Back to the on-screen lesson ·
A height for each point of a region: the domain and range of $f(x, y)$, its graph as a surface, the traces and level curves that let a flat page describe one, and the level surfaces that do the same job for $f(x, y, z)$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the domain and range of a function of two variables, evaluate it at a point, identify its level curves and say which one a given point lies on, read a contour map for steepness rather than for height, and describe the level surfaces of a function of three variables.
Functions of one variable, their graphs as curves in the plane, and domains as sets of numbers. Unit 1 gave you coordinates in space. This lesson joins the two: a function of two variables has a graph, and the graph needs all three axes.
The domain of $f(x, y)$ is the set of points of the plane where the rule makes sense; the range is the set of heights it produces. The graph is the surface $z = f(x, y)$ in space. A level curve is the set of points where $f$ takes one fixed value $c$, and a contour map is several of them drawn together with their heights labelled; the contour interval is the fixed height gap between neighbouring curves. For a function of three variables the same idea gives a level surface instead, and a trace is what you see when you slice a surface with a plane such as $x = 0$.
A function of two variables assigns one number to each point of a region of the plane:
$$f : D \subseteq \mathbb{R}^2 \to \mathbb{R}, \qquad (x, y) \mapsto z.$$
The domain $D$ is where the rule makes sense, and unlike the one-variable case it is a region rather than a set of intervals: $\sqrt{9 - x^2 - y^2}$ needs a closed disc, $\ln(x - y)$ needs an open half-plane, and $1/(xy)$ needs the plane with two lines removed. Sketching the domain first is not busywork; it is the only reliable way to notice that a later integral has to stop somewhere.
The graph is a surface. Plotting $z = f(x, y)$ above each point of the domain produces a surface in space. Reading it directly is hard — paper is flat — so two devices do the reading instead.
Traces. Fix one variable and look at what is left. Setting $y = 0$ in $z = x^2 + y^2$ leaves the parabola $z = x^2$ in the $xz$ plane; setting $z = 4$ leaves the circle $x^2 + y^2 = 4$. A few traces in each direction usually identify a surface completely, and it is how the quadric surfaces of lesson 7 were told apart.
Level curves. The level curve at height $c$ is
$$\{(x, y) \in D : f(x, y) = c\}.$$
It lives in the plane, not in space: it is the horizontal slice $z = c$ pushed flat onto the floor and labelled with its height. Draw several at equally spaced heights and you have a contour map, which carries the whole surface in two dimensions. Because the heights are equally spaced, how close the curves lie is how steep the surface is — and that single reading is what the gradient will make exact in lesson 18.
Three variables and more. $f(x, y, z)$ has no graph anybody can draw — it would need four axes — but it still has level surfaces $f(x, y, z) = c$. For $f = x^2 + y^2 + z^2$ they are spheres; for the temperature in a room they are the surfaces of constant temperature. Nothing about the algebra changes, and the whole of the rest of this course is written for as many variables as the problem has.
Another way: picture
Hold a bowl above a table. Shine a light straight down and dip the bowl into water at several fixed depths; each waterline, traced onto the table below, is a level curve. Rings close together mean the bowl's wall is steep; rings far apart mean it is nearly flat. Nothing about the picture tells you how high the bowl is being held — only how it is shaped.
Another way: steps
To get a feel for an unfamiliar $f(x, y)$:
| $f(x, y)$ | Level curves | What the surface is |
|---|---|---|
| $ax + by$ | parallel lines | a tilted plane |
| $x^2 + y^2$ | circles about the origin | a bowl (paraboloid) |
| $x^2 - y^2$ | hyperbolas, and two lines at height $0$ | a saddle |
| $xy$ | hyperbolas, and the axes at height $0$ | the same saddle, turned |
Two of these deserve a second look. The level curves of $x^2 + y^2$ are circles whose radii go as $\sqrt{c}$, so equally spaced heights give rings that crowd together further out: the bowl steepens. And the level set of $x^2 - y^2$ at height $0$ is a pair of crossing lines — the one height at which a level set is allowed to cross itself, because it passes through the saddle point where the surface is level in two directions at once.
That last observation is worth storing. A contour map is a partition of the domain: each point has exactly one height, so two curves at different heights can never meet. A single curve crossing itself, though, is a signal, and it marks the critical points that lesson 19 will hunt for.
Drawing a level curve in space. A level curve lives in the $xy$ plane. The trace $z = c$ lives on the surface at height $c$. They have the same shape and different addresses, and saying which one you mean prevents an hour of confusion later.
Skipping the domain. $\sqrt{9 - x^2 - y^2}$ is defined only on a disc of radius $3$, and every level curve of it is a circle inside that disc. Starting to sketch before noticing that produces a picture of something that does not exist.
Reading contour spacing as height. Close contours mean steep, not high. A cliff at sea level draws a denser map than a broad plateau at two thousand metres.
Assuming the one-variable picture transfers. A function of one variable is increasing or decreasing at a point; a function of two is increasing in some directions and decreasing in others at the same point. A third coordinate is not a third term bolted onto a two-dimensional formula. The cross product exists only in space, a plane is fixed by a direction rather than by a slope, and a curve in space can bend without ever leaving a surface. Checking a new formula against the plane is worth doing; assuming the plane's formula still governs is how most of the errors in this course begin.
Two confusions account for most early errors here, and both come from squeezing three dimensions onto flat paper.
The first is between the surface and its contour map. The surface $z = f(x, y)$ is a two-dimensional sheet sitting in space; the contour map is a collection of curves drawn in the flat domain, each labelled with a number. The map is a description of the surface, in the way a plan is a description of a building, and asking which quadrant a level curve is in makes sense while asking how high it is does not — its height is written on its label.
The second is reading closeness as height. Contours are drawn at a fixed height interval, so the only information in their spacing is how much horizontal distance one interval of climb takes. Crowded contours mean a steep slope and say nothing about altitude; a sea cliff and a mountain cliff draw the same crowd of lines.
For $f(x, y) = \sqrt{16 - x^2 - y^2}$ the rule needs $16 - x^2 - y^2 \ge 0$, so the domain is the closed disc of radius $4$.
A domain in two variables is a region, not a list of intervals.
Setting $f = c$ gives $x^2 + y^2 = 16 - c^2$: a circle of radius $\sqrt{16 - c^2}$, which exists for $0 \le c \le 4$. So the range is $[0, 4]$.
The level curves also settle the range.
At $c = 0$ the circle has radius $4$ — the rim; at $c = 4$ it shrinks to the single point at the origin — the top. The surface is the upper half of a sphere, and the contours crowd near the rim, which is where a hemisphere is steepest.
The map identified the surface without any drawing in space.
For $f(x, y) = x^2 - y^2$, the level curve at height $9$ is $x^2 - y^2 = 9$: a hyperbola opening left and right.
Positive heights open along the $x$ axis.
At height $-9$ it is $y^2 - x^2 = 9$, a hyperbola opening up and down; at height $0$ it is $y = \pm x$, two crossing lines.
Negative heights open the other way.
So the surface rises along the $x$ axis, falls along the $y$ axis, and is level at the origin in both — a saddle. The origin is a point where the surface is flat and is neither a peak nor a pit, which is the case that makes the second derivative test of lesson 19 necessary.
Crossing contours mark the interesting point.
For $f(x, y) = xy$, the level curve at height $6$ is $xy = 6$: a hyperbola with one branch in the first quadrant and one in the third.
A product is positive when the factors share a sign.
At height $-6$ the branches sit in the second and fourth quadrants, and at height $0$ the level set is the pair of axes themselves.
So the surface climbs into two opposite quadrants and falls into the other two, with the origin level in both axis directions: a saddle again, turned through forty-five degrees. Recognising $xy$ and $x^2 - y^2$ as the same surface in two poses saves re-deriving everything about the second of them.
The level curve of $f(x, y) = 8x + 5y$ at height $40$ is a straight line. Plot the two points where it crosses the axes.
Plot your answer on the grid:
Every level curve of $f(x, y) = x^2 + y^2$ is a circle centred at the origin. Give the radius of each of these three.
| Radius | |
|---|---|
| The level curve at height $16$ | |
| The level curve at height $64$ | |
| The level curve at height $144$ |
For $f(x, y) = 2x^2 + 5y$, what is $f(5, -5)$?
Answer:
On which level curve of $f(x, y) = 3x - 4y$ does the point $(-5, -3)$ lie?
Answer:
The level curve of $f(x, y) = 3x + 9y$ at height $27$ is a straight line. Plot the two points where it crosses the axes.
Plot your answer on the grid:
A walking map draws a contour line every $20$ metres of height. In one place the lines are packed very close together. What does that say about the ground there?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Every level curve of $f(x, y) = x^2 + y^2$ is a circle centred at the origin. Give the radius of each of these three.
| Radius | |
|---|---|
| The level curve at height $9$ | |
| The level curve at height $36$ | |
| The level curve at height $81$ |
You can turn a rule in two variables into a domain, a handful of level curves and a picture of the surface, and you can say what crowded contours mean. Next: what it takes for such a function to have a limit at a point, when there are infinitely many ways to arrive.
10. Your turn: the contour map of a product, step 3