What is $\nabla_{\gamma'_t}\gamma_t'$ equal to? An attempt

Question

Is it correct? Let $\gamma_t$ be a smooth curve on a Riemannian mfd. Then

$\nabla_{\gamma'_t}\gamma_t'=\gamma''_t+(\gamma_t^i)'(\gamma_t^j)'\Gamma_{ij}^k\partial_k$?

Answer 1

Let $\gamma_t$ be a smooth curve on a Riemannian manifold. Then, writing in local coordinates $\gamma'_t=(\gamma_t^i)'\partial_i$ one has that

$\nabla_{\gamma_t'}\gamma_t'=(\gamma_t^i)''\partial_i+(\gamma_t^i)'(\gamma_t^j)'\Gamma_{ij}^k\partial_k$.