Lista 2:

  1. Determine os gradientes nos pontos dados:
\begin{aligned} & g(x, y)=x y^{2}, \quad(2,-1) \\ & g(x, y)=\frac{x^{2}}{2}-\frac{y^{2}}{2}, \quad(\sqrt{2}, 1) \\ & f(x, y)=\sqrt{2 x+3 y}, \quad(-1,2) \\ & f(x, y)=\operatorname{tg}^{-1} \frac{\sqrt{x}}{y}, \quad(4,-2)

Passo 1

O enunciado pede para calcular os vetores gradiente \nabla f ou \nabla g de quatro funções de duas variáveis em pontos específicos. Para isso usaremos a definição de derivada parcial: \displaystyle \frac{\partial f}{\partial x} trata y como constante e \displaystyle \frac{\partial f}{\partial y} trata x como constante. Depois, basta avaliar cada componente no ponto dado.

Passo 2

Para g(x,y)=x\,y^{2} calculamos
\frac{\partial g}{\partial x}=y^{2},\qquad \frac{\partial g}{\partial y}=2\,x\,y.
No ponto (2,-1) temos
\frac{\partial g}{\partial x}(2,-1)=(-1)^{2}=1, \quad \frac{\partial g}{\partial y}(2,-1)=2\cdot2\cdot(-1)=-4.
Logo \nabla g(2,-1)=(1,\,-4).

Passo 3

Para g(x,y)=\tfrac{x^{2}}{2}-\tfrac{y^{2}}{2} temos
\frac{\partial g}{\partial x}=x, \qquad \frac{\partial g}{\partial y}=-y.
No ponto (\sqrt{2},1)
\frac{\partial g}{\partial x}(\sqrt{2},1)=\sqrt{2}, \quad \frac{\partial g}{\partial y}(\sqrt{2},1)=-1.
Logo \nabla g(\sqrt{2},1)=(\sqrt{2},\,-1).

Passo 4

Para f(x,y)=\sqrt{2x+3y} definimos u=2x+3y. Então
\frac{\partial f}{\partial x} =\frac{1}{2\sqrt{u}}\cdot2=\frac{1}{\sqrt{u}}, \quad \frac{\partial f}{\partial y} =\frac{1}{2\sqrt{u}}\cdot3=\frac{3}{2\sqrt{u}}.
Em (-1,2) calculamos u=2(-1)+3\cdot2=4 e \sqrt{u}=2, ficando
\frac{\partial f}{\partial x}(-1,2)=\frac{1}{2} \quad \text{e} \quad \frac{\partial f}{\partial y}(-1,2)=\frac{3}{4}.
Logo \nabla f(-1,2)=\bigl(\tfrac12,\tfrac34\bigr).

Passo 5

Para f(x,y)=\tan^{-1}\!\bigl(\tfrac{\sqrt{x}}{y}\bigr) defina v=\frac{\sqrt{x}}{y}. Então
\frac{\partial f}{\partial v}=\frac{1}{1+v^{2}}, \qquad \frac{\partial v}{\partial x}=\frac{1}{2\sqrt{x}}\cdot\frac{1}{y}, \qquad \frac{\partial v}{\partial y}=-\frac{\sqrt{x}}{y^{2}}.
Assim
\frac{\partial f}{\partial x} =\frac{1}{1+v^{2}}\cdot\frac{1}{2y\sqrt{x}}, \quad \frac{\partial f}{\partial y} =\frac{1}{1+v^{2}}\cdot\Bigl(-\frac{\sqrt{x}}{y^{2}}\Bigr).
No ponto (4,-2) temos \sqrt{x}=2, v=2/(-2)=-1, 1+v^{2}=2. Logo
\frac{\partial f}{\partial x}(4,-2) =\frac{1}{2}\cdot\frac{1}{2\cdot(-2)\cdot2} =-\frac{1}{16}, \quad \frac{\partial f}{\partial y}(4,-2) =\frac{1}{2}\cdot\Bigl(-\frac{2}{4}\Bigr) =-\frac{1}{4}.
Logo \nabla f(4,-2)=\bigl(-\tfrac{1}{16},-\tfrac{1}{4}\bigr).

Resposta

  1. \nabla g(2,-1)=(1,\,-4)
  2. \nabla g(\sqrt{2},1)=(\sqrt{2},\,-1)
  3. \nabla f(-1,2)=\bigl(\tfrac12,\tfrac34\bigr)
  4. \nabla f(4,-2)=\bigl(-\tfrac{1}{16},-\tfrac{1}{4}\bigr)

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