Derivative first fundamental form

Question

Let $X,Y: I \rightarrow T_{\gamma}\Omega$ be vector fields along a curve $\gamma: I \rightarrow \Omega \subset\mathbb{R}^2.$ Now, in our lecture it was claimed that the derivative $\frac{d}{dt} g(X,Y) = D_{\gamma'} g(X,Y),$ (the latter one is a directional derivative) where $g$ is the metric tensor defined by a submanifold $f : \Omega \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3.$ Unfortunately, we did not show this and I don't see how this can be done.

Answer 1

It's just really a matter of notation. Suppose you have a function $f$ defined on a parametrized curve $\gamma\colon I\to \Omega$. What do you mean when you write $\dfrac d{dt} f$? Of course, you mean $\dfrac d{dt} (f\circ\gamma)$, and, by the chain rule, this is the directional derivative $D_{\gamma'} f$, i.e., the directional derivative of $f$ at $\gamma(t)$ in the direction of $\gamma'(t)$.