Matéria: Matemática
Autor: Anônimo
Perguntado 8 anos atrás

$Qual\ o\ vetor\ rotacional\ desse\ campo\ vetorial?\\\\F(x,y,z)\ =\ xz\ i\ +\ xyz\ j-\ y^{2}\ k$

Respostas

respondido por: Lukyo

Dado o campo vetorial

$\mathsf{\mathbf{F}(x,\,y,\,z)=P(x,\,y,\,z)\,\mathbf{i}+Q(x,\,y,\,z)\,\mathbf{j}+R(x,\,y,\,z)\,\mathbf{k}}$

o rotacional de F pode ser calculado pelo seguinte determinante simbólico:

$\mathsf{rot\,\mathbf{F}=\nabla\times \mathbf{F}}\\\\ \mathsf{rot\,\mathbf{F}=det}\begin{bmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\\\ \mathsf{\frac{\partial}{\partial x}}&\mathsf{\frac{\partial}{\partial y}}&\mathsf{\frac{\partial}{\partial z}}\\\\\mathsf{P}&\mathsf{Q}&\mathsf{R} \end{bmatrix}\\\\\\ \mathsf{rot\,\mathbf{F}=\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right)\mathbf{i}-\left(\dfrac{\partial R}{\partial x}-\dfrac{\partial P}{\partial z}\right)\mathbf{j}+\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathbf{k}}$

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Nesta tarefa, temos

$\mathsf{\mathbf{F}(x,\,y,\,z)=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}}\quad\Rightarrow\quad\left\{\!\begin{array}{l} \mathsf{P=xz}\\\\ \mathsf{Q=xyz}\\\\ \mathsf{R=-y^2} \end{array} \right.$

Calculando o rotacional,

$\mathsf{rot\,\mathbf{F}}$

$\mathsf{=\left(\dfrac{\partial}{\partial y}(-y^2)-\dfrac{\partial}{\partial z}(xyz)\right)\mathbf{i}-\left(\dfrac{\partial}{\partial x}(-y^2)-\dfrac{\partial}{\partial z}(xz)\right)\mathbf{j}+\left(\dfrac{\partial}{\partial x}(xyz)-\dfrac{\partial}{\partial y}(xz)\right)\mathbf{k}}$

$\mathsf{=(-2y-xy)\,\mathbf{i}-(0-x)\,\mathbf{j}+(yz-0)\,\mathbf{k}}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{rot\,\mathbf{F}=-y(2+x)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k}}\end{array}}$

Bons estudos! :-)

Anônimo: Ótima resolução !!! Muito obrigado amigo ! =)

respondido por: solkarped

✅ Após resolver os cálculos, concluímos que o rotacional do referido campo vetorial é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\textrm{rot}\:\vec{F} = -y(2 + x)\,\vec{i} + x\,\vec{j} + yz\,\vec{k}\:\:\:}}\end{gathered}$}$

Seja a função:

$\Large\displaystyle\text{$\begin{gathered} F(x, y, z) = xz\,i + xyz\,j - y^{2}\,k\end{gathered}$}$

Organizando o campo vetorial, temos:

$\Large\displaystyle\text{$\begin{gathered} \vec{F}(x, y, z) = (xz)\vec{i} + (xyz)\vec{j} + (-y^{2})\vec{k}\end{gathered}$}$

Sendo F um campo vetorial em R³, podemos dizer que o rotacional de F - denotado por "rot F" - é o produto vetorial entre o operador diferencial e F, isto é:

$\Large\displaystyle\text{$\begin{gathered} \textrm{rot}\:\vec{F} = \nabla\wedge\vec{F}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \bigg(\frac{\partial}{\partial x},\,\frac{\partial}{\partial y},\,\frac{\partial}{\partial z}\bigg) \wedge(X_{F}\vec{i},\,Y_{F}\vec{j},\,Z_{F}\vec{k})\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\X_{F} & Y_{F} & Z_{F}\end{vmatrix}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix}\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\Y_{F} & Z_{F}\end{vmatrix}\vec{i} - \begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\X_{F} & Z_{F}\end{vmatrix}\vec{j} + \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y}\\X_{F} & Y_{F}\end{vmatrix}\vec{k}\end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered} = \left(\frac{\partial Z_{F}}{\partial y} - \frac{\partial Y_{F}}{\partial z}\right)\vec{i} - \left(\frac{\partial Z_{F}}{\partial x} - \frac{\partial X_{F}}{\partial z}\right)\vec{j} + \left(\frac{\partial Y_{F}}{\partial x} - \frac{\partial X_{F}}{\partial y}\right)\vec{k}\end{gathered}$}$