4.1.1 Functions of two variables: surfaces and level curves

A rule with two inputs

A function of two variables takes two numbers, $x$ and $y$ , and returns one number, written $f(x, y)$ . The pair $(x, y)$ is a point of the plane, as in lesson 3.1.1. So $f$ gives a number for each point.

You already know some. A rectangle $w$ cm wide and $h$ cm tall has area $A(w, h) = wh$ cm². Both inputs matter, and neither can be left out.

Take $f(x, y) = x^2 + 2y^2$ . At the point $(1, 2)$ it gives $1 + 8 = 9$ , and at $(2, 1)$ it gives $4 + 2 = 6$ . The order of the inputs matters: $x$ always comes first.

Rule: Function of two variables

(x, y) \mapsto f(x, y)

A rule that gives exactly one number $f(x, y)$ for each point $(x, y)$ where it is defined. The output is in its own units, such as cm² for an area.

A function of one variable fits in a row of values. A function of two variables needs a grid: one column for each $x$ and one row for each $y$ . Here is $x^2 + 2y^2$ on a grid of 25 points.

	$x = -2$	$-1$	0	1	2
$y = 1$	6	3	2	3	6
$y = 0.5$	4.5	1.5	0.5	1.5	4.5
$y = 0$	4	1	0	1	4
$y = -0.5$	4.5	1.5	0.5	1.5	4.5
$y = -1$	6	3	2	3	6

The smallest value, 0, is at the centre $(0, 0)$ . The values grow in every direction from there, faster in $y$ than in $x$ . From the centre, a step of 1 in $y$ climbs to 2, a step of 1 in $x$ only to 1.

x

import numpy as np def f(x, y): return x**2 + 2 * y**2 print(f(1, 2), f(2, 1)) xs = np.array([-2, -1, 0, 1, 2]) ys = np.array([1, 0.5, 0, -0.5, -1]) X, Y = np.meshgrid(xs, ys) print(X[0], Y[:, 0]) print(f(X, Y))

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The domain

The domain of $f$ is the set of points where the rule gives a real number (lesson 1.1.1). For $x^2 + 2y^2$ it is the whole plane.

The same two traps as in one variable can shrink it: a square root of a negative number, and division by zero. Logarithms need a positive input too.

$\sqrt{x - y}$ needs $x - y \ge 0$ , so $y \le x$ . The domain is the half of the plane on or below the line $y = x$ . At $(3, 1)$ the value is $\sqrt{2} \approx 1.4142$ ; at $(1, 2)$ there is none.
$\ln(x + y)$ needs $x + y > 0$ : the points above the line $y = -x$ , not on it. At $(3, -2)$ it is $\ln 1 = 0$ .
$\frac{1}{x^2 + y^2}$ needs $x^2 + y^2 \neq 0$ . That rules out one point, the origin $(0, 0)$ .

So a domain is a region of the plane, often bounded by a line or a curve. Check whether the boundary itself is in it, as with $\le$ against $<$ in lesson 1.1.1.

(2, 2)

import math def root(x, y): return math.sqrt(x - y) def log_sum(x, y): return math.log(x + y) for x, y in [(3, 1), (2, 2), (1, 2)]: try: print((x, y), round(root(x, y), 4)) except ValueError: print((x, y), "ValueError: outside the domain") for x, y in [(3, -2), (1, -2)]: try: print((x, y), log_sum(x, y)) except ValueError: print((x, y), "ValueError: outside the domain")

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The graph is a surface

The graph of a function of one variable is a curve: above each $x$ , a point at height $f(x)$ . With two inputs, put the point $(x, y)$ on a floor and raise a point above it to the height $z = f(x, y)$ . Together these points form a surface.

For $x^2 + 2y^2$ the surface is a bowl. Its bottom sits on the floor at $(0, 0)$ , where $f = 0$ , and its sides rise in every direction, more steeply in $y$ .

The bowl $z = x^2 + 2y^2$ , as a surface and as a map of level curves. The cut at height 4 casts the ellipse $f = 4$ on the floor. The slice $y = 1$ is the parabola $x^2 + 2$ .

A surface is hard to draw and harder to read off exactly. Two simpler pictures carry the same information: slices, which are curves you know from Level 2, and level curves, which are flat maps.

Slices

Hold $y$ fixed at a number $b$ , and $f(x, b)$ is a function of $x$ alone. Its graph is a slice: the curve where the upright plane $y = b$ cuts the surface. Holding $x$ fixed instead gives a slice in the $y$ direction.

For $x^2 + 2y^2$ , the slice $y = 1$ is $x^2 + 2$ : a parabola (lesson 1.2.2) with its lowest point 2 at $x = 0$ . The slice $x = 1$ is $1 + 2y^2$ , a narrower parabola.

moving input	$-2$	$-1$	0	1	2
slice $y = 1$ : $f(x, 1)$	6	3	2	3	6
slice $x = 1$ : $f(1, y)$	9	3	1	3	9

In the first row $x$ moves, and the values are the row $y = 1$ of the grid above. In the second row $y$ moves.

Rule: A slice

\begin{gathered} g(x) = f(x, b) \\[4pt] h(y) = f(a, y) \end{gathered}

Fix one input at a number and let the other move. The result is a function of one variable, and everything in Level 2 applies to it.

Every slice of the bowl in the $x$ direction is the same parabola $x^2$ , lifted by $2b^2$ . Lesson 4.1.2 takes the slopes of slices: those are the partial derivatives.

y = 1

def f(x, y): return x**2 + 2 * y**2 def across(x): # the slice y = 1 return f(x, 1) def along(y): # the slice x = 1 return f(1, y) points = [-2, -1, 0, 1, 2] print([across(x) for x in points], [x**2 + 2 for x in points]) print([along(y) for y in points], [1 + 2 * y**2 for y in points])

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Level curves and the contour map

Cut the surface with a flat plane at height $c$ and look straight down. The cut is a level curve: every point $(x, y)$ where $f(x, y) = c$ .

Draw level curves for several heights, each labelled with its $c$ , and you have a contour map. It works like the contour lines of a hiking map.

For the bowl, $x^2 + 2y^2 = c$ is an ellipse. Setting $y = 0$ gives $x = \pm\sqrt{c}$ , and setting $x = 0$ gives $y = \pm\sqrt{c/2}$ . So $f = 4$ crosses the $x$ -axis at $\pm 2$ and the $y$ -axis at $\pm\sqrt{2} \approx \pm 1.4142$ .

The level $c = 0$ is the single point $(0, 0)$ , and a negative $c$ gives no curve at all: the bowl never goes below the floor.

Rule: Level curve

\{(x, y) : f(x, y) = c\}

All the points where $f$ has the height $c$ . Moving along a level curve leaves $f$ unchanged. Curves for different heights never cross, because each point has only one height.

Reading steepness from the spacing

Walk out along the positive $x$ -axis. The curves $f = 1, 2, 3, 4$ cross it at 1, 1.4142, 1.7321 and 2, with gaps 0.4142, 0.3178 and 0.2679 between them. Each gap is one unit of height, and the gaps shrink, so the climb gets steeper.

Along the $y$ -axis the crossings are at 0.7071, 1, 1.2247 and 1.4142, with gaps 0.2929, 0.2247 and 0.1895. They are closer than along $x$ : the bowl is steeper in the $y$ direction, as the table showed.

So curves close together mean a steep surface, and curves far apart a gentle one. The labels say whether it rises or falls.

Quadratic forms

Lesson 3.7.1 drew level curves of quadratic forms and found that their eigenvalues decide the shape. $x^2 + 2y^2$ is the form of the matrix with diagonal entries 1 and 2, whose eigenvalues are 1 and 2. Both are positive, so the curves are ellipses, with semi-axes $\sqrt{c/1}$ and $\sqrt{c/2}$ .

A contour map works for any function of two variables, quadratic or not. Lessons 4.3.2 and 4.4.2 use these maps to find and reach the lowest points of more general functions.

Set a point with the sliders and read its height; the bold curve through it is its level curve. Move the point along that curve and the height stays the same. Then switch to $x^2 - y^2$ , a saddle met in the next section.

Explore

Each curve joins the points where the function has the same height, like the contour lines of a map. Pick a point and read its height; where the curves crowd together, the surface is steep.

Point x = 2

Point y = 1

Point: (2, 1), f = 6

(2\cos t, \sqrt{2}\sin t)

import math def f(x, y): return x**2 + 2 * y**2 for t in [0, 0.5, 1, 2, 3]: x, y = 2 * math.cos(t), math.sqrt(2) * math.sin(t) print(round(x, 4), round(y, 4), round(f(x, y), 10)) on_x = [math.sqrt(c) for c in [1, 2, 3, 4]] on_y = [math.sqrt(c / 2) for c in [1, 2, 3, 4]] print([round(v, 4) for v in on_x], [round(b - a, 4) for a, b in zip(on_x, on_x[1:])]) print([round(v, 4) for v in on_y], [round(b - a, 4) for a, b in zip(on_y, on_y[1:])])

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Four shapes

Four simple shapes come up again and again in this level. The bowl is the first.

A saddle: $x^2 - y^2$ . Its slice $y = 0$ is $x^2$ , which rises both ways, and its slice $x = 0$ is $-y^2$ , which falls both ways. So the surface is shaped like a horse's saddle, or a mountain pass. Its level curves are hyperbolas: $f = 3$ at $(2, 1)$ , $f = -3$ at $(1, 2)$ , and $f = 0$ on the two crossing lines $y = x$ and $y = -x$ .

A plane: $6 - 2x - 3y$ . Every slice is a straight line: $6 - 2x$ along $y = 0$ , $6 - 3y$ along $x = 0$ . The level curves $2x + 3y = 6 - c$ are parallel straight lines. Equal steps in height give equal gaps: along $y = 0$ the height falls by 1 every 0.5.

A product: $xy$ . The slice $y = 2$ is the line $2x$ , and in general the slice $y = b$ is the line $bx$ , with slope $b$ . The level curves $xy = c$ are hyperbolas, such as $(2, 3)$ and $(4, 1.5)$ on $f = 6$ . They lie in the quadrants where $x$ and $y$ have the same sign when $c > 0$ , and in the others when $c < 0$ .

Level curves of a saddle, a plane and the product $xy$ . Labels give the heights; the dashed lines are where $f = 0$ , with positive and negative heights in different colours.

The product $xy$ is the saddle turned by 45°. Algebra shows it: $xy = \frac{1}{4}\big((x + y)^2 - (x - y)^2\big)$ , a square minus a square.

As a quadratic form, $xy$ has the matrix $\begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix}$ . Its eigenvalues, $\frac{1}{2}$ and $-\frac{1}{2}$ , have both signs, so lesson 3.7.1 predicts hyperbolas.

function	slices	level curves	surface
$x^2 + 2y^2$	parabolas opening up	ellipses	bowl
$x^2 - y^2$	parabolas, up in $x$ and down in $y$	hyperbolas	saddle
$6 - 2x - 3y$	straight lines	parallel, evenly spaced lines	tilted plane
$xy$	straight lines through the origin	hyperbolas in the quadrants	saddle turned 45°

(2, 1)

import numpy as np def saddle(x, y): return x**2 - y**2 def plane(x, y): return 6 - 2 * x - 3 * y def product(x, y): return x * y print(saddle(2, 1), saddle(1, 2), saddle(3, 3), [saddle(x, 0) for x in [-2, 0, 2]], [saddle(0, y) for y in [-2, 0, 2]]) print([plane(x, 0) for x in [0, 0.5, 1, 1.5, 2]], plane(0, 2), plane(3, 0)) print(product(2, 3), product(4, 1.5), product(-2, -3), product(-2, 3), [product(x, 2) for x in [-1, 0, 1, 2]]) x, y = 2, 3 print(((x + y)**2 - (x - y)**2) / 4) print(np.linalg.eigvalsh(np.array([[0, 0.5], [0.5, 0]])), np.linalg.eigvalsh(np.diag([1.0, 2.0])))

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The same ideas reach functions of three or more variables, such as the volume $xyz$ of a box. Their graphs need more dimensions than paper has, but slices and partial derivatives (lesson 4.1.2) still work.

Common mistakes

The order of the inputs matters. For $x^2 + 2y^2$ , $f(1, 2) = 9$ but $f(2, 1) = 6$ .
A level curve's label is a height, not a distance. $x^2 + 2y^2 = 4$ crosses the $x$ -axis at 2, not at 4.
Close curves mean steep, not high. The bowl's crowded outer curves are steep sides; the height is in the labels.
Level curves of different heights never cross. One point has one height. The saddle's two lines where $f = 0$ cross, but they are the same level.
Check the domain. $\sqrt{x - y}$ has no value at $(1, 2)$ , and $\ln(x + y)$ none on the line $y = -x$ .
A slice fixes the other input at a number. The slice $y = 1$ of $x^2 + 2y^2$ is $x^2 + 2$ , a function of $x$ alone.

Functions of two variables

0 of 12 correct

1.
For $f(x, y) = x^2 + 2y^2$ , find $f(-1, 2)$
2.
For $f(x, y) = 3x - xy + y^2$ , find $f(2, -1)$
3.
The domain of $\sqrt{x - y}$ is the region $y \le x$ . With $x = 3$ , what is the largest $y$ in the domain?
4.
Find $\ln(x + y)$ at $(3, -2)$
5.
Write the slice $y = 3$ of $x^2 + 2y^2$ as a function of $x$
6.
The slice $y = b$ of $xy$ is the line $bx$ . What is the slope of the slice $y = -2$ ?
7.
Where does the level curve $x^2 + 2y^2 = 8$ cross the positive $y$ -axis?
8.
For $x^2 + 2y^2$ , the level curve $f = c$ crosses the $x$ -axis at $\sqrt{c}$ . For which $c$ does it cross at $x = 3$ ?
9.
Which level curve of $x^2 - y^2$ passes through $(3, 2)$ ? Give its height $c$
10.
The point $(4, y)$ lies on the level curve $xy = 6$ . Find $y$
11.
For the plane $6 - 2x - 3y$ , where does the level curve $f = 0$ cross the $x$ -axis?
12.
Along the $x$ -axis, how far apart are the level curves $f = 0$ and $f = 1$ of the plane $6 - 2x - 3y$ ?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

f(x, y)

Functions of two variables: surfaces and level curves

A rule with two inputs

Rule: Function of two variables

The domain

The graph is a surface

Slices

Rule: A slice

Level curves and the contour map

Rule: Level curve

Reading steepness from the spacing

Quadratic forms

Four shapes

Common mistakes

Functions of two variables

Summary card

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