91影视

The Parabolic cylinder.

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

The velocity and scale factors are given below.

$$\begin{gathered} h_u=\sqrt{u^2+v^2} \\ {h}_v=\sqrt{u^2+v^2} \\ {h}_z=1 \\ \stackrel{藱}{r}=\stackrel{藱}{u}\sqrt{u^2+v^2}{\stackrel{铃�}{e}}_u\stackrel{藱}{v}\sqrt{u^2+v^2}{\stackrel{铃�}{e}}_u+\stackrel{藱}{z}{\stackrel{铃�}{e}}_z \end{gathered}$$

The lagrangian in the parabolic coordinates are given below.

$\begin{array}{l}L=\frac{1}{2}{m}^2-V\left(uV,z\right)\\ L=\frac{1}{2}m\left[\left(u^2+v^2\right){\stackrel{藱}{u}}^2+\left(u^2+v^v\right){\stackrel{藱}{v}}^2+{\stackrel{藱}{z}}^2\right]-V(u,v,z)\end{array}$

Apply Euler-lagrange equation, the above equation becomes as follows.

$$\begin{gathered} \frac{d}{dt}\left(\frac{鈭�L}{鈭�\stackrel{藱}{u}}\right)=m\left[\stackrel{篓}{u}\left(u^2+v^2\right)+\stackrel{藱}{u}\left(2u\stackrel{藱}{u}+2v\stackrel{藱}{v}\right)\right] \\ =m\left(\stackrel{藱}{u}+\stackrel{藱}{v^2}\right)u-\frac{鈭�V}{鈭�u} \end{gathered}$$

Solve further.

Divide the above equation by respective scalar factors.

The components of acceleration are mentioned below.

$$\begin{gathered} a_u=\stackrel{篓}{u}\sqrt{u^2+v^2}+\frac{u}{\sqrt{u^2+v^2}}{\stackrel{藱}{u}}^2+\frac{2v}{\sqrt{u^2+v^2}}\stackrel{藱}{u}\stackrel{藱}{v}-\frac{u}{\sqrt{u^2+v^2}}{\stackrel{藱}{v}}^2 \\ {a}_v=\stackrel{篓}{v}\sqrt{u^2+v^2}+\frac{u}{\sqrt{u^2+v^2}}{\stackrel{藱}{u}}^2+\frac{2v}{\sqrt{u^2+v^2}}\stackrel{藱}{u}\stackrel{藱}{v}-\frac{u}{\sqrt{u^2+v^2}}{\stackrel{藱}{v}}^2 \\ {a}_z=\stackrel{篓}{z} \end{gathered}$$

The values of components of acceleration are mentioned below.