 Solution

Let $T,n,B$ be the Frenet frame of the original space curve, with $T^{\prime }=kn$, $n^{\prime }=-kT+\tau B$, $B^{\prime }=-\tau n$. Put $w=\sqrt{k^2+{\tau }^2}$. The normal-indicatrix arc-length satisfies $d\bar{s}/ds=w>0$, and its unit tangent is $\overline{T}=(-kT+\tau B)/w$. On the unit sphere its outward normal is $n$, and $\overline{T}\times n=(-kB-\tau T)/w$.

Differentiate $\overline{T}$ in $s$. The derivative of $1/w$ is parallel to $\overline{T}$ and contributes nothing to its projection onto $\overline{T}\times n$. The remaining derivative is $(-k^{\prime }T+{\tau }^{\prime }B-(k^2+{\tau }^2)n)/w$, givingWith the alternative left-normal signed-curvature convention the displayed scalar changes sign, while the area conclusion is unchanged.