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Linear Transformations

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Using [tex]du=.01[/tex], [tex]dv=.01[/tex] find the aroximate area under the transformation of the square bounded by the lines [tex]u=3[/tex], [tex]u=3.01[/tex], [tex]v=5[/tex], [tex]v=5.01[/tex].

    2. Relevant equations

    [tex]T(u,v)=<au+bv, cu+dv>[/tex]
    where [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], and [tex]d[/tex] make a square matrix.


    3. The attempt at a solution

    I am not sure how to approach this problem, as usually there is a function (of the form of the first eqn under relevent equations). I tried finageling, substituting etc.. but could not figure out how to do this. This is for a course in multivariable calculus. any thoughts?
     
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  3. Apr 5, 2007 #2

    Dick

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    Yes. Under a transformation the area of the transformed figure is related to the original area by a formula involving the jacobian of the transformation. What is the jacobian of a linear transformation?
     
  4. Apr 6, 2007 #3

    Thank you for the Reply Dick. The Jacobian, as I understand it, is the determinate of the function X and Y "components" of the [tex]T(u,v)[/tex] function. Evaluating the determinate creates:

    [tex]\frac{dx}{du}\frac{dy}{dv} - \frac{dx}{dv}\frac{dy}{du}[/tex]

    but my misunderstanding comes in to the fact that we do not have [tex]T(u,v)[/tex] (or, specifically [tex]<au+bv, cu+dv>[/tex], [tex]x[/tex] being equal to [tex]au+bv[/tex] and [tex]y[/tex] being equal to [tex]cu+dv[/tex]). Am I right thus far?
     
    Last edited: Apr 6, 2007
  5. Apr 6, 2007 #4

    Dick

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    I'm not clear what your confusion is. Isn't (dx/du)(dy/dv)-(dx/dv)(dy/du)=ad-bc?
     
  6. Apr 6, 2007 #5
    I apologize I guess I wasn't clear. We are given the coordinates in the uv coordinate system, and the first step towards a solution is to translate them into the xy coordinate system. But in all the examples I have done thus far, it is done through a function [tex]T(u,v)=(something, something)[/tex], where the something's are functions of u and v, to which we can set equal to x and y respectively. solving at the critical points allows us to draw the graph in the xy plane.

    Am I looking at this in the wrong way?
     
    Last edited: Apr 6, 2007
  7. Apr 6, 2007 #6

    Dick

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    If T(u,v) isn't your tranformation, what is?
     
  8. Apr 6, 2007 #7

    Dick

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    Thanks for clarifying. I think the idea here is that the square is small in u,v coordinates so the jacobian doesn't vary much over it. So a reasonable approximation to the area of the transformed square would just be to multiply the area of the original square by the jacobian at any point in the square. Say (3,5)?
     
  9. Apr 6, 2007 #8
    wouldn't the jacobian of a point just be zero? ie, dx/du = d3/du = 0 (etc..) ?
     
    Last edited: Apr 6, 2007
  10. Apr 6, 2007 #9

    Dick

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    Dang did it again. No, it's not zero. It's [tex]\frac{dx}{du}\frac{dy}{dv} - \frac{dx}{dv}\frac{dy}{du}[/tex]
    EVALUATED at u=3,v=5. Love stealing tex instead of writing it myself.
     
  11. Apr 6, 2007 #10
    i'm missing something, cause i don't quite get it. Thus far, every time I have computed the jacobian, x was equal to something like [tex]u^2 + v[/tex] and something similar for y. When you say evaluated at u=3, v=5, surly you don't mean with respect to because that doesn't make any sense, so what do you mean?
     
  12. Apr 6, 2007 #11

    Dick

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    If they don't give you an explicit T(u,v) then the best you can say is that the area is approximately 0.01*0.01*jacobian(T)(3,5) or 0.01*0.01*jacobian(T)(3.01,5.01) or any other point in the square. Sorry, gotta sleep now.
     
  13. Apr 6, 2007 #12

    HallsofIvy

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    There is no such thing as "the Jacobian of a point". You mean the Jacobian of the transformation evaluated at a point.


    (I am reminded of students who argue: "The derivative of f(x) at x= a is 0, because f(a) is a constant!")
     
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