Question

Parabolic cylinder.

Short answer

The required values are mentioned below.

$$\begin{gathered} \stackrel{}{\stackrel{..}{S}=\left({\stackrel{..}{q}}_k+{\Gamma }_{ij}^k\stackrel{}{q_i\stackrel{}{q_j}}\right)\frac{\partial S}{\partial q_k}} \\ \frac{{\partial }^{2}S}{\partial q_i\partial q_j}={\Gamma }_{ij}^k\frac{\partial s}{\partial q_k} \end{gathered}$$

Step-by-step solution

Given Information

A parabolic cylinder.

Definition of cylindrical coordinates.

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.

Find all the values.

$\frac{{\partial }^{2}s}{\partial q_i\partial q_j}={\Gamma }_{ij}^k\frac{\partial s}{\partial q_k}$Let velocity is $\stackrel{\mathbf{·}}{s}=\frac{\partial s}{\partial q_i}{\stackrel{\mathbf{.}}{q}}_i$

Derive the formula mentioned above, equation becomes as follows.

$$\begin{gathered} \stackrel{\mathbf{.}\mathbf{.}}{S}=\frac{\partial S}{\partial q_i}\stackrel{\mathbf{.}\mathbf{.}}{q_i}+\frac{d}{dt}\left(\frac{\partial s}{\partial q_i}\right)\stackrel{\mathbf{.}}{q_i} \\ \stackrel{\mathbf{.}\mathbf{.}}{S}=\frac{\partial S}{\partial q_i}\stackrel{\mathbf{.}\mathbf{.}}{q_i}+\frac{{\partial }^{2}s}{\partial q_i\partial q_j}{\stackrel{.}{q}}_i\stackrel{.}{q_j} \end{gathered}$$

$\frac{{\partial }^{2}s}{\partial q_i\partial q_j}$can be written as $\frac{{\partial }^{2}s}{\partial q_i\partial q_j}={\Gamma }_{ij}^k\frac{\partial s}{\partial q_k}$.

Multiply $\frac{\partial s}{\partial q_i}$on both sides of equation mentioned above.

$$\begin{gathered} {\Gamma }_{ij}^k\frac{\partial s}{\partial q_k}.\frac{\partial s}{\partial q_i}=\frac{{\partial }^{2}s}{\partial q_i\partial q_j}.\frac{\partial s}{\partial q_I} \\ {\Gamma }_{ij}^k{g}_{kl}=\frac{{\partial }^{2}s}{\partial q_i\partial q_j}.\frac{\partial s}{\partial q_I} \\ {\Gamma }_{ij}^k=\frac{{\partial }^{2}s}{\partial q_i\partial q_j}.\frac{\partial s}{\partial q_I}{g}_{IK}^{-1} \\ \stackrel{..}{s}=\left({\stackrel{..}{q}}_k+{\Gamma }_{ij}^k\stackrel{.}{q}\stackrel{.}{{}_i{q}_j}\right)\frac{\partial s}{\partial q_k} \end{gathered}$$.

The formula states that $s=\frac{1}{2}\left(u^2-v^2\right)\hat{e}x+uv\hat{e}y+z\hat{e}z$

$$\begin{gathered} \frac{\partial s}{\partial u}=u{\hat{e}}_x+v{\hat{e}}_y \\ {h}_u=-v{\hat{e}}_x+u{\hat{e}}_y \\ {h}_v=\sqrt{u^2+v^2} \end{gathered}$$

Solve further.

$$\begin{gathered} \frac{\partial s}{\partial u}={\hat{e}}_z \\ {h}_z=1 \\ \stackrel{.}{s}=\stackrel{.}{u}\sqrt{u^2+v^2}{\hat{e}}_u+\stackrel{.}{v}\sqrt{u^2+v^2}{\hat{e}}_v+\stackrel{.}{z}{\hat{e}}_z \end{gathered}$$

Calculate second derivative of s

$$\begin{gathered} \frac{{\partial }^{2}s}{\partial u^2}=\frac{u}{u^2+v^2}\frac{\partial s}{\partial u}-\frac{v}{u^2+v^2}\frac{\partial s}{\partial v} \\ {\Gamma }_{uu}^u=\frac{u}{u^2+v^2} \\ {\Gamma }_{uu}^u=-\frac{u}{u^2+v^2} \\ \frac{{\partial }^{2}s}{\partial u\partial v}=\frac{v}{u^2+v^2}\frac{u}{u^2+v^2}\frac{\partial s}{\partial v} \end{gathered}$$

Solve further.

$$\begin{gathered} {\Gamma }_{uu}^u=\frac{v}{u^2+v^2} \\ {\Gamma }_{uv}^u=-\frac{u}{u^2+v^2} \\ \frac{{\partial }^{2}s}{\partial v^2}=-\frac{u}{u^2+v^2}\frac{\partial s}{\partial u}+\frac{v}{u^2+v^2}\frac{\partial s}{\partial v} \end{gathered}$$

Solve further.

$$\begin{gathered} {\Gamma }_{uv}^u=-\frac{u}{u^2+v^2} \\ {\Gamma }_{uv}^v=\frac{u}{u^2+v^2} \\ \frac{{\partial }^{2}s}{\partial u^2}=\frac{u}{u^2+v^2}\frac{\partial s}{\partial u}-\frac{v}{u^2+v^2}\frac{\partial s}{\partial v} \\ {\Gamma }_{uu}^v=\frac{u}{u^2+v^2} \end{gathered}$$

Solve further.

localid="1659406719619" $$\begin{gathered} {\Gamma }_{uu}^v=-\frac{v}{u^2+v^2} \\ \frac{{\partial }^{2}s}{\partial u\partial v}=\frac{u}{u^2+v^2}\frac{\partial s}{\partial u}+\frac{u}{u^2+v^2}\frac{\partial s}{\partial v} \\ {\Gamma }_{uv}^u=\frac{u}{u^2+v^2} \\ {\Gamma }_{uv}^v=\frac{u}{u^2+v^2} \end{gathered}$$

Solve further.

$$\begin{gathered} \frac{{\partial }^{2}s}{\partial v^2}=-\frac{u}{u^2+v^2}\frac{\partial s}{\partial u}+\frac{v}{u^2+v^2}\frac{\partial s}{\partial v} \\ {\Gamma }_{uv}^u=\frac{u}{u^2+v^2} \\ {\Gamma }_{uv}^v=\frac{u}{u^2+v^2} \end{gathered}$$

Acceleration by components is mentioned below.

$$\begin{gathered} \stackrel{..}{s}={\stackrel{..}{s}}_u{\hat{e}}_u+{\stackrel{..}{s}}_v{\hat{e}}_v+{\stackrel{..}{s}}_z{\hat{e}}_z \\ {\stackrel{..}{s}}_u=\left(\stackrel{..}{u}+{\Gamma }_{uu}^u\stackrel{}{u^2+2{\Gamma }_{uv}^u\stackrel{..}{u}\stackrel{..}{v+}{\Gamma }_{vv}^u\stackrel{.}{v^2}}\right)h_u \\ {\stackrel{..}{s}}_u=\left(\stackrel{}{\stackrel{..}{u}+\frac{u}{u^2+v^2}{\stackrel{.}{u}}^2}+\frac{2v}{u^2+v^2}\stackrel{.}{u}\stackrel{.}{v}-\frac{u}{u^2+v^2}\stackrel{.}{v^2}\right)\sqrt{u^2+v^2} \\ {\stackrel{..}{s}}_u={\left(\stackrel{..}{v}+{\Gamma }_{uu}^v\stackrel{.}{u}+2{\Gamma }_{uv}^v\stackrel{.}{u}\stackrel{.}{v}+{\Gamma }_{vv}^v\stackrel{.}{v^2}\right)}^{}{h}_v \end{gathered}$$

Solve further.

$$\begin{gathered} {\stackrel{..}{s}}_u=\stackrel{..}{u}-\frac{v}{u^2+v^2}{\stackrel{.}{u}}^2+\frac{2u}{u^2+v^2}\stackrel{.}{u}\stackrel{.}{v}+\frac{v}{u^2+v^2}\stackrel{.}{v^2} \\ {\stackrel{..}{s}}_z=\stackrel{..}{z} \end{gathered}$$

The required values are mentioned below.

$$\begin{gathered} \stackrel{..}{s}=\left({\stackrel{..}{q}}_k+{\Gamma }_{ij}^k\stackrel{.}{q}{\stackrel{.}{q}}_j\right)\frac{\partial s}{\partial q_k} \\ \frac{{\partial }^{2}s}{\partial q_i\partial q_j}={\Gamma }_{ij}^k\frac{\partial s}{\partial q_k} \end{gathered}$$