Vector Calculus Question (I don't understand)

[der.physika]

Homework Statement

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

Homework Equations

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?

 

[Mark44]

Homework Statement

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

Homework Equations

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?

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.

 

[der.physika]

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.

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??

 

[Mark44]

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.