Regra da Cadeia - Parte Grifada

por **Cleyson007** » Ter Nov 04, 2014 16:47

Se z = f(x,y), onde x = r² + s² e y = 2rs, encontre $\frac{\partial^2 z}{\partial r\partial s}$ .

Alguém me esclarece da passagem grifada em vermelho?

por **Russman** » Ter Nov 04, 2014 22:54

Primeiramente, a resolução separa as derivadas de modo que

$\frac{\partial^2 z }{\partial r \partial s}= \frac{\partial }{\partial r}\left ( \frac{\partial z}{\partial s} \right )$ .

Mas, como
$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$ ,

então

$\frac{\partial }{\partial r}\left ( \frac{\partial z}{\partial s} \right ) = \frac{\partial }{\partial r} \left (\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \right )$

que, pela regra da soma e do produto, fica

$\frac{\partial^2 z }{\partial r \partial s} = \left (\frac{\partial x }{\partial s} \right )\left [ \frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial x} \right )\right ] +\left (\frac{\partial z }{\partial x} \right )\left ( \frac{\partial^2 x }{\partial s \partial r} \rig\right ) +\left (\frac{\partial y }{\partial s} \right )\left [ \frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial y} \right )\right ] +\left (\frac{\partial z }{\partial y} \right )\left ( \frac{\partial^2 y }{\partial s \partial r}\right )$

Agora, como

$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$

então

$\frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial x} \right ) = \frac{\partial^2 z }{\partial x^2}\frac{\partial x }{\partial r}+\frac{\partial^2 z}{\partial x \partial y }\frac{\partial y}{\partial r}$

e

$\frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial y} \right ) = \frac{\partial^2 z }{\partial y^2}\frac{\partial y }{\partial r}+\frac{\partial^2 z}{\partial x \partial y }\frac{\partial x}{\partial r}$ .

Portanto,

$\frac{\partial^2 z }{\partial r \partial s} = \left (\frac{\partial x }{\partial s} \right )\left [ \frac{\partial^2 z }{\partial x^2}\frac{\partial x }{\partial r}+\frac{\partial^2 z}{\partial x \partial y }\frac{\partial y}{\partial r} \right ] +\left (\frac{\partial z }{\partial x} \right )\left ( \frac{\partial^2 x }{\partial s \partial r} \rig\right ) +\left (\frac{\partial y }{\partial s} \right )\left [ \frac{\partial^2 z }{\partial y^2}\frac{\partial y }{\partial r}+\frac{\partial^2 z}{\partial x \partial y }\frac{\partial x}{\partial r}\right ] +\left (\frac{\partial z }{\partial y} \right )\left ( \frac{\partial^2 y }{\partial s \partial r}\right )$

e, enfim,

$\frac{\partial^2 z }{\partial r \partial s} = \frac{\partial^2 z }{\partial x^2} \left (\frac{\partial x }{\partial s} \frac{\partial x }{\partial r} \right ) +\frac{\partial z }{\partial x} \left ( \frac{\partial^2 x }{\partial s \partial r} \rig\right ) + \left \frac{\partial^2 z}{\partial x \partial y } \left ( \frac{\partial x}{\partial s}\frac{\partial y}{\partial r}+\frac{\partial y}{\partial s}\frac{\partial x}{\partial r} \right ) +\frac{\partial z }{\partial y} \left ( \frac{\partial^2 y }{\partial s \partial r}\right ) + \frac{\partial^2 z }{\partial y^2} \left (\frac{\partial y }{\partial s} \frac{\partial y }{\partial r} \right )$

Os termos entre parenteses são calculáveis pois é dada a forma explícita das funções.

por **Cleyson007** » Ter Nov 04, 2014 23:30

Pode me esclarecer essa parte por favor Russman?

Russman escreveu:
que, pela regra da soma e do produto, fica

$\frac{\partial^2 z }{\partial r \partial s} = \left (\frac{\partial x }{\partial s} \right )\left [ \frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial x} \right )\right ] +\left (\frac{\partial z }{\partial x} \right )\left ( \frac{\partial^2 x }{\partial s \partial r} \rig\right ) +\left (\frac{\partial y }{\partial s} \right )\left [ \frac{\partial }{\partial r} \left ( \frac{\partial z }{\partial y} \right )\right ] +\left (\frac{\partial z }{\partial y} \right )\left ( \frac{\partial^2 y }{\partial s \partial r}\right )$

por **Russman** » Qua Nov 05, 2014 01:23

Primeiro, pela regra da soma

$\frac{\partial }{\partial r} \left (\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \right ) = \frac{\partial }{\partial r} \left (\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}\right)+\frac{\partial }{\partial r} \left(\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \right )$

e, depois, pela regra do produto em casa parcela:

$\frac{\partial }{\partial r} \left (\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}\right) = \left (\frac{\partial x}{\partial s} \right )\left [ \frac{\partial }{\partial r}\left (\frac{\partial z}{\partial x} \right ) \right ] + \left (\frac{\partial z}{\partial x} \right ) \left (\frac{\partial^2 x}{\partial r \partial s} \right )$

$\frac{\partial }{\partial r} \left (\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\right) = \left (\frac{\partial y}{\partial s} \right ) \left[ \frac{\partial }{\partial r}\left (\frac{\partial z}{\partial y} \right ) \right] + \left (\frac{\partial z}{\partial y} \right ) \left (\frac{\partial^2 y}{\partial r \partial s} \right )$

Agora basta somar. Mais claro?