I like Serena said:
What is MIT?
Assuming you mean the 2nd equality, they are using that:
\begin{aligned}
\mathbf T(s) &= \mathbf r'(s) \\
\mathbf T'(s) &= \kappa(s) \mathbf n(s) \\
\mathbf n(s) &= \frac{\mathbf r''(s)}{\kappa(s)} \\
\mathbf b(s) &= \mathbf T(s) \times \mathbf n(s) \\
\end{aligned}
Or if you meant the 3rd equality, it is explained on page 2 in your pdf.
Btw, can you please be more specific?
I dislike guessing what someone means.
MIT = Massachusetts Institute of Technology.
As a note, I have shown that \(\frac{1}{\rho} = \frac{\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert} {\lvert\dot{\mathbf{r}}\rvert^3}\)
Here is what I worked out.
\begin{align}
\frac{1}{\tau} &= -\hat{\mathbf{n}}\cdot \frac{d\hat{\mathbf{b}}}{ds}\\
&= -\rho\frac{d^2\mathbf{r}}{ds^2}\cdot \left(\hat{\mathbf{u}}\times
\hat{\mathbf{n}}\right)_t
\end{align}
Now let's write \(\frac{d^2\mathbf{r}} {ds^2}\) as
\begin{align}
\frac{d^2\mathbf{r}}{ds^2} &= \frac{d^2\mathbf{r}} {dt^2}
\left(\frac{dt} {ds}\right)^2\\
&= \frac{1} {v^2}\ddot{\mathbf{r}}
\end{align}
and substitute back into equation above.
\begin{align}
&= -\frac{\rho} {v^2}\ddot{\mathbf{r}} \cdot \frac{\rho} {v^3}\left(\dot{\mathbf{r}}\times
\ddot{\mathbf{r}}\right)_t\\
&= \frac{\lvert\dot{\mathbf{r}}\rvert} {\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert}\dddot{\mathbf{r}} \cdot \frac{1}{\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert} (\dot{\mathbf{r}}\times\ddot{\mathbf{r}})\\
&= ?\\
&= \frac{(\dot{\mathbf{r}} \times\ddot{\mathbf{r}}) \cdot
\dddot{\mathbf{r}}}
{\lvert \dot{\mathbf{r}} \times\ddot{\mathbf{r}} \rvert^2}
\end{align}
So I still have a \(\lvert\dot{\mathbf{r}}\rvert\)
