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

1. Jul 17, 2010

### der.physika

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?
Could someone give me a similar example and solve that?

2. Jul 17, 2010

### 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.

3. Jul 17, 2010

### der.physika

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??

4. Jul 17, 2010

### 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.