Prove that $\frac{dN}{ds}=-\kappa T$

Question

Prove that $\frac{dN}{ds}=-\kappa T$, where $N$ is the oriented normal, and $T$ is the unit tangent vector, and $s$ is arc-length parameter.

Here's what I've got so far from my note and I don't understand some of it:

By definition, suppose $T=<a,b>$, then $N=<-b,a>$, so $T\cdot N=0$.

Differentiate both side with respect to $s$, we get $T\cdot \frac{dN}{ds}+\frac{dT}{ds}\cdot N=0$.

Since by definition of signed curvatuve, $\kappa=\frac{dT}{ds}\cdot N$, we have $T\cdot \frac{dN}{ds}=-\kappa$.

Then, my note says "Since these two vectors are parallel, we have the desired result"

Question1: Why are they parallel? Does it have something to do with the fact that we are talking about planar curves?

Question2: How do get from these two being parallel to the result?

Answer 1

The rate of change of the normal w.r.t. the arc length is parallel to the tangent so: $\widehat{\frac{dN}{ds}}={T}$ (note $T$ is already a unit vector)and the rest follows.

Q1: The hand waving (physicists ;) explanation is that we need only consider the tangent (kissing?) circle with centre at the centre of curvature for the point under consideration. This is easily shown to hold for this tangent circle and so will hold for the curve itself.

Q2: Since $T$ and $\frac{dN}{ds}$ are parallel $\frac{dN}{ds}=\lambda T$ and then $T.\frac{dN}{ds}=\lambda T.T$, but $T$ is a unit vector so $$ T.\frac{dN}{ds}=\lambda T.T=\lambda=-\kappa $$ Hence $\lambda=-\kappa$ and so $\frac{dN}{ds}=-\kappa T$ as required.

Answer 2

We have $N\cdot N=1$ and therefore $\dot N\cdot N= 0$. Since $(T,N)$ is an orthonormal frame this implies $$\dot N=(\dot N\cdot T)\> T\ .\tag{1}$$ (In particular $\dot N$ and $T$ are parallel, as are $\dot T$ and $N$.)

Now from $N\cdot T=0$ we get $\dot N\cdot T+N\cdot \dot T=0$, and by your definition of $\kappa$ this says that $\dot N\cdot T=-\kappa$. From $(1)$ it then follows that $\dot N=-\kappa T$.