MHB Parametrizing Z=y^2-x^2 & f(x,y)=(-x,-y,z)

[mathmari]

Hi!
Which is the parametrization of z= y^2-x^2 , and f(x,y)=(-x,-y,z) ?

 

[mathmari]

Do I have to use x=coshv and y=sinhv, so z=-1??

 

[HallsofIvy]

No, z is to be a variable and there is no reason to believe it would be the constant, -1.

z= x^2- y^2 is a single equation in three variables. It's graph has 3- 1= 2 dimensions so is a surface. A Parameterization must have 2 parameters.

Using your idea of x= cosh(v) and y= sinh(v), since we want "z" rather than -1, multiply by -z: x= -z cosh(v), y= -z sinh(v). If you don't like the idea of using z itself as a parameter (and have already us "v" as a parameter), introduce the paramer u= z. Then x= - u cosh(v),
y= -u sinh(v), z= u.

I don't understand what "f(-x, -y, z)" has to do with this. What is "f"?

 

[mathmari]

I don't understand what "f(-x, -y, z)" has to do with this. What is "f"?

I don't really know...It is given from the exercise :(

 

[mathworker]

$$y=(1+t)^2$$
$$x=(1-t)^2$$
$$z=4t...$$
may be?

 

[I like Serena]

I don't really know...It is given from the exercise :(

Looks to me as if you're supposed to give a parametrization in the form f(x,y).
In that case your parametrization would be:
$$f(x,y)=(-x,-y,y^2-x^2)$$

 

[mathmari]

$$y=(1+t)^2$$
$$x=(1-t)^2$$
$$z=4t...$$
may be?

You mean:
$$y^2=(1+t)^2$$
$$x^2=(1-t)^2$$
$$z=4t$$
Right??

- - - Updated - - -

Looks to me as if you're supposed to give a parametrization in the form f(x,y).
In that case your parametrization would be:
$$f(x,y)=(-x,-y,y^2-x^2)$$

Ok! Thanks! :)

 

[I like Serena]

$$y=(1+t)^2$$
$$x=(1-t)^2$$
$$z=4t...$$
may be?

This is a curve while the original equation is a surface...

 

[mathworker]

You mean:
$$y^2=(1+t)^2$$
$$x^2=(1-t)^2$$
$$z=4t$$
Right??

yes!