A parameterized curve $f(t):I\rightarrow M$ is a geodesic of M iff its acceleration is everywhere perpendicular to M, i.e. \begin{equation} \ddot{f}(t)=g(t)N({f}(t)) \end{equation}
Where $g(t):I\rightarrow \mathbb{R}$. Taking the scalar product of both sides of this equation with $N({f}(t))$ we find ...??
$$g(t)=...$$
How can I find the function $g(t)$ where Find the value of the $g(t)$ and the offset by the equation leads
$f(t):I\rightarrow M$ is geodesic iff it satisfies the differential equation
$$\ddot{f}(t)+\dot{N}({f}(t))N({f}(t))=0$$
I found this solution in one of the books but I did not understand how the derivation and the reasons
the book here Geometrical Dynamics of Complex Systems: A Unified Modelling Approach
Can you help by stating the reasons
Thank you very much anyway