Mutlivariable Calculus: Level Set

[kingwinner]

Q: Find a real-valued function F(x,y,z) whose 0-level set is the image of the map f: R^2 -> R^3 defined by f(u,v) = (2uv, v^2, u+v).

I know that by definition,
f: U C R^n -> R, U is the domain of f.
The level set of value c E R is the set of those points x E U at which f(x)=c.
i.e. c-level set = {x E U | f(x) = c} C R^n

But I don't know how to proceed from here, can someone please help me out?

 

[kingwinner]

I know it's hard, but someone here must know how to solve it...

 

[Dick]

Well, If (x,y,z) lies on your level set surface then v=sqrt(y) and u=z-sqrt(y). What's x? Does that suggest anything?

 

[kingwinner]

Well, If (x,y,z) lies on your level set surface then v=sqrt(y) and u=z-sqrt(y). What's x? Does that suggest anything?

Is this true: F(x,y,z) = F(2uv, v^2, u+v) = 0 ?

Also, why is v=sqrt(y)? Shouldn't it be v=+/-sqrt(y)?

 

[Dick]

Is this true: F(x,y,z) = F(2uv, v^2, u+v) = 0 ?

Also, why is v=sqrt(y)? Shouldn't it be v=+/-sqrt(y)?

You can choose a sign on sqrt(y) if you want. But that's not very important. The point is you can determine u and v in terms of y and z. That means x is determined in terms of y and z. Once you have an expression involving only x,y and z, you can put all the terms on one side and call that F.

 

[kingwinner]

But how do I know which sign of sqrt(y) should I choose? Choosing a different sign gives different answers...

"Once you have an expression involving only x,y and z" <---how can I actually get this?


Thanks a lot!

 

[Dick]

But how do I know which sign of sqrt(y) should I choose? Choosing a different sign gives different answers...

"Once you have an expression involving only x,y and z" <---how can I actually get this?


Thanks a lot!

There are an infinite number of functions that have that level set. You shouldn't be suprised that there isn't just one answer. All you have to do is find one. And I've shown you how to find y and z in terms of u and v, but you still haven't told me what x is in terms of y and z.

 

[kingwinner]

Set
x=2uv
y=v^2
z=u+v

Eliminating u and v, I get
4yz^2-4xy-x^2-4y^2=0

"Once you have an expression involving only x,y and z, you can put all the terms on one side and call that F." <---Why would this be F? I know that I have an expression in terms of x,y,z only, but why would this expression be equal to F?

 

[Dick]

Well, if you define F(x,y,z)=4yz^2-4xy-x^2-4y^2, would you agree F(x,y,z)=0 if (x,y,z) is in the image of f?