Question 10.6.7: Sketching a Hyperboloid of Two Sheets Draw a graph of the qu...
Sketching a Hyperboloid of Two Sheets
Draw a graph of the quadric surface
\frac{x^2}{4}-y^2-\frac{z^2}{2}=1 .Notice that this is the same equation as in example 6.6, except for the sign
of the y \text {-term. } As we have done before, we first look at the traces in the three coordinate planes. The trace in the y z \text {-plane }(x=0) \text { is } defined by
Since it is clearly impossible for two negative numbers to add up to something positive, this is a contradiction and there is no trace in the y z \text {-plane. } That is, the surface does not intersect the y z \text {-plane. } . The traces in the other two coordinate planes are as follows:
x y \text {-plane }(z=0): \frac{x^2}{4}-y^2=1 \text { (hyperbola) }and x z \text {-plane }(y=0): \frac{x^2}{4}-\frac{z^2}{2}=1 \text { (hyperbola) }.
We show these traces in Figure 10.60a. Finally, notice that for x=k, we have that
y^2+\frac{z^2}{2}=\frac{k^2}{4}-1,
so that the traces in the plane x=k \text { are ellipses for } k^2>4 \text {. } It is important to notice here that if k^2<4 \text {, the equation } y^2+\frac{z^2}{9}=\frac{k^2}{4}-1 has no solution. (Why is that?) So, for -2<k<2 \text {, the surface has no trace at all in the plane } x=k \text {, } leaving a gap that separates the hyperbola into two sheets. Putting this all together, we have the surface seen in Figure 10.60b. We call this surface a hyperboloid of two sheets.
We can plot this on a CAS by graphing the two functions z=\sqrt{2\left(\frac{x^2}{4}-y^2-1\right)} and z=-\sqrt{2\left(\frac{x^2}{4}-y^2-1\right)} \text {. (See Figure 10.60c,where we have restricted the } z \text {-range } to -10 \leq z \leq 10, to show the elliptical cross sections.) Notice that this plot shows large gaps between the two halves of the hyperboloid. However, if you have drawn Figure 10.60b yourself, this plotting deficiency won’t fool you.
Alternatively, the parametric plot with x=2 \cosh s, y=\sinh s \cos t and z=\sqrt{2} \sinh s \sin t, \text { for }-4 \leq s \leq 4 \text { and } 0 \leq t \leq 2 \pi, produces the left half of the hyperboloid with its elliptical and hyperbolic traces. The right half of the hyperboloid has parametric equations x=-2 \cosh s, y=\sinh s \cos t \text { and } z=\sqrt{2} \sinh s \sin t, with -4 \leq s \leq 4 \text { and } 0 \leq t \leq 2 \pi . We show both halves in Figure 10.60d.
