Covariant Derivative of a vector field - Parallel Vector Field I'm having trouble to understand the concept of Covariant Derivative of a vector field.
The definition from doCarmo's book states that the Covariant Derivative $(\frac{Dw}{dt})(t), t \in I$ is defined as the orthogonal projection of $\frac{dw}{dt}$ in the tangent plane.
Does that mean that if $w_0 \in T_pS$ is a vector in the tangent plane at point $p$, then its covariant derivative $Dw/dt$ is always zero? Since $dw_0/dt$ will be parallel to the normal $N$ at point $p$.
Is that correct?
If so, then for a vector field to be parallel, then every vector must be in the tangent plane.
Is that also correct?
Could you explain without using tensors and Riemannian Manifolds? Thank you   
 A: The vector fields you are talking about will all lie in the tangent plane. However the (ordinary) derivative of a vector field (in the tangent plane) does not necessary lie in the tangent plane. Remember that the tangent plane may vary from point to point. This (ordinary) derivative does not belong to the intrinsic geometry of a surface, however its projection back onto the tangent plane will again be an intrinsic concept.
Does this answer you concerns ?
A: Consider that the surface is the plane $OXY.$ Consider the curve $(t,0,0)$ and the vector field $V(t)=t\partial_x.$ You have that its covariant derivative  $\frac{dV}{dt}=\partial_x$is not zero. Note that, even being $N$ constant, the length of $V$ changes. This is the reason, in this case, to have non-zero covariant derivative.
Now, when we say that a vector field is parallel we assume it is tangent to the surface. In any case, if you consider that the orthogonal projection is zero without being tangent, think of the above case of the plane and $V=\partial_x+\partial_z.$ 
