At each point on a curve $\mathcal{C}$, the tangent vector is parallel to a non-vanishing vector field $\mathbf{F}$. Show that the curvature $\kappa$ of $\mathcal{C}$ is given by: $$\kappa=\frac{|\mathbf{F}\times (\mathbf{F}\cdot\nabla)\mathbf{F}|}{|\mathbf{F}|^3}\;\;(1)$$
I have done a similar question before: for a curve $\mathbf{r}(t)$ in the $xy$ plane parametrised by $x(t),y(t):$
$$\kappa=\frac{|\mathbf{r}'\times \mathbf{r}''|}{|\mathbf{r'}|^3}\;\;(2)$$
(I can post the proof if anyone is interested)
Given the similarity, I am wondering whether there might be a very efficient/quick way of getting at $(1)$ by cleverly relating it to $(2)$. Is this feasible? I have tried using the same steps as those used to arrive at $(2)$ but I am unsure even of how the definitions carry over in terms of derivatives... Any form of help is appreciated.
First note that (2) is reparametrization invariant. Thus, wlog, $\mathcal{C}$ is an integral curve of $\mathbf{F}$, i.e. given by a solution $\mathbf{r}(t)$ of $\mathbf{r}'=\mathbf{F}(\mathbf{r})$. Take the derivative of the last equation wrt $t$ (use the chain rule), get $\mathbf{r}''= (\mathbf{F}\cdot\nabla)\mathbf{F}$. Plug into (2).