Change of variable for Jacobian: is there a method?

[fatpotato]

Homework Statement

Choose the correct change of variables among the three possible choices to map $\Sigma$ to a rectangular region.

Relevant Equations

None for now

Hello,

This problem comes just prior to introducing change of variables with Jacobian.

Given the following region in the x-y plane, I have to choose (with justification) the correct change of variables associated, for $u\in [0,2]$ and $v \in [0,1]$.

The correct choice here is a), but I do not understand why...Solution only mentions that a possible change of variable for x would be $x = -(y+1) + u(y+1)$, so we get to solution a).

Here is my question: is there a systematic way of finding the appropriate change of variables? How did the instructor suddenly decide that choosing the previous substition would be a good idea to solve the problem? I have no idea where to begin.

Thank you very much for your help.

 

[SammyS]

Homework Statement:: Choose the correct change of variables among the three possible choices to map $\Sigma$ to a rectangular region.
Relevant Equations:: None for now

Hello,

This problem comes just prior to introducing change of variables with Jacobian.

Given the following region in the x-y plane, I have to choose (with justification) the correct change of variables associated, for $u\in [0,2]$ and $v \in [0,1]$.

The correct choice here is a), but I do not understand why...Solution only mentions that a possible change of variable for x would be $x = -(y+1) + u(y+1)$, so we get to solution a).

You can at least verify that $u=0$ and $u=2$ correspond to the left-hand boundary and the right-hand boundary, respectively, for the region in the x-y plane.
Also, $u=1$ corresponds to the vertical line, $x=0$ which passed through the middle of the region.

Here is my question: is there a systematic way of finding the appropriate change of variables? How did the instructor suddenly decide that choosing the previous substitution would be a good idea to solve the problem? I have no idea where to begin.

Thank you very much for your help.

It's hard to guess how the instructor decided on that substitution (whether it was sudden or not). It's also difficult to guess the reason for choosing that particular region for the new variables, except to observe that the new region is in the 1st Quadrant.

Let's examine choice a), with emphasis on the defined mapping for new variable, $u$. - The mapping for variable $v$ is trivial.

Rewrite
$\displaystyle u=\frac{ x+y+1 }{ y+1 }$

$\quad \displaystyle=\frac{ x }{ y+1 } + \frac{ 1 }{ y+1 }$

$\quad \displaystyle=\frac{ x }{ y+1 } +1$

The effect of that final term of 1 is to shift the rectangle 1 unit in the positive $u$ direction.

Let's drop that final $+ 1$ and change the variable name from $u$ to $n$ so as to avoid confusion. We now have:
$\displaystyle n=\frac{ x }{ y+1 }$.

Solving for $x$ gives : $\displaystyle x=n\left(y+1 \right)$.

If you think of the rather unusual practice of graphing $x$ as a function of $y$, we have that this gives a line with slope $n$ and $y$ intercept of $-1$ .

Any such line with slope $n \in [ -1,~1]$ will intersect the trapezoid at all values of $y \in [0,~1]$ and nowhere else.

 

[fatpotato]

Hello,

Thank you for your answer. I did not think about formulating $u$ as a parameter $n$! This gives a nice parametric linear equation and indeed gives insight on the area's border.

It's hard to guess how the instructor decided on that substitution (whether it was sudden or not).

I guess that there are usual tricks to know and to try on typical problems...Going through a Schaum on change of variables with Jacobian matrices, I noticed the prevalence of certain substitutions, like $x=u+v$, $y=u-v$ and the classical polar coordinates substitution.

 

[benorin]

You might begin by noting that $u$ is the same in all three options so all that we must do is to choose the appropriate $v$, since $x$ is bounded by functions of $y$ we know (b) is incorrect. The remaining two give $v=\pm y$ and being as it is given that $v\in \left[ 0,1\right]$ and $0\le y\le 1$ we must have $v=y$ (throwing out the negative) or change of variables (a).

 

[fatpotato]

You might begin by noting that $u$ is the same in all three options so all that we must do is to choose the appropriate $v$, since $x$ is bounded by functions of $y$ we know (b) is incorrect. The remaining two give $v=\pm y$ and being as it is given that $v\in \left[ 0,1\right]$ and $0\le y\le 1$ we must have $v=y$ (throwing out the negative) or change of variables (a).

This is indeed clever, however there is a twist.

I cropped the solution picture: there should be an additional answer d) for "None of the above". I cropped the last choice because it was not in english. But I still agree with you, it is more efficient to only check for the substitution for $v$ in this case.

 

[benorin]

In that case you need only narrow it down to (a) or (d) and then draw out the transformed boundaries to see which of the two options is correct.