# Surfaces and contour maps: functions of two variables

> A hiking map shows a mountain on flat paper. Its contour lines are the level sets of a function of two variables, and their spacing measures steepness.

Series: Multivariable Calculus (https://www.monocurl.com/monocurl-essays/multivariable-calculus/)
Canonical: https://www.monocurl.com/monocurl-essays/multivariable-calculus/functions-of-two-variables/
Keywords: function of two variables, contour map, level curves, level set, graph of a function of two variables, contour lines and steepness, how to read a contour map

[Figure 1, animation: A flooded valley drains. Each time the water has dropped another 100 m, its shoreline is marked on the hill. Every mark is level, and together the marks are the contour lines of the hill.]

A sheet of paper is flat and a mountain is not. Yet a hiker with a good map can tell, before leaving the parking lot, where the summit is, which slope is too steep to climb, and where the easiest crossing between two peaks lies. All of that information is carried by thin brown lines with numbers on them. How does a flat drawing hold a third dimension?

Temperature across a metal plate and pressure on a weather chart also depend on two inputs at once, and the contour map is the standard way to see any such quantity. We will build the map for one hill, then say exactly what was built.

## A height for every point of the map

Start with what a surveyor would collect. Put the origin at a marker post, measure $x$ in kilometers east of it and $y$ in kilometers north, and at each of a grid of positions write down the height of the ground. To keep the numbers small we measure height in hundreds of meters, so a height of 6 means 600 m.

[Figure 2, image: Surveyed heights on the hill, in hundreds of meters, at points half a kilometer apart. The position x is kilometers east of the marker post and y is kilometers north.]

Read along the middle row, where $y = 0$. Going east the heights run 4.9, 4.0, 4.6, 6.0, 7.4, 8.0, 7.1. So the ground dips, climbs to 8.0 at the point $(1, 0)$, and falls again. The rows above and below are each lower than the middle row by exactly 1.

**Pause and try.** Using only the 21 numbers in the figure, where would you guess the summit is? A hiker walks from the west edge to the summit along the middle row. What is the lowest height on that walk, and is the ground at that place lower or higher than the ground just north and south of it?

**Answer.** The largest number is 8.0 at $(1, 0)$, and its neighbors on all four sides are lower, so the summit is at or near $(1, 0)$. On the walk along the middle row the lowest height is 4.0, at $(-1, 0)$. The points north and south of it, $(-1, 1)$ and $(-1, -1)$, are at 3.0. So that place is the low point of the east-west walk and at the same time the high point of the north-south line through it. A place like that is called a pass.

Twenty-one numbers are a sample. The hill itself has a height above *every* point of the map, and for this lesson the hill is given by a formula:

$$
f(x, y) = 6 + 3x - x^3 - y^2.
$$

Check it against the figure: $f(0, 0) = 6$, $f(1, 0) = 6 + 3 - 1 = 8$, and $f(-1, 1) = 6 - 3 + 1 - 1 = 3$. The term $-y^2$ is why the rows at $y = \pm 1$ are lower than the middle row by 1. The map covers $-1.75 \le x \le 2$ and $-1.5 \le y \le 1.5$.

A rule like this, which takes a pair of numbers and returns one number, is the object the rest of the series studies.

**Definition.** Let $D$ be a set of points in the plane. A **function of two variables** with **domain** $D$ is a rule $f$ that assigns to each point $(x, y)$ in $D$ exactly one real number $f(x, y)$. The set of all values $f(x, y)$ is the **range** of $f$.

For the hill, $D$ is the rectangle the map covers, and the range is the set of heights that occur on it, from 1.75 up to 8 at the summit. (The value 1.75 occurs at the eastern corners $(2, \pm 1.5)$ and at $(-1, \pm 1.5)$, due north and south of the pass.) "Exactly one" matters: the ground cannot be at 300 m and at 400 m over the same spot, and we will use that below.

## The graph is a surface

The graph of a function of one variable is a curve in the plane: the input uses one axis and the output the other. Our input already uses two axes, so the output needs a third.

**Definition.** The **graph** of a function $f$ of two variables with domain $D$ is the set of points $(x, y, z)$ in space such that $(x, y)$ is in $D$ and $z = f(x, y)$.

[Figure 3, image: The graph of f(x, y) = 6 + 3x - x^3 - y^2 over the map: above each point (x, y) of the base, the surface is at height z = f(x, y). The summit is the point (1, 0, 8) and the pass is (-1, 0, 4).]

The graph of a function of two variables is a surface, and for our $f$ it is the hill itself, with a summit over $(1, 0)$, a pass over $(-1, 0)$, and ground that rises again toward the west edge, the shoulder of a neighboring hill.

The picture is easy to like and hard to use. Try to read from it the height at $(0.5, 1)$, or to decide whether the north face is steeper than the east face. A drawing of a surface distorts distances and cannot be measured with a ruler. That is the map maker's problem: the graph lives in space, and paper does not.

## Slicing the hill at a fixed height

The map maker's solution is to stop asking "how high is this point?" and ask the reverse question: "which points are at this height?"

Flood the valley until the water surface stands at height 6. The water surface is a horizontal plane, and its edge, the shoreline, is the set of points of the hill that are exactly at height 6. Now drain the water and keep the shoreline as a mark on the ground.

[Figure 4, steps: From a hill to its map. The shoreline at each water level is a curve of constant height, and seen from directly above those curves are the contour lines.]

  - step: Flood the valley to height 6. The shoreline is every point of the hill at exactly that height.
  - step: Drain the water and keep the mark. It stays at height 6 all the way around.
  - step: Do the same at heights 3, 4, 5 and 7.
  - step: Look straight down. Height drops out of view, and the marks are the contour lines of a map.
Seen from directly above, a shoreline is a curve in the $(x, y)$ plane, which paper can show without distortion. The height that the view from above throws away is put back as a label on the curve.

We can find the height-6 curve from the formula. A point $(x, y)$ is on it when $f(x, y) = 6$, that is, when

$$
6 + 3x - x^3 - y^2 = 6, \qquad\text{or}\qquad y^2 = 3x - x^3 = x(3 - x^2).
$$

For $y$ to exist the right side must not be negative. Since $x(3 - x^2) \ge 0$ exactly when $x \le -\sqrt 3$ or $0 \le x \le \sqrt 3 \approx 1.73$, the main part of the curve lies over $0 \le x \le \sqrt 3$. For each such $x$ there are two points, $y = \pm\sqrt{3x - x^3}$, one north of the $x$-axis and one south. At $x = 1$ they are at $y = \pm\sqrt 2 \approx \pm 1.41$. The two halves join at $(0, 0)$ and at $(\sqrt 3, 0)$, so the curve is a closed loop around the summit that passes through the marker post. (The other range, $x \le -\sqrt 3$, contributes a sliver at the far west edge of the map, where the neighboring hill pokes above 6.)

**Pause and try.** Find the points where the height-7 curve crosses the $x$-axis and where it crosses the line $x = 1$. Is the height-7 curve inside or outside the height-6 loop?

**Answer.** On the line $x = 1$ the height is $f(1, y) = 8 - y^2$, which equals 7 when $y = \pm 1$. On the $x$-axis we need $6 + 3x - x^3 = 7$, or $x^3 - 3x + 1 = 0$. A calculator gives $x \approx 0.347$ and $x \approx 1.532$ on the map (check: $3(0.347) - 0.347^3 \approx 1.041 - 0.042 = 0.999$). All four points are inside the height-6 loop, which reaches $y = \pm 1.41$ on $x = 1$ and runs from $x = 0$ to $x = 1.73$ on the axis. Higher ground is nearer the summit, so the 7 curve is nested inside the 6 curve.

These curves have a name that does not depend on hills.

**Definition.** Let $f$ be a function of two variables with domain $D$, and let $c$ be a real number. The **level set** of $f$ at level $c$ is the set of all points $(x, y)$ in $D$ with $f(x, y) = c$. When a level set is a curve it is called a **level curve**, or a **contour line**. A **contour map** of $f$ is a drawing of several level sets, each labelled with its level, usually at equally spaced levels $c, c + \Delta, c + 2\Delta, \dots$; the number $\Delta$ is the **contour interval**.

Notice where each object lives. The graph is a surface in space. A level set is a set of points in the domain, in the plane. The mark on the hillside is where the graph meets the horizontal plane $z = c$, and the level set is that mark seen from directly above.

[Figure 5, image: The same function drawn two ways. Left: its graph, a surface in space. Right: its contour map in the plane, with contour interval 1 (that is, 100 m). The orange curve is the level set at height 6 in both.]

**Key idea.** A contour map shows a function of two variables by drawing its level sets: for each of several heights $c$, the curve of all points $(x, y)$ where $f(x, y) = c$. The graph is the surface $z = f(x, y)$, and a level set is a horizontal slice of that surface seen from above.

## What the lines can and cannot do

Two rules about contour lines are usually given to hikers as folklore. Both are consequences of the definitions, and the first has a one-line proof.

**Theorem.** Let $f$ be a function of two variables. If $c_1 \neq c_2$, the level sets of $f$ at levels $c_1$ and $c_2$ have no point in common.

**Proof.** Suppose a point $(x, y)$ belonged to both. Then $f(x, y) = c_1$ and $f(x, y) = c_2$. But $f$ assigns exactly one number to each point of its domain, so $c_1 = c_2$, contrary to assumption.

So the 300 m line and the 400 m line never meet, and the proof used only the words "exactly one" in the definition of a function.

The theorem says nothing about a level set meeting *itself*, and on our hill one does. Look at level 4, the height of the pass. The equation $f(x, y) = 4$ rearranges to $y^2 = 2 + 3x - x^3$, and the cubic factors:

$$
y^2 = (2 - x)(x + 1)^2, \qquad\text{so}\qquad y = \pm (x + 1)\sqrt{2 - x}.
$$

Both signs give $y = 0$ at $x = -1$. The two branches pass through the same point $(-1, 0)$ with different directions, and the level set is shaped like an X there.

[Figure 6, image: The level set at height 4 (purple) crosses itself at the pass. It divides the nearby ground into four sectors: higher ground to the east and west, lower ground to the north and south. Contours at 3.5 and 4.5 show how the neighboring level sets avoid the crossing.]

Stand at the pass. To the east and to the west the ground rises; to the north and south it falls away. The level-4 set is the boundary between "higher than me" and "lower than me", and at a pass that boundary has to separate four regions, which is why it crosses. A contour line that crosses itself marks a pass (mathematicians say **saddle point**).

Level sets also need not be curves at all. At level 8 the level set of our $f$ on the map is the single point $(1, 0)$, the summit. At level 9 it is empty, since no ground is that high. And a flat plateau, a region where $f$ is constant, is a level set with area. "Level curve" describes the usual case; "level set" is always correct.

## Why close lines mean steep ground

The second piece of folklore is that closely spaced contour lines mean steep ground. This one can be made exact. Walk in a straight line across the map. Each time you pass from one contour line to the next, your height has changed by one contour interval. If the lines are far apart, you walked a long way for that change; if they are close, you got it in a few steps.

[Figure 7, animation: A walk due east along the line y = 0. Each orange dot on the map (left) is a place where the route crosses a contour line, and the same crossings are marked on the height profile of the route (right). Where the dots crowd together on the map, the profile is steep.]

On this route the height is $p(x) = f(x, 0) = 6 + 3x - x^3$. West of the summit the route crosses the 6 line at $x = 0$ and the 7 line at $x \approx 0.347$. East of the summit it crosses the 7 line again at $x \approx 1.532$ and the 6 line at $x = \sqrt 3 \approx 1.732$.

**Pause and try.** On which side of the summit is the hill steeper along this route, and by roughly what factor? Use only the four crossing points just listed.

**Answer.** On the west side the height rises by 1 over a distance of $0.347$, an average slope of $1 / 0.347 \approx 2.9$. On the east side it falls by 1 over a distance of $1.732 - 1.532 = 0.200$, an average slope of about $-5.0$. The east side is steeper by a factor of about $1.7$. In ordinary units, a slope of 5 here means 500 m of height per kilometer of map distance, a 50% grade.

That computation is the method: the change in height between adjacent lines is the contour interval $\Delta$, the map distance between them is $d$, and their quotient is the average slope. Here is the precise statement, including what it says about the slope at individual points.

**Theorem.** **(Contour spacing and slope.)** Let $f$ be a function of two variables, and let $P$ and $Q$ be two points a distance $d > 0$ apart such that the straight segment from $P$ to $Q$ lies in the domain of $f$. Let $\mathbf{u}$ be the unit vector pointing from $P$ to $Q$, and let $\varphi(t) = f(P + t\mathbf{u})$ for $0 \le t \le d$ be the height profile of the straight walk from $P$ to $Q$. Suppose $f(P) = c$ and $f(Q) = c + \Delta$.

- The average rate of change of height along the walk is $\Delta / d$.
- If $\varphi$ is continuous on $[0, d]$ and differentiable on $(0, d)$, then at some point of the walk the slope $\varphi'(t)$ is exactly $\Delta / d$.
- Consequently, let $M > 0$, and suppose that every straight walk in the domain has a profile as in (2) with $|\varphi'(t)| \le M$ throughout. Then any two points that are joined by a segment in the domain and whose levels differ by $\Delta$ are at least $|\Delta| / M$ apart.

**Proof.** For (1), the average rate of change of $\varphi$ over $[0, d]$ is by definition $\dfrac{\varphi(d) - \varphi(0)}{d} = \dfrac{f(Q) - f(P)}{d} = \dfrac{\Delta}{d}$.

Part (2) is the mean value theorem of single-variable calculus applied to $\varphi$: a function continuous on $[a, b]$ and differentiable on $(a, b)$ has, at some point strictly between, a derivative equal to its average rate of change over $[a, b]$. We are citing that theorem; it is proved in every calculus text from the [extreme value theorem](@calculus/optimization).

For (3), take $P$ and $Q$ joined by a segment in the domain, with levels differing by $\Delta$. By (2) there is a point of the segment where the slope is $\Delta / d$, and by hypothesis $|\Delta / d| \le M$. Hence $d \ge |\Delta| / M$.

The segment from $P$ to $Q$ is your straight route between two contour lines. The function $\varphi$ is the height profile, the curve in the right panel of the last figure, and $\varphi'$ is its [derivative](@calculus/what-is-a-derivative), the slope of the profile. Part (3) is the hiker's rule with a number in it: on ground nowhere steeper than $M$, contour lines one interval apart are never closer than $\Delta / M$. Turned around, lines closer than that prove the ground is steeper than $M$ somewhere between them.

Check part (2) on the east side of our route. The average slope between the 7 line and the 6 line was $-5.0$. The profile has $p'(x) = 3 - 3x^2$, and $3 - 3x^2 = -5$ at $x = \sqrt{8/3} \approx 1.633$, which does lie between $1.532$ and $1.732$.

One caution. The quotient $\Delta / d$ is the slope *along the route you measured*. A route that cuts across the lines at a slant covers more map distance between them and gives a smaller slope, and a route along a contour line gives slope zero. The steepest slope comes from measuring straight across the lines. Making "the slope at a point in a given direction" precise is the subject of [the next lesson](@multivariable-calculus/partial-derivatives-and-gradient).

## Three maps to know by sight

The same construction applies to any function of two variables. Three simple maps are worth recognizing.

[Figure 8, image: Contour maps of three functions on the square -2 \le x, y \le 2, each with its levels labelled. A plane gives evenly spaced parallel lines, a bowl gives circles that crowd together as the bowl steepens, and a saddle gives hyperbolas around a level set that crosses itself.]

**Worked example.** **The level sets of the plane $f(x, y) = 2x + y$.**

The level set at level $c$ is the line $y = c - 2x$. Every level gives a line of slope $-2$, so the level sets are parallel. To find their spacing, measure straight across them, along the perpendicular line $y = x/2$. It meets level $c$ at $\left(\tfrac{2c}{5}, \tfrac{c}{5}\right)$, so levels $c$ and $c + 2$ are $\sqrt{(4/5)^2 + (2/5)^2} = 2/\sqrt 5 \approx 0.89$ apart, whatever $c$ is. The map is a set of evenly spaced parallel lines, and the slope straight across them is $\Delta / d = 2 / (2/\sqrt 5) = \sqrt 5 \approx 2.24$ everywhere, as it should be for a plane.

**Worked example.** **The level sets of the bowl $f(x, y) = x^2 + y^2$, and what their spacing says.**

The equation $x^2 + y^2 = c$ has no solutions when $c < 0$, the single solution $(0, 0)$ when $c = 0$, and for $c > 0$ describes the circle of radius $\sqrt c$ about the origin. So the levels $1, 2, 3, 4$ are circles of radius $1$, $1.414$, $1.732$, $2$. The gaps between them are $0.414$, $0.318$, $0.268$: equal steps in height take shorter and shorter steps outward.

By part (1) of the theorem the average slopes of a walk straight outward are $1/0.414 \approx 2.4$, then $1/0.318 \approx 3.1$, then $1/0.268 \approx 3.7$. The bowl gets steeper as you leave the center.

*Check.* Along the positive $x$-axis the height is $x^2$, whose derivative $2x$ goes from 2 to 4 between $x = 1$ and $x = 2$. Each of the three averages lies between the derivative values at the ends of its gap, for instance $2.4$ between $2(1) = 2$ and $2(1.414) = 2.83$.

The third map, $f(x, y) = xy$, is a pass with nothing else around it: its level set at 0 is the pair of coordinate axes, crossing at the origin as the level-4 set crosses at the pass of our hill. Problem 4 works it out.

## Reading errors to avoid

**Common mistake.** **Reading closeness as height.** Students often look at a map and say the hill is highest where the lines are densest. Density of lines says the ground is *steep* there. On our hill the lines are densest on the east flank, around $x = 1.9$, where the height is only about 5; around the summit, the highest ground on the map, there is a wide area with no lines at all, because the ground there is nearly flat. Height is read from the labels. Steepness is read from the spacing.

**Common mistake.** **Calling the map the graph.** The graph of $f$ is the surface in space. The contour map lives in the domain, and both of its axes are inputs. A test: a vertical line meets a one-variable graph at most once, but it can cross the same contour several times, because on the map neither of $x$ and $y$ is a function of the other.

**Check yourself.** Two adjacent contour lines on a map with a contour interval of 100 m are 250 m apart, measured straight across on the map. What is the average slope of the ground between them along that line?
Answer: 0.4, a 40% grade Rise over run is $100 / 250 = 0.4$: the height changes by one contour interval over 250 m of map distance.

We began with a mountain that would not fit on paper. The step that made it fit was to turn the question around, from "what is the height here?" to "where is the height equal to $c$?", because the answers to the second question are sets of points in the plane. The same reversal works for any function of two variables, and later lessons lean on it: [the direction of steepest climb](@multivariable-calculus/partial-derivatives-and-gradient) and [the highest point along a trail](@multivariable-calculus/lagrange-multipliers) can both be read from how the level curves sit.

**Recap.** - A function of two variables assigns one number $f(x, y)$ to each point $(x, y)$ of its domain. Its graph is the surface $z = f(x, y)$ in space.
- The level set at level $c$ is the set of points in the domain where $f(x, y) = c$. A contour map draws the level sets for equally spaced levels and labels them.
- Level sets at different levels never meet, because $f$ has one value at each point. A single level set can cross itself, and does so at a pass.
- A level set can be a curve, a point (a summit), empty, or a whole region.
- If two points at map distance $d$ have heights differing by $\Delta$, the average slope of the straight walk between them is $\Delta / d$, and somewhere on that walk the slope equals $\Delta / d$ exactly. Close lines mean steep ground.

## Problems

Problems 1 and 2 repeat the moves of the lesson. Problems 3 to 5 ask you to combine them or explain something. The last two go further. The hill is $f(x, y) = 6 + 3x - x^3 - y^2$ throughout.

**Problem 1.** Compute the height of the hill at $(2, 1)$, at $(-1, -1)$ and at $(0.5, 0.5)$. Which two of these points lie on the same level set? Is $(0.5, 0.5)$ inside or outside the level curve at height 7 that surrounds the summit?

**Solution.** $f(2, 1) = 6 + 6 - 8 - 1 = 3$. $f(-1, -1) = 6 - 3 + 1 - 1 = 3$. $f(0.5, 0.5) = 6 + 1.5 - 0.125 - 0.25 = 7.125$. The first two points are both at height 3, so both belong to the level set at level 3 (they lie on different pieces of it, one near the east edge and one south of the pass). The third point is at height $7.125 > 7$. The ground inside the 7 loop is the ground higher than 7, so $(0.5, 0.5)$ is inside it, though only just.

**Problem 2.** Let $f(x, y) = 3x - 4y$. Describe the level set at level $c$. Find the distance between the level sets at levels $c$ and $c + 5$, measured straight across them, and the slope of the plane in that direction.

**Solution.** The level set is the line $3x - 4y = c$, that is, $y = \tfrac34 x - \tfrac c4$. These lines all have slope $\tfrac34$, so measure along the perpendicular through the origin, in the direction of the unit vector $\left(\tfrac35, -\tfrac45\right)$. Its point at distance $t$ from the origin is $\left(\tfrac{3t}{5}, -\tfrac{4t}{5}\right)$, where $f = \tfrac{9t}{5} + \tfrac{16t}{5} = 5t$. So level $c$ is met at $t = c/5$ and level $c + 5$ at $t = c/5 + 1$: the lines are exactly 1 apart for every $c$. The slope straight across is $\Delta / d = 5 / 1 = 5$.

**Problem 3.** Consider the east-west line $y = 1$ on the hill. Show that the height along it is $7 - (x - 1)^2(x + 2)$. Deduce that this line meets the level curve at height 7 at exactly one point of the map, and that it never enters the region above 7. What does the line look like relative to that level curve?

**Solution.** Along $y = 1$, $f(x, 1) = 5 + 3x - x^3$. Then $7 - f(x, 1) = x^3 - 3x + 2$. Since $x = 1$ is a root, divide: $x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)^2(x + 2)$. So $f(x, 1) = 7 - (x - 1)^2(x + 2)$. On the map $x \ge -1.75$, so $x + 2 > 0$, and $(x - 1)^2 \ge 0$; hence $f(x, 1) \le 7$, with equality only at $x = 1$. The line touches the 7 curve at the single point $(1, 1)$ and stays on the low side of it everywhere else. It is tangent to the level curve there: it grazes it without crossing. (A path that just touches a level curve at its highest point is the central picture in the lesson on [Lagrange multipliers](@multivariable-calculus/lagrange-multipliers).)

**Problem 4.** A student's sketch of a contour map shows the 300 m line and the 400 m line crossing at a point $A$. Explain why this cannot be the contour map of a function. What feature of real terrain might the student have been trying to draw, and what does a correct map of it look like?

**Solution.** At $A$ the height would be both 300 m and 400 m, but a function assigns exactly one value to each point of its domain, so level sets at different levels are disjoint. Real terrain that tempts this drawing is a cliff. At a vertical cliff the ground changes height by 100 m over almost no map distance, and the contour lines run together into what looks like a single line, without crossing. An overhang, where the ground really does have two heights above one map position, is not the graph of a function, and a contour map cannot show it.

**Problem 5.** Let $f(x, y) = xy$. (a) Describe the level set at level $c$ for $c > 0$, $c = 0$ and $c < 0$. (b) For $c > 0$, find the point of the level set nearest the origin in the first quadrant and its distance from the origin. (c) Use (b) to decide whether the surface gets steeper or gentler as you walk away from the origin along the line $y = x$.

**Solution.** (a) For $c \ne 0$ the level set is the hyperbola $y = c/x$, which has two branches: in the first and third quadrants if $c > 0$, in the second and fourth if $c < 0$. For $c = 0$ the equation $xy = 0$ holds when $x = 0$ or $y = 0$, so the level set is the two coordinate axes, which cross at the origin. (b) By symmetry in $x$ and $y$ the nearest point is on the line $y = x$: there $x^2 = c$, so the point is $(\sqrt c, \sqrt c)$ at distance $\sqrt{2c}$. (Indeed the squared distance of $(x, c/x)$ from the origin is $x^2 + c^2/x^2 = (x - c/x)^2 + 2c \ge 2c$.) (c) Levels $1, 2, 3, 4$ meet the line $y = x$ at distances $1.414$, $2$, $2.449$, $2.828$ from the origin. The gaps are $0.586$, $0.449$, $0.379$, so the average slopes are $1.71$, $2.23$, $2.64$: the surface gets steeper. (Along $y = x$ at distance $s$ from the origin the height is $s^2/2$, with derivative $s$, in agreement.)

**Problem 6.** The functions $f(x, y) = x^2 + y^2$ and $g(x, y) = \sqrt{x^2 + y^2}$ both have circles about the origin as level sets. (a) Draw, or describe, the contour maps of each at levels $1, 2, 3, 4$. (b) What does the difference between the two maps say about the two surfaces? (c) Show in general that if $g(x, y) = \varphi(f(x, y))$ for some function $\varphi$ of one variable, then every level set of $f$ is contained in a level set of $g$.

**Solution.** (a) For $f$ the circles have radii $1$, $1.414$, $1.732$, $2$. For $g$ the level set at level $c > 0$ is the circle of radius $c$, so the radii are $1, 2, 3, 4$, evenly spaced. (b) Even spacing means the slope of $g$ straight outward is the same everywhere, $\Delta/d = 1$: its graph is a cone. The crowding circles of $f$ mean its graph steepens outward: a bowl. The shapes of the level curves are identical, and the labels and spacing carry the difference. (c) Suppose $P$ and $Q$ are in the same level set of $f$, so $f(P) = f(Q) = c$. Then $g(P) = \varphi(f(P)) = \varphi(c) = \varphi(f(Q)) = g(Q)$, so $P$ and $Q$ are in the same level set of $g$, the one at level $\varphi(c)$. Here $g = \sqrt{f}$, with $\varphi(t) = \sqrt t$.

**Problem 7.** For the hill, the lesson showed that the level set at height 4 is $y = \pm(x + 1)\sqrt{2 - x}$. Find the slopes of the two branches at the pass $(-1, 0)$, and hence the angle of each of the four sectors into which they divide the ground near the pass. Which sectors are the higher ground?

**Solution.** Let $y = (x + 1)\sqrt{2 - x}$. By the [product rule](@calculus/product-and-chain-rule), $\dfrac{dy}{dx} = \sqrt{2 - x} + (x + 1)\cdot\dfrac{-1}{2\sqrt{2 - x}}$. At $x = -1$ the second term vanishes and the slope is $\sqrt 3$. The other branch is the negative of this one and has slope $-\sqrt 3$. A line of slope $\sqrt 3$ makes an angle of $60^\circ$ with the $x$-axis, and one of slope $-\sqrt 3$ makes $-60^\circ$. So the two sectors that contain the $x$-axis, to the east and west, each span $120^\circ$, and the two that contain the north and south directions each span $60^\circ$. On the $x$-axis near the pass, $f(x, 0) = 4 + (2 - x)(x + 1)^2 > 4$ for $x \neq -1$, so the east and west sectors are the higher ground, and the narrower north and south sectors are the lower ground.