# Inverse mapping theorem & local inverses

1. Nov 29, 2007

### kingwinner

[Related concepts: Inverse mapping theorem, transformations and coordinate systems]

1) For each of the following transformations (u,v) = f(x,y), (i) compute det Df, (ii) find formulas for the local inverses of f when they exist.
a) u=x^2, v=y/x
b) u=(e^x) cos y, v=(e^x) sin y

I got part (i) easily, but I don't understand part (iii) at all.

Here are the answers from the solutions manual:
For 1a)(iii), the answer is f^-1 (u,v) = (sqrt u, v (sqrt u) )
x= ln(u^2 + v^2) / 2
y is given up to multiples of 2pi by arctan(v/u) when u>0, pi/2 - arctan(u/v) when v>0, pi + arctan(v/u) when u<0, 3pi/2 - arctan(u/v) when v<0

But they didn't show any of the steps, nor do they show me how they arrive at the answers. Can someone please explain this part? How exactly can I find the formula for the local inverse? I am terribly confused...

Thanks a million!

2. Nov 29, 2007

### Hurkyl

Staff Emeritus
You have two equations. Solve for x and y.

3. Nov 30, 2007

### Despondent

The first one should be fairly easy. In the second one what you need to do is eliminate x so that you can solve for y in terms of u and v only and then eliminate y so that you can solve for x in terms of u and v only.

If you're stuck on solving for x and y, then just consider u^2 + v^2 and u/v.

4. Nov 30, 2007

### HallsofIvy

Staff Emeritus
The crucial point is what Hurkyl said: solve u= x2, v= y/x for x and y in terms of u and v. Try it. If you have difficulty get back to us.

5. Nov 30, 2007

### kingwinner

1a)(iii) Solving for x and y, I get x=+/-sqrt u, y=+/-v (sqrt u), but the answer says that f^-1 (u,v) = (sqrt u, v (sqrt u) ), they only take the positive signs, this is what I don't understand...

How can I do so?

Thanks!

Last edited: Nov 30, 2007
6. Nov 30, 2007

### HallsofIvy

Staff Emeritus
A function does not have two values! Since you say you got part i easily, I presume you showed that there is a local inverse as long as x is not 0. You can guarantee a local inverse by restricting x to be positive or negative. If you restrict x to be positive, the "local" inverse function is $x= \sqrt{u}$, $y= v\sqrt{u}$. If you restrict x to be negative, the "local" inverse function is $x= -\sqrt{u}$, $y= -v\sqrt{u}$. Each part has a unique inverse function- you can't write them together.

7. Nov 30, 2007

### kingwinner

Yes, I showed that there is a local inverse as long as x is not 0

So now there are two answers to (iii) $x= \sqrt{u}$, $y= v\sqrt{u}$ and $x= -\sqrt{u}$, $y= -v\sqrt{u}$.
Now how can I decide which one to take? If I just take one of them, would it still be a complete answer? Which one is the correct answer that the question is actually looking for?

Last edited: Nov 30, 2007
8. Dec 1, 2007

### kingwinner

Can someone please explain this part? I really want to understand it...Thanks!