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

In summary, the problem asks you to solve for x, y, find the Jacobian, and find what a parallelogram would be in the "u,v" coordinate system. You solved correctly for x and y, but the Jacobian is NOT -2. How did you get that?
  • #1
VinnyCee
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0
Solve u = 2x - 3y & v = -x + y in terms of x & y, then find Jacobian.

Here is the problem:

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

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

Here is what I have:

[tex]y = -u\;-\;2v\;and\;x = -u\;-\;3v[/tex] I am not sure of these answers.

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

[tex]y = x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y = x\;+\;1[/tex]
[tex]-2v\;-\;u = -3v\;-\;u\;\;\;\;\;-2v\;-\;u = -3v\;-\;u\;+\;1[/tex]
[tex]v = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;v = 1[/tex]

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

Please help!
 
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  • #2
Did anyone check these?

Please, this is the last problem and probably the toughest. Thank you all in advance.
 
  • #3
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 [tex]x = -3,\;\;x = 0,\;\;y = x\;and\;y = x\;+\;1[/tex]."

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
How I got the Jacobian

I used

[tex]\frac{\partial x}{\partial u}\;\frac{\partial y}{\partial v}\;-\;\frac{\partial y}{\partial u}\;\frac{\partial x}{\partial v}[/tex]

with the earlier equations solved for [tex]x[/tex] and [tex]y[/tex] and got

[tex](-1)(-1) - (-1)(-3) = 1 - 3 = -2[/tex]

What do I do now? put the limits into an integral with what as the integrand? Rectangular, cylindrical of spherical?
 
  • #5
[tex]y = -u\;-\;2v\;and\;x = -u\;-\;3v[/tex]
so [tex]\frac{\partial y}{\partial x}= -2[/tex], 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!
 
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Related to Solve u = 2x - 3y & v = -x + y in terms of u & v, then find Jacobian.

1. What is the meaning of "solve u = 2x - 3y & v = -x + y in terms of u & v"?

When we are asked to solve a system of equations in terms of certain variables, it means that we need to express each variable in the system (in this case, u and v) in terms of the other variables. This will allow us to rewrite the equations and solve for the desired variables.

2. How do we express u and v in terms of x and y?

To express u and v in terms of x and y, we need to manipulate the given equations to isolate the desired variables. In this case, we can rewrite the first equation as x = (u + 3y)/2 and the second equation as y = v + x. These expressions can then be substituted into the original equations to solve for u and v.

3. What is the Jacobian and why do we need to find it?

The Jacobian is a mathematical concept that represents the rate of change of a multivariable function. In this case, it is used to calculate the partial derivatives of u and v with respect to x and y. We need to find the Jacobian in order to properly solve the system of equations and express u and v in terms of x and y.

4. How do we find the Jacobian in this problem?

To find the Jacobian in this problem, we first need to calculate the partial derivatives of u and v with respect to x and y. This can be done by taking the derivative of each equation with respect to x and y, respectively. Then, we can use these partial derivatives to construct the Jacobian matrix, which represents the rate of change of u and v with respect to x and y.

5. Can the Jacobian be used to solve other types of equations?

Yes, the Jacobian can be used to solve a variety of mathematical problems, including systems of equations and optimization problems. It is a useful tool in many areas of mathematics and science, and is often used to analyze the behavior of complex systems.

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