Let $S$ be a surface in $R^3$ with induced Riemannian metric. let $\gamma : I\to S$ be a smooth curve on $S$ and $V$ a vector field tangent to $S$ along $\gamma$ . $V$ can be thought of as a smooth function $V : I \to R^3$ Show that $V$ is parallel if and only if $V'(t)$ is orthogonal to $T_{\gamma(t)}M$, where $V'(t)$ is the usual time derivative of $V$ .
I know that a vector field $V$ along a curve $\gamma$ is parallel if $ \frac{DV}{dt}=0$ i.e the co variant derivative of $V$ is zero. Then I am trying to solve one direction by using the definition of $\frac{DV }{dt}$ but I got stuck there.I need some help.
You can write the vector field in local coordinates as $V=\sum_{j=1}^3 v_j\partial_j$. Thus $$\frac{DV}{dt}=(\overline{\nabla}_\overline{c'} \overline{V})^T=\left(\sum_k(\frac{dv_k}{dt}+\sum_{i,j}\Gamma_{i j}^k v_j \frac{dx_i}{dt})\partial_k \right)^T=\left(\sum_{k=1}^3\frac{dv_k}{dt}\partial_k\right)^T=\left(\frac{DV}{dt}\right)^T$$
The first equality is given by exercise $3$ from chap $2$ of manfredo and the third equality comes from the fact that the Christoffel symbols are zero in Euclidean spaces.
Now, note that you can write
$$\frac{dV}{dt}=(\frac{dV}{dt})^T+(\frac{dV}{dt})^N=\frac{DV}{dt}+(\frac{dV}{dt})^N$$
$(\Rightarrow)$ Supose that $V$ is parallel. The above equation says that $\frac{dV}{dt}=(\frac{dV}{dt})^N$ (has only the normal component) which is perpendicular to $T_{c(t)}M$.
$(\Leftarrow)$ Now, suppose that $\frac{dV}{dt}$ is perpendicular to $T_{c(t)}M$.
As $\frac{DV}{dt}\in T_{c(t)}M$, it follows that
$$\langle \frac{dV}{dt},\frac{DV}{dt} \rangle =0 \iff \langle \frac{DV}{dt}+(\frac{dV}{dt})^N, \frac{DV}{dt} \rangle =0 \iff \langle \frac{DV}{dt} ,\frac{DV}{dt} \rangle =0 \iff \frac{DV}{dt}=0$$