# Solve u = 2x - 3y & v = -x + y in terms of u & v, then find Jacobian.

1. Apr 14, 2005

### VinnyCee

Solve u = 2x - 3y & v = -x + y in terms of x & y, then find Jacobian.

Here is the problem:

Solve the system $$u = 2x\;-\;3y,\;\;v = -x\;+\;y$$ for x and y in terms of u and v. Then find the Jacobian $$\frac{\partial\left(x,\;y\right)}{\partial\left(u,\;v\right)}$$.

Find the image under the trasformation $$u = 2x\;-\;3y,\;\;v = -x\;+\;y$$ of the parallelogram $$R$$ in the xy-plane with boundries $$x = -3,\;\;x = 0,\;\;y = x\;and\;y = x\;+\;1$$

Here is what I have:

$$y = -u\;-\;2v\;and\;x = -u\;-\;3v$$ I am not sure of these answers.

Assuming those are correct, I get the Jacobian to be $$-2$$. This is where I am stuck, how do I do the second part? I can get the $$v-values$$ to be $$0\;and\;1$$ here,

$$y = x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y = x\;+\;1$$
$$-2v\;-\;u = -3v\;-\;u\;\;\;\;\;-2v\;-\;u = -3v\;-\;u\;+\;1$$
$$v = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;v = 1$$

But when I try and do the last part for the u values, I get $$u = u\;+\;3$$, which is impossible becaus then $$0 = 3$$.

Last edited: Apr 14, 2005
2. Apr 14, 2005

### VinnyCee

Did anyone check these?

Please, this is the last problem and probably the toughest. Thank you all in advance. :surprised

3. Apr 14, 2005

### HallsofIvy

Staff Emeritus
Yes, you solved correctly for x and y, but the Jacobian is NOT -2. How did you get that?

"Find the image under the trasformation of the parallelogram in the xy-plane with boundries $$x = -3,\;\;x = 0,\;\;y = x\;and\;y = x\;+\;1$$."

The boundaries do NOT neccessarily convert to "u= constant" or "v= constant".

The line x= -3, y anything, converts to u= 2(-3)-3y, v= -(-3)+ y= 3+ y so y= v- 3.
Then u= -6- 3(v- 3)= -6- 3v+ 9= -3v+ 3. One boundary is u= -3v+ 3.

The line x= 0 converts to u= 2(0)- 3y, v= -(0)+ y so y= v. Then u= -3(v).
Another boundary is u= -3v. (Which is, of course, parallel to the previous line.)

The line y= x converts to u= 2x- 3(x)= -x, v= -x+ (x)= 0. One boundary is v= 0.
(Since x can be anything, u can be anything.)

The line y= x+1 converts to u= 2x-3(x+1)= 2x-3x- 3= -x- 3, v= -x+ (x+1)= 1.
The last boundary is v= 1 (since x can be anything, u can be anything). That line is, of course, parallel to v= 0.

4. Apr 14, 2005

### VinnyCee

How I got the Jacobian

I used

$$\frac{\partial x}{\partial u}\;\frac{\partial y}{\partial v}\;-\;\frac{\partial y}{\partial u}\;\frac{\partial x}{\partial v}$$

with the earlier equations solved for $$x$$ and $$y$$ and got

$$(-1)(-1) - (-1)(-3) = 1 - 3 = -2$$

What do I do now? put the limits into an integral with what as the integrand? Rectangular, cylindrical of spherical?

5. Apr 14, 2005

### HallsofIvy

Staff Emeritus
$$y = -u\;-\;2v\;and\;x = -u\;-\;3v$$
so $$\frac{\partial y}{\partial x}= -2$$, not -1!

I don't know! What does the problem ask? Your original post just said "solve for x, y, find the Jacobian, find what a parallelogram would be in the "u,v" coordinate system". You didn't say anything about an integral!

I presume you are to do whatever integral you are given in uv- coordinates- that's not any of those!

Last edited: Apr 14, 2005