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The equation of a hyperbolic paraboloid to derive the corner points of rectangle

bugatti79

Member
Feb 1, 2012
71
Hi Folks,


I have come across some text where [tex]f(x,y)=c_1+c_2x+c_3y+c_4xy[/tex] is used to define the corner points


[tex]f_1=f(0,0)=c_1[/tex]
[tex]f_2=f(a,0)=c_1+c_2a[/tex]
[tex]f_3=f(a,b)=c_1+c_2a+c_3b+c_4ab[/tex]
[tex]f_4=f(0,b)=c_1+c_3b[/tex]


How are these equations determined? F_1 to F_4 starts at bottom left hand corner and rotates counter clockwise.


Thanks
Note: This was posted last week on MHB with no response.
The equation of a hyperbolic paraboloid to derive the corner points of rectangle
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hi Folks,


I have come across some text where [tex]f(x,y)=c_1+c_2x+c_3y+c_4xy[/tex] is used to define the corner points


[tex]f_1=f(0,0)=c_1[/tex]
[tex]f_2=f(a,0)=c_1+c_2a[/tex]
[tex]f_3=f(a,b)=c_1+c_2a+c_3b+c_4ab[/tex]
[tex]f_4=f(0,b)=c_1+c_3b[/tex]


How are these equations determined? F_1 to F_4 starts at bottom left hand corner and rotates counter clockwise.


Thanks
Note: This was posted last week on MHB with no response.
The equation of a hyperbolic paraboloid to derive the corner points of rectangle
Hi bugatti79! :)

I think you meant MHF instead of MHB. ;-)

Your f(x,y) is defined on a square domain of the form [0,a] x [0,b].
The corner points are the corner points of this domain.
 

bugatti79

Member
Feb 1, 2012
71
Hi bugatti79! :)

I think you meant MHF instead of MHB. ;-)

Your f(x,y) is defined on a square domain of the form [0,a] x [0,b].
The corner points are the corner points of this domain.
Hello Sir,

Yes I stand corrected regarding MHF. :)
Ok but just wondering how those equations were derived particularly the third one with the xy product term :)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
If you take a general hyperbolic paraboloid and rotate it appropriately, you can get its form to look like your $f(x,y)$. It's entirely analogous to being able to take a certain hyperbola, rotate it, and obtain the equation $xy=1$. I haven't worked it out, but there probably is a standard-form hyperbolic paraboloid equation looking something like
$$ \frac{z}{c}= \frac{y^{2}}{b^{2}}- \frac{x^{2}}{a^{2}},$$
which, when rotated using rotation matrices, gives you the form of your $f(x,y)$.
 

bugatti79

Member
Feb 1, 2012
71
If you take a general hyperbolic paraboloid and rotate it appropriately, you can get its form to look like your $f(x,y)$. It's entirely analogous to being able to take a certain hyperbola, rotate it, and obtain the equation $xy=1$. I haven't worked it out, but there probably is a standard-form hyperbolic paraboloid equation looking something like
$$ \frac{z}{c}= \frac{y^{2}}{b^{2}}- \frac{x^{2}}{a^{2}},$$
which, when rotated using rotation matrices, gives you the form of your $f(x,y)$.
Hi Ackbach,
Good to see you are still around! :)

That is interesting. I can see now why this expression is required. Each corner needs to movable while the other three are fixed thus resembling the shape of hyperbolic parabola from a flat surface.

I googled information on rotation of a hyperbolic parabola to a flat surface and vica versa, no luck. Perhaps some one know of a link?

Thanks guys
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Good to see you around, too! I remember you having the most interesting problems in the old country - quantum mechanics and such-like.

Well, I just got my info from the wiki on hyperbolic paraboloids. There's the general equation right away, and then a little info on rotation in the "Multiplication Table" section. To be clear: you're not rotating the hyperbolic paraboloid into the flat surface. A hyperbolic paraboloid is not flat in a Euclidean sense! If I had to make a guess, I'd say the rectangle you're dealing with is some region of interest for other reasons.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Here's a wiki link to quadric surfaces.
Not to be confused with quartics or quadratics.

The link includes the method how to analyze a quadric surface.
In your case you have a hyperbolic paraboloid that has been rotated around the z-axis by 45 degrees.
 

bugatti79

Member
Feb 1, 2012
71
Here's a wiki link to quadric surfaces.
Not to be confused with quartics or quadratics.

The link includes the method how to analyze a quadric surface.
In your case you have a hyperbolic paraboloid that has been rotated around the z-axis by 45 degrees.
Hi,

Thanks for that. Where did you establish that regarding the rotation of 45degrees?

Ed
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hi,

Thanks for that. Where did you establish that regarding the rotation of 45degrees?

Ed
I'll give a summarized analysis.

Your quadric equation is $c_1 + c_2x + c_3y + c_4xy - z = 0$.
This can be written as the following quadric equation in matrix form (see quadric on wiki):
$$\begin{bmatrix}x & y & z\end{bmatrix} \begin{bmatrix}0 & c_4/2 & 0 \\ c_4/2 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} + \begin{bmatrix}c_2 \\ c_3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}x \\ y \\ z\end{bmatrix} = 0$$

The eigenvalues of the matrix are $\dfrac {c_4} 2, -\dfrac {c_4} 2, 0$ and their corresponding eigenvectors are $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.

Those eigenvectors are diagonals, showing a rotation of 45 degrees.


The "regular" (unrotated and untranslated) form of your quadric surface is:
$$\frac {c_4}{2} x^2 - \frac {c_4}{2}y^2 - C z = 0$$
where the constant C is yet to be determined (but it's 1).

As you can see in the article, this is classified as a hyperbolic paraboloid.
 
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bugatti79

Member
Feb 1, 2012
71
I'll give a summarized analysis.

Your quadric equation is $c_1 + c_2x + c_3y + c_4xy - z = 0$.
This can be written as the following quadric equation in matrix form (see quadric on wiki):
$$\begin{bmatrix}x & y & z\end{bmatrix} \begin{bmatrix}0 & c_4/2 & 0 \\ c_4/2 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} + \begin{bmatrix}c_2 \\ c_3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}x \\ y \\ z\end{bmatrix} = 0$$

The eigenvalues of the matrix are $\dfrac {c_4} 2, -\dfrac {c_4} 2, 0$ and their corresponding eigenvectors are $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.

Those eigenvectors are diagonals, showing a rotation of 45 degrees.


The "regular" (unrotated and untranslated) form of your quadric surface is:
$$\frac {c_4}{2} x^2 - \frac {c_4}{2}y^2 - C z = 0$$
where the constant C is yet to be determined (but it's 1).

As you can see in the article, this is classified as a hyperbolic paraboloid.


I have just stumbled across this lovely applet Applet: Hyperbolic paraboloid coefficients - Math Insight

So far how I understand it is that we start off with the quadric surface [tex]f(x,y)=c_1+c_2x+c_3y+c_4xy[/tex], rotate it 45 degrees as Serena showed which gives us the non degenerate form. If we set the A abd B coefficients to 0 then z=0 and thus the surface becomes flat (Use applet)...
 
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