Activity 6.5.1. Slicing the Hyperboloids.
Use the same slicing method as in the previous subsections to study the two surfaces
\begin{equation*}
\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1
\qquad \text{and} \qquad
\frac{z^2}{16} - \frac{x^2}{4} - \frac{y^2}{9} = 1
\end{equation*}
and discover why the first is called a hyperboloid of one sheet while the second is called a hyperboloid of two sheets. Pay particular attention to the traces in the horizontal planes \(z = c\text{:}\) they are what tells the two surfaces apart.
(a)
Both surfaces can be obtained from equation (6.1). Find the values of \(A, B, C, D,\) and \(E\) that produce each surface. Compared with the ellipsoid, what changed in the signs of the coefficients?
Solution.
For the first surface \(A = 1/4\text{,}\) \(B = 1/9\text{,}\) \(C = -1/16\text{,}\) \(D = 0\text{,}\) and \(E = 1\text{.}\) For the second surface \(A = -1/4\text{,}\) \(B = -1/9\text{,}\) \(C = 1/16\text{,}\) \(D = 0\text{,}\) and \(E = 1\text{.}\) For the ellipsoid all three squared terms were positive; here exactly one squared term is negative for the first surface and exactly two are negative for the second. Keep this count in mind as you work through the traces.
(b)
For the surface \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1\text{,}\) find the trace in the plane \(z = c\) for an arbitrary constant \(c\text{,}\) and the traces in the coordinate planes \(x = 0\) and \(y = 0\text{.}\) For which values of \(c\) does the plane \(z = c\) actually intersect the surface? Identify each trace as an ellipse, parabola, or hyperbola.
Solution.
Slicing with the plane \(z = c\) gives
\begin{equation*}
\frac{x^2}{4} + \frac{y^2}{9} = 1 + \frac{c^2}{16}
\end{equation*}
Since the right-hand side is positive for every value of \(c\text{,}\) every horizontal plane meets the surface in an ellipse. The smallest one, \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) in the plane \(z = 0\text{,}\) is the “waist” of the surface, and the ellipses grow as \(|c|\) increases. In the coordinate planes we find hyperbolas:
\begin{equation*}
\frac{y^2}{9} - \frac{z^2}{16} = 1 \; (x = 0)
\qquad
\frac{x^2}{4} - \frac{z^2}{16} = 1 \; (y = 0)
\end{equation*}
The traces are shown in Figure 6.13, and Figure 6.14 lets you slice the surface yourself.
Diagram Exploration Keyboard Controls
Instructions.
Use the buttons to slice the surface \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1\) with planes \(x=c\text{,}\) \(y=c\text{,}\) or \(z=c\text{,}\) and drag the slider to vary \(c\text{.}\) The equation of each cross-section is displayed above the figure. Drag the figure to view it from a different angle, or press the “Rotate” button to spin it automatically.
The animation in Figure 6.15 shows these slices being taken one at a time.
(c)
Repeat the analysis for the surface \(\frac{z^2}{16} - \frac{x^2}{4} - \frac{y^2}{9} = 1\text{.}\) For which values of \(c\) does the plane \(z = c\) intersect the surface this time? What happens for \(c\) between those values?
Solution.
Slicing with the plane \(z = c\) now gives
\begin{equation*}
\frac{x^2}{4} + \frac{y^2}{9} = \frac{c^2}{16} - 1
\end{equation*}
and this time the sign of the right-hand side matters. For \(|c| \lt 4\) the right-hand side is negative, so the plane misses the surface entirely; there is no trace at all. For \(c = \pm 4\) the trace is the single point \((0, 0, \pm 4)\text{,}\) and only for \(|c| \gt 4\) do we get ellipses, which grow as \(|c|\) increases. In the coordinate planes we again find hyperbolas, both opening up and down:
\begin{equation*}
\frac{z^2}{16} - \frac{y^2}{9} = 1 \; (x = 0)
\qquad
\frac{z^2}{16} - \frac{x^2}{4} = 1 \; (y = 0)
\end{equation*}
The traces are shown in Figure 6.16, and Figure 6.17 lets you watch the plane \(z = c\) pass through the gap without touching the surface.
Diagram Exploration Keyboard Controls
Instructions.
Use the buttons to slice the surface \(\frac{z^2}{16} - \frac{x^2}{4} - \frac{y^2}{9} = 1\) with planes \(x=c\text{,}\) \(y=c\text{,}\) or \(z=c\text{,}\) and drag the slider to vary \(c\text{.}\) The equation of each cross-section is displayed above the figure. Drag the figure to view it from a different angle, or press the “Rotate” button to spin it automatically.
The animation in Figure 6.18 shows these slices being taken one at a time.
(d)
Using only the traces in the planes \(z = c\text{,}\) explain why the first surface is called a hyperboloid of one sheet and the second a hyperboloid of two sheets. Can you predict the number of sheets directly from the signs in the equation?
Solution.
The horizontal traces tell the two surfaces apart. For the first surface the plane \(z = c\) produces an ellipse for every value of \(c\text{:}\) the ellipses stack on top of one another without interruption, so the surface is a single connected piece — one sheet. For the second surface there is no trace at all when \(-4 \lt c \lt 4\text{:}\) the surface has a gap around the origin and splits into two separate pieces, one with \(z \geq 4\) and one with \(z \leq -4\) — two sheets.
The signs in the equation predict this without any graphing. With the equation written with \(1\) on the right-hand side, count the negative squared terms: one negative term gives a hyperboloid of one sheet, and two negative terms give a hyperboloid of two sheets. The axis of the hyperboloid is the axis of the variable that appears with the minority sign: for the first surface the \(z\)-term is the lone negative, and for the second the \(z\)-term is the lone positive, so both hyperboloids have the \(z\)-axis as their axis.
