= Solution
Put $T=\alpha'(s)$. Since $s$ is <arc length>, $|T|=1$, so $T'\perp T$. The <curvature of a space curve> is
$$
k(s)=|T'(s)|=|\alpha''(s)|.
$$
Where $k(s)\ne0$, define the principal normal and binormal by
$$
N=\frac{T'}k,
\qquad
B=T\times N,
$$
and define the <torsion of a space curve> by
$$
\tau=-B'\cdot N.
$$
Thus torsion requires nonzero curvature, so that $N$ and $B$ are defined.
Differentiating the orthonormality relations for the frame $(T,N,B)$ shows that its derivative matrix is skew-symmetric. Since $T'=kN$, differentiating $B=T\times N$ and using the definition of $\tau$ gives the <Frenet-Serret formulas>
$$
\boxed{
T'=kN,
\qquad
N'=-kT+\tau B,
\qquad
B'=-\tau N.}
$$
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