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Homework Help: Vector Calculus Question (I don't understand)

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Let D* = [0,1]x[0,1] and define T on D* by T(u,v)=(-u^2+4u, v). Find the image D. Is T one-to one

    2. Relevant equations



    3. The attempt at a solution

    I have no idea... I don't know how to do it.
    The solution is [0,3] x [0,1]... yes it is one to one.
    am I supposed to say det A does not equal zero or use the u=u' v=v' approach?
    and... how do I find D? i don't know how to go about this.
    Could someone give me a similar example and solve that?
     
  2. jcsd
  3. Jul 17, 2010 #2

    Mark44

    Staff: Mentor

    This seems pretty straightforward to me, as the image vector doesn't have u and v tangled together. For fixed u, T maps v to v. For fixed v, T maps u to -u^2 + 4u, and this graph is a parabola.

    This is kind of a simple-minded way to look at this problem, but I think it will work.
     
  4. Jul 17, 2010 #3
    Okay... so I took your advice.

    so I thought about what you said

    so T(u,v) maps u => -u^2 + 4u
    and v to v

    so for u [0,1] is the interval... so that means [0, -1+4] = [0,3]

    for v [0,1] goes to [0,1]

    [0,3] x [0,1] so that means... it maps the square into a rectangle.

    But... how do I show that the mapping is one-to one??
     
  5. Jul 17, 2010 #4

    Mark44

    Staff: Mentor

    Let's call the outputs (w, z), so that T(u, v) = (w, z), with w = -u2 + 4u and z = v. Show that (w1, z1) = (w2, z2) ==> (u1, v1) = (u2, v2).

    It's also helpful to look at the portion of the parabola for which 0 <= u <= 1. Quadratic functions aren't normally one-to-one, but if the domain is limited in the right way, the limit domain version can be one-to-one.
     
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