Distance between the geodesic and the $z$-axis Let $\gamma$ be a unit-speed curve on the helicoid $$\sigma (u,v)=(u\cos v, u\sin v, v)$$ I have shown that $$\dot u^2+(1+u^2)\dot v^2=1$$ and that if $\gamma$ is a geodesic on $\sigma$ then $$\dot v=\frac{a}{1+u^2}$$ where $a$ is a constant. 
I have also find the geodesics corresponding to $a = 0$ and $a = 1$.  
Suppose that a geodesic $\gamma$ on $\sigma$ intersects a ruling at a point $p$ a distance $D > 0$ from the $z$-axis, and that the angle between $\gamma$ and the ruling at $p$ is $α$, where $0 < α < \frac{\pi}{2}$. 
How could we show that the geodesic intersects the $z$-axis if $D > \cot α$, but that if $D < \cot α$ its smallest distance from the $z$-axis is  $\sqrt{D^2 \sin^2 α − \cos^2 α}$ ? 
 A: This is what I have, which is incomplete and maybe incorrect, since my solution doesn't agree with the problem.
First, $\sigma$ is a $u$-Clairaut patch, i.e. $F = 0$ and $E,G$ only depend on on $u$. By theorem, if $\gamma$ is a geodesic on a $u$-Clairaut patch, and $\phi(t)$ gives the angle between $\gamma$ and $\sigma_u$, then $c = \sqrt{G}\sin\phi$ is constant along $\gamma.$ So, $\gamma$ must stay in the region where $G\geq c^2$. The curves on which $G=c^2$ (if they exist) are called the barrier curves, and $\gamma$ either hits a barrier curve tangentially if the barrier curve is not a geodesic or spirals towards it asymptotically if it is. Since $G = \sigma_v\cdot\sigma_v = u^2+1,$ and $D = |u_0|$ where $p = \sigma(u_0,v_0)$, we have $c = \sqrt{D^2+1}\sin\alpha.$
Also, a $v$-parameter curve $\sigma(u_0,v)$ on a $u$-Clairaut patch is a geodesic if and only if $G_u(u_0) = 0$, which here is only if $u_0 = 0$. If $D>\cot\alpha$, then $c^2>1$, so $u_0\neq 0$ and the barrier curves are not geodesics, so $\gamma$ hits the barrier curve, at which point it will have its minimum distance to the $z$-axis. By definition of the barrier curve, at this point we have $u^2+1 = (D^2+1)\sin^2\alpha$, so the distance to the $z$-axis there is $|u| = \sqrt{D^2\sin^2\alpha-\cos^2\alpha}.$
When $D<\cot\alpha$, we have $c^2<1$, and, since $G\geq 1$ everywhere, there is no barrier curve, so I'm not sure what happens. :(
