Prove that $\frac{d^2 t}{d s^2} = - \frac{\alpha ' (t) \cdot \alpha '' (t)}{|| \alpha ' (t) ||^4}$ I have to prove that if $\alpha(t)$ is a regular curve in space and $\beta(s)$ is its reparametrization of unit speed then $\frac{d^2 t}{d s^2} = - \frac{\alpha ' (t) \cdot \alpha '' (t)}{|| \alpha ' (t) ||^4}$.
Attempt:
I know that the RHS is:
\begin{gather*}
\alpha'(t)=\left(\frac{ds}{dt}\right)T(t) \\
\alpha''(t)= \left(\frac{ds}{dt}\right)T'(t)+\left(\frac{d^2s}{dt^2}\right)T(t)
\end{gather*}
where $T(t)$ is the unit tangent vector. When doing $\alpha'(t)\cdot \alpha''(t)$ we get
\begin{equation*}
\alpha'(t) \cdot \alpha ''(t) = \left(\frac{ds}{dt}\right) \left(\frac{d^2s}{dt^2}\right) T(t) \cdot T(t) = ||\alpha'(t)||\left(\frac{d^2s}{dt^2}\right)
\end{equation*}
And so
\begin{equation*}
-\frac{\alpha'(t)\cdot \alpha''(t)}{||\alpha'(t)||^4} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^3}
\end{equation*}
On the other hand, the LHS is:
\begin{gather*}
\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{||\alpha'(t)||} = (||\alpha'(t)||)^{-1} \\
\Rightarrow \frac{d^2 t}{ds^2} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^2}
\end{gather*}
So there is a factor of $||\alpha'(t)||$ I am missing. Is the statement wrong or where am I wrong? Thank you in advance.
 A: 
On the other hand, the LHS is:
  \begin{gather*}
\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{||\alpha'(t)||} = (||\alpha'(t)||)^{-1} \\
\Rightarrow \frac{d^2 t}{ds^2} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^2}
\end{gather*}
  So there is a factor of $||\alpha'(t)||$ I am missing. Is the statement wrong or where am I wrong? Thank you in advance.

The ($\Rightarrow$) step is incorrect: 
$$
\frac{d}{ds}(\frac{ds}{dt})\neq\frac{d^2s}{dt^2}.
$$
Note that
$$
LHS=\frac{d}{ds}(\frac{dt}{ds})=\frac{d}{ds}(\frac{1}{ds/dt})
=\frac{-1}{(ds/dt)^2}\cdot\frac{d}{ds}(\frac{ds}{dt})
=\frac{-1}{\|\alpha'(t)\|^2}\cdot \frac{d}{ds}(\frac{ds}{dt})\tag{1}
$$
Now, 
$$
\frac{d}{ds}(\frac{ds}{dt})=\frac{d}{dt}(\frac{ds}{dt})\cdot{\frac{dt}{ds}}
=\frac{d^2s}{dt^2}\frac{1}{\|\alpha'(t)\|}\tag{2}
$$
Combining (1) and (2) you get the desired LHS, which is equal to what you calculated for the RHS.
A: I think the introduction of $T(t)$ etc. confuses things somewhat.
It can be done as follows:
$\dfrac{ds}{dt} = \Vert \alpha'(t) \Vert = \langle \alpha'(t), \alpha'(t) \rangle^{1/2}; \tag 1$
$\dfrac{dt}{ds} = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}; \tag 2$
$\dfrac{d^2 t}{ds^2} = \dfrac{d}{ds} \dfrac{dt}{ds} = \dfrac{d}{ds} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2} = \dfrac{dt}{ds} \dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}; \tag 3$
$\dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2} =  -\dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-3/2} \dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle$
$=  -\dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-3/2} \left ( 2 \langle \alpha'(t), \alpha''(t) \rangle \right ) = -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}}; \tag 4$
finally,
$\dfrac{d^2 t}{ds^2} = \dfrac{dt}{ds} \left (  -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}} \right ) =  \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\left (  -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}} \right )$
$= -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^2} = -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\Vert \alpha'(t) \Vert^4}. \tag 5$
