Question

If u=y2/2x,v=x/2+y2/2x find d(u,v)/d(x,y)

Answer 1

Given:

The given values are $u=\frac{y^2}{2x}$ and $v=\frac{x}{2} +\frac{y^2}{2x}$.

To find:

We need to find the value of $\frac{\partial(u,v)}{\partial(x,y)}$.

Solution:

We know that, $\frac{\partial(u,v)}{\partial(x,y)}$ is known as Jacobian.

The value of Jacobian is

$=\frac{\partial(u,v)}{\partial(x,y)}=det \ \ \[\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\\end{matrix}\]$

Where det represents determinant of the matrix.

First find the values of $\frac{\partial u}{\partial x} , \frac{\partial u}{\partial y} , \frac{\partial v}{\partial x} \ \text{and}\ \frac{\partial v}{\partial y}$.

$\frac{\partial u}{\partial x} =\[-\frac{{{y}^{2}}}{2{{x}^{2}}}\]$

$\frac{\partial u}{\partial y} =\[\frac{{{2y}}}{2{{x}}}\]$

    $=\[\frac{{{y}}}{{{x}}}\]$

$\frac{\partial v}{\partial x} \[=\frac{1}{2}-\frac{{{y}^{2}}}{2{{x}^{2}}}\]$

$\frac{\partial v}{\partial y} =0+\[\frac{{{2y}}}{2{{x}}}\]$

    $=\[\frac{{{y}}}{{{x}}}\]$

Now substitute the above values in Jacobian.

$\Rightarrow\frac{\partial(u,v)}{\partial(x,y)}=det \ \ \[\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\\end{matrix}\]$

$\[\Rightarrow\frac{\partial(u,v)}{\partial(x,y)}=det\ \ \begin{matrix} -\frac{{{y}^{2}}}{2{{x}^{2}}} & \frac{y}{x} \\ \frac{1}{2}-\frac{{{y}^{2}}}{2{{x}^{2}}} & \frac{y}{x} \\\end{matrix}\]$

Now, find the determinant of above value.

$\Rightarrow\frac{\partial(u,v)}{\partial(x,y)}=-\frac{{{y}^{2}}}{2{{x}^{2}}}\times \frac{y}{x}-\frac{y}{x}\times \left( \frac{1}{2}-\frac{{{y}^{2}}}{2{{x}^{2}}} \right)$

$\Rightarrow \frac{\partial(u,v)}{\partial(x,y)}=-\frac{{{y}^{2}}}{2{{x}^{2}}}\times \frac{y}{x}-\frac{y}{x}\times \frac{1}{2}-\frac{y}{x}\times \left( -\frac{{{y}^{2}}}{2{{x}^{2}}} \right)$

$\Rightarrow \frac{\partial(u,v)}{\partial(x,y)}=-\frac{{{y}^{3}}}{2{{x}^{3}}}-\frac{y}{2x}+\frac{{{y}^{3}}}{2{{x}^{3}}}$

Rearranging the terms,

$\Rightarrow \frac{\partial(u,v)}{\partial(x,y)}=-\frac{{{y}^{3}}}{2{{x}^{3}}}+\frac{{{y}^{3}}}{2{{x}^{3}}}-\frac{y}{2x}$

$& \Rightarrow\frac{\partial(u,v)}{\partial(x,y)}=-\frac{y}{2x}$

Therefore, the value $\frac{\partial(u,v)}{\partial(x,y)}$ of is $-\frac{y}{2x}$.

Answer 2

SOLUTION :

GIVEN

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

TO DETERMINE

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

FORMULA TO BE IMPLEMENTED

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

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

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

EVALUATION

Here

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

So

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

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

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

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

So

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

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

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

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

FINAL RESULT

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

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