Proving the Frenet-Serret formulae

Question

I have been asked to prove:

$$ \frac{d\mathbf{N}}{ds}=-\kappa\mathbf{T}+\tau\mathbf{B} $$

I have been given $\mathbf{N}=\mathbf{B} \times \mathbf{T}$

$$ \frac{d\mathbf{B}}{ds}=-\tau\mathbf{N} $$

To start I said: $$ \frac{d\mathbf{N}}{ds}=\frac{d(\mathbf{B} \times \mathbf{T})}{ds} = (\frac{d\mathbf{B}}{ds} \times \mathbf{T}) +( \mathbf{B} \times \frac{d\mathbf{T}}{ds} )$$

I can sub in the things I know so: $$ \frac{d\mathbf{N}}{ds}=(\tau\mathbf{N} \times \mathbf{T}) +(\mathbf{B} \times \frac{d\mathbf{T}}{ds})$$

But now I am having trouble finding $\frac{d\mathbf{T}}{ds}$,

So far I have tried using the fact that:

$$ \mathbf{T}'(t)=|\mathbf{T}'(t)|\mathbf{N}(t)$$

and the chain rule:

$$\mathbf{T}'(t)=\frac{d\mathbf{T}}{ds} \frac{{ds}}{dt} $$

But when I combine them I get:

$$\frac{d\mathbf{T}}{ds} \frac{{ds}}{dt} =|\mathbf{T}'(t)|\mathbf{N}(t)$$

Which becomes:

$$\frac{d\mathbf{T}}{ds}\frac{{ds}}{dt} =\kappa\mathbf{N}(t)$$

Where I can't seem to cancel the $\frac{{ds}}{dt}$ or to find a minus in there somewhere? Not sure if I am being stupid or have just made a mistake somewhere? If anyone could help that would be much appreciated!

Answer 1

HINT

We have that

$$\frac{d\mathbf{N}}{ds}\perp \mathbf{N} \implies \frac{d\mathbf{N}}{ds}=a\mathbf{T}+b\mathbf{B}$$

then from

$$\mathbf{N} \cdot \mathbf{T}=0 \implies \frac{d(\mathbf{\mathbf{N} \cdot \mathbf{T}})}{ds}=\frac{d\mathbf{\mathbf{N} }}{ds}\cdot \mathbf{T}+ \mathbf{N}\cdot \frac{d\mathbf{\mathbf{T} }}{ds}=a+\kappa=0$$

then use also that

$$\mathbf{N} \cdot \mathbf{B}=0 \implies \dots$$

Answer 2

Not sure I get your question right, but it appears you are asking for why $\frac{{\rm d}t}{{\rm d}s}=\kappa n$ ($t$: tangential vector, $n$: normal vector, $\kappa$: curvature, $s$: arc length).

Pick any point $s_0$ on the curve $c(s)$. In the vicinity of $s_0$ the image $$c(s)\approx c(s_0) + c'(s_0)(s-s_0) + \frac{c''(s_0)}{2} \, (s-s_0)^2 \tag{1}$$ for the approximation of $c(s)$ about $s_0$ is 2-dimensional with the two basis vectors $c'=t$ and $c''=\kappa n$. It is easily seen that $n$ is normal to $t$ and for now $\kappa$ is just an arbitrary proportionality constant. It therefore suffices to consider this part of the curve in 2d-space by rotating and shifting the coordinate system, so that the center of the curvature circle of radius $R$ becomes the origin and $c(s_0)$ points in $x$-direction. By expanding the circle (with arbitrary parameter $t\in (0,2\pi)$) about the point $c(s_0)$ corresponding to $t=0$, we have $$R\begin{pmatrix} \cos(t) \\ \sin(t) \end{pmatrix}\approx R \begin{pmatrix} 1 - \frac{t^2}{2} \\ t \end{pmatrix} \, .$$ On the other hand by the definition of the arc length equation (1) in components reads $$c(s) \approx \begin{pmatrix} R - \frac{\kappa}{2} (s-s_0)^2 \\ s-s_0 \end{pmatrix} \, .$$ Hence by comparison $s-s_0=Rt$ from the $y$-component and therefore $$R-\frac{Rt^2}{2}=R-\frac{\kappa R^2t^2}{2} \quad \Rightarrow \quad \kappa=\frac{1}{R}$$ from the $x$-component, which is just the definition of the curvature.