Simple question about geodesic in Riemann manifold I was read my book of Riemannian Geometry and the book says the follow: " A parameterized curve $\gamma:I\to M$ is a geodesic in $t_{0}\in I$ if $\dfrac{D}{dt}\left(\dfrac{d\gamma}{dt}\right)=0$ in the point $t_{0}$ ". I don't understand what's mean this, because the intuition behind geodesic is the curve join two point, why the book uses this definition?Thanks!. 
 A: The one you gave is the standard definition for a parametrized curve $\gamma:I\rightarrow M$ to be geodesic at a point $t_0\in I$. The formal meaning of the definition is that the covariant derivative $\frac{D}{dt}$ of the tangent vector field to the curve is $0$ at $t_0$. We say that a parametrized curve is a parametrized geodesic if it is geodesic at every $t_0\in I$. If $C\subset M$ is a regular connected curve, we say that $C$ is a geodesic of M if for every point $p\in C$ the parametrization of C in a coordinate neighbourhood of $p$ by the arc length is a parametrized geodesic.
Intuitively speaking, a curve $\gamma$ on M is a geodesic if the tangent vector to $\gamma$ is 'always the same' when viewed in different tangent spaces. The intuitive meaning you talk about is that a geodesic connecting two points is the shortest path from one to the other. This is not wrong, but it's a definition related to calculus of variations, while in the study of Riemannian geometry it is more common to use the definition you found on the book.
A: The term 'geodesic in $t_0$' seems quite uncommon to me, but with the definition you wrote down it's not wrong. Usually one talks about geodesic (not mentioning points) and this means that the equation you wrote down for $t_0$ is valid for every $t_0$. 
There are other definitions (e.g. locally shortest curve joining two points), and of course they are all equivalent to each other, but that has to be verified, of course.
A: In ${\bf R}^3$, consider a surface $S$ If $p,\ q\in S$, and if
$f(u,0)=p,\ f(u,1)=q$ where $f(u,v)=(x,y,z)(u,v)$ is a
parametrization around $p,\ q$, then $f(u,\ )$ is a curve from $p$
to $q$ for all $u$. Here we assume that $f(0,\ )$ is a shortest
curve from $p$ to $q$. So if $l(u)$ is a length of a curve $f(u,\
)$, then $$ l(u)=\int_0^1 |f_v| dv $$ $$ l'(0)=\int \frac{
(f_v,f_{uv}) }{ |f_v| } $$
That is $f(0,\ )$ is shortest iff $(f_v,f_{uv})=0$ Here if we
reparametrize so that $|f_v(0,v)|=C,\ (f_u,f_v)=0$, then $$
0=\frac{d}{dv} (f_u,f_v)- (f_u,f_{vv}) $$
Hence we have ${\rm Proj}_{T_{f(0,v)} S} f_{vv} =0 $ which is
equivalent to $\frac{D}{dt}\gamma'=0$ where $\gamma (v)=f(0,v)$.
That is geodesic iff locally shortest curve.
