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In the following notes: , the author (on page 5) says that $T$ and $N$ are related by the Frenet formula $\partial_x N = -s'(x)\kappa T$ where $N,T$ are given by some curve parameterized by $x\mapsto \gamma(x)$ and $s$ is the arclength parameter.

I'm not quite sure how the author derived this formula. Any help would be much appreciated.

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2 Answers

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$T$ and $N$ form an orthonormal frame, so $$ T \cdot T = 1, \quad T \cdot N = 0, \quad N \cdot N = 1. $$ Differentiating gives $T \cdot T' = 0$, $N \cdot N' = 0$, and $ T \cdot N' = - N \cdot T' $. But $T' = s'\kappa N$ by the chain rule and the definition of the curvature, so it follows that $N'$ is parallel to $T$ and $T \cdot N' = -s'\kappa$, whence the formula you ask about.

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I would say it like this:

The standard Frenet-Serret equation for $N' = \partial_s N$ is:

$N' = \partial_s N = -\kappa T; \tag 1$

here $T$ and $N$ are the unit tangent an normal vectors, respectively, and $\kappa$ is the curvature of the given curve $\gamma$, of which $s$ is the arc-length. If the curve is parametrized by some other parameter $x$, $x \to \gamma(x)$, then we may compute $\partial_x N$ in terms of $s(x)$ by means of the chain rule:

$\partial_x N = \partial_s N \dfrac{ds(x)}{dx}; \tag 2$

inserting this into (1) yields

$\partial_x N = \partial_s N \dfrac{ds(x)}{dx} = -s'(x)\kappa T, \tag 3$

the desired result.

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