Radial geodesics in a graph of a function I'm trying to figure out how to prove the following claim:
Suppose that $S$ is the graph of a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and every plane containing the $z$-axis intersects $S$ along a geodesic of $S$. Suppose in addition that $f(0)=0$ and $\nabla f(0)=0$. Then, $\nabla f$ is a radial vector field.
I've tried to use some kind of Gauss lemma, in order to verify that  the level sets of $f$ are circles, but I could not make any progress. if someone could help me, I'd appreciate. This is not homework. Thanks in advance.
 A: The most direct (and tedious) way to do this is probably by computing the Christoffel symbols, etc, but here's a geometric argument. I'll use cylindrical coordinates $(r,\theta,z)$ and the fact that the covariant derivative in a submanifold is just the orthogonal projection of the ambient covariant derivative. 
If the radial curve $\gamma_\theta : t \mapsto (r(t), \theta, f(r(t),\theta))$ is a geodesic, then its covariant acceleration in $S$ is zero; so its acceleration in $\mathbb R^3$ must be orthogonal to $S$ - i.e. $T_p S = \ddot \gamma_\theta |_p^\perp$ whenever $p$ is on $\gamma_\theta$. Since this curve stays in the plane given by the constant $\theta$, its acceleration must have no $\theta$ component - that is $\langle \partial_\theta, \ddot \gamma_\theta\rangle = 0$, and so $\partial_\theta \in T_p S$. (I used here that the cylindrical coordinate system has orthogonal coordinate vectors.)
This implies $\nabla f$ has no component in the $\theta$ direction - one way to see this formally is noting that $S$ is a level set of $z - f(x,y)$ and thus $T_p S$ is the set of vectors $v$ such that $\langle v , \partial_z - \nabla f\rangle = 0$. Thus $\partial_\theta \in T_pS \implies \langle \nabla f , \partial_\theta \rangle = 0$.
Since every point on $S$ lies on some radial curve, this shows $\nabla f$ is everywhere radial.
