Question

Sendo f(x, y , z) = (0, cos xz, - sen xy) um campo vetorial, determine o seu rotacional.

Answer 1

• Seu rotacional será:

$\large\displaystyle\begin{gathered}rot F=(-x\cos(xy)+ x\sin(xz) ) \vec{i}+y\cos (xy)\ \vec{j}-z\sin(xz) \vec{k}\end{gathered}$

Desejamos encontrar o rotacional do campo vetorial f(x, y , z) = (0, cos xz, - sen xy).

Considere um campo vetorial: $\large\displaystyle\begin{gathered} f(x,y,z)=A\cdot \vec{i}+B\cdot \vec{j}+C\cdot \vec{k}\end{gathered}$. Seu rotacional será definido por: $\large\displaystyle\begin{gathered} rot F=\vec{\nabla}\times F\end{gathered}$. Sendo que:

                $\large\displaystyle\begin{gathered}rot F=\left[\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\A&B&C\end{array}\right] \end{gathered}$

Sendo dado no enunciado o campo vetorial f(x, y , z) = (0, cos xz, - sen xy). Logo, seu rotacional será dado por:

       $\large\displaystyle\begin{gathered}rot F=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\0&cosxz&-senxy\end{array}\right| \end{gathered}$

• Para resolver, irei utilizar a regra de sarrus. Logo:

$\large\displaystyle\begin{gathered}rot F=\left|\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\0&cosxz&-senxy\end{array}\right| \left|\begin{array}{ccc}i&j\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} \\0&cosxz\end{array}\right| \end{gathered}$

$\large\displaystyle\begin{gathered}rot F=\left(\frac{\partial}{\partial y}(-\sin xy)\vec{i} +\frac{\partial }{\partial z}(0)\vec{j}+\frac{\partial}{\partial x}(\cos xz)\vec{k}\right)-...\end{gathered}$

         $\large\displaystyle\begin{gathered}...\left(\frac{\partial}{\partial y}(0)\vec{k}+ \frac{\partial}{\partial z}(\cos xz)\vec{i}+\frac{\partial}{\partial x}(-\sin xy)\vec{j}\right)\end{gathered}$

$\large\displaystyle\begin{gathered}rot F=\left((-x\cos (xy))\vec{i}+(-z\sin (xz))\vec{k}\right)-...\end{gathered}$

             $\large\displaystyle\begin{gathered}...-\left( (-x\sin(xz))\vec{i}+(-y\cos (xy))\vec{j}\right)\end{gathered}$

$\large\displaystyle\begin{gathered}rot F=-x\cos(xy) \vec{i}-z\sin(xz) \vec{k}+ x\sin(xz) \vec{i}+y\cos (xy)\ \vec{j}\end{gathered}$

Podemos então colocar em evidência os termos. Ficando por final:

$\large\displaystyle\begin{gathered}rot F=-x\cos(xy) \vec{i}-z\sin(xy) \vec{k}+ x\sin(xz) \vec{i}+y\cos (xy)\ \vec{j}\end{gathered}$

$\large\displaystyle\begin{gathered}rot F=(-x\cos(xy)+ x\sin(xz) ) \vec{i}-z\sin(xz) \vec{k}+y\cos (xy)\ \vec{j}\end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\therefore \boxed{\boxed{\green{rot F=(-x\cos(xy)+ x\sin(xz) ) \vec{i}+y\cos (xy)\ \vec{j}-z\sin(xz) \vec{k}}}}\ \checkmark\end{gathered} }$

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