Derive $\kappa(t)=\frac{\lvert \boldsymbol{a}(t) \times \boldsymbol{v}(t) \rvert}{v^3(t)}$ directly instead of proving it. Let $\boldsymbol{r}(t)$ be a parametrised curve. Then,
$$\boldsymbol{\hat{T}}=\frac{\boldsymbol{r'}(t)}{\lvert \boldsymbol{r'}(t) \rvert }$$
$$\boldsymbol{\hat{N}}=\frac{\boldsymbol{\hat{T}'}(t)}{\lvert \boldsymbol{\hat{T}'}(t) \rvert } $$
$$\frac{d\boldsymbol{\hat{T}}}{ds}=\frac{\lvert \boldsymbol{\hat{T'}}(t) \rvert}{v(t)} \boldsymbol{\hat{N}}=\kappa(t)\boldsymbol{\hat{N}}$$
$$v(t)=\frac{ds}{dt}$$
$$\kappa(t) \equiv \frac{\lvert \boldsymbol{\hat{T'}}(t) \rvert}{v(t)} $$
$$\kappa(t)=\frac{\lvert \boldsymbol{a}(t) \times \boldsymbol{v}(t) \rvert}{v^3(t)}$$
I have seen some proofs of the last formula. But nevertheless I wonder whether it is possible to derive this formula directly. I think I need to clarify my intention: Suppose you know the definition of curvature as in the third equation but you don't know the formula for curvature as in the last equation. Then, someone asks you to give a formula for the curvature in terms of $\boldsymbol{r}(t)$ and its derivatives. How would you proceed, and how would you find this last equation? I think, it has to come from somewhere...
Well, I have got another, closely linked, question: Is there a formula for $$\frac{d\boldsymbol{\hat{T}}}{ds}$$ in terms of $\boldsymbol{r}(t)$ and its derivatives?
 A: Given $$\frac{|d\boldsymbol{\hat{T}}|}{ds}= \frac{|d\boldsymbol{\hat{T}}|}{dt} \frac{dt}{ds}=\kappa(s)$$
Since we know $\boldsymbol{r}'(t)=\boldsymbol{T}(t)=\frac{ds}{dt} \boldsymbol{\hat{T}}(t)$
It is easy to see if we differentiate both sides with respect to $t$, we will get some relationship on the curvature.
$$\boldsymbol{r}''(t)=\frac{d^2 s}{dt^2}\boldsymbol{\hat{T}}(t)+\kappa  v^2\boldsymbol{\hat{N}}$$
We see that the second derivative of the position vector is related to the curvature in the above formula. To make it more clean, we want to eliminate the unit tangent vector, we can do this by crossing it with the tangent vector $\boldsymbol{r}'(t)$
$$\boldsymbol{r}'(t)\times\boldsymbol{r}''(t)=\frac{d^2 s}{dt^2}\boldsymbol{r}'(t)\times\boldsymbol{\hat{T}}(t)+\kappa  v^2\boldsymbol{r}'(t)\times\boldsymbol{\hat{N}}$$
Since the cross product of parallel vectors is the zero vector,
$$\boldsymbol{r}'(t)\times\boldsymbol{r}''(t)=\kappa v^3 \boldsymbol{\hat{B}}(t)$$
By taking the norm of both sides,
$$\kappa(t)=\frac{||\boldsymbol{r}'(t)\times\boldsymbol{r}''(t)||}{v(t)^3}$$
For your second question, using our expression above,
$$\boldsymbol{r}''(t)=v'(t)\boldsymbol{\hat{T}}(t)+v(t)\frac{d\boldsymbol{\hat{T}}(t)}{dt}$$
Rearranging,
$$\frac{d\boldsymbol{\hat{T}}(t)}{ds}=\frac{v(t)\boldsymbol{r}''(t)-v'(t)\boldsymbol{r}'(t)}{v(t)^3}$$
