Functions of several variables — graphs, level curves, level surfaces: Visual walkthrough

We derive one central result:

Level curves of f (x, y) = x^{2} + y^{2} are concentric circles of radius c .

Everything below earns that sentence, symbol by symbol.

Step 1 — What a two-input machine even is

WHAT. A function $f (x, y)$ is a machine with two dials and one readout. You choose a spot on the flat floor — a pair of numbers $(x, y)$ — and the machine hands back a single number $z$ , the height.

WHY start here. Before we can slice a surface we must agree the surface exists. Every symbol we will use is born in this picture: $x$ = how far east, $y$ = how far north, $z$ = how high.

PICTURE. Look at the figure. The flat grid is the floor (the $x y$ -plane). One chosen point $(x, y)$ is marked; a vertical stick of length $z = f (x, y)$ rises from it. The tip of that stick is one point of the surface. (We use the plain letters $x, y, z$ throughout — the marked point is just some choice of them.)

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

height, the readout z = the machine f (east dial x, north dial y)

Step 2 — Grow the whole surface: the paraboloid

WHAT. Do Step 1 at every floor point at once. Collect all the stick-tips. That cloud of tips is the graph — a surface floating over the floor.

WHY. A single stick is one number. The surface is the whole function seen in one glance. We need the full bowl before we can cut it.

PICTURE. The orange surface is the bowl $z = x^{2} + y^{2}$ . Notice it touches the floor at exactly one point (the origin, where both dials are $0$ ) and flares upward and outward. The steepness grows as you climb — remember that, it returns in Step 6.

z = x^{2} + y^{2} ⟹ z \geq 0, z = 0 ⟺ (x, y) = (0, 0)

Here $x^{2} \geq 0$ and $y^{2} \geq 0$ always, so their sum can never dip below zero — that is why the bowl never pokes below the floor.

Step 3 — Cut a horizontal slice at height $c$

WHAT. Pick a height, call it $c$ (just a chosen number, like " $c = 4$ metres"). Push a perfectly flat sheet of glass through the bowl at that height. Where the glass meets the surface, we get a curve floating in 3D — this curve is called the trace at height $c$ .

WHY the letter $c$ ? $c$ stands for constant: on this whole curve the height never changes, it is frozen at $c$ . We use one symbol so the whole family (all heights) shares one formula.

WHY a horizontal cut and not any other? Because we want places of equal value. Equal height means equal readout $z = c$ . A tilted cut would mix different heights and destroy that meaning.

PICTURE. The violet ring is where the glass sheet $z = c$ kisses the orange bowl. It lives up in the air at height $c$ — it is not on the floor yet.

the surface’s height z = frozen height of the glass c on the trace

Step 4 — Force the two heights to agree (the key algebra)

WHAT. On the trace, two facts hold at the same time: the point is on the surface ( $z = x^{2} + y^{2}$ ) and it is at the glass height ( $z = c$ ). Set them equal.

WHY. This is the only step that turns geometry into an equation. "Being on both" means the same $z$ from both descriptions, so we substitute one into the other and $z$ disappears — leaving a relation between just $x$ and $y$ .

PICTURE. The figure shows the two coloured height-labels ( $z$ from the bowl, $z$ from the glass) meeting and cancelling, leaving a flat equation in $x, y$ alone.

height from the bowl x^{2} + y^{2} = height from the glass c

Each symbol now: $x^{2} + y^{2}$ is the machine's output at the floor point; $c$ is the constant we chose. The equation says "which floor points make the machine read exactly $c$ ?" — that question is exactly the level curve defined at the top of the page.

Step 5 — Recognise the shape: a circle of radius $c$

WHAT. Compare $x^{2} + y^{2} = c$ with the circle you already know, $x^{2} + y^{2} = r^{2}$ (all points a fixed distance $r$ from the origin).

WHY this comparison and not another? Because $x^{2} + y^{2}$ is literally the squared distance from the floor point $(x, y)$ to the origin (Pythagoras on the floor: the two legs are $x$ and $y$ ). So " $x^{2} + y^{2} = c$ " reads "squared distance $= c$ ", i.e. "distance $= c$ " — a circle.

PICTURE. The right triangle on the floor has legs $x$ and $y$ and hypotenuse the straight line to the origin; its length squared is $x^{2} + y^{2}$ . Freezing that length gives the circle.

x^{2} + y^{2} = c ⟹ match r^{2} = c r = radius; needs c \geq 0 c

We take the positive root because a radius is a length, never negative. That single choice quietly rules out the degenerate cases — which we now face head-on.

Step 6 — Every case: $c > 0$ , $c = 0$ , $c < 0$ , and the crowding

WHAT. A good derivation must survive every value of $c$ . Run all three.

$c > 0$ : $r = c$ is a genuine positive radius → an honest circle. Why: a real positive length can be a radius.
$c = 0$ : $x^{2} + y^{2} = 0$ forces $x = 0$ and $y = 0$ (two squares can only sum to zero if both are zero) → the level "curve" shrinks to a single point, the origin. Why: the glass touches the bowl only at its lowest point.
$c < 0$ : $x^{2} + y^{2}$ can never be negative, so no floor point works → the level set is empty. Why: the glass sheet is below the bowl and touches nothing.

WHY the rings crowd outward. Take equally spaced heights $c = 1, 2, 3, 4$ . Their radii are $1, 2, 3, 4 = 1, 1.41, 1.73, 2$ . The gaps between radii shrink: $0.41, 0.32, 0.27$ . So equal steps in height give circles that bunch together as you go out — and bunched contours mean steep ground. This is exactly the bowl getting steeper as you climb, now visible on the flat map.

PICTURE. The floor map shows nested circles labelled $c = 1, 2, 3, 4$ ; watch the shrinking gaps. A tiny dot marks $c = 0$ ; a faint crossed-out ring reminds you $c < 0$ is empty.

case c > 0 c = 0 c < 0 level set circle, r = c point (0, 0) empty why positive length is a valid radius two squares sum to 0 only if both are 0 x^{2} + y^{2} \geq 0 can never equal a negative

Step 7 — Read the map backwards (sanity check)

WHAT. Stand on the flat contour map alone (no bowl drawn) and rebuild the height at any point. If you are at $(x, y)$ , find which circle passes through you — its label is your height.

WHY. A derivation is trustworthy only if it is reversible. Map → surface must undo surface → map.

PICTURE. A hiker-dot at $(1.5, 0)$ sits on the ring $c = 2.25$ , because $1. 5^{2} + 0^{2} = 2.25$ . The vertical arrow reconstructs the stick of Step 1.

floor point (x, y) ⟼ label of its ring = x^{2} + y^{2} = z

We recovered the readout of Step 1 — the loop closes. ✅

The one-picture summary

Everything at once: the orange bowl, three glass sheets cutting violet traces at $c = 1, 2, 3$ , and their magenta shadows landing as nested circles on the floor — the gaps visibly shrinking outward.

Recall Feynman retelling — the whole walk in plain words

Picture a machine that, standing on any spot of a flat floor, tells you a height by squaring how far east you are, squaring how far north, and adding them. Do that everywhere and the heights make an orange bowl sitting on the floor (Steps 1–2). Now shove a flat glass sheet through the bowl at some fixed height $c$ ; where glass meets bowl you get a ring floating in the air — a trace (Step 3). Ask "which floor spots gave that height?" — set the bowl's height equal to the glass's height, the height cancels, and you're left with $x^{2} + y^{2} = c$ (Step 4). Since $x^{2} + y^{2}$ is just squared distance from the middle (Pythagoras), that says "all spots the same distance $c$ from the centre" — a circle (Step 5). Try every height: positive gives a real circle, zero gives just the centre dot, negative gives nothing because you can't add two squares and get a negative (Step 6). Drop all the air-rings straight down to the floor and you get a contour map whose circles bunch up as you go out — bunched circles mean the bowl is getting steep (Step 6 again). Finally, read the map backwards: stand anywhere, see which ring you're on, and its label is your height — the machine, rebuilt (Step 7).

Recall

Why do we cut horizontally to get level curves?

Horizontal means equal height

z = c

, i.e. equal output — exactly the points that share one value.

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

, what disappeared and why?

The height

z

— because the point lies both on the surface (

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

) and on the glass (

z = c

), so we set them equal.

Why is the radius

c

and not

c

x^{2} + y^{2}

is squared distance from the origin, so distance is its square root.

What happens at

c = 0

and

c < 0

c = 0

gives the single point

(0, 0)

;

c < 0

gives the empty set (a sum of squares is never negative).

Trace vs level curve?

The trace floats at height

c

in 3D; the level curve is its shadow on the floor with

c

as a label only.

Why do the circles crowd outward?

Equal height-steps give radii

1, 2, 3, \dots

whose gaps shrink — crowding signals steepness.

Connections

Parent topic — this walkthrough builds its central "slice-and-project" claim from zero.
Gradient vector — points across the rings toward steepest ascent; crowding here previews its size.
Partial derivatives — slopes of the bowl along the $x$ - and $y$ -directions.
Quadric surfaces — the bowl $z = x^{2} + y^{2}$ is the paraboloid, a named quadric.
Directional derivatives — how height changes as you walk across the contour map.