= Solution
For an <arc-length parametrization>, the <unit tangent vector> is $t=r'(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The <curvature of a space curve> is $\kappa=|t'|$; where $\kappa>0$, its <principal normal vector> is $n=t'/\kappa$, so $t'=\kappa n$. The <binormal vector> $b=t\times n$ completes the positively oriented <Frenet frame>.
Differentiate $b\cdot b=1$ to get $b'\cdot b=0$. Also
$$
\frac d{ds}(b\cdot t)=0
\quad\Longrightarrow\quad b'\cdot t=-b\cdot t'=-\kappa b\cdot n=0.
$$
Thus $b'$ has only an $n$ component. Define the <torsion of a space curve> by $\tau=-b'\cdot n$, giving $b'=-\tau n$.
Next, $n'\cdot n=0$, and differentiating the other inner products gives $n'\cdot t=-\kappa$ and $n'\cdot b=\tau$. Hence the <Frenet-Serret formulas> are
$$
\boxed{t'=\kappa n,\qquad b'=-\tau n,\qquad n'=-\kappa t+\tau b.}
$$
These statements require enough smoothness, for example a $C^3$ <curve> with positive curvature on the interval under discussion. If $\kappa=0$, the <unit tangent vector> is still defined, but the standard <principal normal vector>, <binormal vector> and <torsion of a space curve> need not be; their existence cannot be inferred just from smoothness of the <curve>.
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