Consider first \(c=0\text{.}\) The level curve for \(c=0\) is the set of all points \((x,y)\) such that \(0=\sqrt{1-\frac{x^2}9-\frac{y^2}4}\text{.}\) Squaring both sides gives us \begin{equation*} \frac{x^2}9+\frac{y^2}4=1, \end{equation*} an ellipse centered at \((0,0)\) with horizontal major axis of length 6 and minor axis of length 4. Thus for any point \((x,y)\) on this curve, \(f(x,y) = 0\text{.}\)

Now consider the level curve for \(c=0.2\) \begin{align*} 0.2 \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4}\\ 0.04 \amp = 1-\frac{x^2}9-\frac{y^2}4\\ \frac{x^2}9+\frac{y^2}4 \amp =0.96\\ \frac{x^2}{8.64}+\frac{y^2}{3.84} \amp =1. \end{align*}

This is also an ellipse, where \(a = \sqrt{8.64}\approx 2.94\) and \(b=\sqrt{3.84}\approx 1.96\text{.}\)

In general, for \(z=c\text{,}\) the level curve is: \begin{align*} c \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4}\\ c^2 \amp = 1-\frac{x^2}9-\frac{y^2}4\\ \frac{x^2}9+\frac{y^2}4 \amp =1-c^2\\ \frac{x^2}{9(1-c^2)}+\frac{y^2}{4(1-c^2)} \amp =1, \end{align*} ellipses that are decreasing in size as \(c\) increases. A special case is when \(c=1\text{;}\) there the ellipse is just the point \((0,0)\text{.}\)

The level curves are shown in Figure 12.1.10(a). Note how the level curves for \(c=0\) and \(c=0.2\) are very, very close together: this indicates that \(f\) is growing rapidly along those curves.

In Figure 12.1.10(b), the curves are drawn on a graph of \(f\) in space. Note how the elevations are evenly spaced. Near the level curves of \(c=0\) and \(c=0.2\) we can see that \(f\) indeed is growing quickly.

We begin by setting \(f(x,y)=c\) for an arbitrary \(c\) and seeing if algebraic manipulation of the equation reveals anything significant. \begin{align*} \frac{x+y}{x^2+y^2+1} \amp = c\\ x+y \amp = c(x^2+y^2+1).\\ \end{align*} We recognize this as a circle, though the center and radius are not yet clear. By completing the square, we can obtain: \begin{align*} \left(x-\frac{1}{2c}\right)^2+\left(y-\frac1{2c}\right)^2\amp =\frac{1}{2c^2}-1, \end{align*} a circle centered at \(\big(1/(2c),1/(2c)\big)\) with radius \(\sqrt{1/(2c^2)-1}\text{,}\) where \(\abs{c}\lt 1/\sqrt{2}\text{.}\) The level curves for \(c=\pm 0.2,\ \pm 0.4\) and \(\pm0.6\) are sketched in Figure 12.1.14(a). To help illustrate “elevation,” we use thicker lines for \(c\) values near 0, and dashed lines indicate where \(c\lt 0\text{.}\)

There is one special level curve, when \(c=0\text{.}\) The level curve in this situation is \(x+y=0\text{,}\) the line \(y=-x\text{.}\)

In Figure 12.1.14(b) we see a graph of the surface. Note how the \(y\)-axis is pointing away from the viewer to more closely resemble the orientation of the level curves in (a).

Seeing the level curves helps us understand the graph. For instance, the graph does not make it clear that one can “walk” along the line \(y=-x\) without elevation change, though the level curve does.

\(\ds f(3,0,2) = \frac{3^2+2+3\sin(0) }{3+2(0)-2} = 11\text{.}\)

As the domain of \(f\) is not specified, we take it to be the set of all triples \((x,y,z)\) for which \(f(x,y,z)\) is defined. As we cannot divide by \(0\text{,}\) we find the domain \(D\) is \begin{equation*} D = \{(x,y,z)\ |\ x+2y-z\neq 0\}. \end{equation*}

We recognize that the set of all points in \(\mathbb{R}^3\) that are not in \(D\) form a plane in space that passes through the origin (with normal vector \(\la 1,2,-1\ra\)).

We determine the range \(R\) is \(\mathbb{R}\text{;}\) that is, all real numbers are possible outputs of \(f\text{.}\) There is no set way of establishing this. Rather, to get numbers near 0 we can let \(y=0\) and choose \(z \approx -x^2\text{.}\) To get numbers of arbitrarily large magnitude, we can let \(z\approx x+2y\text{.}\)

We can (mostly) answer this question using “common sense.” If energy (say, in the form of light) is emanating from the origin, its intensity will be the same at all points equidistant from the origin. That is, at any point on the surface of a sphere centered at the origin, the intensity should be the same. Therefore, the level surfaces are spheres.

We now find this mathematically. The level surface at \(I=c\) is defined by \begin{align*} c \amp = \frac{1}{x^2+y^2+z^2}.\\ \end{align*} A small amount of algebra reveals \begin{align*} x^2+y^2+z^2 \amp = \frac1c. \end{align*}

Given an intensity \(c\text{,}\) the level surface \(I=c\) is a sphere of radius \(1/\sqrt{c}\text{,}\) centered at the origin.

Figure 12.1.20 gives a table of the radii of the spheres for given \(c\) values. Normally one would use equally spaced \(c\) values, but these values have been chosen purposefully. At a distance of 0.25 from the point source, the intensity is 16; to move to a point of half that intensity, one just moves out 0.1 to 0.35 — not much at all. To again halve the intensity, one moves 0.15, a little more than before.

Note how each time the intensity if halved, the distance required to move away grows. We conclude that the closer one is to the source, the more rapidly the intensity changes.