Question

Sobre as questões abaixo, calcule o o gradiente de campo escalar: (PARTE 2)

Answer 1

Resposta:

Veja abaixo.

Explicação passo-a-passo:

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• Essa tarefa é sobre gradiente.

• O "vetor" gradiente, como o nome diz, é uma função vetorial que aponta na direção de maior crescimento de uma função escalar.

• Por definição, o gradiente de uma função f, denotado por grad f é dado por:

        $grad\,f=\vec {\nabla}f=\bigg(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},...} \bigg)$

Sem mais delongas, bora para a solução!

Solução:

13)

$f(x,y)=arc\,tg\,xy$

$\dfrac{\partial f}{\partial x}=\dfrac{1}{1+(xy)^2}\cdot y=\dfrac{y}{1+(xy)^2}$

$\dfrac{\partial f}{\partial y}=\dfrac{1}{1+(xy)^2}\cdot x=\dfrac{x}{1+(xy)^2}$

$\therefore \vec{\nabla}f=\bigg(\dfrac{y}{1+x^2y^2}\,,\dfrac{x}{1+x^2y^2}}\bigg)$

14)

$f(x,y)=\dfrac{2x}{x-y}=\dfrac{u(x,y)}{v(x,y)}$

$\dfrac{\partial f}{\partial x}=\dfrac{u'\cdot v-v'\cdot u}{v^2}=\dfrac{2(x-y)-(-y)\cdot 2x}{(x-y)^2}\\\\\therefore \dfrac{\partial f}{\partial x}=\dfrac{2(x-y)+2xy}{(x-y)^2}$

$\dfrac{\partial f}{\partial y}=\dfrac{u'\cdot v-v'\cdot u}{v^2}=\dfrac{0\cdot(x-y)-(-1)\cdot 2x}{(x-y)^2}\\\\\therefore \dfrac{\partial f}{\partial y}=\dfrac{2x}{(x-y)^2}$

$\therefore \vec{\nabla}f=\bigg(\dfrac{2(x-y)+2xy}{(x-y)^2}\,,\dfrac{2x}{(x-y)^2}}\bigg)$

15)

$f(x,y,z)=2xy+yz^2+ln\,z$

$\dfrac{\partial f}{\partial x}=2y$

$\dfrac{\partial f}{\partial y}=2x+z^2$

$\dfrac{\partial f}{\partial z}=2yz+\dfrac{1}{z}$

$\therefore \vec{\nabla}f=\bigg(2y}\,,2x+z^2}\,,2yz+\dfrac{1}{z}}\bigg)$

16)

$f(x,y,z)=\sqrt{\dfrac{x+y}{z}}=\dfrac{(x+y)^{\frac{1}{2}}}{z^{\frac{1}{2}}}=\dfrac{u}{v}$

$\dfrac{\partial f}{\partial x}=\dfrac{u'\cdot v-v'\cdot u}{v^2}=\dfrac{\frac{1}{2}(x+y)^{-\frac{1}{2}}\cdot1-0\cdot(x+y)^{\frac{1}{2}}}{(z^{\frac{1}{2}})^2}}$

$\therefore \dfrac{\partial f}{\partial x}=\dfrac{1}{2z\sqrt{x+y}}$

$\dfrac{\partial f}{\partial y}=\dfrac{1}{2z\sqrt{x+y}}$

$\dfrac{\partial f}{\partial z}=\dfrac{u'\cdot v-v'\cdot u}{v^2}=\dfrac{0\cdot z^{\frac{1}{2}}-\frac{1}{2}z^{-\frac{1}{2}}\cdot(x+y)^{\frac{1}{2}}}{(z^{\frac{1}{2}})^2}}$

$\dfrac{\partial f}{\partial z}=-\dfrac{1}{2}z^{-\frac{3}{2}}\cdot (x+y)^{\frac{1}{2}}=-\dfrac{\sqrt{x+y}}{2z^{\frac{3}{2}}}$

$\therefore \vec{\nabla}f=\bigg(\dfrac{1}{2z\sqrt{x+y}}\,,\dfrac{1}{2z\sqrt{x+y}}\,,-\dfrac{\sqrt{x+y}}{2z^{\frac{3}{2}}}\bigg)$

17)

$f(x,y,z)=ze^{x^2-y}=u\cdot v$

$\dfrac{\partial f}{\partial x}=u'\cdot v+v'\cdot u=0\cdot e^{x^2-y}+2xe^{x^2-y}\cdot z$

$\therefore \dfrac{\partial f}{\partial x}= 2xze^{x^2-y}$

$\dfrac{\partial f}{\partial y}=u'\cdot v+v'\cdot u=0\cdot e^{x^2-y}-1\cdot e^{x^2-y}\cdot z$

$\therefore \dfrac{\partial f}{\partial y}=-ze^{x^2-y}$

$\dfrac{\partial f}{\partial z}=u'\cdot v+v'\cdot u=1\cdot e^{x^2-y}+0\cdot z$

$\therefore \dfrac{\partial f}{\partial z}=e^{x^2-y}$

$\therefore \vec{\nabla}f=\bigg(2xze^{x^2-y},-ze^{x^2-y}\,,e^{x^2-y}\bigg)$

Continue aprendendo com o link abaixo:

Vetor gradiente

https://brainly.com.br/tarefa/32024990

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