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

  1. Apr 14, 2005 #1
    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!
     
    Last edited: Apr 14, 2005
  2. jcsd
  3. Apr 14, 2005 #2
    Did anyone check these?

    Please, this is the last problem and probably the toughest. Thank you all in advance. :surprised
     
  4. Apr 14, 2005 #3

    HallsofIvy

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    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.
     
  5. Apr 14, 2005 #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?
     
  6. Apr 14, 2005 #5

    HallsofIvy

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    [tex]y = -u\;-\;2v\;and\;x = -u\;-\;3v[/tex]
    so [tex]\frac{\partial y}{\partial x}= -2[/tex], 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
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