Question

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

Answer 1

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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