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Math Help - f: R^2 to R^2 function, derivatives and inverses

  1. #1
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    f: R^2 to R^2 function, derivatives and inverses

    I would be greatful for your comments.


    Question.

    A function f:R^2->R^2 is defined by

    f(x,y)=(u,v)^T=(x-y,x^2-y^2).

    (a) Find the set of critical points of f.

    Roughly sketch the u and v contours in the (x,y)-plane together with the critical points of f. Inlcude and label the contours u=0 and v=0.

    (b) Write down the derivative of the function f and the derivative of the local inverse at a non-critical point (x,y).

    Deduce an expression for \frac{\partial{x}}{\partial{u}} and evaluate it at the point (x,y)=(2,1).

    (c) Find, if possible, an appropriate formula for the local inverse function at each of the points (x,y)=(2,1) and (1,1) justifying your answers. Carefully state the region in the (u,v)-plane on which the local inverse is defined and its image on the (x,y)-plane.

    On separate diagrams, sketch the u and v contours through each of the given points. Comment on the relationship between the contours in each diagram.

    (d) Find \frac{\partial{x}}{\partial{u}} directly from the local inverse function you found in part (c).

    Evaluate it at the image point of (x,y)=(2,1) in the (u,v)-plane and verify that this agrees with the result in part (b).

    (e) Write down the equations of the tangent flat to f at the point (x,y)=(2,1).


    I'll post my work in a separate post.
    Last edited by Volga; March 23rd 2011 at 06:36 AM.
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  2. #2
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    EDITED: The first attempt didn't really solve anything, so I deleted it. Please take a look at the below post.
    Last edited by Volga; March 23rd 2011 at 06:38 AM.
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  3. #3
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    Attempt at solution # 2.


    (a) critical points and sketch of u, v contours on (x,y)-plane.

    The contour of u is a straight line of the type y=x-u, and u can take any value on the real line -> it is a series of straight lines with slope of 1.

    Contour v=x^2-y^2 are hyperbolas with different distance from the origin (varies with v).

    The critical point of f are the points where the determinant of Jacobian is zero:

    If Jacobian is

    |1 -1|
    |2x -2y|

    then |J|=-2y+2x=2(x-y)=0 iff x=y. So the critical point is an infinite set represented by a line y=x.


    (b) Derivative

    The derivative of a R^2->R^2 function is a matrix (Jacobian) calculated as follows:

    |\frac{\partial{u}}{\partial{x}}\frac{\partial{u}}  {\partial{y}}|
    |\frac{\partial{v}}{\partial{x}}\frac{\partial{v}}  {\partial{y}}|

    which is

    |1 -1|
    |2x -2y|

    The derivative of the local inverse is the inverse Jacobian:

    |\frac{-2y}{2(x-y)} \frac{1}{2(x-y)}|
    |\frac{-2x}{2(x-y)} \frac{1}{2(x-y)}|

    At point (2,1) the value of this J^{-1} is \frac{-2*1}{2(2-1)}=-1


    (c) formula for local inverse

    I solve for x and y the system of equations

    u=x-y
    v=x^2-y^2

    and I get x=\frac{u^2+v}{2u}, y=\frac{v-u^2}{2u}

    At (2,1) it is 2=\frac{u^2+v}{2u}, 1=\frac{v-u^2}{2u}

    We want x>0, y>0
    (a) u^2+v<0, u<0, v-u^2<0
    (b) u^2+v>0, u>0, v-u^2>0

    Solving the system of equations, I got a pair:
    u=1, v=3 so (1,3) is the value of f at the point (2,1). For the local inverse to exist, each (x,y) must have only one pair (u,v). Therefore, I think the relevant area is (b): v>u^2 (v>0), u>0

    At the point (1,1) the local inverse does not exist because at this point the |J|=0 and the Jacobian matrix is not invertible.


    (d)

    \frac{\partial{x}}{\partial{u}}=\frac{1}{2}\frac{(  u^2+v)'-(u^2+v)u'}{u^2}=\frac{1}{2}\frac{u^2-v}{u^2}=\frac{-2}{2}=-1 as in (b)



    (e) tangent flat equation at (2,1)

    (u,v)^T=f(2,1)+f'(2,1)(x-2, y-1)^T

    f'(2,1):

    |1 -1|
    |4 2|

    f(2,1)=(1,3)^T

    therefore, via some matrix multiplication we get the equation

    u=1+(x-2)-(y-1)
    v=3+4(x-2)+2(y-1)


    I would really appreciate your comments! Thank you!
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