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Parametrization

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,149
Hi!!!
Which is the parametrization of z= y^2-x^2 , and f(x,y)=(-x,-y,z) ???
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,149
Do I have to use x=coshv and y=sinhv, so z=-1??
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
No, z is to be a variable and there is no reason to believe it would be the constant, -1.

[tex]z= x^2- y^2[/tex] 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

Well-known member
MHB Site Helper
Apr 14, 2013
4,149

mathworker

Active member
May 31, 2013
118
\(\displaystyle y=(1+t)^2\)
\(\displaystyle x=(1-t)^2\)
\(\displaystyle z=4t.....\)
may be?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,893
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,149
\(\displaystyle y=(1+t)^2\)
\(\displaystyle x=(1-t)^2\)
\(\displaystyle z=4t.....\)
may be?
You mean:
\(\displaystyle y^2=(1+t)^2\)
\(\displaystyle x^2=(1-t)^2\)
\(\displaystyle z=4t\)
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!!!! :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,893
\(\displaystyle y=(1+t)^2\)
\(\displaystyle x=(1-t)^2\)
\(\displaystyle z=4t.....\)
may be?
This is a curve while the original equation is a surface...
 

mathworker

Active member
May 31, 2013
118