Show that$\mathbf{F}\mathbf{=}{\mathbf{y}}^{\mathbf{2}}\mathbf{zsinh}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{xz}\mathbf{\right)}\mathbf{i}\mathbf{+}\mathbf{2}{\mathbf{ycosh}}^{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{xz}\mathbf{\right)}\mathbf{j}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{xsinh}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{xz}\mathbf{\right)}\mathbf{k}$ is conservative, and find a scalar potential $\mathbf{\phi }$such that $\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{\varphi }$.

It has been given that$\mathrm{F}={\mathrm{y}}^2\mathrm{zsinh}\left(2\mathrm{xz}\right)\mathrm{i}+2{\mathrm{ycosh}}^2\left(\mathrm{xz}\right)\mathrm{j}+{\mathrm{y}}^2\mathrm{xsinh}\left(2\mathrm{xz}\right)\mathrm{k}$

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines its magnitude.

For $\mathrm{F}$to be conservative, the condition required is $\mathrm{\nabla }\times \mathrm{F}=0$.So check this condition.

$\mathrm{\nabla }\times \mathrm{F}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{x}}& \frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}& \frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{z}}\\ {\mathrm{y}}^2\mathrm{zsinh}2\mathrm{xz}& 2{\mathrm{ycosh}}^2\mathrm{xz}& {\mathrm{y}}^2\mathrm{xsinh}2\mathrm{xz}\end{array}\right|$

Solve the matrix as shown below.

$=\left(\right(2\mathrm{yxsinh}2\mathrm{xz})-(2\mathrm{yx}(2\cosh \mathrm{xzsinh}\mathrm{xz})\left)\right)\mathrm{i}$

$\begin{array}{r}-\left(\left(y^2\sinh 2xz+2y^{2}xz\cosh 2xz\right)-\left(y^2\sinh 2xz+2y^{2}xz\cosh 2xz\right)\right)j\\ +\left(\right(2yz(2\cosh xz\sinh xz))-(2yx\sinh 2xz\left)\right)k\end{array}$

If you use the identity $2\cosh \mathrm{xzsinh}\mathrm{xz}=\sinh 2\mathrm{xz},$the first and third components will vanish. on addition, the second component will vanish directly.

$\mathrm{\nabla }\times \mathrm{F}=0$and role="math" localid="1657350380054" $\mathrm{F}$is conservative.

Now, we need to find a scalar potential $\mathrm{\varphi }$such that $\mathrm{F}=-\mathrm{\nabla }\mathrm{\varphi }$

$\frac{\mathrm{\partial }\mathrm{\varphi }}{\mathrm{\partial }\mathrm{x}}=-{\mathrm{F}}_{\mathrm{x}}$

$=-{\mathrm{y}}^2\mathrm{zsinh}2\mathrm{xz}$

$\frac{\mathrm{\partial }\mathrm{\varphi }}{\mathrm{\partial }\mathrm{x}}=-{\mathrm{F}}_{\mathrm{y}}$

$=-2{\mathrm{ycosh}}^2\mathrm{xz}$

$\frac{\mathrm{\partial }\mathrm{\varphi }}{\mathrm{\partial }\mathrm{z}}=-{\mathrm{F}}_{\mathrm{z}}$

$=-{\mathrm{y}}^2\mathrm{xsinh}2\mathrm{xz}$

Start by the $y$-component, integrating the equation of the $\mathrm{y}$-component with respect to $\mathrm{y}$results.

$\mathrm{\varphi }=\int -2{\mathrm{ycosh}}^2\mathrm{xzdy}$

$=-{\mathrm{y}}^2{\cosh }^2\mathrm{xz}+\mathrm{g}(\mathrm{x},\mathrm{z})$

Here, $\mathrm{g}$is an arbitrary function in $\mathrm{x}$and $\mathrm{z}$only. Now, substitute in the $\mathrm{z}$-equation as shown below.

$\frac{\mathrm{\partial }\mathrm{\varphi }}{\mathrm{\partial }\mathrm{x}}=-{\mathrm{y}}^2\mathrm{zsinh}2\mathrm{xz}+\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{x}}=-{\mathrm{y}}^2\mathrm{zsinh}2\mathrm{xz}$

$\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{x}}=0\to \mathrm{g}(\mathrm{x},\mathrm{Z})=\mathrm{g}\left(\mathrm{Z}\right)$

Deduce that $\mathrm{g}$is a function of $\mathrm{z}$only. Now, substitute in $\mathrm{z}$-equation as shown below.

$\frac{\mathrm{\partial }\mathrm{\varphi }}{\mathrm{\partial }\mathrm{z}}=-{\mathrm{y}}^2\mathrm{xsinh}2\mathrm{xz}+\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{z}}=-{\mathrm{y}}^2\mathrm{xsinh}2\mathrm{xz}$

$\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{z}}=0\to \mathrm{g}\left(\mathrm{Z}\right)=\mathrm{C}$

Here, $\mathrm{c}$is an arbitrary constant. Since this constant won't matter in determining set it to be zero.

$\mathrm{\varphi }=-{\mathrm{y}}^2{\cosh }^2\mathrm{xz}$

Hence, the solution is $\mathrm{\varphi }=-{\mathrm{y}}^2{\cosh }^2\mathrm{xz}$.