Question

If u+v=ex cosy and u-v=ex sin y the value of j(u,v/x,y)is

Answer 1

Given:

Given that, $\[u-v={{e}^{x}}\text{ siny}\]$ and $\[u+v={{e}^{x}}\text{ cosy}\]$.

To find:

We need to find the value of $\[J\left( \frac{u,v}{x,y} \right)\]$.

Solution:

In vector calculus, the Jacobean matrix  of a vector-valued function in several variables is that the matrix of all its first-order partial derivatives. When this matrix is square, that is, when the function takes the identical number of variables as input because the number of vector components of its output, its determinant is observed because the Jacobian determinant. Both the matrix and (if applicable) the determinant are often observed simply because the Jacobian in literature.

We know that, $\[J\left( \frac{u,v}{x,y} \right)\text{ = }\frac{\partial \left( u,v \right)}{\partial \left( x,y \right)}\text{ = }\left| \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\\end{matrix} \right|\]$.

But we have to find $u$ and $v$ first using given equation,

$\[u-v={{e}^{x}}\text{ siny}\]$......(1)

$\[u+v={{e}^{x}}\text{ cosy}\]$.......(2)

Adding equation 1 and 2, we get

$\[u\text{ = }\frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y+\sin y \right)\]$

Subtracting equation 2 and 1, we get

$v= \frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y-\sin y \right)$

When differentiating the above two equation, we will get the values of

$\[\frac{\partial u}{\partial x},\[\frac{\partial v}{\partial x},\[\frac{\partial u}{\partial y},\[\frac{\partial u}{\partial y}.$

$& \frac{\partial u}{\partial x}\text{= }\frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y+\sin y \right) \\ & \\ & \frac{\partial v}{\partial x}\text{= }\frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y-\sin y \right) \\ & \\ & \frac{\partial u}{\partial y}\text{= }\frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y-\sin y \right) \\ & \\ & \frac{\partial u}{\partial y}\text{= }\frac{\text{ }{{\text{e}}^{x}}}{2}\left( -\cos y-\sin y \right)$

Substituting these values in the above formula, we get

$\[J\left( \frac{u,v}{x,y} \right)\text{ = }\frac{\partial \left( u,v \right)}{\partial \left( x,y \right)}\text{ = }\left| \begin{matrix} \frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y+\sin y \right) & \frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y-\sin y \right) \\ \frac{\text{ }{{\text{e}}^{x}}}{2}\left( \cos y-\sin y \right) & \frac{\text{ }{{\text{e}}^{x}}}{2}\left( -\cos y-\sin y \right) \\\end{matrix} \right|\]$

                           $\[=\left( \frac{{{e}^{x}}}{2} \right)\left( \frac{{{e}^{x}}}{2} \right)\left\{ -{{\left( \cos y+\sin y \right)}^{2}}-{{\left( \cos y-\sin y \right)}^{2}} \right\}\]$

After expanding & solving, we get

                            $\[=\text{ -}\frac{{{e}^{2}}x}{2}\]$