TNB Frame
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.3: “Arc Length and Curvature” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature |
| Textbook used in class | Stewart, Calculus, Section 13.3: “Arc Length and Curvature” (Examples 6, 7) |
Quick Reference
Unit tangent: $\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$
Principal unit normal: $\mathbf{N}(t) = \dfrac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}$
Binormal: $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$
Torsion formula: $\tau = -\dfrac{d\mathbf{B}}{ds} \cdot \mathbf{N}$, or computationally: $$\tau = \frac{(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}'''}{|\mathbf{r}' \times \mathbf{r}''|^2}$$
Motivation
A particle moving through space traces a curve that can bend (change direction in the tangent-normal plane) and also twist (rotate the plane itself). The TNB frame is a moving coordinate system that rides along with the particle and captures both behaviors.
Think of a roller coaster car: $\mathbf{T}$ points forward (along the track), $\mathbf{N}$ points toward the center of each curve (it is why riders feel pushed sideways on turns), and $\mathbf{B}$ points up relative to the track. As the coaster goes through a corkscrew, $\mathbf{B}$ rotates -- that rotation is torsion.
Key Concepts
1. The Unit Tangent Vector T
$$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$$
$\mathbf{T}$ always points in the direction of motion and has magnitude 1. Because $|\mathbf{T}| = 1$, its derivative $\mathbf{T}'$ is always perpendicular to $\mathbf{T}$ (differentiating $\mathbf{T} \cdot \mathbf{T} = 1$ gives $2\mathbf{T} \cdot \mathbf{T}' = 0$).
2. The Principal Unit Normal N
$$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}$$
$\mathbf{N}$ points in the direction $\mathbf{T}'$ -- that is, toward the center of the osculating circle, perpendicular to the direction of motion. It is the direction in which the curve turns.
3. The Binormal Vector B
$$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$$
$\mathbf{B}$ is perpendicular to both $\mathbf{T}$ and $\mathbf{N}$, and $|\mathbf{B}| = 1$ because $\mathbf{T}$ and $\mathbf{N}$ are orthonormal. The plane spanned by $\mathbf{T}$ and $\mathbf{N}$ at each point is called the osculating plane -- the plane in which the curve lies locally. $\mathbf{B}$ is normal to this plane.
4. Torsion
Torsion $\tau$ measures how fast the binormal $\mathbf{B}$ rotates with respect to arc length -- equivalently, how fast the curve twists out of its osculating plane. A curve with $\tau = 0$ everywhere lies in a fixed plane (a plane curve). A helix has constant nonzero torsion.
Worked Example
Find $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ for the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$. (Stewart 13.3, Example 6.)
Step 1. Compute $\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$ and $|\mathbf{r}'(t)| = \sqrt{2}$.
Unit tangent: $$\mathbf{T}(t) = \frac{1}{\sqrt{2}}\langle -\sin t, \cos t, 1\rangle.$$
Step 2. Differentiate $\mathbf{T}$. $$\mathbf{T}'(t) = \frac{1}{\sqrt{2}}\langle -\cos t, -\sin t, 0\rangle, \qquad |\mathbf{T}'(t)| = \frac{1}{\sqrt{2}}.$$
Principal unit normal: $$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} = \langle -\cos t, -\sin t, 0\rangle.$$
$\mathbf{N}$ points toward the $z$-axis (the center of the helix) -- which matches intuition.
Step 3. Compute the binormal. $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \frac{1}{\sqrt{2}}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin t & \cos t & 1 \\ -\cos t & -\sin t & 0\end{vmatrix}$$ $$= \frac{1}{\sqrt{2}}\langle 0\cdot\cos t - 1\cdot(-\sin t),\; 1\cdot(-\cos t)-(-\sin t)\cdot 0,\; (-\sin t)(-\sin t) - \cos t(-\cos t)\rangle$$ $$= \frac{1}{\sqrt{2}}\langle \sin t,\, -\cos t,\, 1\rangle.$$
All three vectors are unit vectors and mutually perpendicular. The frame rotates smoothly as $t$ increases.
$\mathbf{N}$ points away from the center of curvature. The principal normal $\mathbf{N}$ points TOWARD the center of curvature (toward the inside of the bend), not away from it. This is because $\mathbf{N} = \mathbf{T}'/|\mathbf{T}'|$ and $\mathbf{T}'$ points in the direction that $\mathbf{T}$ is turning toward, which is always the inside of the curve.
Common Misconceptions
the principal normal vector $\mathbf{N}$ points away from the center of curvature.
This is the concept-image-conflicts-definition error. The principal normal $\mathbf{N} = \mathbf{T}'/|\mathbf{T}'|$ points in the direction that $\mathbf{T}$ is turning, which is always toward the inside of the curve, toward the center of curvature. For a particle rounding a curve, $\mathbf{N}$ points inward. The intuition “normal means outward” comes from surface normals, a different context; in the TNB frame, $\mathbf{N}$ always points toward the center of the osculating circle.
Leveled Practice
Problem 1. For the helix in the worked example, find the torsion $\tau$ using the formula $\tau = (\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}''' / |\mathbf{r}' \times \mathbf{r}''|^2$.
Show answer
From the arc length example: $\mathbf{r}' = \langle -\sin t, \cos t, 1\rangle$, $\mathbf{r}'' = \langle -\cos t, -\sin t, 0\rangle$.
$\mathbf{r}''' = \langle \sin t, -\cos t, 0\rangle$.
From before: $\mathbf{r}' \times \mathbf{r}'' = \langle \sin t, -\cos t, 1\rangle$ and $|\mathbf{r}' \times \mathbf{r}''|^2 = 2$.
$(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}''' = \sin t \cdot \sin t + (-\cos t)(-\cos t) + 1 \cdot 0 = \sin^2 t + \cos^2 t = 1$.
$\tau = \dfrac{1}{2}$.
The helix has constant torsion $\tau = 1/2$, just as it has constant curvature $\kappa = 1/2$.