Given a covariant derivative operator $\nabla$ in a manifold $M$, if a vector field $X(t)$ is only defined along a curve $\alpha$, it is shown in @malament2012topics page 55 that there exist an operator $D_{\alpha}$ such that if $X$ were the restriction to $\alpha$ of a vector field $\tilde{X}$ defined on an open set then $(D_{\alpha} X)(t)=(\nabla_{\alpha'(t)}\tilde{X} )(\alpha(t))$.
Idea of the proof
Suppose a vector field $V(t)$ defined over a curve $\alpha(t)$. Suppose, too, that in a local chart we have
$$ V(t)=V_1(t) \partial x_1+V_2(t) \partial x_2 $$and
$$ \alpha'(t)=A_1(t) \partial x_1+A_2(t) \partial x_2 $$We cannot calculate $\nabla_{\alpha'(t)} V(t)$ because we need $V$ to be defined in an open set. But if there exists $D_{\alpha}$ satisfying "what it must satisfy", it will make:
$$ D_{\alpha}(V(t))=D_{\alpha}(V_1(t) \partial x_1+V_2(t) \partial x_2)= $$ $$ =D_{\alpha}(V_1(t))\partial x_1+V_1(t)\cdot D_{\alpha}(\partial x_1)+ $$ $$ +D_{\alpha}(V_2(t))\partial x_2+V_2(t)\cdot D_{\alpha}(\partial x_2) $$and this way, it should be
$$ D_{\alpha}(V(t))=\frac{d V_1(t)}{dt}\partial x_1+V_1(t)\cdot \nabla_{\alpha'(t)}(\partial x_1)+ $$ $$ +\frac{d V_2(t)}{dt}\partial x_2+V_2(t)\cdot \nabla_{\alpha'(t)}(\partial x_2) $$And this expression can be used to define $D$.
$\blacksquare$
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Author of the notes: Antonio J. Pan-Collantes
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