Question

If z = log [(x ^ 2 - y ^ 2) / (x ^ 2 + y ^ 2)] then . x (partial x /partial x )+y (partial z/ partial y) =0​

Answer 1

Answer:

If z = log [(x ^ 2 - y ^ 2) / (x ^ 2 + y ^ 2)] then . x (partial x /partial x )+y (partial z/ partial y) =0

Answer 2

Appropriate Question :-

$\rm \: If \: z = log\bigg[\dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \bigg], \: prove \: thatx\dfrac{\partial z}{\partial x}+ y\dfrac{\partial z}{\partial y}= 0$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: z = log\bigg[\dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \bigg]$

can be rewritten as

$\rm \: {e}^{z} = \dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} }$

$\rm \: {e}^{z} =\dfrac{1 - \dfrac{ {y}^{2} }{ {x}^{2} } }{1 + \dfrac{ {y}^{2} }{ {x}^{2} } }$

$\rm\implies \: {e}^{z} \: is \: a \: homogenous \: function \: of \: degree \: 0$

We Know, Euler's Theorem :- It states that if u is a homogeneous function of x and y of degree n, then

$\boxed{ \rm{ \: x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}= nu \: \: }}$

So, By Euler's Theorem, we have

$\rm \: x\dfrac{\partial }{\partial x} {e}^{z} + y\dfrac{\partial }{\partial y} {e}^{z} = 0 \times {e}^{z}$

$\rm \: x{e}^{z}\dfrac{\partial z}{\partial x}+ y{e}^{z}\dfrac{\partial z}{\partial y} = 0$

$\rm \:{e}^{z}\bigg( x\dfrac{\partial z}{\partial x}+ y\dfrac{\partial z}{\partial y}\bigg)= 0$

$\rm\implies \: x\dfrac{\partial z}{\partial x}+ y\dfrac{\partial z}{\partial y}= 0$

$\rule{190pt}{2pt}$

Additional Information :-

If u is a homogeneous function of x and y of degree n, then

$\boxed{ \rm{ \: x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}= nu \: \: }}$

$\boxed{ \rm{ \: x\dfrac{\partial^{2} u}{\partial {x}^{2} }+ y\dfrac{\partial^{2} u}{\partial {y} \: \partial x}= (n - 1)\dfrac{\partial u}{\partial x} \: \: }}$

$\boxed{ \rm{ \: {x}^{2} \dfrac{\partial^{2} u}{\partial {x}^{2} }+ 2xy\dfrac{\partial^{2} u}{\partial {y} \: \partial x} + {y}^{2} \dfrac{ {\partial }^{2} u}{\partial {y}^{2} } = n(n - 1)u \: \: }}$

$\boxed{ \rm{ \: x\dfrac{\partial^{2} u}{\partial x \: \partial y }+ y\dfrac{\partial^{2} u}{\partial {y}^{2} }= (n - 1)\dfrac{\partial u}{\partial y} \: \: }}$