Question

if u=x+y+z ,y+z=uv,z=uvw,s.t d(x,y,z)/d(u,v,w)=u*u*v

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Answer 1

Step-by-step explanation:

I hope it will be help full for

Answer 2

Answer:    

$\frac{\partial (x,y,z) }{\partial (u,v,v)} = u^{2}v$ or d(x,y,z)/d(u,v,w)=u*u*v

Step-by-step explanation:

Given Equations are, u= x+ y + z --(1)

                                   y + z = u × v, --(2)

                                   z= u × v × w --(3)

The Question here indicates the use of Jacobian concept:

                                 $\frac{\partial (x,y,z)}{\partial (u,v,w)} = \left[\begin{array}{ccc}\frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} &\frac{\partial x}{\partial w} \\\frac{\partial y}{\partial u} &\frac{\partial y}{\partial v} &\frac{\partial y}{\partial w} \\\frac{\partial z}{\partial u} &\frac{\partial z}{\partial v} &\frac{\partial z}{\partial w} \end{array}\right]$

In the given variables, the variables are depended as denoted :

                                  (x,y,z) --> (u,v,w)

By solving the given equations we can write x in terms of u ,v, w .

(1) - (2) ⇒ x= u- u × v

From (2) and (3) we write, uv= y+uvw ⇒ y= u× v-(u ×v× w)

and z= u× v× w

Let us substitute the derived x, y ,z values in the Jacobian formula :

$\frac{\partial (x,y,z)}{\partial (u,v,w)} = \left[\begin{array}{ccc}\frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} &\frac{\partial x}{\partial w} \\\frac{\partial y}{\partial u} &\frac{\partial y}{\partial v} &\frac{\partial y}{\partial w} \\\frac{\partial z}{\partial u} &\frac{\partial z}{\partial v} &\frac{\partial z}{\partial w} \end{array}\right]$

$\frac{\partial x}{\partial u}$ = $\frac{\partial (u- u \times v)}{\partial u}$ = 1-v

$\frac{\partial x}{\partial v}$ = $\frac{\partial (u- u\times v)}{\partial v}$= -u

$\frac{\partial x}{\partial w}$ = $\frac{\partial u-u \times v}{\partial w}$=0

$\frac{\partial y}{\partial u}$ = $\frac{\partial (u\times v)-(u \times v \times w)}{\partial u}$ = v- v× w

$\frac{\partial y}{\partial v}$ =$\frac{\partial (u\times v)-(u \times v \times w)}{\partial v}$ =u-  u× w

$\frac{\partial y}{\partial w}$ = $\frac{\partial (u\times v)-(u \times v \times w)}{\partial w}$ = - u× v

$\frac{\partial z}{\partial u}$ =$\frac{\partial u \times v \times w}{\partial u}$ = v× w

$\frac{\partial z}{\partial v}$ = $\frac{\partial u \times v \times w}{\partial v}$ = u× w

$\frac{\partial z}{\partial w}$ = $\frac{\partial u \times v \times w}{\partial w}$ = u× v

Now substitute all the values in the Jacobian matrix;

   =       $\left[\begin{array}{ccc}(1-v)&-u&0\\v- v\times w &u- u\times w&- u\times v\\v\times w &u\times w & u\times v\end{array}\right]$

   = (1-v) [ (u-uw)(uv) - (-uv) (uw)] -(-u) [ v-vw(uw) - (u-uw)(vw)]

  =  (1-v) [$u^{2}v$ - $u^{2}v w$ + $u^{2}vw$] +u[vuw -$w^{2}uv$] -(uvw +$w^{2}vu$)]

=$u^{2}v -u^{2}v w+u^{2}vw -u^{2}v^{2} +u^{2}v^{2}w -u^{2}v^{2}w +u^{2}v}w- u^{2}w^{2}v- u^{2} vw + w^{2}vu^{2}$

  = $u^{2}v$

Therefore,    $\frac{\partial (x,y,z)}{\partial (u,v,w)}= u^{2}v$

Hence proved.