Geometrical properties of a surface

Question

Actually I am from physics background, so I need a help in understanding the geometrical properties of a surface, Suppose If an expression for a surface is given such as

$$Ax^2+By^2+Cz^2+Dxy+Eyz+Fxz+G=0$$

How can I find the geometrical properties of such a 3D surface? In 2D, I know how to find the maxima, minima, eccentricity of conic sections, If I change the 3D surface into 2D plot by introducing contours? will it be useful in finding the properties of 3D surface?

Answer 1

There are only a few basic shapes that an equation in this form can take on.

An elipsoid, a hyperboloid in one sheet, and a hyperboloid in two sheets.

If you opened it up a littlle bit wiht

$Ax^2 + By^2 +Cz^2 + Dxy + Eyz+Fxz+Gx+Hy+Iz + K = 0$ paraboloids and hyperbolic paraboloids enter the picture.

I would suggest you become familiar with the more simplified equations first

$Ax^2 + By^2 +Cz^2 = K$ and $Ax^2 + By^2 = z$

And look at what you get with positive, negative and zero values for $A,B$ and $C.$

If you now linear algebra:

Build this matrix

\begin{bmatrix} A &\frac D2 &\frac F2\\ \frac D2 & B &\frac E2\\ \frac F2&\frac E2&C\end{bmatrix}

And the eigenvalues will determine the shape (same as A,B and C when we didn't have those pesky cross terms). The eigenvectors will determine the orientation.

That said, contours are certainly useful. You can also use partial derivatives to find the critical points (peaks, saddles, etc.)

Answer 2

This equation is a quadratic form, and as such can be characterized using linear algebra. If you are not familiar with the methods used (matrix diagonalization in particular), a bit more background information may be required.

The equation can first be rewritten in matrix form.

$$-\frac AG x^2-\frac BGy^2-\frac CG z^2-\frac DGxy-\frac EGyz-\frac FGxz=1$$

let $A=-\frac 1{2G}\begin{bmatrix}2A&D&F\\D&2B&E\\F&E&2C\end{bmatrix}$ and $\mathbf v=\begin{bmatrix}x\\y\\z\end{bmatrix}$. The above equation can then be written as:

$$\mathbf v^TA\mathbf v=1$$

Since $A$ is a symmetric matrix, it is (orthogonally) diagonalizable; which is to say $A=P^TDP$, where $D$ is a diagonal matrix and $P$ is an orthogonal matrix. Plugging in this expansion gives:

$$(P\mathbf v)^TD(P\mathbf v)=1$$

Let $\begin{bmatrix}a\\b\\c\end{bmatrix}=\mathbf w=P\mathbf v$, and let $\lambda_1$, $\lambda_2$, and $\lambda_3$ be the diagonal entries of $D$ (they are also the eigenvalues of $A$).

$$\mathbf w^TD\mathbf w=1$$

$$\lambda_1a^2+\lambda_2b^2+\lambda_3c^2=1$$

Basically, diagonalization transforms the original surface into one that has no cross terms, and useful symmeties about the $a$, $b$, and $c$ axes. This diagonalized quadratic form is much easier to characterize.

-If the eigenvalues $\lambda_1$, $\lambda_2$, and $\lambda_3$ are all positive, the equation describes an ellipsoid.

-If two are positive and one is zero, it describes an elliptical cylinder.

-If one is positive and two are zero, it describes two parallel planes.

-If two are positive and one is negative, it describes a hyperboloid of one sheet.

-If one is positive and two are negative, it describes a hyperboloid of two sheets.

-If one is positive, one is negative, and one is zero, it describes two sheets whose cross section is a hyperbola.

All of the other cases have no real solutions.

Let $V$ be the solution set of $\mathbf v^TA\mathbf v=1$ and $W$ be the solution set of $\mathbf w^TD\mathbf w=1$. Because $\mathbf v=P\mathbf w$ and $P$ is orthogonal, $V$ is just $W$ rotated and/or reflected about the origin, and has the same shape. Because of this, any distances or angles computed from $W$ are equivalent for $V$, and any vector quantities computed from $W$ are left multiplied by $P$ to obtain the corresponding vector in $V$.

If there is anything in this outline that requires more explanation or sources, let me know.