O vetor gradiente de f(x,y)= x² + y² em (1,1)?

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\[\begin{align} {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &={\partial (x^2+y^2) \over \partial x}\Bigg|_{(x,y)=(1,1)} \\ &=2x|_{(x,y)=(1,1)} \\ &=2\cdot 1 \\ &= 2 \,\,\,\,(I) \end{align}\]

E o valor de \({\partial f(x,y) \over \partial y}\) no ponto \((1,1)\) é:

\[\begin{align} {\partial f(x,y) \over \partial y}\Bigg|_{(x,y)=(1,1)} &={\partial (x^2+y^2) \over \partial y}\Bigg|_{(x,y)=(1,1)} \\ &=2y|_{(x,y)=(1,1)} \\ &=2\cdot 1 \\ &= 2 \,\,\,\,(II) \end{align}\]

Portanto, pelas equações \((I)\) e \((II)\), o vetor gradiente correspondente é:

\[\begin{align} \nabla f(1,1) &= \Bigg({\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)}, {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)}\Bigg) \\ &=(2,2) \end{align}\]

Concluindo, o vetor gradiente é \(\boxed{\nabla f(1,1)=(2,2)}\).