Question

Calcule a derivada direcional de f(x,y,z) = xyz, no ponto (1,1,3) e na direção i + j + k

Answer 1

Primeiro deixamos o vetor unitario:

$\vec{v}=i+j+k\to\hat{v}=\dfrac{i+j+k}{\sqrt3}$

Agora eh soh aplicar a formula:

$\hat{v}=ai+bj+ck\\\\\\D_{\vec{v}}f(x,y,z)=f_x\cdot a+f_y\cdot b+f_z\cdot c$

Logo:

$f_x=yz\\\\f_y=xz\\\\f_z=xy\\\\a=b=c=\dfrac{1}{\sqrt3}\\\\\\D_{\vec{v}}f(x,y,z)=\dfrac{yz+xz+xy}{\sqrt3}\\\\\\D_{\vec{v}}f(1,1,3)=\dfrac{(1)(3)+(1)(3)+(1)(1)}{\sqrt3}=\dfrac{7}{\sqrt3}=\boxed{\dfrac{7\sqrt3}{3}}$

Answer 2

✅ Após resolver os cálculos, concluímos que a derivada direcional da referida função polinomial a partir do ponto "P" na direção do versor de "w" é:

               $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf D_{\hat{w}} f(1, 1, 3) = \frac{7\sqrt{3}}{3}\:\:\:}}\end{gathered} }$

Sejam os dados:

                         $\Large\begin{cases} f(x, y, z) = xyz\\P(1, 1, 3)\\\vec{w} = \vec{i} + \vec{j} + \vec{k}\end{cases}$

Para calcularmos a derivada direcional da função a partir do ponto "P" na direção do vetor "w", devemos realizar os seguintes passos:

• Calcular o vetor gradiente da função.

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

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

                                  $\Large\displaystyle\begin{gathered} = yz\,\vec{i} + xz\,\vec{j} + xy\,\vec{k}\end{gathered}$

                                  $\Large\displaystyle\begin{gathered} = (yz, xz, xy)\end{gathered}$

          $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{\nabla} f(x, y, z) = (yz, xz, xy)\end{gathered}$

• Obter o vetor gradiente da função aplicado ao ponto "P".

           $\Large\displaystyle\begin{gathered} \vec{\nabla} f(1, 1, 3) = (1\cdot3, 1\cdot3, 1\cdot1) = (3, 3, 1)\end{gathered}$

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

• Calcular o versor do vetor "w".

                  $\Large\displaystyle\text{ \begin{gathered} \hat{w} = \frac{\vec{w}}{\parallel \vec{w}\parallel}\end{gathered} }$

                       $\Large\displaystyle\text{ \begin{gathered} = \frac{(1, 1, 1)}{\sqrt{1^{2} + 1^{2} + 1^{2}}}\end{gathered} }$  

                       $\Large\displaystyle\text{ \begin{gathered} = \bigg(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\bigg)\end{gathered} }$

           $\Large\displaystyle\text{ \begin{gathered} \therefore\:\:\:\hat{w}= \bigg(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\bigg)\end{gathered} }$

• Obter a derivada direcional da função a partir do ponto "P" na direção do versor de "w".

        Observe que a derivada direcional pode ser calculada a partir do produto escalar entre o vetor gradiente aplicado ao ponto "P" com o versor de "w", ou seja:

        $\Large\displaystyle\begin{gathered} \bf(I)\end{gathered}$          $\Large\displaystyle\text{ \begin{gathered} D_{\hat{w}} f(x, y, z) = \vec{\nabla} f(x, y, z)\cdot \hat{w}\end{gathered} }$

        Substituindo os dados na equação "I", temos:

         $\Large\displaystyle\text{ \begin{gathered} D_{\hat{w}} f(1, 1, 3) = \vec{\nabla} f(1, 1, 3)\cdot\hat{w}\end{gathered} }$

                                     $\Large\displaystyle\text{ \begin{gathered} = (3, 3, 1)\cdot\bigg(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\bigg)\end{gathered} }$

                                      $\Large\displaystyle\text{ \begin{gathered} = 3\cdot\frac{1}{\sqrt{3}} + 3\cdot\frac{1}{\sqrt{3}} + 1\cdot\frac{1}{\sqrt{3}}\end{gathered} }$

                                      $\Large\displaystyle\text{ \begin{gathered} = \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}}\end{gathered} }$

                                      $\Large\displaystyle\text{ \begin{gathered} = \frac{7}{\sqrt{3}}\end{gathered} }$

                                      $\Large\displaystyle\text{ \begin{gathered} = \frac{7\sqrt{3}}{3}\end{gathered} }$

✅ Portanto, a derivada direcional procurada é:

          $\Large\displaystyle\text{ \begin{gathered} D_{\hat{w}} f(1, 1, 3) = \frac{7\sqrt{3}}{3}\end{gathered} }$

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

Saiba mais:

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