TITLE: Vector calculus and Frenet-Serret equations
QUESTION [5 upvotes]: I have shown the first two equality and I am working on the showing the 1st equals the 3rd.
\begin{alignat*}{4}
  \frac{1}{\rho}\hat{\mathbf{{n}}} &= \frac{d\hat{\mathbf{{u}}}}{ds}
  &{}= \frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}}
  &{}= \left((\dot{\mathbf{r}}\cdot\dot{\mathbf{r}})\ddot{\mathbf{r}} -
  (\dot{\mathbf{r}}\cdot\ddot{\mathbf{r}})\dot{\mathbf{r}}\right)
  \frac{1}{\dot{r}^4}
\end{alignat*}

$$
\frac{1}{\rho}\hat{\mathbf{{n}}} = \frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}}
$$
We know that $\mathbf{v} = \frac{ds}{dt}\frac{dr}{ds}$ where $\dot{s} = v$ and $\hat{\mathbf{u}} = \frac{dr}{ds}$.
So $\mathbf{v} = v\hat{\mathbf{u}}\iff \dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt}$.
Then $\frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}} = \frac{1}{v^2}\frac{d\mathbf{v}}{dt}$.
I know that
$$
\frac{d\mathbf{v}}{dt} = \frac{dv}{dt}\hat{\mathbf{u}} + \frac{v^2}{\rho}\hat{\mathbf{n}}.
$$
Then $\frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}} = \frac{1}{v^2}\frac{dv}{dt}\hat{\mathbf{u}} + \frac{1}{\rho}\hat{\mathbf{n}}$.
Therefore, $\frac{1}{v^2}\frac{dv}{dt}\hat{\mathbf{u}} = 0$ but how do I show that this is $0$?

REPLY [2 votes]: There is a mistake in the computation of $\dot {\hat {\mathbf u}}$.
Using the product rule, you have$$\dot {\hat {\mathbf u}}=-\frac {\dot v}{v^2}\mathbf v + \frac 1{v}\dot {\mathbf v}=-\frac {\dot v}{v}\hat{\mathbf u} + \frac 1{v}\dot {\mathbf v}$$and therefore
$$\frac {\dot {\hat {\mathbf u}}}{v}=\frac 1{v^2}(-\dot v\, \hat{\mathbf u}+\dot {\mathbf v})=\frac 1{v^2}\left(-\dot v\, \hat{\mathbf u}+\dot v\, \hat{\mathbf u}+ \frac {v^2}{\rho}\hat {\mathbf n}\right)==\frac 1{\rho}\hat {\mathbf n}$$