# Is this a ruled surface?

I'd like to know if the contour surfaces of the following $f$ are ruled:

$f(x, y, z)$ is a 3-dim scalar field (real-valued function) defined in the 1st octant $x > 0,\, y > 0,\,z > 0$ that exhibits the following scaling property:

$$f(x, y, z) = f\left(~ \frac{z_0}z \, x, \frac{z_0}z \, y,\, z_0 ~\right) \qquad \text{or} \qquad f(x, y, z) = f(\, sx,\, sy,\, s z) \tag*{s \equiv \frac{z_0}z}$$ That is, for any point, the value of $f$ can also be found on a reference plane $z = z_0$ by scaling radially $(x, y) \mapsto (sx, sy)$. Of course, any plane $z = z_0 > 0$ can act as the reference plane, and each contour surface $f(x, y, z) = p$ is in fact "swept out" by rays originating from the origin and passing through $(x, y, z)$.

This reminds me of ruled surface for the following reason:

given a contour surface $f(x, y, z) = p$, there is a level curve on the reference plane $g(x, y)=f(x, y, z_0)=p$ which can be parametrized into $g\left(\, x(u),\, y(u) \, \right) \to \beta(u)$ as in $$f = \alpha(u) + v \beta(u) \qquad \text{with v = z}$$ where $(u, v)$ are the parameters, with $\alpha(u) = \{0, 0, 0\}$ being a trivial base curve (directrix) and $\beta(u) = \{ x(u), y(u), z_0 \}$ the director. There is also the alternative parametrization of interpolation/extrapolation of two non-intersecting curves on the contour surface: just take two planar level curves at two different heigh $z = z_1$ and $z = z_2$.

However, I'm not sure if the base curve is allowed to be degenerate. In addition, I have trouble constructing a parametrization with rulings orthogonal to the rays.

Formally,my question are:

1. Does the scaling property above implies it to be a ruled surface?
2. If it is ruled, is it necessary that there exists a parametrization with two orthogonal rulings?

I have very little knowledge about even intro level differential geometry, and I basically would like to know if my field $~f~$ belongs to this type of function (ruled surface) so that I should go ahead and learn the relevant tools and study it that way.

p.s.

I don't know if this is relevant, but the field $f$ and its 1st derivatives are all continuous, but there are some boundaries of regions where the 2nd derivatives are not continuous with infinite jumps. The boundaries are like $z = y$ and $z = \sqrt{x^2 + y^2}$.

$\newcommand{\Orthant}{\mathcal{O}}$If $f$ is a $C^{1}$ real-valued function in the positive orthant $\Orthant$, and if $$f(x, y, z) = f(sx, sy, sz),\quad s > 0, \tag{1}$$ for all $(x, y, z)$ in $\Orthant$, then:

1. Yes, each level set of $f$ is a ruled surface (provided the level set is a surface).

2. No, a ruled surface need not even admit two families of rulings; that is, there may exist only one ruling through each point of the surface.

Notes:

1. A ruled surface with constant directrix is sometimes called a (generalized) cone. If it matters, note that you can instead parametrize your levels by $$\beta(u) + v\beta(u) = (1 + v)\beta(u),$$ using a "coordinate slice" as both the directrix and director.

2. The coordinate slices of the levels are ("usually") not rulings. The only rulings guaranteed by the homogeneity property (1) are sections of the level set by a plane through the origin.

• Thanks for the pointer! Just one terminology to make sure: I've read that only some surfaces are doubly-ruled and I can understand you're saying that the construction of my surface by the `rays' doesn't guarantee double ruling. I'm mostly wondering about the contour curves for parameter $v$... are these not called rulings? – Lee David Chung Lin Nov 25 '16 at 16:11
• In the context of ruled surfaces, "rulings" are Euclidean lines. Contour curves with $v$ constant are not (generally) lines. :) – Andrew D. Hwang Nov 25 '16 at 16:23