Multivariable-Calculus · Unit 7 · Video 3 · Interactive Practice

Contour Plots Show Which Way a Surface Goes: Circles, Saddles and Mountain Passes

IKey Formulas

Formula	Name	What it tells you
$f(x, y) = 1 - (x^2 + y^2) = c \iff x^2 + y^2 = 1 - c$	Level curves of the dome	Circle of radius $\sqrt{1 - c}$ : levels $1, 0, -1, -2, -3$ sit at $r = 0, 1, \sqrt{2}, \sqrt{3}, 2$
$1,\ \sqrt{2} - 1 \approx 0.41,\ \sqrt{3} - \sqrt{2} \approx 0.32,\ 2 - \sqrt{3} \approx 0.27$	Gaps between levels	Horizontal distance for a height change of 1: a shorter gap means a steeper surface
$y^2 - x^2 = 0 \iff y = \pm x$	Level through the saddle	The level curve through a saddle point crosses itself

Key Insight: The curves say where each level lies, the spacing says how steep the surface is, and the labels say whether $f$ rises or falls from one curve to the next.

IIVisualization 1 — Same Circles, Dome or Bowl

$1 - x^2 - y^2$ and $x^2 + y^2$ share every level circle; only the labels and the profile differ.

Function

IIIVisualization 2 — The Mountain Pass of $y^2 - x^2$

From the saddle point, which headings go uphill, which go downhill, and which stay level?

IVVisualization 3 — Which Way Does $f$ Go?

From a point on the two-mountain map, predict whether $f$ rises or falls as $x$ or $y$ increases.

x increases ⇒ f y increases ⇒ f

💡 The map tells you which way $f$ goes, not how fast: measuring the rate with $y$ held fixed, or with $x$ held fixed, needs the partial derivatives.

VQuiz Questions

Problem 1 · Radius of a Level Curve

Given: $f(x, y) = 4 - x^2 - y^2$ — find the radius of the level curve $f = -5$ .

✅ Correct!

4 - (x^2 + y^2) = -5

gives

x^2 + y^2 = 9

, a circle of radius

3

❌ Close, but…

9

r^2

, not

r

: the circle

x^2 + y^2 = 9

has radius

\sqrt{9}

❌ Check the sign. Solving

4 - (x^2 + y^2) = -5

gives

x^2 + y^2 = 4 - (-5) = 9

, not

4 - 5

❌ Not quite. Set

f = -5

and solve for

x^2 + y^2

first.

Show solution

Write $f = 4 - (x^2 + y^2)$ and set it equal to the level:

4 - (x^2 + y^2) = -5 \quad\Longrightarrow\quad x^2 + y^2 = 4 + 5 = 9.

This is the circle centred at the origin with radius $r = \sqrt{9} = 3$ .

Problem 2 · Evenly Spaced Circles

Given: A contour plot shows a point at the origin labelled $0$ and circles of radius $1, 2, 3, 4$ labelled $1, 2, 3, 4$ — which describes the graph of $f$ ?

✅ Correct! The height climbs by

1

over every gap of

1

, so the steepness never changes:

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

, a cone.

❌ Close, but… The paraboloid's level circles

x^2 + y^2 = c

have radii

\sqrt{c}

, whose gaps shrink (

1, 0.41, 0.32, 0.27

). These gaps are all equal.

❌ Read the labels. They increase from

0

at the centre to

4

outside, so

f

rises as you move out.

❌ Check the spacing. Steeper further out would need the circles to crowd together; here every gap is

1

Show solution

Labels: they go $0, 1, 2, 3, 4$ outward, so $f$ increases away from the origin (a bowl-like shape, not a dome).

Spacing: each gap between neighbouring levels is $1$ , so the height changes by $1$ over every horizontal distance of $1$ : the steepness is constant.

The level $f = c$ is the circle of radius $c$ , i.e. $\sqrt{x^2 + y^2} = c$ . So $f(x, y) = \sqrt{x^2 + y^2}$ , a cone. For comparison, $x^2 + y^2$ puts level $c$ at radius $\sqrt{c}$ , and its gaps shrink.

Problem 3 · A Stretched Saddle

Given: $f(x, y) = x^2 - 4y^2$ , which has a saddle point at the origin — find the level curve through the saddle and what $f$ does along the $y$ -axis.

What is the level curve $f = 0$ ?

Moving away from the origin along the $y$ -axis, $f$ …

✅ Correct! The zero level is the crossing pair

y = \pm \tfrac{1}{2}x

, and

f(0, y) = -4y^2 \le 0

, so the ground falls to the north and south of the pass.

❌ Close, but…

x^2 = 4y^2

gives

y^2 = \tfrac{x^2}{4}

, so

y = \pm \tfrac{x}{2}

— you divided the wrong way.

❌ Not quite. A single point is the level curve at a peak or pit. At a saddle,

x^2 - 4y^2 = 0

factors as

(x - 2y)(x + 2y) = 0

: two lines.

❌ Check the level curve. Solve

x^2 - 4y^2 = 0

for

y

❌ Check the direction. On the

y

-axis

x = 0

, so

f(0, y) = -4y^2

Show solution

Level through the saddle:

x^2 - 4y^2 = 0 \;\Longrightarrow\; x^2 = 4y^2 \;\Longrightarrow\; y^2 = \tfrac{x^2}{4} \;\Longrightarrow\; y = \pm\tfrac{1}{2}x.

Two lines crossing at the origin: the level curve crosses itself, as it must at a saddle.

Along the $y$ -axis: $f(0, y) = -4y^2 \le 0$ , which is $0$ at the origin and negative elsewhere, so $f$ decreases moving north or south. (Along the $x$ -axis, $f(x, 0) = x^2 \ge 0$ increases.)

Problem 4 · Reading Direction off the Map

Given: $f(x, y) = y - x^2$ , whose level curves are the parabolas $y = x^2 + c$ . Start at $(x_0, y_0) = (-1, 2)$ , on the level $f = 1$ — decide how $f$ changes for a small step in each direction.

If $x$ increases with $y$ fixed, $f$ …

If $y$ increases with $x$ fixed, $f$ …

✅ Correct! Moving right from

x = -1

shrinks

x^2

, so

f = y - x^2

grows; moving up grows

y

, so

f

grows too.

❌ Watch the sign of $x_0$ . At

x = -1

, increasing

x

brings it closer to

0

, so

x^2

gets smaller: e.g.

f(-0.9, 2) = 2 - 0.81 = 1.19 > 1

❌ Check the $x$ -step. Compare

f(-1, 2) = 1

with

f(-0.9, 2)

❌ Check the $y$ -step. Compare

f(-1, 2) = 1

with

f(-1, 2.1)

Show solution

At the start, $f(-1, 2) = 2 - 1 = 1$ .

Increase $x$ (step right): $f(-0.9, 2) = 2 - 0.81 = 1.19$ . Moving right from $x = -1$ heads toward the bottom of the parabola, where $x^2$ is smaller, so we cross toward higher levels: $f$ increases.

Increase $y$ (step up): $f(-1, 2.1) = 2.1 - 1 = 1.1$ . Moving up crosses into the region above $y = x^2 + 1$ , where the levels are higher: $f$ increases.

The map fixes the direction of change but not its size; the rate needs partial derivatives.

Solved: 0 / 4

Contour Plots Show Which Way a Surface Goes: Circles, Saddles and Mountain Passes

IKey Formulas

IIVisualization 1 — Same Circles, Dome or Bowl

IIIVisualization 2 — The Mountain Pass of y2−x2y^2 - x^2y2−x2

IVVisualization 3 — Which Way Does fff Go?

VQuiz Questions

IIIVisualization 2 — The Mountain Pass of $y^2 - x^2$

IVVisualization 3 — Which Way Does $f$ Go?