91影视

The given equation is$\mathrm{V}=\mathrm{i}\left({\mathrm{ze}}^{\mathrm{zy}}+\mathrm{xsin}鈦�\mathrm{zx}\right)+\mathrm{jxcos}鈦�\mathrm{xz}鈭�\mathrm{kzsin}鈦�\mathrm{zx}$

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.

In this problem find vector whose curl gives the vector

$\mathrm{V}=\mathrm{i}\left({\mathrm{ze}}^{\mathrm{zy}}+\mathrm{xsin}鈦�\mathrm{zx}\right)+\mathrm{jxcos}鈦�\mathrm{xz}鈭�\mathrm{kzsin}鈦�\mathrm{zx}$

The components of vector are given in terms of the derivatives of$\mathrm{A}.$

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{z}}}{\mathrm{鈭�}\mathrm{y}}鈭�\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{y}}}{\mathrm{鈭�}\mathrm{z}}={\mathrm{ze}}^{\mathrm{zy}}+\mathrm{xsin}鈦�\mathrm{Zx}$

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{x}}}{\mathrm{鈭�}\mathrm{z}}鈭�\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{z}}}{\mathrm{鈭�}\mathrm{x}}=x\cos 鈦�\mathrm{xz}$

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{y}}}{\mathrm{鈭�}\mathrm{x}}鈭�\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{x}}}{\mathrm{鈭�}\mathrm{y}}=\mathrm{Zsin}鈦�\mathrm{zx}$

Observe the second and third equations.

role="math" localid="1657347819231" ${\mathrm{A}}_{\mathrm{x}}=\sin 鈦�\mathrm{xz}$

${\mathrm{A}}_{\mathrm{y}}=\cos 鈦�\mathrm{xz}$

By substituting those solutions into the first equation, we get,

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{z}}}{\mathrm{鈭�}\mathrm{y}}鈭�(鈭�\mathrm{xsin}鈦�\mathrm{zx})={\mathrm{ze}}^{\mathrm{yz}}+\mathrm{xsin}鈦�\mathrm{zx}$

Solve further.

$\frac{\mathrm{鈭�}{\mathrm{A}}_z}{\mathrm{鈭�}\mathrm{y}}={\mathrm{zee}}^{\mathrm{yz}}$

The solution can be written as ${\mathrm{A}}_{\mathrm{z}}={\mathrm{e}}^{\mathrm{zy}}$

Then the total solution becomes $\mathrm{V}=\mathrm{i}\left({\mathrm{ze}}^{\mathrm{zy}}+\mathrm{xsin}鈦�\mathrm{zx}\right)+\mathrm{jxcos}鈦�\mathrm{xz}鈭�\mathrm{kzsin}鈦�\mathrm{zx}$.

Here,$\mathrm{u}$ is an arbitrary function.

Hence, the vector field derived is $\mathrm{A}=\mathrm{isin}鈦�\mathrm{zx}+\mathrm{jcos}鈦�\mathrm{zx}+{\mathrm{ke}}^{\mathrm{zy}}+\mathrm{鈭�}\mathrm{u}$.