Question

calcule o vetor gradiente da funçao

Answer 1

Olá,

Temos a seguinte função:

$\tt \: f (x, y) = y \cdot \: ln \: x$

Vamos calcular as derivadas parciais de f:

$\tt \: \dfrac{ \partial f}{ \partial x} = y. \: \dfrac{1}{x}$

$\tt \: \dfrac{ \partial f}{ \partial y} = 1 \cdot \: ln \: x$

Vamos calcular as derivadas parciais em (1, -3):

$\tt \: \dfrac{ \partial f}{ \partial x} (1, -3) = - 3 \cdot \frac{1}{1} \\ \\ \tt \: \dfrac{ \partial f}{ \partial x} (1, -3) = - 3$

$\tt \: \dfrac{ \partial f}{ \partial y} (1, -3) = 1 \cdot ln \: 1 \\ \\ \tt \: \dfrac{ \partial f}{ \partial y} (1, -3) = 1 \cdot 0 \\ \\ \tt \: \dfrac{ \partial f}{ \partial y} (1, -3) = 0$

Vetor gradiente:

$\tt \nabla f (x, y) = \dfrac{ \partial f}{ \partial x} (x, y) \vec{i} + \dfrac{ \partial f}{ \partial y} (x, y) \vec{j}$

Substituindo os valores:

$\tt \nabla f (1, - 3) = \: - 3 \cdot\vec{i} + 0 \cdot\vec{j} \\ \\ \tt \nabla f (1, - 3) = \: - 3\vec{i}$

Resposta: d)

Answer 2

✅ Após resolver os cálculos, concluímos que o vetor gradiente da superfície aplicado ao ponto "P" é:

         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\ \vec{\nabla} f(1, -3) = (-3, 0)\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

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

Seja os dados:

               $\Large\begin{cases}s: f(x, y) = y\,\ln\,x\\P = (1, -3)\end{cases}$

Para resolver esta questão, devemos:

• Calcular o vetor gradiente da superfície:

          $\Large\displaystyle\begin{gathered} \vec{\nabla} f(x, y) = \langle f_{x}(x, y),\,f_{y}(x, y)\rangle\end{gathered}$

                               $\Large\displaystyle\begin{gathered} = \frac{\partial f}{\partial x}\,\vec{i} + \frac{\partial f}{\partial y}\,\vec{j}\end{gathered}$

                               $\Large\displaystyle\begin{gathered} = y\cdot\frac{1}{x}\,\vec{i} + \ln\,x\,\vec{j}\end{gathered}$

                               $\Large\displaystyle\begin{gathered} = \frac{y}{x}\,\vec{i} + \ln\,x\,\vec{j}\end{gathered}$

          $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{\nabla} f(x, y) = \frac{y}{x}\,\vec{i} + \ln\,x\,\vec{j}\end{gathered}$

• Determinar o vetor gradiente aplicado ao ponto P:

           $\Large\displaystyle\begin{gathered} \vec{\nabla} f(1, -3) = -\frac{3}{1}\,\vec{i} + \ln\,1\,\vec{j}\end{gathered}$

                                   $\Large\displaystyle\begin{gathered} = -3\,\vec{i} + 0\,\vec{j}\end{gathered}$

                                   $\Large\displaystyle\begin{gathered} = (-3, 0)\end{gathered}$

            $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{\nabla} f(1, -3) = (-3, 0)\end{gathered}$

✅ Portanto, o vetor gradiente aplicado ao ponto "P" é:

                     $\Large\displaystyle\begin{gathered} \vec{\nabla} f(1, -3) = (-3, 0)\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$