Derivative of a parallel

Question

I'm fully aware of the fact that there is something simple here that I am just not seeing and it's not been four hours of trying to substitute every potential rewrite of $n'(t)$ but can someone kindly explain to me the following. Given the arc-length-parametrised curve $γ(t)$ and its parallel $p(t)$ s.t. $$p(t)=γ(t)+dn(t)$$ $$\text{ and } n(t) \text{is the unit normal vector to } γ(t)$$ Why does this hold? $$p′(t)=γ′(t)+dn′(t)=(1−k(t)d)γ′(t)$$ $$ \text{where } k(t) \text{ is the curvature of } γ(t)$$

Answer 1

Expanding both sides of the last equality shows that the equality is equivalent to (suppressing parameters in most places throughout) $$n' = - k \gamma' .$$ But this follows from the Frenet-Serret equations for planar curves: \begin{align*} {\bf T}' &= \phantom{-k {\bf T}+}k n \\ n' &= -k {\bf T} \end{align*} Here $\bf T$ is the unit tangent vector; since $\gamma$ is parameterized by arc length, $\gamma' = {\bf T}$.

If you're only familiar with the more commonly seen Frenet-Serret equations for space curves, you can recover the plane curve version just by viewing $\gamma$ as the space curve $t \mapsto (\gamma(t), 0)$; since the curve is planar, the torsion is $\tau \equiv 0$.