How level sets associated with a quadratic expression can switch from one to two components?

Question

So I have two equations:

$$ x^2 + y^2 - z^2 = 1/2 $$ and $$ x^2 + y^2 - z^2 = -1/2, $$ which represent single-sheeted hyperboloid, and a two-sheeted hyperboloid, respectively. How do you explain how this different sign has this effect on the level surfaces?

Answer 1

The best way is to look at level surfaces at constant $z$. At $z=1/2$ you get $$x^2+y^2=3/4$$ for the first one, and $$x^2+y^2=-1/4$$ for the second one. The first one therefore has a circle as the level surface, while the second has no solution, which disconnects the graph. For larger values of $z$ you do get a circle for both, but the fact that there are values of $z$ for which no solution exists makes the second graph disconnected, into 2 sheets.

Answer 2

In normal form, the constant is always positive. So in normal form, you have \begin{align*} x^2 + y^2 - z^2 &= 1/2 &\text{and} \\ -x^2 + -y^2 + z^2 &= 1/2 \text{.} \end{align*} Then count the minus signs to see how many sheets there are.

To expose the geometry, arrange for all variable's signs to be positive. For $$ x^2 + y^2 = z^2 + 1/2 $$ the left-hand side is a circle and the right-hand side is a radius lower bounded by $\sqrt{1/2}$, so the upper and lower barts are joined by a "tube" of radius $/sqrt{1/2}$. For $$ x^2 + y^2 = z^2 - 1/2 $$ the left-hand side is a circle and the right-hand side only gives realizable radii when $|z| \geq \sqrt{1/2}$.

Answer 3

Indeed, a graphical explanation is very convincing. Look at the (rather fundamental) figure :

on which we see 3 "level surfaces" of function :

$$w=f(x,y,z)=x^2+y^2-z^2$$

for values $w=-1/2, 0,1/2$ representing resp. an hyperboloid with 1 sheet, a cone which is a transition surface, and an hyperboloid with 2 sheets.

In fact, level surfaces generalize level curves to dimension 3.

Answer 4

Note that the expression $x^2+y^2-z^2$ has rotational symmetry about the $z$-axis, which can be seen by introducing polar coordinates:

• $x=r\cos t, y= r\sin t \Rightarrow r^2-z^2 = \pm\frac{1}{2}$

This means, any intersection of a plane containing the $z$-axis with $x^2+y^2-z^2 = \pm \frac{1}{2}$ produces the curve $r^2-z^2 = \pm\frac{1}{2}$ with $r\geq 0$.

Factoring shows that we have in both cases a hyperbolic curve:

• $(r-z)(r+z) = \pm\frac{1}{2} \Leftrightarrow r+z = \pm\frac{1}{2(r-z)}$

Now, just check the shape of the two curves (for example plotting on an $r$-$z$-coordinate system):

• $r^2-z^2 =\frac{1}{2} \Leftrightarrow z = \pm\sqrt{r^2-\frac{1}{2}} \Rightarrow$ only one branch (the seeming two branches are connected at $z =0 \Leftrightarrow r=\frac{1}{\sqrt{2}}$) $\Rightarrow$ one sheet
• $r^2-z^2 =-\frac{1}{2} \Leftrightarrow z = \pm\sqrt{r^2+\frac{1}{2}} \Rightarrow$ two branches $\Rightarrow$ two sheets