Recipe

Find curvature, the unit tangent, and the principal normal

Use when: you have r(t) and need the curvature κ, the unit tangent T, or the principal normal N at a point

Goal: κ, T(t), and N(t), built from r(t)'s first two derivatives

Taught in Curvature and the normal vector · a later section extends this recipe with more cases

1. 1

   Differentiate each component of \mathbf r(t) twice.

   Example

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

   Why: both derivatives feed directly into the cross-product form of curvature below, and no other information is needed.

2. 2

   Take the magnitude of \mathbf r'(t).

   Example

   |\mathbf r'(t)|=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2, constant for every t.

   Why: speed appears cubed in curvature's denominator, and appears again, undivided, when \mathbf T(t) is built in Step 4.

3. 3

   Cross \mathbf r'(t) with \mathbf r''(t), take that vector's magnitude, and divide by |\mathbf r'(t)|^3.

   Example

   \mathbf r'\times\mathbf r''=\langle\sin t,-\cos t,1\rangle, magnitude \sqrt2, so \kappa=\dfrac{\sqrt2}{(\sqrt2)^3}=\dfrac{\sqrt2}{2\sqrt2}=\dfrac12.

   ⚡ Shortcut

   Skip finding \mathbf T(t) at all when only \kappa is needed. The cross-product form never requires normalizing \mathbf r'(t) first, unlike the definition \kappa=|\mathbf T'(t)|/|\mathbf r'(t)|, which does.

   Why: the cross-product form is algebraically identical to |\mathbf T'(t)|/|\mathbf r'(t)| but skips normalizing \mathbf r'(t) into \mathbf T(t) first, which is often the messier step.

4. 4

   Divide \mathbf r'(t) by its own magnitude from Step 2.

   Example

   \mathbf T(t)=\dfrac{\langle-\sin t,\cos t,1\rangle}{\sqrt2}; at t=\pi/2: \mathbf T=\left\langle-\dfrac{\sqrt2}2,0,\dfrac{\sqrt2}2\right\rangle.

   Why: \mathbf T(t) is the direction of travel with speed divided out, and Step 2 already found exactly the number to divide by.

5. 5

   Differentiate \mathbf T(t), then divide the result by its own magnitude.

   Example

   \mathbf T'(t)=\dfrac{\langle-\cos t,-\sin t,0\rangle}{\sqrt2}, magnitude \dfrac1{\sqrt2}, so \mathbf N(t)=\langle-\cos t,-\sin t,0\rangle; at t=\pi/2: \mathbf N=\langle0,-1,0\rangle.

   • Not yet: the binormal B and torsion τ — covered in TNB, torsion, and tangential/normal acceleration

   Why: \mathbf N(t) has to be a unit vector, so the direction \mathbf T is turning toward needs the same kind of normalizing that built \mathbf T itself in Step 4.