Question

Em um ponto (x,y,z) do espaço, a interpretação física do rotacional do campo vetorial F esta relacionada a tendencia do campo F de produzir rotação naquele ponto. Sabendo disso, considere o campo F (x,y,z)=(2x³+4Z²) i + (X² - Y)J + (Z)K, e assinale a alternativa que contenha o seu rotacional

Answer 1

Resposta:  $\mathrm{rot}(\mathbf{F})=(0,\,8z,\,2x).$

Explicação passo a passo:

Calcular o rotacional do campo vetorial no $\mathbb{R}^3$

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

com

    $\begin{cases}~P(x,\,y,\,z)=2x^3+4z^2\\ ~Q(x,\,y,\,z)=x^2-y\\ ~R(x,\,y,\,z)=z \end{cases}$

O rotacional de $\mathbf{F}$ é dado por

    $\mathrm{rot}(\mathbf{F})=\nabla\times \mathbf{F}\\\\\\=\left(\dfrac{\partial}{\partial x},\,\dfrac{\partial}{\partial y},\,\dfrac{\partial}{\partial z}\right)\times \Big(P(x,\,y,\,z),\,Q(x,\,y,\,z),\,R(x,\,y,\,z)\Big)$

Podemos escrever o cálculo acima na forma de um determinante simbólico:

    $=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\\ P&Q&R \end{vmatrix}$

    $=\begin{vmatrix}\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\\ Q&R \end{vmatrix}\mathbf{i}-\begin{vmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial z}\\\\ P&R \end{vmatrix}\mathbf{j}+\begin{vmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\\\\ P&Q \end{vmatrix}\mathbf{k}$

    $=\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right)\!\mathbf{i}-\left(\dfrac{\partial R}{\partial x}-\dfrac{\partial P}{\partial z}\right)\!\mathbf{j}+\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\!\mathbf{k}$

Substituindo as coordenadas $P,\,Q$ e $R$ do campo vetorial e calculando as derivadas parciais indicadas, temos

    $=\left(\dfrac{\partial}{\partial y}(z)-\dfrac{\partial}{\partial z}(x^2-y)\right)\!\mathbf{i}-\left(\dfrac{\partial}{\partial x}(z)-\dfrac{\partial}{\partial z}(2x^3+4z^2)\right)\!\mathbf{j}+\left(\dfrac{\partial}{\partial x}(x^2-y)-\dfrac{\partial}{\partial y}(2x^3+4z^2)\right)\!\mathbf{k}$

    $=(0-0)\mathbf{i}-(0-8z)\mathbf{j}+(2x-0)\mathbf{k}\\\\\\=(0)\mathbf{i}+(8z)\mathbf{j}+(2x)\mathbf{k}\\\\\\=(0,\,8z,\,2x)\quad\longleftarrow\quad\mathsf{resposta.}$

Dúvidas? Comente.

Bons estudos!