Question

dentre as alternativas abaixo, selecione a que contenha o gradiente da função f(x,y)=y-x2,

Answer 1

$\displaystyle \vec{\nabla}f=\frac{\partial}{\partial x}y-x^2\hat{i}+\frac{\partial}{\partial y}y-x^2\,\hat{j}=\boxed{2x\,\hat{i}+1\hat{j}}$

Answer 2

✅ Tendo resolvido os cálculos, concluímos que o vetor gradiente é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\: \vec{\nabla} f(x, y)= (-2x,\,1) \:\:\:}}\end{gathered} }$

Seja a função em duas variáveis:

               $\Large\displaystyle\begin{gathered} f(x, y) = y - x^{2}\end{gathered}$

Calculando o vetor gradiente da referida função que representa um superfície no espaço:

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

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

                        $\Large\displaystyle\begin{gathered} = 2\cdot(-1)\cdot x^{2 - 1}\,\vec{i} + 1\cdot y^{1 - 1}\,\vec{j}\end{gathered}$

                        $\Large\displaystyle\begin{gathered} = -2x\,\vec{i} + 1\,\vec{j}\end{gathered}$

                        $\Large\displaystyle\begin{gathered} = (-2x,\,1)\end{gathered}$

✅ Portanto, o vetor gradiente é:

    $\Large\displaystyle\begin{gathered} \vec{\nabla} f(x, y)= (-2x,\,1)\end{gathered}$

   

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

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