Meridians on surface of revolution A curve $\alpha(t)=(r(t),z(t))$ in the $(r,z)$-plane, where $r(t)>0$, is rotated around the $z$-axis. We can parameterise it with $x(t,\phi)=(r(t) \cos\phi,r(t) \sin\phi,z(t))$ for $t \in (a,b)$ and $\phi \in (0,2\pi)$.
Now given the second fundamental form $(L_{ij})$ 
$$\frac{1}{\sqrt{\dot r^2+ \dot z^2}}  \begin{pmatrix} \dot r \ddot z- \dot z \ddot r & 0\\ 0& r \dot z \end{pmatrix}$$
I've got to prove that $\det(L_{ij})=0 $ only then if every meridian is a straight line. Well, calculating the determinant, I get the equation $\dot z \ddot r= \dot r \ddot z$. Now I don't know how to go on. I'd be grateful for any help.

Hints: Using primes instead of dots to denote derivatives with respect to $t$, you have
$$
r' z'' - r'' z' = 0\quad\text{for all $t$.}
\tag{1}
$$


*

*If $r'z'$ is non-vanishing in some interval, (1) is equivalent to
$$
\left[\log\left(\frac{z'}{r'}\right)\right]' = \frac{z''}{z'} - \frac{r''}{r'} = 0.
$$

*If $r'z' = 0$, either $r' = 0$ or $z' = 0$.
Continuity and the first bullet point show that $r'z'$ cannot vanish on a set with empty interior if the profile curve $\alpha$ is regular. 
On the other hand, if $r'z'$ vanishes identically on some interval and $\alpha$ is regular, then either $r'$ or $z'$ vanishes identically.
(These claims leave non-trivial details to check, so I trust this outline doesn't spoil all your fun.)
