What is the restriction on this variation? Shortest distance between two points... The shortest path from Manila to Acapulco by sea called a rumb line. The distance is $$\int^A_M\sqrt{\vec dx}^2 = \int^A_M\sqrt{\vec v^2}dt, \quad \vec v= \dot {\vec x}$$
Now wevary the path, $\vec x(t)$ becomes $\vec x + d \vec x$ and the variation of the path length is
$$\int^A_M\frac{\vec v \cdot \vec dv}{\sqrt{\vec v^2}}dt$$
This has to be zero for all possible variations of the path. This implies a restriction on $d\vec v$. What is the restriction on the variation?
 A: You have the following functional
$$D[\vec{x}(t)]=\int_{t_{A}}^{t_{B}}\Big(\frac{\vec{x}(t)}{dt}\Big)^{1/2}dt$$
You have to minimize that one by requiring
$$\frac{\delta{D}[\vec{x}(t)]}{\delta\vec{x}(t)}=0$$
subject to the Dirichlet boundary conditions
$$x(t_{A})=Akapulko$$
$$x(t_{M})=Manila$$
The variational derrivative is defined as
$$\frac{\delta{D}[\vec{x}(t)]}{\delta\vec{x}(t)}=\lim_{\epsilon\rightarrow0}\frac{d{D}[\vec{x}(t)+\epsilon\vec{y}(t)]}{d\epsilon}$$
For the Dirichlet boundary conditions to be applied you require
$$\delta\vec{x}(t_{A})=\vec{x}(t_{A})+\epsilon\vec{y}(t_{A})=Akapulko$$
and
$$\delta\vec{x}(t_{M})=\vec{x}(t_{M})+\epsilon\vec{y}(t_{M})=Manila$$
From this you can see that you have the following restrictions on the variation of the position vector, mainly
$$\vec{y}(t_{M})=\vec{y}(t_{A})=0$$
I.e. the variation is equal to zero on the boundary. This imply exactly no condition on the derrivatives of the variation. If you have had the Neumann boundary conditions, you then would have them, for sure!)
