Derivative along a curve Suppose $M$ is a hypersurface of the sphere $S^n \subset \mathbb{R}^{n+1}$, and denote the riemannian connections of $M$, $S^n$ and $\mathbb{R}^{n+1}$ by $\nabla, \overline{\nabla}$ and $\tilde{\nabla}$, respectively.
Given a differentiable curve $\alpha : (-\varepsilon, \varepsilon) \to M$ with $\alpha(0) = p$ and $\alpha'(0) = v \in T_p M$, how can I "explicitly" calculate $\overline{\nabla}_v \alpha'$ in terms of the well known derivatives in $\mathbb{R}^{n+1}$? Is it true that $\tilde{\nabla}_v \alpha' = \alpha''(0)$, so that
$$ \overline{\nabla}_v \alpha' = \operatorname{proj}_{T_p S^n} (\alpha''(0)) = \alpha''(0) - \langle \alpha''(0), p \rangle p \,\text{ ?} $$
I am having trouble understanding covariant derivatives along curves. Any help will be appreciated.
 A: (1) LC connection is a connection satisfying the following : $$X(Y,Z)=(\nabla_XY,Z)+(\nabla_XZ,Y)$$
$$ [X,Y]=\nabla_XY - \nabla_YX$$
where $X,\ Y,\ Z$ are vector field on a neighborhood $U$
Here LC connection is uniquely determined by functions
$\Gamma_{ij}^k$ on $U$ where $$ \nabla_{E_i} E_j =\Gamma_{ij}^k E_k
$$ and $E_i$ is coordinate vector field
So $$ \nabla_XY= (X^i E_i(Y^m) + X^i Y^j\Gamma_{ij}^m )E_m\ \ast $$
(2) Consider a curve $c$ on $U$ where $c(0)=p\in U,\ c'(0)=X(p)$
Hence $$X^i E_i(Y^m)  =\frac{d}{dt} Y^m(c(t))   $$
That is to obtain $\nabla_XY$ at $p$, we need $X(p)$ and $Y(c(t)),\
|t|<\epsilon$ only That is if $Y$ is just a vector field along $c$ then
the covariant derivative is $\frac{D}{dt} Y = c'(0)Y +[c'(0)]^i
Y^j\Gamma_{ij}^m E_m $
(3) $ f: {\bf R}^n\rightarrow ({\bf R}^n,g_0) $ where $g_0$ is
canonical metric is identity so that it is coordinate map. That is
$E_i:=df\ e_i=e_i$ Hence $$ g_0(E_i,E_j) =f^\ast g_0
(e_i,e_j)=g_0(e_i,e_j)=\delta_{ij} $$
Hence $\Gamma_{ij}^k=0$ so that $\ast$ implies that $\frac{D}{dt}
Y=c'(0)Y $ So first answer is right.
(4) Also second is right : Covariant derivative on $S^2$ by using
projection is in fact LC connection on $(S^2,g)$ where $g$ is
induced from $g_0$
A: Yes, your formula is true. In fact, given a vector field $w(t)$ along $\gamma$, projecting $dw/dt$ onto tangent plane is exactly how classical differential geometry define the covariant derivative of the vector field $w(t)$ along curve $\gamma$ (One can check Do Carmo's Differential geometry of curves and surface). In your case, you specifically choose tangent field $w(t)=\alpha'(t)$ .
As a matter of fact, it is not only true for $n$ dimensional surface embedded into $\mathbb R^{n+1}$, but also true for $n$ dimensional surface embedded into $\mathbb R^m$, for any $m>n$. One can study more in this lecture notes, which I really love. The related topic is in section 1.8, 
