Parameterizing a Hyperboloid: A Challenge!

[*Helix*]

NEEDD HELP! parameterizing a surface given by an implicit function (HYPERBOLOID?!)

Homework Statement

I need to parameterize the surface given by the following implicit equation:

x² + y² + z² + 2xy + 2xz - 2yz = 1

Homework Equations

I tried using all sorts of subsititutions, getting one varible in terms of the other two, using cylindrical and spherical coordinate systems, etc..and I still cannot come up with parametric equations for x y and z!

The Attempt at a Solution

I'm just hoping someone can point me in the right direction, or give me a good hint to crack out this answer. thanks!

 

[Dick]

You probably know how to parametrize a standard hyperboloid, right? Ok, so write that as (x+y)^2+z^2+2*z*(x-y)=1. Suggests you want to do a change of variables like u=(x+y), v=(x-y). Now what? Is it starting to look like a usual hyperboloid form?

 

[*Helix*]

Hi Dick, your suggestion looked goods, I have come up with the following, I just need a little help with grasping how the parameters work. let me know what you think of this solution so far:

if we let
u = x - y
v = x + y
then:
u² = x² - 2xy + y²

Substituting the change of varibles into the orginal equation:
x² + y² + z² + 2xy + 2xz - 2yz = 1
and simplifying we have:

z² +(2u)z + u² = 1
or
z² +(2u)z + (u² - 1) = 0

z is in quadratic form which yields a solution, after simplifying:

z = -u +/- sqrt(1)

so equations for x y and z are:
x = (u+v)/2
y = (v-u)/2
z = -u

now its interesting..when I graph x y and z strictly as these equations, I get a plane that slices the surface exactly in half... how do I get this equation in standard form for my surface.. also, I know that a standard form of parametric equations for a hyperboloid of one sheet are:
x= a*cosh(u).*cos(k)
y= b*cosh(u).*sin(k)
z = c*sinh(u)

where u in this sense is I'm guessing a vector space, and k is an angle measured between 0 and 2 pi. and a b and c are scalers. (ie. if they all equal one, then its a hyperboloid made up of circles)...how would I relate what I found to these parametric equations?

 

[Dick]

Hi Dick, your suggestion looked goods, I have come up with the following, I just need a little help with grasping how the parameters work. let me know what you think of this solution so far:

if we let
u = x - y
v = x + y
then:
u² = x² - 2xy + y²

Substituting the change of varibles into the orginal equation:
x² + y² + z² + 2xy + 2xz - 2yz = 1
and simplifying we have:

z² +(2u)z + u² = 1

You are getting things all mixed up. u^2 has a -2xy in it's expansion and the original form had +2xy. Try that again and keep track of the signs better this time.

 

[*Helix*]

well, using v and u in the equation, you end up with ( u^2 - v^2 + 1) which is (-4xy +1) under the radical...what happens to z??

 

[Dick]

Don't put anything under the radical. You are just trying to introduce new variables so it looks like a standard hyperboloid. Your expression z^2+2zu+u^2=1 is wrong. One of those u's should be a v. Which one? Now complete the square in z and introduce another variable w.

 

[*Helix*]

so I should end up with z² +z(2u) + v² = 1 ...right? And I'm to complete the square in z and use another varible?

 

[*Helix*]

when I complete the square in z, assuming I have the right u and v now, I get z = -u plus or minus the square root of (u² - v² + 1) ..this is what I was getting before...hmmm

 

[Dick]

so I should end up with z² +z(2u) + v² = 1 ...right? And I'm to complete the square in z and use another varible?

Don't go all soft on me here. Yes, you have z^2+2zu+v^2=1. I would make that z^2+2zu+u^2-u^2+v^2=1. Or (z+u)^2-u^2+v^2=1. Now how about putting w=z+u? Don't you see where this is going? w^2-u^2+v^2=1, now parametrize the hyperboloid and then go back to x,y,z coordinates.

 

[*Helix*]

nice..thanks..I at least get this part!

we let:
v=x+y
u=x-y
w=z-u

then w² + v² -u² = 1

which we see is hyperboloid with an axis of symmetry around u?!

anddd...expanding all that out in terms of x y and z, we get the original equation
...which is sweet
so x = (v+u)/2 y = (v-u)/2 and z = (w - u)

now is this "standard" form? and as for parameterizing it..and those funky hyperbolic cosine functions and stuff...I'd have no idea how to relate what we found here, back to that..if that's what you're talking about parameterizing it.
I can visualize like a "mapping" of the surface...its slanted and pointed out along a line running through the origin
thanks for your help so far! hope to hear from you soon!

 

[Dick]

You don't understand how the parametric form you gave before parametrizes (x/a)^2+(y/b)^2-(z/c)^2=1?? Show me how it does. Substitute the 'funky' stuff into the equation (x/a)^2+(y/b)^2-(z/c)^2=1 and show me you get 1. Do it. Then apply the same thing to the u,v,w equation.

 

[*Helix*]

Well, I know the standard parametric form for when a hyperboloid is around the z axis. That is:

x=a*cosh(u)*cos(v)
y=b*cosh(u)*sin(v)
z=c*sinh(u)

v in this sense is between 0 and 2*PI
and if a=b=c, then it is a circular type hyperboloid, with the apex in the xy plane as a circle..am I right?
I'm not sure what "u" does, and how it relates, though when I graph the regular function in Matlab, it's basically a matrix of values along a line, which I guessing is the z axis in this case.
I'm getting confused about actually how I'd plot this is Matlab. I need to do that too..Using parameters only, NOT the implicit function
I know I could just write this:
(z+x-y)² + (x+y)² -(x-y)² = 1
then
(z+x-y) = a*cosh(u)*cos(v)
(x+y) =b*cosh(u)*sin(v)
(x-y)=c*sinh(u)

z' = a*cosh(u)*cos(v) + (y-x)
y' = b*cosh(u)*sin(v) -(x)
x' = c*sinh(u) +(y)

(still not sure about that "u")
am I on the right path? let me know.

 

[LCKurtz]

Well, I know the standard parametric form for when a hyperboloid is around the z axis. That is:

x=a*cosh(u)*cos(v)
y=b*cosh(u)*sin(v)
z=c*sinh(u)

v in this sense is between 0 and 2*PI
and if a=b=c, then it is a circular type hyperboloid, with the apex in the xy plane as a circle..am I right?
I'm not sure what "u" does, and how it relates, though when I graph the regular function in Matlab, it's basically a matrix of values along a line, which I guessing is the z axis in this case.
I'm getting confused about actually how I'd plot this is Matlab. I need to do that too..Using parameters only, NOT the implicit function
I know I could just write this:
(z+x-y)² + (x+y)² -(x-y)² = 1
then
(z+x-y) = a*cosh(u)*cos(v)
(x+y) =b*cosh(u)*sin(v)
(x-y)=c*sinh(u)

So what would a, b, and c be for your equation?

And if you don't want the implicit function, why not let Matlab solve those three equations for x, y, and z? That would give you x, y, and z in terms of two parameters u and v.

 

[Dick]

Don't let Matlab solve anything. You should do it yourself. It's not even hard. You've got w^2+v^2-u^2=1. Use your standard parametrization to write that in terms of two parameters. I wouldn't call them u and v just for the sake of clarity. Now since you know u,v and w in terms of x,y and z. Just solve for x, y and z in terms of the parameters. Then use Matlab to plot x, y and z. I don't know why this is seeming so difficult.

 

[*Helix*]

I'm going to work on it. my final solution is due monday morning :( ..so we'll see what i can come up with..I'll post very soon!