We already have a definition of geodesics which lie in surfaces in $\RR^3$, namely curves whose geodesic curvature vanishes. Recall that \begin{equation} \kappa_g \,ds = d\TT\cdot\NN \end{equation} where $\TT$ is the unit tangent vector to the curve, and $\NN$ is a particular choice of normal vector to the curve. Thus, for $\kappa_g$ to vanish, we must have \begin{equation} \dot\TT \cdot \NN = 0 \end{equation} Since $\TT$ is a unit vector, we know that \begin{equation} \dot\TT \cdot \TT = 0 \end{equation} Since we are in a two-dimensional surface, $\dot\TT$ only has two components. Since the $\TT$-component vanishes, the other component vanishes if and only if the vector itself is zero. Thus, \begin{equation} \kappa_g = 0 \Longleftrightarrow \dot\TT=0 \end{equation} This is almost, but not quite, our new definition of geodesic, which says that $\dot\vv=0$. But \begin{equation} \vv = v\,\TT \end{equation} where $v=|\vv|$ is the speed, so we have \begin{equation} \dot\vv = \dot{v}\,\TT + v\,\dot\TT \end{equation} Since $\dot\TT\perp\TT$, these components are independent, and \begin{equation} \dot\vv = 0 \Longleftrightarrow \dot{v}=0 \hbox{ and } \dot\TT=0 \end{equation} Our new definition of geodesic is therefore equivalent to the old definition, provided we traverse the curve at constant (but not necessarily unit) speed. 1)