: Occurs at $t = \pi/2, 3\pi/2$ (vertices $(0, \pm b)$):
$$\kappa_{min} = \frac{ab}{(a^2)^{3/2}} = \frac{ab}{a^3} = \mathbf{\frac{b}{a^2}}$$

Part (c): Radii of Curvature at Vertices

  • At major vertices $(\pm a, 0)$: $\rho = \frac{1}{\kappa_{max}} = \mathbf{\frac{b^2}{a}}$.
  • At minor vertices $(0, \pm b)$: $\rho = \frac{1}{\kappa_{min}} = \mathbf{\frac{a^2}{b}}$.
Solved Problem Example 2.3: Complete Frenet-Serret Apparatus and Lancret Ratio for a Circular Helix

For the general circular helix:

$$\vec{r}(t) = \langle a\cos t, \; a\sin t, \; bt \rangle \quad (a > 0, \; b > 0)$$

(a) Compute the complete Frenet-Serret apparatus: $\vec{T}(t), \vec{N}(t), \vec{B}(t)$, curvature $\kappa$, and torsion $\tau$. (b) Prove that both curvature and torsion are constant along the entire helix. (c) Verify Lancret's Theorem by demonstrating that the ratio $\tau / \kappa$ is constant, and determine the angle that the tangent vector makes with the $z$-axis.

Part (a): Derivation of Frenet-Serret Vectors and Invariants

1. Velocity and Unit Tangent:

$$\vec{r}'(t) = \langle -a\sin t, \; a\cos t, \; b \rangle, \qquad |\vec{r}'(t)| = \sqrt{a^2 + b^2}$$
$$\mathbf{\vec{T}(t) = \frac{1}{\sqrt{a^2+b^2}} \langle -a\sin t, \; a\cos t, \; b \rangle}$$

2. Principal Normal:

$$\vec{T}'(t) = \frac{1}{\sqrt{a^2+b^2}} \langle -a\cos t, \; -a\sin t, \; 0 \rangle$$
$$|\vec{T}'(t)| = \frac{a}{\sqrt{a^2+b^2}}$$
$$\mathbf{\vec{N}(t) = \frac{\vec{T}'}{|\vec{T}'|} = \langle -\cos t, \; -\sin t, \; 0 \rangle}$$

Notice that $\vec{N}$ points horizontally toward the $z$-axis!

3. Curvature:

$$\mathbf{\kappa = \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|} = \frac{a/\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = \frac{a}{a^2 + b^2}}$$

4. Binormal Vector:

$$\vec{B}(t) = \vec{T}(t) \times \vec{N}(t) = \frac{1}{\sqrt{a^2+b^2}} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -a\sin t & a\cos t & b \\ -\cos t & -\sin t & 0 \end{vmatrix}$$
$$\vec{B}(t) = \frac{1}{\sqrt{a^2+b^2}} \left[ \hat{i}(b\sin t) - \hat{j}(-b\cos t) + \hat{k}(a\sin^2 t + a\cos^2 t) \right]$$
$$\mathbf{\vec{B}(t) = \frac{1}{\sqrt{a^2+b^2}} \langle b\sin t, \; -b\cos t, \; a \rangle}$$

5. Torsion:

Differentiate $\vec{B}(t)$ with respect to $t$:

$$\vec{B}'(t) = \frac{1}{\sqrt{a^2+b^2}} \langle b\cos t, \; b\sin t, \; 0 \rangle = -\frac{b}{\sqrt{a^2+b^2}} \langle -\cos t, \; -\sin t, \; 0 \rangle = -\frac{b}{\sqrt{a^2+b^2}} \vec{N}(t)$$

By the third Frenet-Serret relation, $\frac{d\vec{B}}{dt} = \frac{ds}{dt} \frac{d\vec{B}}{ds} = \sqrt{a^2+b^2}(-\tau \vec{N})$. Equating:

$$\sqrt{a^2+b^2}(-\tau \vec{N}) = -\frac{b}{\sqrt{a^2+b^2}} \vec{N} \implies \mathbf{\tau = \frac{b}{a^2 + b^2}}$$

Part (b): Constancy of Invariants

$$\kappa = \frac{a}{a^2+b^2} = \text{const}, \qquad \tau = \frac{b}{a^2+b^2} = \text{const}$$

Since neither $\kappa$ nor $\tau$ contains the parameter $t$, both invariants are constant everywhere. $\blacksquare$


Part (c): Lancret's Theorem and Axis Incline Evaluate the ratio of torsion to curvature:

$$\frac{\tau}{\kappa} = \frac{b / (a^2+b^2)}{a / (a^2+b^2)} = \frac{b}{a} = \text{constant}$$

By Lancret's Theorem (1806), a space curve is a general helix (its tangent vector makes a constant angle with a fixed direction) if and only if the ratio $\tau / \kappa$ is constant.

The angle $\phi$ between $\vec{T}$ and the $z$-axis ($\hat{k}$) is:

$$\cos\phi = \vec{T} \cdot \hat{k} = \frac{b}{\sqrt{a^2+b^2}} = \text{constant}$$

Thus, $\phi = \arccos\left(\frac{b}{\sqrt{a^2+b^2}}\right)$ is strictly constant along the entire curve. $\blacksquare$