If $M$ is a surface on $\mathbb{R}^3$ then $\frac{D}{dt}V = \frac{d}{dt}V?$

Question

If $M$ is a surface on $\mathbb{R}^3$, then I can imagine that a vector field along a curve $c:I \to M$ as a function $V : I \to \mathbb{R}^3$ such that $V(t) \in T_{c(t)}M.$

Then is it true that $\frac{D}{dt} V(t) = \frac{d}{dt}V(t)$, where $\frac{D}{dt}$ is the covariant derivative?

If yes, why?

Answer 1

You must mean a vector field along a curve in $M$. In that case, you can see the field as $V\colon I \to \Bbb R^3$ with the condition $V(t) \in T_{\alpha(t)}M$ for all $t$, where $\alpha: I \to M$ is the curve. We actually have $$\frac{{\rm D}V}{{\rm d}t}(t) = \left(\frac{{\rm d}V}{{\rm d}t}(t)\right)^{\tan}, $$since although $\Bbb R^3$ is flat, $M$ need not be, and ${\rm D}$ derives from $M$'s Levi-Civita connection, and not $\Bbb R^3$'s.