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According to the Hamilton principle, a variational problem for a functional based on a single function, the Lagrangian, determines the dynamics of a physical system. The Lagrangian may contain all physical information about the system and the forces operating on it.

Given a surface $\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$and subject to no other forces, moves along a geodesic of the surface.

Also given that the potential energy is constant.

Since the geodesics acceleration vector$\stackrel{篓}{\mathrm{r}}$is normal to the surface.

Thus,

$\stackrel{篓}{\mathrm{r}}鈥�\mathrm{dr}=0$,

Where denotes the small displacement tangent to surface $\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$.

As we know that $\mathrm{L}=\mathrm{T}-\mathrm{V}$.

Thus, from the Hamilton鈥檚 principle:

And by the Newton鈥檚 second law $\mathrm{m}\stackrel{篓}{\mathrm{r}}=-鈭�\mathrm{V}$

Let assume that potential energy is due to constraint $\mathrm{f}=0$.

Therefore, the Gradient of $\mathrm{V}$is normal to the surface.

And$鈭�\mathrm{V}鈥�\mathrm{dr}=0$if $\mathrm{dr}$is a displacement tangent to surface.

Then,

$$\begin{gathered} m\stackrel{篓}{r}鈥�dr=鈥�鈭�V路dr;鈭�V鈥�dr=0 \\ \stackrel{篓}{r鈥�dr=0} \end{gathered}$$

So, the geodesics acceleration vector is normal to the surface and gives localid="1664354192539" $\stackrel{篓}{\mathrm{r}}鈥�\mathrm{dr}=0$.