Answer to Question #268880 in Calculus for Bhuvana

Question

Find the Jacobian ∂(x, y, z)/∂(u, v, w).

(a) u = xy, v = y, w = x + z (b) u = x + y + z, v = x + y − z, w = x − y + z

Answer 1

Solution;

(a)

Given;

u=xy

v=y

w=x+z

Now;

$x=\frac uy=\frac uv$

$y=v$

$z=w-x=w-\frac uv$

$\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} &\frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{vmatrix}$ =$\begin{vmatrix} \frac1v & \frac u{v^2}&0 \\ 0 & 1&0\\ -\frac1v&-\frac{u}{v^2}&1 \end{vmatrix}$ =$\frac1v(1-0)-\frac{u}{v^2}(0-0)+0(0+\frac1v)$

Hence;

$\frac{\partial(x,y,z)}{\partial(u,v,w)}=\frac1v$

(b)

u=x+y+z

v=x+y-z

w=x-y+z

Now;

$v+w=x+y-z+x-y+z=2x$

Hence;

$x=\frac{v+w}{2}$

Also;

$u-v=x+y+z-x-y+z=2z$

Hence;

$z=\frac{u-v}{2}$

Also;

$y=x+z-w$ =$\frac{v+w}{2}+\frac{u-v}{2}-w$

Hence;

$y=\frac{u-w}{2}$

Therefore;

$\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} &\frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{vmatrix}=\begin{vmatrix} 0 &\frac 12&\frac12\\ \frac12 & 0&-\frac{1}{2}\\ \frac12&-\frac12&0 \end{vmatrix}$=$0(0+\frac14)-\frac12(0+\frac14)+\frac12(-\frac14-0)$

Therefore;

$\frac{\partial(x,y,z)}{\partial(u,v,w)}=-\frac14$