Parabolic cylinder.

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.

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

The lagrangian in the parabolic coordinates are given below.

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

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

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{u}}\right)=m\left[\ddot{u}\left(u^2+v^2\right)+\dot{u}\left(2u\dot{u}+2v\dot{v}\right)\right] \\ =m\left(\dot{u}+\dot{v^2}\right)u-\frac{\partial V}{\partial u} \end{gathered}$$

Solve further.

Divide the above equation by respective scalar factors.

The components of acceleration are mentioned below.

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

The values of components of acceleration are mentioned below.