# The TNB components of the jerk vector

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1. Nov 19, 2015

### clustro

It can be found in any advanced calculus textbook the proof that, for a "well-behaved" space curve, the acceleration vector can be decomposed into components along the tangent and normal unit vectors. The acceleration vector is always orthogonal to the binormal vector.

The decomposition is written as:

$\bf{\vec{a}}=\frac{d^2s}{dt^2}\bf{\vec{T}}+\kappa(\frac{ds}{dt})^2\bf{\vec{N}}$

Where s is the arc-length, t is the time, $\kappa$ is the curvature, and T and N are the tangent and normal unit vectors.

I read somewhere that, while there is never an acceleration in the binormal direction, the jerk vector will in general have components in all three directions of the TNB frame.

I just wanted to make sure I am doing this differentiation correctly. The primes indicate differentiation with respect to time.

$\bf\vec{a}'=s'''\bf{\vec{T}} + s''\bf\vec{T}'+[\kappa (s')^2]'\bf\vec{N}+\kappa(s')^2\bf\vec{N}'$

From the Frenet-Serret formulas and the chain rule:

$\bf\vec{T}'=\frac{d\bf\vec{T}}{ds}\frac{ds}{dt}=\kappa s'\bf\vec{N}$
$\bf\vec{N}'=\frac{d\bf\vec{N}}{ds}\frac{ds}{dt}=s'(-\kappa\bf\vec{T} + \tau\bf\vec{B})$

Where B is the binormal vector and $\tau$ is the torsion.

Plugging in the expressions for $\bf\vec{T}'$ and $\bf\vec{N}'$ into the jerk vector expression, combining like terms, and simplifying yields:

$\bf\vec{a}'=(s'''-\kappa^2(s')^3)\bf\vec{T} + (3\kappa s's'' + \kappa' s'^2)\bf\vec{N}+\kappa\tau s'^3 \bf\vec{B}$

I could not find any other derivation of this to check my work against. Does this look like the correct derivation of the TNB components of the jerk vector?

Am I correct in assuming that $\kappa$ is a function of time? It seems like it would be, since if I vary my speed along the curve, the curvature at my current point will change according to the speed I have chosen.

2. Nov 24, 2015