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.

Assume that : $$\dot r^2 + \dot z^2=1 \Rightarrow \dot r \ddot r + \dot z \ddot z =0$$
$$det(L_{ij})=0\Rightarrow (\dot r \ddot z-\dot z \ddot z)r\dot z = 0 $$
If $\dot z = 0$ we have :
$$\dot z=0 \Rightarrow \dot r^2=1 \Rightarrow \dot r=\pm 1 $$
If $\dot z\neq 0$ we have :
$$\dot r \ddot z - \dot z \ddot r=0 \Rightarrow \dot r =0 \  \vee \ \dot r \neq 0  $$
$$\dot r \neq 0 \Rightarrow \ddot z = \dot z(\frac{\ddot r}{\dot r ^2} ) \Rightarrow \ddot z =\dot z(\frac{- \dot z \ddot z}{\dot r ^2})\Rightarrow \ddot z= 0$$
sinc if $\ddot z \neq 0 \Rightarrow \dot r^2 + \dot z ^2 =0$ Contradiction 
So if $\dot r \neq 0 \Rightarrow \ddot z =0 \Rightarrow \dot z =0 \Rightarrow \dot r = c$
$$ $$
Note : You can assume that : $\dot r^2 + \dot z ^2 = c$
