I am following a proof in my vector calculus book, but I am getting stuck.
Let $T(s)$ be the the unit tangent vector at $s$ and let $k(s) = \|(T’(s))\|$ -- here $T’(s)$ is orthogonal to $T(s)$ -- and let $N(s)$ be the unit vector such that $T’(s)=k(s) \cdot N(s)$.
See photo
Now, in the proof it says
differentiating $N(s) \cdot T(s) =0$ gives $N’(s)\cdot T(s)+N(s)\cdot T’(s) =0$.
Hence $N’(s) \cdot T(s)=-k(s)$. But I don’t see how this is derived.
$N(s)\cdot T(s)=0 \\ \Rightarrow \frac{d}{dt}\left(N(s)\cdot T(s)\right)=0 \\ \Rightarrow \frac{dN(s)}{dt}\cdot T(s)+N(s)\cdot \frac{dT(s)}{dt}=0 \\ \Rightarrow \frac{dN(s)}{dt}\cdot T(s)=-N(s)\cdot \frac{dT(s)}{dt}$
but you are told that $\frac{dT(s)}{dt}=k(s)N(s)$ so
$\frac{dN(s)}{dt}\cdot T(s)=-N(s)\cdot \left(k(s)N(s)\right)=-k(s)\left(N(s)\cdot N(s)\right)$
and $N(s)\cdot N(s)=1$ because $N(s)$ is a unit vector, so
$\frac{dN(s)}{dt}\cdot T(s)=-k(s)$