In a Frenet-Serret frame, what are $\Delta\vec T$ and $(\vec a\vec\nabla)\vec T$ Given a Frenet-Serret frame $(\vec T(t), \vec N(t), \vec B(t))$ defined by a curve $\vec \gamma(t)$ with
$$\begin{array}{rcl}
   |\tfrac{d}{dt}\vec\gamma(t)| &\equiv& 1,
\\ \vec T(t) &:=& \tfrac{d}{dt}\vec\gamma(t),
\\ \kappa(t) &:=& |\tfrac{d}{dt}\vec T(t)|,
\\ \kappa(t)\vec N(t) &:=& \tfrac{d}{dt}\vec T(t),
\\ \vec B(t) &:=& \vec T(t)\times\vec N(t),
\\ \tfrac{d}{dt}\vec B(t) &=:& -\tau\vec N(t)
\end{array}$$
one can define a local coordinate system $$\vec x(t,n,b) := \vec\gamma(t) + n\vec N(t) + b\vec B(t)$$ (where the range of $n,b$ is defined such that this yields an injective function, assuming $\kappa,\tau\neq0$). $(\vec T,\vec N ,\vec B)$ are assumed to be independent of $n,b$, i.e. for a fixed $t$ the $\vec x(t,n,b)$ span a 2D plane perpendicular to $\vec T(t)$ with Cartesian coordinates $(n,b)$.
Then what are the derivatives of these unit vectors, especially

what are $(\vec a\vec\nabla)\vec T$ (for a vector $\vec a$, i.e. directional derivative of $\vec T$) and $\Delta\vec T$ (i.e. Laplacian of $\vec T$) expressed in these coordinates?

 A: Alright, let's just follow the Wikipedia article on curvilinear coordinates:
From $\vec x(t,n,b) = \vec\gamma(t) + n\vec N(t) + b\vec B(t)$ one obtains
$$\begin{array}{rcl}
  \vec h_t &=& \partial_t\vec x = \underbrace{\vec\gamma'}_{=\vec T} + n\underbrace{\vec N'}_{=-\kappa\vec T + \tau\vec B} + b\underbrace{\vec B'}_{=-\tau\vec N}
\\ &=& (1-n\kappa)\vec T - n\tau\vec N + b\tau\vec B
\\\Rightarrow h_t &=& \sqrt{(1-n\kappa)^2+(n^2+b^2)\tau^2},
\\\vec h_n &=& \partial_n\vec x = \vec N \Rightarrow h_n=1,
\\\vec h_b &=& \partial_b\vec x = \vec B \Rightarrow h_b=1.
\end{array}$$
Note how for non-vanishing torque $\tau\neq0$ this yields $\vec h_t\,\not\|\,\vec T$. Then, following these formulas, one obtains
$$(\vec a\vec\nabla)\vec T = \frac{\vec a\vec h_t}{h_t^2}\underbrace{\partial_t\vec T}_{=\kappa\vec N}$$
and
$$\begin{array}{rcl}
  \Delta\vec T &=& \left(\frac1{h_t}\partial_t\right)^2 \vec T = \frac1{h_t}\partial_t\left(\frac{\kappa(t)}{h_t}\vec N\right)
\\ &=& \frac1{h_t}\left(\frac\kappa{h_t}\right)'\vec N + \frac{\kappa}{h_t^2}\left(-\kappa\vec T + \tau\vec B\right)
\\ \text{with}\quad h_t' &=& \frac{(1-n\kappa)n\kappa'+(n^2+b^2)\tau\tau'}{h_t}
\\ \Rightarrow \frac1{h_t}\left(\frac\kappa{h_t}\right)' &=& \frac{\kappa' h_t^2 - \kappa[(1-n\kappa)n\kappa' + (n^2+b^2)\tau\tau']}{h_t^4}
\\ &=& \color{red}{\frac{\kappa'[(n^2+b^2)\tau+2\kappa^2-3\kappa+1] - \tau'\kappa\tau(n^2+b^2)}{h_t^4}}
\end{array}$$
The last line still needs to be fixed for the $\kappa\to n\kappa$ error in $h_t$.
A: \begin{array}{rcl}
  \vec h_t &=& \partial_t\vec x = \underbrace{\vec\gamma'}_{=\vec T} + n\underbrace{\vec N'}_{=-\kappa\vec T + \tau\vec B} + b\underbrace{\vec B'}_{=-\tau\vec N}
\\ &=& (1-\kappa)\vec T - n\tau\vec N + b\tau\vec B
\\\Rightarrow h_t &=& \sqrt{(1-\kappa)^2+(n^2+b^2)\tau^2},
\\\vec h_n &=& \partial_n\vec x = \vec N \Rightarrow h_n=1,
\\\vec h_b &=& \partial_b\vec x = \vec B \Rightarrow h_b=1.
\end{array}
should be
\begin{array}{rcl}
  \vec h_t &=& \partial_t\vec x = \underbrace{\vec\gamma'}_{=\vec T} + n\underbrace{\vec N'}_{=-\kappa\vec T + \tau\vec B} + b\underbrace{\vec B'}_{=-\tau\vec N}
\\ &=& (1-\kappa n)\vec T - n\tau\vec N + b\tau\vec B
\\\Rightarrow h_t &=& \sqrt{(1-\kappa n)^2+(n^2+b^2)\tau^2},
\\\vec h_n &=& \partial_n\vec x = \vec N \Rightarrow h_n=1,
\\\vec h_b &=& \partial_b\vec x = \vec B \Rightarrow h_b=1.
\end{array}
