Question

Calcule o rotacional do campo vetorial F (x, y,z) = xy² z4 i + (2x²y + z) j + (y³ z²) k e assinale a alternativa correta.

Answer 1

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⠀⠀☞ Tendo calculado as derivadas parciais pudemos encontrar o rotacional do campo vetorial que está de acordo com a opção e). ✅

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• ⠀⠀ O rotacional de um campo vetorial F é o campo vetorial resultante do produto do operador diferencial ∇ por F, ou seja, rot F = ∇ × F. Temos que:

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$\blue{\Large\sf~~\begin{cases}\sf~M = xy^2z^4\\\\ \sf~N = 2x^2y + z\\\\ \sf~P = y^3z^2 \end{cases}}$

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⠀⠀Como já explícito no enunciado, basta então calcularmos as seis derivadas parciais à seguir:

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$\blue{\Large\sf~~\begin{cases}\sf~\dfrac{\partial~M}{\partial y}~\overrightarrow{\sf i}~~=~~\dfrac{\partial~(xy^2z^4)}{\partial y}~\overrightarrow{\sf i}~~=~~2xyz^4\\\\\sf~\dfrac{\partial~M}{\partial z}~\overrightarrow{\sf i}~~=~~\dfrac{\partial~(xy^2z^4)}{\partial z}~\overrightarrow{\sf i}~~=~~4xy^2z^3 \\\\\sf~\dfrac{\partial~N}{\partial x}~\overrightarrow{\sf j}~~=~~\dfrac{\partial~(2x^2y + z)}{\partial x}~\overrightarrow{\sf j}~~=~~4xy \\\\\sf~\dfrac{\partial~N}{\partial z}~\overrightarrow{\sf j}~~=~~\dfrac{\partial~(2x^2y + z)}{\partial z}~\overrightarrow{\sf j}~~=~~1 \\\\\sf~\dfrac{\partial~P}{\partial x}~\overrightarrow{\sf k}~~=~~\dfrac{\partial~(y^3z^2)}{\partial x}~\overrightarrow{\sf k}~~=~~0 \\\\\sf~\dfrac{\partial~P}{\partial z}~\overrightarrow{\sf k}~~=~~\dfrac{\partial~(y^3z^2)}{\partial y}~\overrightarrow{\sf k}~~=~~3y^2z^2 \end{cases}}$

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⠀⠀Com os valores das derivadas parciais podemos agora realizar as operações para encontrar o nosso rotacional:

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$\Large\blue{\sf \left(\dfrac{\partial P}{\partial y} - \dfrac{\partial N}{\partial z}\right) \overrightarrow{\sf i} + \left(\dfrac{\partial M}{\partial z} - \dfrac{\partial P}{\partial x}\right) \overrightarrow{\sf j} + \left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right) \overrightarrow{\sf k}}$

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$\Large\blue{\sf (3y^2z^2 - 1) \overrightarrow{\sf i} + (4xy^2z^3 - 0) \overrightarrow{\sf j} + (4xy - 2xyz^4) \overrightarrow{\sf k}}$

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⠀⠀O que nos leva à opção E). ✌  

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$\Large\green{\boxed{\rm~~~\red{E)}~\gray{rot~F}~\pink{=}~\blue{ (3y^2z^2... }~~~}}$ ✅

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$\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$

⠀⠀☀️ Leia mais sobre rotacional e divergente de campos vetoriais:

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✈ https://brainly.com.br/tarefa/6628979

$\bf\large\red{\underline{\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

⠀⠀⠀⠀☕ $\Large\blue{\text{\bf Bons~estudos.}}$

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($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$

Answer 2

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

      $\large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\textrm{rot}\:\vec{F} = (3y^{2}z^{2} - 1)\vec{i} + (4xy^{2}z^{3})\vec{j} + (4xy - 2xyz^{4})\vec{k}\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

                    $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:E\:\:\:}}\end{gathered} }$

Seja:

 $\Large\displaystyle\begin{gathered} F(x, y, z) = (xy^{2}z^{4}) i + (2x^{2}y + z)\vec{j} + (y^{3}z^{2})k\end{gathered}$

Organizando o campo vetorial, temos:

 $\Large\displaystyle\begin{gathered} \vec{F}(x, y, z) = (xy^{2}z^{4}) i + (2x^{2}y + z)\vec{j} + (y^{3}z^{2})k\end{gathered}$

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:

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

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

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

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

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

                   $\Large\displaystyle\begin{gathered} = (3y^{2}z^{2} - 1)\vec{i} - (0 - 4xy^{2}z^{3})\vec{j} + (4xy - 2xyz^{4})\vec{k}\end{gathered}$

                   $\Large\displaystyle\begin{gathered} = (3y^{2}z^{2} - 1)\vec{i} + (4xy^{2}z^{3})\vec{j} + (4xy - 2xyz^{4})\vec{k}\end{gathered}$      

✅ Portanto, a resposta é:

     $\Large\displaystyle\begin{gathered} \textrm{rot}\:\vec{F} = (3y^{2}z^{2} - 1)\vec{i} + (4xy^{2}z^{3})\vec{j} + (4xy - 2xyz^{4})\vec{k}\end{gathered}$

Saiba mais:

1. https://brainly.com.br/tarefa/52158446
2. https://brainly.com.br/tarefa/1295497
3. https://brainly.com.br/tarefa/38763398