A quadric surface is defined by the set of all points which satisfy a value for a given equation, of three variables, of degree 2.
A famous, through boring, example of a quadric surface is the sphere of a certain radius! For example, a sphere of radius 1 has the following form.
\[x^2 + y^2 + z^2 = 1\]
Referring back to the first statement, this means that the sphere is just all the points x, y, and z such that the sum of their squares is 1. This is a similar to a situation to a level curve.
For a level curve, we have a function \(f(x,y)\) that maps points from the plane, also known as \(\mathbb{R}^2\), to real space , which can be referred to as \(\mathbb{R} ^3\). We then find all values of \(f(x,y)\) that have a value, \(k\).
This is to say,
\[f(x,y)=k\]
For a quadric surface, the motivation is the same. We now have a function \(F(x,y,z)\) which takes points from real space, (\(\mathbb{R}^3\)), and maps them to 4D space (\(\mathbb{R}^4\)).
However, that means nothing since we can't see 4D space, but we can see all points in 3D space that satisfy a value!
So, we can have:
\[F(x,y,z)=k\]
It's the same principle as a level curve, and we now have a level surface!
I chose to look at a two-sheet hyperboloid. I find them quite interesting, because there is a space where no points can satisfy the criteria to define the surface, and so the surface is made of two separate parts.
(make surface in geogebra and paste here)
The two-sheet hyperboloid has the (rather ugly) general form:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = -1\]
For the sake of simplicity, and because of my love of palindromes, I chose \(a=2\), \(b=3\), and \(c=2\)
So, my surface has the form:
\[\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{4} = -1\]
My surface also has the following shape:
However, it's evident that if we solve for z, we will not get a continuous equation.
\[z = \pm \sqrt{4(\frac{x^2}{4} + \frac{y^2}{9} +1)}\]
Unfortunately, this means the surface has to be parametrized so that it is completely continuous. The 2-sheet hyperboloid is defined by three parametric equations:
\[r(u,v)= x= a \sinh(u) \cos(v) \\ y = b \sinh(u) \sin(v) \\ z = c \cosh(u)\]
Given my constants, my surface will have the following form:
\[ x= 2 \sinh(u) \cos(v) \\ y = 3 \sinh(u) \sin(v) \\ z = 2 \cosh(u)\]
We can also slice a quadric surface with a plane to get a 2D intersection of the surface. A plane is defined as:
\[Ax + By + Cz + D= 0\]
The plane I will use to size my surface is defined by the normal vector\(<0,1,1>\)
My plane is thus:
\[y+z=0\]
The intersection forms a hyperbola, since it intersects my hyperboloid surface at 45 degrees.
The equation of the resulting hyperbola is below, along with its graph
\[\frac{ x^2}{4} - \frac{5}{36}y^2 = -1\]
Interestingly, if we take the plane \(z+5=0\), we get an ellipse that has the equation
\[\frac{x^2}{4} + \frac{y^2}{9} = \frac{21}{4}\]
Due to the inherent symmetry of the 2-sheet hyperboloid, I only need to render the upper half of it to generate a 3D representation. My print will span 48 millimeters in the x direction, and span roughly 30 millimeters in the y direction. It will be roughly 30 millimeters tall. Below I give the image of the 3D rendering of the surface and the intersection cut out.
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