Level curves

The two main ways to visualize functions of two variables is via graphs and level curves. Both were introduced in an earlier learning module.

For your convenience, that learning module page is reproduced here:

Level curves: for a function $z=f (x,\,y) :\, D \subseteq {\mathbb R}^2 \to {\mathbb R}$ the level curve of value $c$ is the curve $C$ in $D \subseteq {\mathbb R}^2$ on which $f\Bigl|_{C} = c\, $.

Notice the critical difference between a level curve $C$ of value $c$ and the trace on the plane $z = c$: a level curve $C$ always lies in the $xy$-plane, and is the set $C$ of points in the $xy$-plane on which $f(x,\,y) = c$, whereas the trace lies in the plane $z = c$, and is the set of points $(x,\,y,\, c)$ with $(x,\,y)$ in $C$.

    By combining the level curves $f(x,\,y) = c$ for equally spaced values of $c$ into one figure, say $c = -1, \,0,\, 1,\, 2,\, \ldots \,,$ in the $xy$-plane, we obtain a contour map of the graph of $z=f(x,\,y)$. Thus the graph of $z = f(x,\,y)$ can be visualized in two ways,

  one as a surface in $3$-space, the graph of $z = f(x,\,y)$,

  the other as a contour map in the $xy$-plane, the level curves of value $c$ for equally spaced values of $c$.

Problem: Describe the contour map of a plane in $3$-space. Solution: The equation of a plane in $3$-space is $$Ax + By + Cz \ = \ D\,,$$ so the horizontal plane $z= c$ intersects the plane when $$Ax + By +Cc \ = \ D\,.$$

For each $c$, this is a line with slope $-A/B$ and $y$-intercept $y = (D-Cc)/B$. Since the slope does not depend on $c$, the level curves are parallel lines, and as $c$ runs over equally spaced values these lines will be a constant distance apart. Consequently, the contour map of a plane consists of equally spaced parallel lines. (Does this make good geometric sense?)