Multivariable-Calculus · Unit 7 · Video 2 · Interactive Practice

Level Curves: Reading a Surface from a Flat Map

IKey Formulas

Formula	Name	What you need
$\{(x, y) : f(x, y) = c\}$	Level curve of $f$ at level $c$	A fixed constant $c$ ; a contour plot draws these for regular values $c = \dots, -1, 0, 1, 2, \dots$
$z = c \ \cap\ z = f(x, y)$ , dropped onto the $xy$ -plane	Horizontal slice	Along the level curve $f = c$ the graph sits at height $c$ , above the plane or below it
$f(x, y) = -y$ : $\ f = c \iff y = -c$	Contours of the plane $z = -y$	Evenly spaced horizontal lines; the labels increase as you move down the page
$\text{steepness} \approx \dfrac{\Delta c}{\text{gap between neighbouring contours}}$	Spacing ↔ steepness	Close contours: steep graph; far-apart contours: gentle graph

Key Insight: A contour plot is the graph read from directly above. A point lying between the curves labelled $1$ and $2$ has $1 < f < 2$ , with no formula and no measuring.

IISlicing the Graph with the Plane $z = c$

The level curve $f = c$ of $f(x, y) = 4 - \tfrac{x^2}{4} - y^2$ lies directly beneath the slice of the graph at height $c$ .

Height c 2.00

💡 In the sketch the $x$ -axis comes toward you and the $y$ -axis points right, so the map is the floor turned a quarter turn: in both pictures the ellipse is longer along $x$ .

IIISpacing Tells Steepness

Neighbouring contours one unit of $f$ apart sit a distance $1/\text{steepness}$ apart: crowded lines mean a steep graph.

Function

IVReading a Map with No Formula

On a topographic map the contours alone bound the altitude at every point, and the profile along any line follows from them.

VQuiz Questions

Problem 1 · A Level Curve of the Plane

Given: $f(x, y) = -y$ — find the level curve $f = 3$ .

✅ Correct!

-y = 3

exactly when

y = -3

: a horizontal line three units below the

x

-axis, labelled

3

❌ Watch the sign. On

y = 3

the value is

f = -3

. The curve labelled

3

sits below the

x

-axis: the labels increase as you move down.

❌ Not quite.

f = -y

does not involve

x

at all, so its level curves cannot be vertical lines

x = \text{const}

; they run parallel to the

x

-axis.

❌ Not quite.

z = 3

is the slicing plane in space. The level curve is the set of points of the

xy

-plane where

f = 3

, directly beneath that slice.

Show solution

A level curve is the set of points $(x, y)$ where $f(x, y)$ equals the given constant:

f(x, y) = 3 \iff -y = 3 \iff y = -3.

Every $x$ works, so the level curve is the whole horizontal line $y = -3$ . It is the slice of the plane $z = -y$ at height $3$ , dropped onto the $xy$ -plane.

Problem 2 · Circles, and a Level with No Curve

Given: $f(x, y) = x^2 + y^2$ — find the level curves $f = 4$ and $f = -1$ .

What is the level curve $f = 4$ ?

What is the level curve $f = -1$ ?

✅ Correct!

x^2 + y^2 = 4

is the circle of radius

\sqrt{4} = 2

, and since

x^2 + y^2 \ge 0

the plane

z = -1

never meets the graph.

❌ Take the square root.

x^2 + y^2 = r^2

is the circle of radius

r

, so

x^2 + y^2 = 4

has radius

\sqrt{4} = 2

, not

4

❌ Not quite. Squares do not cancel like that:

(2, 0)

lies on

x + y = 2

with

f = 4

, but so does

(1, 1)

with

f = 2

❌ Not quite.

z = 4

is the slicing plane. The level curve is the curve in the

xy

-plane beneath that slice.

❌ Not quite. Solve

x^2 + y^2 = 4

for the set of points

(x, y)

❌ Not quite. Can a sum of two squares of real numbers ever equal

-1

Show solution

Level $4$ : $x^2 + y^2 = 4 = 2^2$ is the set of points at distance $2$ from the origin, the circle of radius $2$ .

Level $-1$ : for real $x, y$ , $x^2 + y^2 \ge 0$ , so $x^2 + y^2 = -1$ has no solutions. The graph is a bowl whose lowest point is at height $0$ ; the horizontal plane $z = -1$ lies entirely below it, so the slice is empty and there is nothing to draw on the contour plot at that level.

Problem 3 · A Steeper Plane

Given: $g(x, y) = -2y$ , with contours drawn at $c = \dots, -1, 0, 1, 2, \dots$ — find the level curve $g = 1$ and the gap between neighbouring contours.

What is the level curve $g = 1$ ?

What is the gap between neighbouring contours?

✅ Correct! The contours

y = -\tfrac{c}{2}

are half as far apart as those of

-y

, because

z = -2y

is twice as steep.

❌ Divide, don't multiply.

-2y = 1

gives

y = -\tfrac{1}{2}

; on

y = -2

the value is

g = 4

❌ Watch the sign. On

y = \tfrac{1}{2}

the value is

g = -2 \cdot \tfrac{1}{2} = -1

❌ Backwards. A steeper graph reaches the next level sooner, so its contours are closer together, not farther apart.

❌ Not quite. A gap of

1

belongs to

f = -y

. Here

g

changes by

2

per unit of

y

, so one unit of

g

takes only half a unit of

y

❌ Not quite. Solve

-2y = 1

for

y

;

g

does not depend on

x

❌ Not quite. Compare the positions of the contours

g = 0

and

g = 1

Show solution

Level curve: $g = c \iff -2y = c \iff y = -\tfrac{c}{2}$ . For $c = 1$ this is the horizontal line $y = -\tfrac{1}{2}$ .

Gap: neighbouring levels $c$ and $c + 1$ sit at

y = -\tfrac{c}{2} \quad\text{and}\quad y = -\tfrac{c + 1}{2}, \qquad \text{gap} = \tfrac{1}{2}.

Moving up by $\tfrac{1}{2}$ in $y$ lowers $g$ by $1$ , so the steepness is $\dfrac{\Delta c}{\text{gap}} = \dfrac{1}{1/2} = 2$ : the contours of $z = -2y$ are twice as crowded as those of $z = -y$ because the plane is twice as steep.

Problem 4 · Which Graph Has Evenly Spaced Rings?

Given: a contour plot whose level curves $f = 1, 2, 3, 4$ are circles centred at the origin, evenly spaced one unit apart — find the function and describe its graph.

Which function has this contour plot?

What does the even spacing say about its graph?

✅ Correct!

\sqrt{x^2 + y^2} = c

is the circle of radius

c

, so the rings sit at radii

1, 2, 3, 4

; equal gaps mean constant steepness, and the graph is a cone.

❌ Check the radii.

x^2 + y^2 = c

has radius

\sqrt{c}

1, \sqrt{2}, \sqrt{3}, 2

. Those gaps shrink outward, the signature of a bowl that steepens.

❌ Not quite. Its level curves

\tfrac{x^2}{4} + y^2 = 4 - c

are ellipses twice as long in

x

as in

y

, not circles, and there is no level curve

f = 4

beyond the single summit point.

❌ Not quite. The level curves of

-y

are horizontal lines, not circles.

❌ Not quite. Solve

f = c

for each candidate and compare the radii with

1, 2, 3, 4

❌ Not quite. Equal gaps for equal steps in

c

mean the height grows by the same amount for each unit of distance travelled outward.

Show solution

Radii of the level curves:

$\sqrt{x^2 + y^2} = c \iff x^2 + y^2 = c^2$ : radius $c$ , so radii $1, 2, 3, 4$ , evenly spaced. ✓
$x^2 + y^2 = c$ : radius $\sqrt{c}$ , so radii $1, 1.41, 1.73, 2$ , crowding outward.
$4 - \tfrac{x^2}{4} - y^2 = c$ : ellipses, not circles.
$-y = c$ : horizontal lines.

Graph: the height $z = \sqrt{x^2 + y^2}$ is the distance from the origin, so walking outward in any direction the height rises by $1$ for every unit travelled. The gap between neighbouring contours is always $1$ , the steepness is always $\dfrac{1}{1} = 1$ , and the surface is a cone with its tip at the origin.

Solved: 0 / 4

Level Curves: Reading a Surface from a Flat Map

IKey Formulas

IISlicing the Graph with the Plane z=cz = cz=c

IIISpacing Tells Steepness

IVReading a Map with No Formula

VQuiz Questions

IISlicing the Graph with the Plane $z = c$