Curvature, Torsion, and the Moving Frame, multivariable and vector calculus lesson

Drive a mountain road and you feel two things. The road bends left and right, and it also climbs and twists out of any flat plane. Curvature measures the first, torsion the second. Both are read off from a small right-handed frame of three unit vectors that rides along the curve.

Key terms

Curvature

Definition Let $\mathbf{r}$ be a regular $C^2$ curve with unit tangent $\mathbf{T} = \mathbf{r}'/\lvert \mathbf{r}' \rvert$ and arc length $s$ . The curvature is

\kappa = \left\lvert \frac{d\mathbf{T}}{ds} \right\rvert = \frac{\lvert \mathbf{T}'(t) \rvert}{\lvert \mathbf{r}'(t) \rvert}.

For a curve in $\mathbb{R}^3$ , or a plane curve written with third coordinate $0$ , this equals

\kappa(t) = \frac{\lvert \mathbf{r}'(t) \times \mathbf{r}''(t) \rvert}{\lvert \mathbf{r}'(t) \rvert^3}.

Curvature is a property of the traced curve. It is unchanged by any $C^2$ reparametrization with nonzero derivative, including one that reverses orientation.

In plain words Walk along the curve at unit speed and watch the direction arrow $\mathbf{T}$ . Curvature is how fast that arrow turns, in radians per unit of length.

Example The circle $\mathbf{r}(t) = \langle R\cos t, R\sin t \rangle$ with $R > 0$ has speed $R$ and $\mathbf{T}(t) = \langle -\sin t, \cos t \rangle$ , so $\lvert \mathbf{T}'(t) \rvert = 1$ and $\kappa = 1/R$ . Walk $1$ meter along a circle of radius $2$ meters and the tangent turns by $0.5$ radian, while on a circle of radius $10$ meters it turns by only $0.1$ radian. For a graph $y = f(x)$ with $f$ twice differentiable, the cross product formula with $\mathbf{r}(x) = \langle x, f(x), 0 \rangle$ gives

\kappa = \frac{\lvert f''(x) \rvert}{\left(1 + f'(x)^2\right)^{3/2}},

so the parabola $y = x^2$ has $\kappa = 2$ at the origin.

Principal unit normal

Definition Let $\mathbf{r}$ be regular and $C^2$ , and suppose $\kappa(t) \ne 0$ , which is the same as $\mathbf{T}'(t) \ne \mathbf{0}$ . The principal unit normal is

\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\lvert \mathbf{T}'(t) \rvert}.

Differentiating $\mathbf{T} \cdot \mathbf{T} = 1$ gives $\mathbf{T} \cdot \mathbf{T}' = 0$ , so $\mathbf{N}$ is a unit vector perpendicular to $\mathbf{T}$ . In terms of arc length, $\dfrac{d\mathbf{T}}{ds} = \kappa \mathbf{N}$ .

In plain words The normal points the way the curve is turning, toward the inside of the bend. It does not flip when you reverse direction, because the curve still bends the same way.

Example For the helix $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangle$ , the speed is $5$ and

\mathbf{T}(t) = \left\langle -\tfrac{3}{5}\sin t, \ \tfrac{3}{5}\cos t, \ \tfrac{4}{5} \right\rangle.

Then $\mathbf{T}'(t) = \left\langle -\tfrac{3}{5}\cos t, -\tfrac{3}{5}\sin t, 0 \right\rangle$ has length $\tfrac{3}{5}$ , so $\mathbf{N}(t) = \langle -\cos t, -\sin t, 0 \rangle$ . It is horizontal and points straight at the $z$ -axis. The curvature is $\kappa = \frac{3/5}{5} = \frac{3}{25}$ .

Binormal

Definition Where $\mathbf{T}$ and $\mathbf{N}$ are both defined, the binormal is the cross product

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

Since $\mathbf{T}$ and $\mathbf{N}$ are perpendicular unit vectors, $\mathbf{B}$ is a unit vector perpendicular to both, and its direction follows the right-hand rule. Equivalently, $\mathbf{B} = \dfrac{\mathbf{r}' \times \mathbf{r}''}{\lvert \mathbf{r}' \times \mathbf{r}'' \rvert}$ .

In plain words The tangent and normal span the flat plane in which the curve is bending right now. The binormal sticks straight out of that plane.

Example For the helix above,

\mathbf{B}(t) = \left\langle \tfrac{4}{5}\sin t, \ -\tfrac{4}{5}\cos t, \ \tfrac{3}{5} \right\rangle.

At $t = 0$ this is $\langle 0, -\tfrac{4}{5}, \tfrac{3}{5} \rangle$ , and it checks against $\mathbf{T}(0) \times \mathbf{N}(0) = \langle 0, \tfrac{3}{5}, \tfrac{4}{5} \rangle \times \langle -1, 0, 0 \rangle$ .

The TNB frame

Definition At each point of a regular $C^2$ curve with $\kappa \ne 0$ , the vectors $\mathbf{T}, \mathbf{N}, \mathbf{B}$ form a right-handed orthonormal frame, called the TNB or Frenet frame. If the curve is $C^3$ , the frame changes along the curve according to the Frenet-Serret formulas,

\begin{aligned} \frac{d\mathbf{T}}{ds} &= \kappa \mathbf{N}, \\[0.4em] \frac{d\mathbf{N}}{ds} &= -\kappa \mathbf{T} + \tau \mathbf{B}, \\[0.4em] \frac{d\mathbf{B}}{ds} &= -\tau \mathbf{N}, \end{aligned}

where $\tau$ is the torsion defined below.

In plain words Picture a tiny set of three perpendicular arrows riding along the curve. Curvature and torsion are exactly the rates at which this frame rotates as you move one unit of length.

Example For the helix at $t = 0$ the frame is

\begin{gathered} \mathbf{T} = \left\langle 0, \tfrac{3}{5}, \tfrac{4}{5} \right\rangle, \quad \mathbf{N} = \langle -1, 0, 0 \rangle, \\[0.4em] \mathbf{B} = \left\langle 0, -\tfrac{4}{5}, \tfrac{3}{5} \right\rangle. \end{gathered}

Each has length $1$ , every pairwise dot product is $0$ , and $\mathbf{T} \times \mathbf{N} = \mathbf{B}$ . At every $t$ , the derivative $d\mathbf{T}/ds = \mathbf{T}'/5 = \langle -\tfrac{3}{25}\cos t, -\tfrac{3}{25}\sin t, 0 \rangle = \tfrac{3}{25}\mathbf{N}$ , matching $\kappa = \tfrac{3}{25}$ .

Normal and osculating planes

Definition At a point $P = \mathbf{r}(t_0)$ of a regular curve, the normal plane is the plane through $P$ perpendicular to $\mathbf{T}(t_0)$ , so it contains $\mathbf{N}(t_0)$ and $\mathbf{B}(t_0)$ whenever they are defined. If also $\kappa(t_0) \ne 0$ , the osculating plane is the plane through $P$ spanned by $\mathbf{T}(t_0)$ and $\mathbf{N}(t_0)$ , so its normal vector is $\mathbf{B}(t_0)$ .

In plain words The normal plane is the cross section you would cut to slice the curve straight across. The osculating plane is the plane the curve almost lies in near $P$ .

Example For the helix at $P = (3, 0, 0)$ , the normal vector $\langle 0, 3, 4 \rangle$ (a multiple of $\mathbf{T}$ ) gives the normal plane $3y + 4z = 0$ , and the normal vector $\langle 0, -4, 3 \rangle$ (a multiple of $\mathbf{B}$ ) gives the osculating plane $-4y + 3z = 0$ . Both pass through $(3, 0, 0)$ .

Osculating circle

Definition At a point $P = \mathbf{r}(t_0)$ of a regular $C^2$ curve with $\kappa(t_0) \ne 0$ , the osculating circle is the circle in the osculating plane with radius of curvature $\rho = 1/\kappa(t_0)$ and center

\mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)}\, \mathbf{N}(t_0).

It passes through $P$ , has the same tangent line there, and has the same curvature as the curve at $P$ .

In plain words It is the circle that hugs the curve best at $P$ . A tight bend gives a small circle, and a gentle bend gives a big one.

Example For $y = x^2$ at the origin, $\kappa = 2$ and $\mathbf{N} = \langle 0, 1 \rangle$ , so the osculating circle has radius $\tfrac{1}{2}$ and center $(0, \tfrac{1}{2})$ . Its equation is

x^2 + \left(y - \tfrac{1}{2}\right)^2 = \tfrac{1}{4}.

Torsion

Definition Let $\mathbf{r}$ be a regular $C^3$ curve with $\kappa \ne 0$ . The torsion $\tau$ is the number with $d\mathbf{B}/ds = -\tau \mathbf{N}$ , that is $\tau = -\dfrac{d\mathbf{B}}{ds} \cdot \mathbf{N}$ . In any regular parametrization,

\tau = \frac{(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}'''}{\lvert \mathbf{r}' \times \mathbf{r}'' \rvert^2}.

Torsion is signed. With this sign convention, a helix that turns counterclockwise as it rises in a right-handed coordinate system has $\tau > 0$ . A curve on an interval with $\kappa \ne 0$ everywhere lies in a single plane exactly when $\tau = 0$ everywhere.

In plain words Torsion measures how fast the osculating plane tips over as you move along. A flat curve never twists out of its plane, so its torsion is zero.

Example For the twisted cubic $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ at $t = 0$ , the derivatives are $\mathbf{r}' = \langle 1, 0, 0 \rangle$ , $\mathbf{r}'' = \langle 0, 2, 0 \rangle$ and $\mathbf{r}''' = \langle 0, 0, 6 \rangle$ . Then $\mathbf{r}' \times \mathbf{r}'' = \langle 0, 0, 2 \rangle$ , so

\kappa = \frac{2}{1^3} = 2, \qquad \tau = \frac{12}{4} = 3.

The frame there is $\mathbf{T} = \mathbf{i}$ , $\mathbf{N} = \mathbf{j}$ , $\mathbf{B} = \mathbf{k}$ . For the helix $\langle 3\cos t, 3\sin t, 4t \rangle$ the same formula gives $\tau = \tfrac{4}{25}$ at every point.

Nonzero speed and curvature conditions

Definition Each object in this lesson needs its own hypotheses at the point in question. Each row of the table adds its condition to all the rows above it.

Object	Needs
$\mathbf{T}$ , normal plane	$C^1$ and $\mathbf{r}' \ne \mathbf{0}$
$\kappa$	also $C^2$
$\mathbf{N}$ , $\mathbf{B}$ , osculating plane and circle	also $\kappa \ne 0$
$\tau$ and the Frenet-Serret formulas	also $C^3$

At a point where $\kappa = 0$ , the curvature itself is defined and equals $0$ , but $\mathbf{N}$ , $\mathbf{B}$ and $\tau$ are undefined. At a point where $\mathbf{r}' = \mathbf{0}$ , even $\mathbf{T}$ may fail to exist.

In plain words You cannot say which way a curve is turning where it is not turning, and you cannot say which way it points where the parametrization stops.

Example A straight line $\mathbf{r}(t) = \langle 1 + t, 2t, 3 \rangle$ has constant $\mathbf{T}$ , so $\kappa = 0$ everywhere and no principal normal exists anywhere. The plane curve $\mathbf{r}(t) = \langle t, t^3, 0 \rangle$ has

\kappa(t) = \frac{6\lvert t \rvert}{(1 + 9t^4)^{3/2}},

which is zero only at $t = 0$ . There the curve has an inflection point. The normal points one way for $t < 0$ and the other way for $t > 0$ , and it is undefined at $t = 0$ .

Check yourself

Find the curvature of $y = x^2$ at $(1, 1)$ and the radius of its osculating circle there.

Show the answer

With $f'(1) = 2$ and $f''(1) = 2$ ,

\kappa = \frac{2}{(1 + 4)^{3/2}} = \frac{2}{5\sqrt{5}} \approx 0.179.

The radius of curvature is $\frac{5\sqrt{5}}{2} \approx 5.59$ .

Find the curvature and torsion of the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ .

Show the answer

The derivatives are $\mathbf{r}' = \langle -\sin t, \cos t, 1 \rangle$ , $\mathbf{r}'' = \langle -\cos t, -\sin t, 0 \rangle$ and $\mathbf{r}''' = \langle \sin t, -\cos t, 0 \rangle$ . The cross product is $\mathbf{r}' \times \mathbf{r}'' = \langle \sin t, -\cos t, 1 \rangle$ , with length $\sqrt{2}$ , and its dot product with $\mathbf{r}'''$ is $1$ . So

\kappa = \frac{\sqrt{2}}{(\sqrt{2})^3} = \frac{1}{2}, \qquad \tau = \frac{1}{2}.

Is the principal unit normal of $\mathbf{r}(t) = \langle t, t^3, 0 \rangle$ defined at $t = 0$ ? What is its torsion where it is defined?

Show the answer

No. The curvature is $6\lvert t \rvert/(1 + 9t^4)^{3/2}$ , which is $0$ at $t = 0$ , so $\mathbf{N}$ is undefined there. Where $t \ne 0$ the torsion is $0$ , because $\mathbf{r}' \times \mathbf{r}'' = \langle 0, 0, 6t \rangle$ is perpendicular to $\mathbf{r}''' = \langle 0, 6, 0 \rangle$ , as it must be for a curve in the plane $z = 0$ .

Curvature, Torsion, and the Moving Frame

Key terms

Curvature

Principal unit normal

Binormal

The TNB frame

Normal and osculating planes

Osculating circle

Torsion

Nonzero speed and curvature conditions

Check yourself

How this lesson connects