- Thread starter
- #1

- Apr 14, 2013

- 4,951

Hi!!!

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

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

- Thread starter mathmari
- Start date

- Thread starter
- #1

- Apr 14, 2013

- 4,951

Hi!!!

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

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

- Thread starter
- #2

- Apr 14, 2013

- 4,951

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

- Jan 29, 2012

- 1,151

[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

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"?

- Thread starter
- #4

- Apr 14, 2013

- 4,951

I don't really know...It is given from the exerciseI don't understand what "f(-x, -y, z)" has to do with this. What is "f"?

- May 31, 2013

- 119

\(\displaystyle y=(1+t)^2\)

\(\displaystyle x=(1-t)^2\)

\(\displaystyle z=4t.....\)

may be?

\(\displaystyle x=(1-t)^2\)

\(\displaystyle z=4t.....\)

may be?

- Admin
- #6

- Mar 5, 2012

- 9,805

Looks to me as if you're supposed to give a parametrization in the form f(x,y).I don't really know...It is given from the exercise

In that case your parametrization would be:

$$f(x,y)=(-x,-y,y^2-x^2)$$

- Thread starter
- #7

- Apr 14, 2013

- 4,951

You mean:\(\displaystyle y=(1+t)^2\)

\(\displaystyle x=(1-t)^2\)

\(\displaystyle z=4t.....\)

may be?

\(\displaystyle y^2=(1+t)^2\)

\(\displaystyle x^2=(1-t)^2\)

\(\displaystyle z=4t\)

Right??

- - - Updated - - -

Ok! Thanks!!!!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:

$$f(x,y)=(-x,-y,y^2-x^2)$$

- Admin
- #8

- Mar 5, 2012

- 9,805

This is a curve while the original equation is a surface...\(\displaystyle y=(1+t)^2\)

\(\displaystyle x=(1-t)^2\)

\(\displaystyle z=4t.....\)

may be?

- May 31, 2013

- 119

yes!!You mean:

\(\displaystyle y^2=(1+t)^2\)

\(\displaystyle x^2=(1-t)^2\)

\(\displaystyle z=4t\)

Right??