Question

seja $f(x,y) = ln(4 -x^{2} -y^{2} )$ . O vetor gradiente de f(-1,2) corresponde a alternativa:


1) Grad(-1,2) = (-2,4)
2) Grad(-1,2) = (4,-2)
3) Grad(-1,2) = (-4,-2)
4) Grad(-1,2) = (-2,-4)

Answer 1

$\texttt{Olhar para baixo}$

$f(x,y) = \ln(4 - {x}^{2} - {y}^{2} ) \\ \\ \dfrac{ \partial f}{ \partial x} ( - 1,2) = \ln(4 - {x}^{2} - {y}^{2} ) = - \cfrac{2x}{4 - {x}^{2} - {y}^{2} } \\ \\ \dfrac{ \partial f}{ \partial x}(-1;2) =- \dfrac{2(-1)}{4-(-1)^2-2^2}= \dfrac{\partial f}{\partial x} (-1,2) =-\dfrac{-2}{4-1-4}=-\dfrac{-2}{-1} \\ \\ \boxed{ \dfrac{ \partial f}{ \partial x}(-1;2) =-2}$

$\dfrac{ \partial f}{ \partial y} ( - 1,2) = \ln(4 - {x}^{2} - {y}^{2} ) = - \cfrac{2y}{4 - {x}^{2} - {y}^{2} } \\ \\ \dfrac{ \partial f}{ \partial y}(-1;2) =- \dfrac{2(2)}{4-(-1)^2-2^2}= \dfrac{\partial f}{\partial y} (-1,2) =-\dfrac{4}{4-1-4}=- \dfrac{4}{-1}\\ \\ \boxed{ \dfrac{ \partial f}{ \partial y}(-1;2) =4}$

$\rule{12cm}{0.1mm}$

$1) \texttt{Grad}(-1,2)= (-2,4)\checkmark$