Functions of several variables — graphs, level curves, level surfaces: Question bank

First, pin down "trace" vs "level curve" (visual anchor)

Figure — Functions of several variables — graphs, level curves, level surfaces

Look at the amber ring (the trace) hovering at height $c$ on the bowl, and the cyan ring (its level curve) flattened onto the ground directly beneath it. Notice too how the ground rings crowd together outward — that crowding is the steepness cue tested repeatedly below.

True or false — justify

A level curve

f (x, y) = c

lives in

R^{3}

at height

z = c

False. The trace lives in 3D at height

c

, but the level curve is that trace's shadow projected onto the

x y

-plane, so it lives in 2D —

c

is only a label.

Equally spaced values of

c

always give equally spaced level curves.

False. Spacing depends on the function: for

x^{2} + y^{2}

the radius is

c

, so equal

c

-steps produce curves that crowd together outward.

The level set of

f (x, y, z) = c

is generally a curve.

False. One equation in three unknowns removes one degree of freedom, leaving a 2D surface, not a curve.

Crowded level curves mean the terrain is nearly flat there.

False. Crowding means the value changes fast over small horizontal distance — that's steep, the opposite of flat.

The graph of

z = f (x, y)

and the level curve

f (x, y) = c

live in the same space.

False. The graph is a surface in

R^{3}

; the level curve is a set in the 2D domain

R^{2}

Every value of

c

produces a non-empty level set.

False. For

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

any

c < 0

gives the empty set, since a sum of squares can never be negative.

Two different level curves of the same function can cross at a point.

False in general. A crossing would mean one input point

(x, y)

has two different outputs

c_{1} \neq = c_{2}

, impossible for a single-valued function.

A function of two variables can have a level "curve" that is a single point.

True. For

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

the level set at

c = 0

is just the origin — a degenerate curve where

\nabla f = 0

Spot the error

f (x, y) = 9 - x^{2} - y^{2}

has domain all of

R^{2}

since squaring is always defined."

Error: the square root needs a non-negative inside, so we need

9 - x^{2} - y^{2} \geq 0

, giving only the disk

x^{2} + y^{2} \leq 9

"For

f = x^{2} - y^{2}

, setting

c = 0

gives no level set because the hyperbola disappears."

Error:

c = 0

gives

x^{2} = y^{2}

, i.e. the two crossing lines

y = \pm x

— the degenerate case, not an empty set.

"Level surfaces of

x^{2} + y^{2} + z^{2} = c

are spheres for every

c

Error: only for

c > 0

. At

c = 0

it is the single point at the origin, and for

c < 0

it is empty.

"To graph

f (x, y, z)

I just need to plot points in 3D."

Error: the graph of

f (x, y, z)

needs 4 axes (3 inputs + 1 output) and cannot be drawn; we use level surfaces in ordinary 3D instead.

"The level curves of

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

get farther apart as you move outward because

c

grows."

Error: radius

= c

grows slower than

c

, so equal

c

-jumps make circles crowd closer together outward.

"A trace and a level curve are the same object."

Error: the trace is the actual slice sitting in 3D at height

z = c

; the level curve is its projection down into the

x y

-plane.

"The level set of any

f

R^{n}

is always a smooth surface of dimension

n - 1

Error: this needs

\nabla f \neq = 0

on the set (implicit function theorem). At a critical value where

\nabla f = 0

the level set may degenerate — e.g. to a point or crossing lines — and fail to be a smooth

(n - 1)

-manifold.

Why questions

Why can we directly graph

f (x, y)

but not

f (x, y, z)

A graph needs one axis per input plus one for the output;

n = 2

needs 3 axes (drawable) while

n = 3

needs 4 axes (not drawable).

Why do cartographers' contour maps and level curves work by the same idea?

Both slice a height surface at constant heights and project those slices onto flat ground, labelling each by its height.

Why does the level set of

f

R^{n}

generically have dimension

n - 1

, and when can this fail?

One equation

f = c

removes one degree of freedom, so where

\nabla f \neq = 0

the implicit function theorem gives a smooth

(n - 1)

-manifold; at critical values (

\nabla f = 0

) this can break down into points, crossings, or worse.

Why can a level curve never be labelled by two different values

c

Because

f

is single-valued: each point

(x, y)

maps to exactly one output, so it belongs to exactly one level curve.

Why does the spacing of level curves encode steepness rather than the shape alone?

Steepness is (change in height)/(horizontal distance); fixing equal height steps, closer curves mean less horizontal distance per step, hence a steeper slope.

Why are concentric circles the signature of the paraboloid

x^{2} + y^{2}

Fixing

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

forces constant distance

c

from the origin, and the radii grow with

c

, producing nested rings of a bowl.

Edge cases

What is the level set of

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

when

c = 0

A single point, the origin — the smallest, degenerate member of the family of circles, and the critical value where

\nabla f = 0

What happens to the level set of

f = x^{2} + y^{2} + z^{2}

c \to 0^{+}

The spheres shrink toward the origin, collapsing to the single point at

c = 0

(radius

c \to 0

What is the level set of a constant function

f (x, y) = 5

c = 5

and at

c = 3

c = 5

it is the entire domain (every point outputs 5); at

c = 3

it is empty, since no point outputs 3.

For the saddle

f = x^{2} - y^{2}

, how do the level sets differ across

c > 0

c = 0

, and

c < 0

c > 0

: hyperbolas opening left/right (crossing the

x

-axis);

c = 0

: the two degenerate crossing lines

y = \pm x

;

c < 0

: hyperbolas opening up/down (crossing the

y

-axis) — a

9 0^{\circ}

-rotated family.

Why do the

c > 0

and

c < 0

saddle hyperbolas point in perpendicular directions?

For

c > 0

x^{2} - y^{2} = c

needs

x^{2} \geq c

so real points exist only away from the

y

-axis (opens sideways); for

c < 0

y^{2} - x^{2} = ∣ c ∣

forces the same condition on

y

instead (opens vertically).

What is the highest-value level curve of

f = 9 - x^{2} - y^{2}

, and what shape is it?

c = 3

(its maximum),

9 - x^{2} - y^{2} = 9

forces

x^{2} + y^{2} = 0

, so the level set is the single origin point.

Can a level set be a curve that is unbounded (goes to infinity)?

Yes. For the saddle

x^{2} - y^{2} = c

with

c > 0

, the hyperbola branches extend infinitely far — level sets need not be closed loops.

Connections

Gradient vector — perpendicular to every level set; $\nabla f = 0$ marks the degenerate critical values
Partial derivatives — build the slopes that the crowding reflects
Directional derivatives — zero when you walk along a level curve
Quadric surfaces — the paraboloids and saddles whose level sets these traps probe
Limits and continuity of multivariable functions — approaching a point across level sets