Question

$\{\begin{array}{l}x=u-v\\ y=2\sqrt{uv}\end{array}$

Short answer

The required values are mentioned below.

$$\begin{gathered} \dot{\mathrm{S}}=\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{u}}{\hat{\mathrm{e}}}_{\mathrm{u}}+\sqrt{1+\frac{\mathrm{u}}{\mathrm{u}}}\dot{\mathrm{v}}{\hat{\mathrm{e}}}_{\mathrm{v}} \\ \mathrm{L}=\frac{1}{2}\mathrm{m}\left(\left(1+\frac{\mathrm{v}}{\mathrm{u}}\right){\dot{\mathrm{u}}}^2+\left(1+\frac{\mathrm{u}}{\mathrm{v}}\right){\dot{\mathrm{v}}}^2\right)-\mathrm{V}(\mathrm{u},\mathrm{v}) \\ -\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{u}}=\mathrm{m}\left\{\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}\dot{\mathrm{u}}-\frac{\mathrm{v}}{{\mathrm{u}}^2\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{v}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{v}}}^2\right\} \\ -\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{v}}=\mathrm{m}\left\{\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{v}}-\frac{1}{2\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{u}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{v}}}^2\right\}=-\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{v}} \end{gathered}$$

Step-by-step solution

Given Information

The equations are mentioned below.

$\{\begin{array}{l}\mathrm{x}=\mathrm{u}-\mathrm{v}\\ \mathrm{y}=2\sqrt{\mathrm{uv}}\end{array}$

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 the value.

The equations are mentioned below.

$\{\begin{array}{l}\mathrm{x}=\mathrm{u}-\mathrm{v}\\ \mathrm{y}=2\sqrt{\mathrm{uv}}\end{array}$

The formula for position vector s is $\mathrm{s}=(\mathrm{u}-\mathrm{v}){\hat{\mathrm{e}}}_{\mathrm{x}}+2\sqrt{\mathrm{uv}}{\hat{\mathrm{e}}}_{\mathrm{y}}$.

Compute the velocity.

$$\begin{gathered} \dot{\mathrm{s}}=\frac{\partial \mathrm{s}}{\partial \mathrm{u}}\dot{\mathrm{u}}+\frac{\partial \mathrm{s}}{\partial \mathrm{v}}\dot{\mathrm{v}} \\ \frac{\partial \mathrm{s}}{\partial \mathrm{u}}=1.{\hat{\mathrm{e}}}_{\mathrm{x}}+\sqrt{\frac{\mathrm{v}}{\mathrm{u}}}{\hat{\mathrm{e}}}_{\mathrm{y}} \\ \frac{\partial \mathrm{s}}{\partial \mathrm{v}}=-1.{\hat{\mathrm{e}}}_{\mathrm{x}}+\sqrt{\frac{\mathrm{u}}{\mathrm{v}}}{\hat{\mathrm{e}}}_{\mathrm{y}} \end{gathered}$$

Orthogonal metric tensors are diagonal hence, the scalar vectors are given below.

$$\begin{gathered} {\mathrm{h}}_{\mathrm{u}}=\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}} \\ {\mathrm{h}}_{\mathrm{v}}=\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}} \end{gathered}$$

Express the velocity in terms of u and v.

$\dot{\mathrm{s}}=\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{u}}{\hat{\mathrm{e}}}_{\mathrm{u}}+\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}\dot{\mathrm{v}}{\hat{\mathrm{e}}}_{\mathrm{v}}$

Express the Lagrangian in terms of u and v.

$$\begin{gathered} \mathrm{L}=\frac{1}{2}\mathrm{m}{\dot{\mathrm{s}}}^2-\mathrm{V}(\mathrm{u},\mathrm{v}) \\ \mathrm{L}=\frac{1}{2}\mathrm{m}\left(\left(1+\frac{\mathrm{v}}{\mathrm{u}}\right){\dot{\mathrm{u}}}^2+\left(1+\frac{\mathrm{u}}{\mathrm{v}}\right){\dot{\mathrm{v}}}^2\right)-\mathrm{V}(\mathrm{u},\mathrm{v}) \end{gathered}$$

Apply Euler-lagrange equation.

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{u}}}\right)=\frac{\partial \mathrm{L}}{\partial \mathrm{u}} \\ \mathrm{m}\left\{\left(1+\frac{\mathrm{v}}{\mathrm{u}}\right)\ddot{\mathrm{u}}-\frac{\mathrm{v}}{{\mathrm{u}}^2}\dot{{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}}\mathrm{u}}\dot{\mathrm{v}}+\frac{1}{2\mathrm{v}}{\dot{\mathrm{v}}}^2\right\}=-\frac{\partial \mathrm{V}}{\partial \mathrm{u}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{v}}}\right)=\frac{\partial \mathrm{L}}{\partial \mathrm{v}} \\ \mathrm{m}\left\{\left(1+\frac{\mathrm{u}}{\mathrm{v}}\right)\ddot{\mathrm{v}}-\frac{\mathrm{v}}{2\mathrm{u}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{v}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{u}}{2{\mathrm{v}}^2}{\dot{\mathrm{v}}}^2\right\}=-\frac{\partial \mathrm{V}}{\partial \mathrm{v}} \end{gathered}$$

Find the components of acceleration.

$$\begin{gathered} \mathrm{m}\left\{\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}\dot{\mathrm{u}}-\frac{\mathrm{v}}{{\mathrm{u}}^2\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{v}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{v}}}^2\right\}=-\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{u}} \\ \mathrm{m}\left\{\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{v}}-\frac{1}{2\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{u}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{v}}}^2\right\}=-\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{v}} \end{gathered}$$

Hence, the required values are mentioned below.

$$\begin{gathered} \dot{\mathrm{S}}=\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{u}}{\hat{\mathrm{e}}}_{\mathrm{u}}+\sqrt{1+\frac{\mathrm{u}}{\mathrm{u}}}\dot{\mathrm{v}}{\hat{\mathrm{e}}}_{\mathrm{v}} \\ \mathrm{L}=\frac{1}{2}\mathrm{m}\left(\left(1+\frac{\mathrm{v}}{\mathrm{u}}\right){\dot{\mathrm{u}}}^2+\left(1+\frac{\mathrm{u}}{\mathrm{v}}\right){\dot{\mathrm{v}}}^2\right)-\mathrm{V}(\mathrm{u},\mathrm{v}) \\ -\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{u}}=\mathrm{m}\left\{\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}\dot{\mathrm{u}}-\frac{\mathrm{v}}{{\mathrm{u}}^2\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{v}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{v}}{\mathrm{u}}}}{\dot{\mathrm{v}}}^2\right\} \\ -\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{v}}=\mathrm{m}\left\{\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}\dot{\mathrm{v}}-\frac{1}{2\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{u}}}^2+\frac{1}{\mathrm{u}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}\dot{\mathrm{u}}\dot{\mathrm{v}}+\frac{\mathrm{u}}{2\mathrm{v}\sqrt{1+\frac{\mathrm{u}}{\mathrm{v}}}}{\dot{\mathrm{v}}}^2\right\}=-\frac{1}{{\mathrm{h}}_{\mathrm{u}}}\frac{\partial \mathrm{V}}{\partial \mathrm{v}} \end{gathered}$$