= Solution
For fixed $u$, varying $v$ traces a unit circle in the radial-vertical plane centred at
$$
(2\cos u,2\sin u,u).
$$
Thus $\Sigma_1$ is a unit tube whose centre winds helically around the $z$-axis. Since
$$
\sigma(t,\pi)=(\cos t,\sin t,t)=\gamma(t),
$$
the given curve is its inner helix.
Take $\Sigma_2$ to be the circular cylinder $x^2+y^2=1$. Along $v=\pi$,
$$
\sigma_u=(-\sin u,\cos u,1),
\qquad
\sigma_v=(0,0,-1),
$$
so their span is the tangent plane of that cylinder. The two surfaces are therefore tangent along $\gamma$. Moreover,
$$
\ddot\gamma(t)=(-\cos t,-\sin t,0),
$$
which is normal to the cylinder, so $\gamma$ is a geodesic of $\Sigma_2$. Part (ii) now proves that it is a geodesic of $\Sigma_1$. This is the <cylindrical-helix tangency construction>.
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