Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - composite transformation

  1. #1
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    composite transformation

    I am going through this past papers and I am utterly confused with the question- looked at all the notes and examples but simply can't understand what to do for part (b) Please explain. I have sucessfully done part (a) and answer is:

    A= (-2 - -2, -1+2
    3-2 2-2) = (0 1
    1 0)

    And a = (-2
    2)

    Hence g(x) = (0 1 + (-2
    1 0)x 2)

    this is the whole question:
    In this question, f and g are both affine transformations. The
    transformation f is reflection in the line y = 1, and the transformation g
    maps the points (0, 0), (1, 0) and (0, 1) to the points (−2, 2), (−2, 3) and
    (−1, 2), respectively.
    (a) Determine g (in the form g(x) = Ax + a, where A is a 22 matrix
    and a is a vector with two components).
    (b) Express f as a composite of three transformations: a translation,
    followed by reflection in a line through the origin, followed by a
    translation. Hence determine f (in the same form as you found g in
    part (a)).
    (c) Use the expressions that you found for f and g in parts (a) and (b) to
    calculate f(g(x)), and hence find the affine transformation f ◦ g in the
    same form as you found g in part (a).

    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    for part (b) use the translation T(x,y) = (x,y) - (0,1), to move the the line y = 1 so that it now goes through the origin. then you may apply the reflection

    R(x,y) = (x,-y), and finally apply the reverse translation (you should be able to easily write down what this is). can you see how to put f in the required form?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    what angle would I consoder

    Hi
    Thanks Deveno
    What confuses me as I had been given a similar problem where x=2 - so after resolving part (a) as I did in this one, my tutor gave me follwoing method which we are expected to follow: can you help me make sense out of it:

    this is what he wrote: Since x=2 makes an angle pi/2 with the positive x- axis, it follows that refelction in this line is qpi/2
    so I need to start my part b like this- which I never understood at the time and I am still confused- can you please help to get this correct.


    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    Quote Originally Posted by vidhi96 View Post
    Hi
    Thanks Deveno
    What confuses me as I had been given a similar problem where x=2 - so after resolving part (a) as I did in this one, my tutor gave me follwoing method which we are expected to follow: can you help me make sense out of it:

    this is what he wrote: Since x=2 makes an angle pi/2 with the positive x- axis, it follows that refelction in this line is qpi/2
    so I need to start my part b like this- which I never understood at the time and I am still confused- can you please help to get this correct.
    what do you mean by q? the line y = 1 makes an angle 0 with the x-axis. now, normally a reflection across a line of angle θ through the origin is:

    [ cos(2θ) sin(2θ)]
    [sin(2θ) -cos(2θ)], but when θ = 0, this matrix has a particularly simple form.

    the form above is derived by seeing that a reflection about y = θx, can be viewed as a rotation of -θ, followed by a reflection across the x-axis, followed by a rotation of angle θ:

    [cosθ -sinθ][ 1 0][ cosθ sinθ]
    [ sinθ cosθ][0 -1][-sinθ cosθ]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    oh

    Quote Originally Posted by Deveno View Post
    what do you mean by q? the line y = 1 makes an angle 0 with the x-axis. now, normally a reflection across a line of angle θ through the origin is:

    [ cos(2θ) sin(2θ)]
    [sin(2θ) -cos(2θ)], but when θ = 0, this matrix has a particularly simple form.

    the form above is derived by seeing that a reflection about y = θx, can be viewed as a rotation of -θ, followed by a reflection across the x-axis, followed by a rotation of angle θ:

    [cosθ -sinθ][ 1 0][ cosθ sinθ]
    [ sinθ cosθ][0 -1][-sinθ cosθ]
    sorry I should have mentioned- notation r= represents rotation and q is for reflection
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    are you still confused as to how to proceed?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    confused and frustrated

    see attached file- that's what he is expecting and I am completely lost
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    what i have written as T(x,y) would be t(0,-1). and what i have written as R would be q0.

    so the entire transformation is f = t(0,1) o q0 o t(0,-1).

    t(0,-1) moves the line y = 1 "down 1" so that now we have a line going through the origin. q0 reflects the plane about the x-axis (y = 0),

    which keeps all the x's the same, but exchanges y with -y. then t(0,1) moves the reflected image "back up 1". on an arbitrary point (x,y), we have:

    f(x,y) = t(0,1)(q0(t(0,-1)(x,y))) =

    [...([x]....[ 0])].....[0]
    [q0([y] + [-1])] + [1] =

    [1 .0][..x.].....[0]
    [0 -1][y-1] + [1] =

    [1 .0][x]...[1 .0][..0]....[0]
    [0 -1][y] - [0 -1][-1] + [1] =

    [1 .0][x]....[0]
    [0 -1][y] + [2], so that our matrix A is just the matrix for q0 and the vector a is (0,2).

    as a check, let's verify that f leaves points on the line y = 1 unchanged.

    f(x,1) = (x,-1) + (0,2) = (x,1). see? now the point (x,1+k) should go to (x,1-k), we can verify this, too:

    f(x,1+k) = (x,-1-k) + (0,2) = (x,1-k), as expected.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    thanks

    Quote Originally Posted by Deveno View Post
    what i have written as T(x,y) would be t(0,-1). and what i have written as R would be q0.

    so the entire transformation is f = t(0,1) o q0 o t(0,-1).

    t(0,-1) moves the line y = 1 "down 1" so that now we have a line going through the origin. q0 reflects the plane about the x-axis (y = 0),

    which keeps all the x's the same, but exchanges y with -y. then t(0,1) moves the reflected image "back up 1". on an arbitrary point (x,y), we have:

    f(x,y) = t(0,1)(q0(t(0,-1)(x,y))) =

    [...([x]....[ 0])].....[0]
    [q0([y] + [-1])] + [1] =

    [1 .0][..x.].....[0]
    [0 -1][y-1] + [1] =

    [1 .0][x]...[1 .0][..0]....[0]
    [0 -1][y] - [0 -1][-1] + [1] =

    [1 .0][x]....[0]
    [0 -1][y] + [2], so that our matrix A is just the matrix for q0 and the vector a is (0,2).

    as a check, let's verify that f leaves points on the line y = 1 unchanged.

    f(x,1) = (x,-1) + (0,2) = (x,1). see? now the point (x,1+k) should go to (x,1-k), we can verify this, too:

    f(x,1+k) = (x,-1-k) + (0,2) = (x,1-k), as expected.
    Hey this is great it makes lot more sense. I will resolve the whole thing. Thanks a lot.
    you are brilliant
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2011
    Posts
    60
    for part c)
    for the last bit
    f o g i got
    f o g ={1 0} * {0 1} X + {1 0} * {-2} + {0}
    {0 -1} {1 0} {0 -1} {2} {2}

    = {0 1} X + {-2}
    {-1 0} {0}

    so AM threw with part c. I will appriciate if you can guide me for part d

    d) Use your answer to part (c) to determine any points (x, y) that are
    left unchanged by the transformation f ◦ g, or to show that there are
    no such points. (To do this, find any values of x and y for which
    f(g(x, y)) = (x, y).)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    i agree with your answer for (c). good work.

    if we explicity calculate f(g(x,y)), we get f(g(x,y)) = (y-2, -x).

    so if (x,y) = (y-2,-x) we get 2 equations in x and y. can you solve these?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    hey

    Quote Originally Posted by Deveno View Post
    i agree with your answer for (c). good work.

    if we explicity calculate f(g(x,y)), we get f(g(x,y)) = (y-2, -x).

    so if (x,y) = (y-2,-x) we get 2 equations in x and y. can you solve these?
    Hi follwoing this I get two equations:
    x=y-2 and y=-x
    solving them I get x=-1 and y=1

    is this correct?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    you can check this: calculate f(g(-1,1)). is it equal to (-1,1)?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    May 2011
    Posts
    3

    found this thread

    Hello Deveno,
    I found this thread (and MHF) this weekend as I was searching for some help in the subject and this thread really helped me.
    I've got to the conclusion that f(g(x)) = f o g.
    Some places would write f(g(x)) = f o g (x).
    Am I right or did I loose the plot?
    Thank you
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    yes. to evaluate (fog) at the vector x, one evaluates g at the vector x, and then evaluates f at the vector g(x).

    ...g.....f.....
    X-->Y-->Z (here X is the space where the x's live, Y is the "target space" of g (which could be the same as X, just "re-arranged"), and Z is the "target space" of f).

    g(x) = y
    f(y) = z
    fog(x) = f(g(x)) = f(y) = z

    in the above problem, X = Y = Z = R^2, the real 2-dimensional plane, and f and g are affine mappings.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Composite Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 27th 2009, 08:41 AM
  2. composite functions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 15th 2009, 07:09 PM
  3. composite
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: January 27th 2009, 06:46 PM
  4. Composite Functions! Help! :|
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 21st 2008, 08:41 PM
  5. Composite f, g, and/or h
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 25th 2007, 03:14 PM

Search Tags


/mathhelpforum @mathhelpforum