Recipe

Find the full TNB frame, curvature, and torsion

Use when: you have r(t) and need the complete moving frame T, N, B together with κ and τ, not just T and N alone

Goal: all three frame vectors plus the two turning rates that describe the curve's local shape completely

Taught in TNB, torsion, and tangential/normal acceleration · extends Find curvature, the unit tangent, and the principal normal

Quick recipe

Find $\mathbf r'(t)$ , $\mathbf r''(t)$ , and $\mathbf r'''(t)$
Build $\mathbf T(t)$ and $\mathbf N(t)$ exactly as before
Cross $\mathbf T(t)$ with $\mathbf N(t)$ to get $\mathbf B(t)$
Compute torsion with the scalar triple product
Report the full frame together

1
Differentiate one more time than the curvature-only recipe needs.

Example

$\mathbf r(t)=\langle\cos t,\sin t,t\rangle$ : $\mathbf r'=\langle-\sin t,\cos t,1\rangle$ , $\mathbf r''=\langle-\cos t,-\sin t,0\rangle$ , $\mathbf r'''=\langle\sin t,-\cos t,0\rangle$ .

Why: torsion's formula needs a third derivative that curvature alone never required.
2
Divide $\mathbf r'(t)$ by its magnitude for $\mathbf T(t)$ ; differentiate $\mathbf T(t)$ and normalize for $\mathbf N(t)$ , Steps 2 through 5 of the curvature recipe.

Example

$|\mathbf r'|=\sqrt2$ ; $\mathbf T(t)=\langle-\sin t,\cos t,1\rangle/\sqrt2$ ; $\mathbf N(t)=\langle-\cos t,-\sin t,0\rangle$ .

Why: $\mathbf B$ is defined from $\mathbf T$ and $\mathbf N$ together, so both have to already be in hand.
3
Compute $\mathbf B(t)=\mathbf T(t)\times\mathbf N(t)$ .

Example

At $t=\pi/2$ : $\mathbf T=\left\langle-\dfrac{\sqrt2}2,0,\dfrac{\sqrt2}2\right\rangle$ , $\mathbf N=\langle0,-1,0\rangle$ , so $\mathbf B=\mathbf T\times\mathbf N=\left\langle\dfrac{\sqrt2}2,0,\dfrac{\sqrt2}2\right\rangle$ .

Why: $\mathbf B$ is the unique unit vector completing a right-handed frame with $\mathbf T$ and $\mathbf N$ , exactly what crossing two perpendicular unit vectors produces.
4
Compute $\tau=\dfrac{(\mathbf r'\times\mathbf r'')\cdot\mathbf r'''}{|\mathbf r'\times\mathbf r''|^2}$ .

Example

$\mathbf r'\times\mathbf r''=\langle\sin t,-\cos t,1\rangle$ , so $|\mathbf r'\times\mathbf r''|^2=2$ ; then $(\mathbf r'\times\mathbf r'')\cdot\mathbf r'''=\sin^2t+\cos^2t+0=1$ , giving $\tau=1/2$ .
- $\tau=0$ everywhere — the curve is planar, lying entirely in one flat plane.
- $\tau$ constant and nonzero — the curve twists at a steady rate, the way a helix does.
Why: torsion measures how much $\mathbf r'''$ pushes out of the plane spanned by $\mathbf r'$ and $\mathbf r''$ , exactly what the triple product with $\mathbf r'\times\mathbf r''$ detects.
5
State $\mathbf T$ , $\mathbf N$ , $\mathbf B$ , $\kappa$ , and $\tau$ as one complete answer.

Why: together these five quantities are everything the Frenet-Serret framework describes about the curve's local shape, and reconstructing the curve up to position and rotation needs no more than this.

Common mistakes

Computing $\mathbf r'\times\mathbf r''$ from one parametrization and $\mathbf r'''$ from a different one, after an unnoticed algebra change partway through.
Forgetting to square $|\mathbf r'\times\mathbf r''|$ in torsion's denominator.
Assuming $\mathbf B$ always points a particular direction, when it is set entirely by $\mathbf T\times\mathbf N$ , whichever way that happens to point.
Redoing $\mathbf r'\times\mathbf r''$ from scratch for torsion after already finding it for curvature, risking a sign slip instead of reusing the same computation.