I am a bit confused with the geometric interpretation of the geodesics. I am new to this subject so there may be mistakes in what is to follow. I know there is a lot of background with connections, vector spaces etc with which I am fine, but the example that got me confused is the following:
Let $c(t)=(t^3,t^3)$ a curve in $\mathbb R^2$ (with the euclidean metric, and the euclidean connection defined by the metric). Then you can show formally that this curve/line is not a geodesic. However, its trace, meaning the line $y=t$ is a geodesic (every line in $\mathbb R^2$ is geodesic with the euclidean metric).
Geometrically geodesics are supposed to be the shortest paths in the manifold, at least locally (although when the manifold is the whole space $\mathbb R^2$ then globally), or in other terms the curves in which the covariant derivative along the curve is $0$. With that geometric interpretation I would expect that the above curve would indeed be a geodesic as its trace is nothing more than a line. It 's easilly shown however that it 's not which got me thinking that the geodesics are somehow dependent of the velocity. What I have in mind is the following question:
Can geodesics be described not as the shortest but as the fastest paths on a manifold? When the velocity is stable, meaning $||c'(t)||=\lambda$ for every t, then obviously shortest=fastest. This however is not true when the velocity is constantly changing, as in $c(t)=(t^3,t^3)$. Thus in my naive interpretation, that curve may be the shortest but not the fastest. Is that interpretation at all valid? and if not is there any other geometric intuition/interpretation behind the example I described?