A curve parametrized by arc length 
Let $C$ be a plane curve parametrized by arc length by $\alpha(s)$,
  $T(s)$ (unit tangent vector) and $N(s)$ (unit normal vector). Prove
  that $$\frac{d}{ds} N(s)=-\kappa(s)T(s).$$

I know that since $C$ is a curve parametrized by arc length then the tangent vector $\alpha'(s)$ has unit length which equals $T(s) = \frac{\alpha'(s)}{||\alpha'(s)||}$, thus $N(s) = \frac{T'(s)}{||T'(s)||}$. 
How can I prove this? . 
 A: Since $N(s)$ and $T(s)$ are normal for all $s$, we have
$\langle T, N \rangle = 0$
for all $s$.  Thus
$0 = \langle T, N \rangle' = \langle T', N \rangle + \langle T, N' \rangle$,
whence
$\langle T', N \rangle = -\langle T, N' \rangle$;
by definition, $T' = \kappa N$; therefore
$\langle T, N' \rangle = -\kappa$.
Now since $\langle N, N' \rangle = 0$ we must have 
$N' = -\kappa T$,
since $T$ and $N$ form an orthorormal basis for the tangent space at any point they are defined.
QED!
Thirteen minutes single-finger 'droid typing!
A: Let's express $\frac{d\mathbf{N}}{ds}$ as a linear combination of the tangent, normal, and binormal vectors
$$\frac{d\mathbf{N}}{ds}=a(s)\mathbf{T}+b(s)\mathbf{N}+c(s)\mathbf{B}.$$
These are all unit vectors perpendicular to each other, and the scalar functions $a,b,c$ are arbitrary scalar functions.  So if we take the dot product with respect to the tangent vector $\mathbf{T}$
$$\frac{d\mathbf{N}}{ds}\cdot\mathbf{T}=a(s)\mathbf{T}\cdot\mathbf{T}+b(s)\mathbf{N}\cdot\mathbf{T}+c(s)\mathbf{B}\cdot\mathbf{T}=a(s)+0+0=a(s)$$
Now we have that $$a(s)=\frac{d\mathbf{N}}{ds}\cdot\mathbf{T}=-\mathbf{N}\cdot\frac{d\mathbf{T}}{ds}=\mathbf{N}\cdot\kappa\mathbf{N}=-\kappa$$
