Do Problem $\mathbf{2}$in spherical coordinates.

An ordinary first-order differential equation that is linear in the independent variable and unknown function but not solved for the derivative is termed as Lagrange's equation.

A function that equals the difference between potential and kinetic energy and characterises the state of a dynamic system in terms of position coordinates and their time derivatives is termed as Lagrangian.

Given in the problem 2, Potential field is$\mathrm{V}\left(\mathrm{r},\mathrm{\theta },\mathrm{\varphi }\right)$ and mass is .

To obtain the kinetic energy, the velocity in spherical coordinate should be mentioned.

The velocity vector will be given by:

Then the kinetic energy will be given by:

According to the definition of the Lagrangian,

So, the lagrangian is

Since the lagrangian has three degrees of freedom:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{r}}}-\frac{\partial \mathrm{L}}{\partial \mathrm{r}}=0$

Then the calculation of the required derivation will be:

Then the Euler equation will be:

The first Lagrange equation will be:

The Euler equation of $\mathrm{\theta }$degree of freedom will be given by:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{\theta }}}-\frac{\partial \mathrm{L}}{\partial \mathrm{\theta }}=0$

Then the calculation of the required derivation will be:

Then the Euler equation will be:

Now dividing the above Euler equation by localid="1664365691504" $\mathrm{r}$, then the result will be:

So, the second Lagrange equation is

Now for the $\mathrm{\varphi }$degree of freedom, the Euler equation will be given by:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{\varphi }}}-\frac{\partial \mathrm{L}}{\partial \mathrm{\varphi }}=0$

Then the required derivative will be:

Then the Euler equation will be:

Divide the above Euler equation by$\mathrm{r}\mathrm{and}\sin \mathrm{\theta }$, the result will be

So, the third Lagrange’s equation is