Suppose that $V$ is a connected open subset of $M$, and, furthermore, suppose that the shape operator vanishes throughout $V$, i.e, for every $p \in M$ and $v \in T_pM$, $S_p(v)=0$. Show then that $V$ must be flat, i.e., it's part of a plane.
So far, I know it's enough to show that the normal vector is the same for any two points $p,q$ in $M$. Since $V$ is open and connected, there exists a smooth curve $\gamma:[0,1]\rightarrow M$ with $\gamma(0)=p$ and $\gamma(1)=q$. So we'll define $f:[0,1] \rightarrow R$ by $f(t)=n(\gamma(t))$ and differentiate to get $f'(t)=dn_\gamma(\gamma'(t))$. Now, if we can justify that this last bit is $0$, we're done. My question is, is it $0$ and why?