Let $\alpha:I=[t_0,t_1] \to \mathbb{R}^3$, $\alpha=\alpha(t)$ is a regular curve not parametrized by arc length and $\beta: J=[s_0,s_1]\to \mathbb{R}^3$, $\beta=\beta(s)$ a reparametrization by arc, where $s=s(t)$ is calculated from $t_0$. Let $t=t(s)$ be the inverse function and $\frac{d\alpha}{dt}=\alpha',\frac{d^2\alpha}{dt^2}=\alpha''$, etc. I want to prove that
$$\frac{d^2t}{ds^2}=-\frac{\alpha'\cdot \alpha''}{||\alpha'||^4}.$$
We can define $s:[t_0,t_1]\to [s_0,s_1]$ and it's a difeomorphism as
$$s=s(t)=\int_{t_0}^t ||\alpha'(u)||du,$$
that is to say, $s(t)$ measure the arc length. Then, it's easy to see that
$$||\alpha'(t)||^{-1}=\frac{dt}{ds}.$$
I try to develop $\alpha''(t)$, but I don't get anything similar.
$$\alpha(t)=s'(t)^2\boldsymbol{t}'_\beta(s(t))+s''(t)\boldsymbol{t}_\beta(s(t))$$
Can anyone help me?
I would be tempted to square everything, and write $$ \left( \frac{dt}{ds}\right)^2 = \frac 1 {\alpha'(t(s)) . \alpha'(t(s))}.$$ This gets rid of the horrible square roots.
Now differentiate both sides, using the product rule and the chain rule: $$ 2 \frac{dt}{ds} \times \frac{d^2 t}{ds^2} = - \frac {2\alpha'(t(s)).\alpha''(t(s)) \times dt/ds}{ (\alpha'(t(s)) . \alpha'(t(s)))^2}.$$
I leave you to finish off...