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

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 $\mathrm{鈭�}脳\mathrm{F}=0$.So check this condition.

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

Solve the matrix as shown below.

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

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

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

$\mathrm{鈭�}脳\mathrm{F}=0$and role="math" localid="1657350380054" $\mathrm{F}$is conservative.

Now, we need to find a scalar potential $\mathrm{蠒}$such that $\mathrm{F}=鈭�\mathrm{鈭�}\mathrm{蠒}$

$\frac{\mathrm{鈭�}\mathrm{蠒}}{\mathrm{鈭�}\mathrm{x}}=鈭�{\mathrm{F}}_{\mathrm{x}}$

$=鈭�{\mathrm{y}}^2\mathrm{zsinh}鈦�2\mathrm{xz}$

$\frac{\mathrm{鈭�}\mathrm{蠒}}{\mathrm{鈭�}\mathrm{x}}=鈭�{\mathrm{F}}_{\mathrm{y}}$

$=鈭�2{\mathrm{ycosh}}^2鈦�\mathrm{xz}$

$\frac{\mathrm{鈭�}\mathrm{蠒}}{\mathrm{鈭�}\mathrm{z}}=鈭�{\mathrm{F}}_{\mathrm{z}}$

$=鈭�{\mathrm{y}}^2\mathrm{xsinh}鈦�2\mathrm{xz}$

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

$\mathrm{蠒}=鈭�鈭�2{\mathrm{ycosh}}^2鈦�\mathrm{xzdy}$

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

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

$\frac{\mathrm{鈭�}\mathrm{蠒}}{\mathrm{鈭�}\mathrm{x}}=鈭�{\mathrm{y}}^2\mathrm{zsinh}鈦�2\mathrm{xz}+\frac{\mathrm{鈭�}\mathrm{g}}{\mathrm{鈭�}\mathrm{x}}=鈭�{\mathrm{y}}^2\mathrm{zsinh}鈦�2\mathrm{xz}$

$\frac{\mathrm{鈭�}\mathrm{g}}{\mathrm{鈭�}\mathrm{x}}=0鈫�\mathrm{g}(\mathrm{x},\mathrm{Z})=\mathrm{g}\left(\mathrm{Z}\right)$

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

$\frac{\mathrm{鈭�}\mathrm{蠒}}{\mathrm{鈭�}\mathrm{z}}=鈭�{\mathrm{y}}^2\mathrm{xsinh}鈦�2\mathrm{xz}+\frac{\mathrm{鈭�}\mathrm{g}}{\mathrm{鈭�}\mathrm{z}}=鈭�{\mathrm{y}}^2\mathrm{xsinh}鈦�2\mathrm{xz}$

$\frac{\mathrm{鈭�}\mathrm{g}}{\mathrm{鈭�}\mathrm{z}}=0鈫�\mathrm{g}\left(\mathrm{Z}\right)=\mathrm{C}$

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

$\mathrm{蠒}=鈭�{\mathrm{y}}^2{\cosh }^2鈦�\mathrm{xz}$

Hence, the solution is $\mathrm{蠒}=鈭�{\mathrm{y}}^2{\cosh }^2鈦�\mathrm{xz}$.