# Find the Jacobian of the transformation T: (u,v) \rightarrow (x(u,v), y(u,v)) when x = u +...

## Question:

Find the Jacobian of the transformation

{eq}T: (u,v) \rightarrow (x(u,v), y(u,v)) {/eq}

when

{eq}x = u + 2v, \quad y = u + 4v {/eq}

## Jacobian:

The Jacobian of a transformation can refer to either a matrix or the determinant of a matrix, but more often we want the determinant. It is a square matrix composed of all the partials of the transformation:

{eq}\begin{align*} \frac{\partial(x,y)}{\partial(u,v)} &= \left| \begin{matrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{matrix} \right| \\\\ &= \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \end{align*} {/eq}

Jacobians get a lot of use when working with iterated integrals, where they are used as sort of a scaling factor for the transformation.

Let's get the partials first. We have

{eq}\begin{align*} \frac{\partial x}{\partial u} &= \frac{\partial }{\partial u} \left( u+2v \right) \\ &= 1 \end{align*} {/eq}

and

{eq}\begin{align*} \frac{\partial x}{\partial v} &= \frac{\partial }{\partial v} \left( u+2v \right) \\ &= 2 \end{align*} {/eq}

And also

{eq}\begin{align*} \frac{\partial y}{\partial u} &= \frac{\partial }{\partial u} \left( u+4v \right) \\ &= 1 \end{align*} {/eq}

and

{eq}\begin{align*} \frac{\partial y}{\partial v} &= \frac{\partial }{\partial v} \left( u+4v \right) \\ &= 4 \end{align*} {/eq}

So the Jacobian is

{eq}\begin{align*} \frac{\partial(x,y)}{\partial(u,v)} &= \left| \begin{matrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{matrix} \right| \\\\ &= \left| \begin{matrix} 1 & 2 \\ 1 & 4 \end{matrix} \right| \\ &= 4-2 \\ &= 2 \end{align*} {/eq}