# Thread: do Carmo question - differential geometry....

1. ## do Carmo question - differential geometry....

Let P = {(x,y,z) : x=y}, and let a function x:U->R^3 be given by x(u,v) = (u+v,u+v, uv) where U = {(u,v) : u>v}

Is x a parameterization of P?

I'm really stuck on this question and I feel it is easier than I am making it out to be.

The answer is Yes, (as it says this in the back of the book) - i'm not really sure how to be rigorous in proving it though.

I see that in the function x, x = y. so this is a start. I think What the question is asking is that given any (x,y,z) can you find a u and v such that x = y = u+v and uv = z??

I have got something like x = u + v, z = uv and then have solved for u and v in terms of x and z. Is this all we need to show? I end up getting u and v are derived from the quadratic formula.

$u=\frac{x\pm \sqrt{x^2 - 4z}}{2}$

the same for v...

this would mean that x^2 needs to be > 4z... does this matter?

now if u>vwould mean you would need to take the positive sqare root for u and negative for v. is this right?

If u = v, then it is possible we might run in to problems.

2. ## Re: do Carmo question - differential geometry....

The point $(0,0,1)$ belongs to $P$ , however the system $\begin{Bmatrix} u+v=0\\uv=1\end{matrix}$ has no real solutions. Perhaps there is a typo in your book.

3. ## Re: do Carmo question - differential geometry....

Originally Posted by mathswannabe
Let P = {(x,y,z) : x=y}, and let a function x:U->R^3 be given by x(u,v) = (u+v,u+v, uv) where U = {(u,v) : u>v}

Is x a parameterization of P?

I'm really stuck on this question and I feel it is easier than I am making it out to be.

The answer is Yes, (as it says this in the back of the book) - i'm not really sure how to be rigorous in proving it though.

I see that in the function x, x = y. so this is a start. I think What the question is asking is that given any (x,y,z) can you find a u and v such that x = y = u+v and uv = z??

I have got something like x = u + v, z = uv and then have solved for u and v in terms of x and z. Is this all we need to show? I end up getting u and v are derived from the quadratic formula.

$u=\frac{x\pm \sqrt{x^2 - 4z}}{2}$

the same for v...

this would mean that x^2 needs to be > 4z... does this matter?

now if u>vwould mean you would need to take the positive sqare root for u and negative for v. is this right?

If u = v, then it is possible we might run in to problems.