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

[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.

Please help!

 

[VinnyCee]

Did anyone check these?

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

 

[HallsofIvy]

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.

 

[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?

 

[HallsofIvy]

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

What do I do now? put the limits into an integral with what as the integrand?

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!

Rectangular, cylindrical of spherical?

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