Question

if u=y2/2 x v=x/2+y 2/2x find ∂(u v)/ ∂(x y)​

Answer 1

$\displaystyle \sf{ } \: \: \: \frac{ \partial(u,v)}{\partial(x,y)} = - \frac{y}{2x}$

Given :

$\displaystyle \sf{}u = \frac{ {y}^{2} }{2x} \: \: \: and \: \: v = \frac{x}{2} + \frac{ {y}^{2} }{2x}$

To find :

$\displaystyle \sf{ } \frac{ \partial(u,v)}{\partial(x,y)}$

Concept :

If u and v are functions of two independent variables x and y then

$\displaystyle \sf{ } \frac{ \partial(u,v)}{\partial(x,y)}$

$\displaystyle \sf = \begin{vmatrix} \frac{ \partial u}{ \partial x} &\frac{ \partial u}{ \partial y} \\ \frac{ \partial v}{ \partial x} & \frac{ \partial v}{ \partial y} \end{vmatrix}$

This is called JACOBIAN of u and v with respect to x and y

Solution :

Step 1 of 3 :

Write down the given functions

Here it is given that

$\displaystyle \sf{}u = \frac{ {y}^{2} }{2x} \: \: \: and \: \: v = \frac{x}{2} + \frac{ {y}^{2} }{2x}$

Step 2 of 3 :

Find the partial derivatives of u and v

$\displaystyle \sf{}u = \frac{ {y}^{2} }{2x} \: \: \: and \: \: v = \frac{x}{2} + \frac{ {y}^{2} }{2x}$

So

$\displaystyle \sf{} \frac{ \partial u}{ \partial x} = - \frac{ {y}^{2} }{2 {x}^{2} }$

$\displaystyle \sf{} \frac{ \partial u}{ \partial y} = \frac{y}{x}$

$\displaystyle \sf{} \frac{ \partial v}{ \partial x} = \frac{1}{2} - \frac{ {y}^{2} }{2 {x}^{2} }$

$\displaystyle \sf{} \frac{ \partial v}{ \partial y} = \frac{y}{x}$

Step 3 of 3 :

Find the Jacobian

The Jacobian

$\displaystyle \sf{ } \frac{ \partial(u,v)}{\partial(x,y)}$

$\displaystyle \sf = \begin{vmatrix} \frac{ \partial u}{ \partial x} &\frac{ \partial u}{ \partial y} \\ \\\frac{ \partial v}{ \partial x} & \frac{ \partial v}{ \partial y} \end{vmatrix}$

$= \displaystyle \sf \begin{vmatrix} - \frac{ {y}^{2} }{2 {x}^{2} } & \frac{y}{x} \\ \\ \frac{1}{2} - \frac{ {y}^{2} }{2 {x}^{2} } & \frac{y}{x} \end{vmatrix}$

$= \displaystyle \sf \begin{vmatrix} - \frac{ {y}^{2} }{2 {x}^{2} } & \frac{y}{x} \\ \\ \frac{1}{2} & 0 \end{vmatrix} \: \sf{} \: \: using \: \: \: R_2'= R_2 - R_1$

$= \displaystyle \sf{} - \frac{y}{2x}$

Thus we have

$\boxed{\displaystyle \sf{ } \: \: \: \frac{ \partial(u,v)}{\partial(x,y)} = - \frac{y}{2x} \: \: \: }$

Answer 2

Answer:

Step-by-step explanation: