Functions of several variables — graphs, level curves, level surfaces: Worked examples

The scenario matrix

Every level-set problem falls into one of these cells. The worked examples below are tagged with the cell(s) they hit — together they cover the whole table.

#	Cell (what makes it special)	Example that hits it
A	$c > 0$ gives a nice bounded curve (circle/ellipse)	Ex 1, Ex 6
B	$c = 0$ degenerate (point / crossing lines)	Ex 1, Ex 2
C	$c < 0$ empty set (impossible value)	Ex 1, Ex 6
D	Sign of $c$ flips the shape family (hyperbola flips axis)	Ex 2
E	Unbounded level curve (line, parabola, whole strip)	Ex 3
F	Domain restriction bites (log, root, division)	Ex 4, Ex 7
G	Level surface in 3D ( $n = 3$ )	Ex 5
H	Real-world word problem (a contour map)	Ex 6
I	Exam twist — recognise shape without a clean formula	Ex 7

Definitions we lean on (from the parent):

Ex 1 — Circles: all three signs of $c$ (cells A, B, C)

Worked example

$f (x, y) = x^{2} + y^{2}$ , sketch level curves for $c = - 4, 0, 1, 4, 9$ . Forecast: you have seen this bowl. Guess: which of these five values give a real curve, which give a single point, which give nothing? Write your guess before reading on.

Step 1. Set $f = c$ : $x^{2} + y^{2} = c$ . Why this step? By definition a level curve collects all inputs sharing the output $c$ ; setting the formula equal to $c$ is literally that condition.

Step 2. Compare with the circle template $x^{2} + y^{2} = r^{2}$ . Why this step? We want the shape's name and size. Matching to a known template gives radius $r = c$ instantly.

Step 3. Split by the sign of $c$ — this is where "all cases" matters:

$c = 9$ : circle radius $9 = 3$ . (cell A)
$c = 4$ : circle radius $2$ . (cell A)
$c = 1$ : circle radius $1$ . (cell A)
$c = 0$ : $x^{2} + y^{2} = 0$ forces $x = y = 0$ — a single point, the origin. (cell B, degenerate)
$c = - 4$ : a sum of two squares can never be negative, so no points — the empty set. (cell C)

Why split by sign? Because $c$ is only real for $c \geq 0$ , and $c = 0$ collapses the circle to a point. The formula secretly changes character at $c = 0$ .

Step 4. Notice the spacing: radii are $1, 2, 3$ for $c = 1, 4, 9$ — equal jumps in $c$ but the radii grow like $c$ , so rings crowd outward. Why note this? Crowded rings $=$ steep terrain (the "crowding = steepness" fact stated at the top of this page).

Figure 1 — Concentric level circles of $f (x, y) = x^{2} + y^{2}$ : green ring $c = 1$ (radius 1), blue ring $c = 4$ (radius 2), orange ring $c = 9$ (radius 3), the red centre dot for the degenerate $c = 0$ point, and a corner note that $c = - 4$ draws nothing.

Look at the figure: the three coloured rings (green $c = 1$ , blue $c = 4$ , orange $c = 9$ ) are the real circles; the red dot at the centre is the $c = 0$ point; and the note in the corner reminds you $c = - 4$ draws nothing. Trace outward and watch the rings bunch up — that visual crowding is the point of Step 4.

Verify: plug $(3, 0)$ into $f$ : $3^{2} + 0^{2} = 9$ ✓ (on the $c = 9$ ring). Plug $(0, 0)$ : $0$ ✓ (the $c = 0$ point). And $4 = 2$ ✓. Why verify? Plugging a point back into $f$ confirms it truly sits on the ring we claimed, catching any arithmetic slip in the radius.

Ex 2 — The saddle: sign of $c$ flips the axis (cells B, D)

Worked example

$f (x, y) = x^{2} - y^{2}$ , level curves for $c = 4$ , $c = 0$ , $c = - 4$ . Forecast: one of these three is not a smooth curve. Which, and what does it become?

Step 1. $c = 4$ : $x^{2} - y^{2} = 4$ , i.e. $\frac{x ^{2}}{4} - \frac{y ^{2}}{4} = 1$ . Why divide by 4? To reach the standard hyperbola form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , whose $+$ sits on $x^{2}$ ⇒ it opens left–right (crosses the $x$ -axis at $x = \pm 2$ ). (cell D)

Step 2. $c = - 4$ : $x^{2} - y^{2} = - 4 \Rightarrow \frac{y ^{2}}{4} - \frac{x ^{2}}{4} = 1$ . Why flip signs? Multiplying by $- 1$ moves the $+$ onto $y^{2}$ ⇒ the hyperbola now opens up–down (crosses the $y$ -axis at $y = \pm 2$ ). The sign of $c$ chose the axis. (cell D)

Step 3. $c = 0$ : $x^{2} - y^{2} = 0 \Rightarrow (x - y) (x + y) = 0 \Rightarrow y = x or y = - x$ . Why factor? A difference of squares factors; each factor $= 0$ is a straight line. Two crossing lines — the degenerate hyperbola (its own asymptotes). (cell B)

Figure 2 — Level curves of the saddle $f (x, y) = x^{2} - y^{2}$ : the blue hyperbola ( $c = 4$ ) opens left–right through $x = \pm 2$ , the orange hyperbola ( $c = - 4$ ) opens up–down through $y = \pm 2$ , and the two dashed red lines $y = x$ , $y = - x$ are the degenerate $c = 0$ case, which are also the shared asymptotes.

Look at the figure: the blue hyperbola opens left–right ( $c = 4$ ), the orange one opens up–down ( $c = - 4$ ), and the two dashed red lines through the origin are the $c = 0$ case. Notice the blue and orange curves both hug the red dashed lines far from the origin — those lines are their shared asymptotes, which is exactly why $c = 0$ is those lines.

Verify: on $c = 4$ test $(2, 0)$ : $4 - 0 = 4$ ✓. On $c = - 4$ test $(0, 2)$ : $0 - 4 = - 4$ ✓. On $c = 0$ test $(3, 3)$ (line $y = x$ ): $9 - 9 = 0$ ✓. Why verify? Testing one point per level value confirms each point lands on the correct branch (and that the $c = 0$ point really lies on a line, not a curve).

Ex 3 — An unbounded level curve (cell E)

Worked example

$f (x, y) = y - x^{2}$ , level curves for $c = - 1, 0, 2$ . Forecast: solving for $y$ should reveal a familiar 1-variable curve, just shifted. Guess the shape.

Step 1. Set $y - x^{2} = c$ and solve for $y$ : $y = x^{2} + c$ . Why solve for $y$ ? Because then each level curve is the graph of a plain 1-variable function — instantly recognisable.

Step 2. Read off: every level curve is an upward parabola $y = x^{2}$ shifted vertically by $c$ . These parabolas extend forever — they are unbounded (no largest $x$ ). (cell E)

$c = 2$ : parabola with vertex $(0, 2)$ .
$c = 0$ : vertex at origin.
$c = - 1$ : vertex $(0, - 1)$ .

Step 3. Do any values of $c$ give the empty set? No. For every real $c$ , picking $x = 0, y = c$ already satisfies it. So there is no "impossible" cell here — unlike Ex 1. Why check? A subtraction (not a sum of squares) can hit any real value, so no $c$ is forbidden.

Figure 3 — Three identical upward parabolas $y = x^{2} + c$ stacked vertically: green ( $c = - 1$ , vertex at $(0, - 1)$ ), blue ( $c = 0$ , vertex at origin), orange ( $c = 2$ , vertex at $(0, 2)$ ), with the red test point $(1, 3)$ sitting on the orange curve. The arms never turn back — the curves are unbounded.

Look at the figure: three parabolas of identical shape stacked vertically — green ( $c = - 1$ ) lowest, blue ( $c = 0$ ) in the middle, orange ( $c = 2$ ) highest. The red dot marks the test point $(1, 3)$ sitting on the orange curve. See how none of the arms ever turn back: that is the "unbounded" claim made visible.

Verify: point $(1, 3)$ on the $c = 2$ curve: $y = x^{2} + 2 = 1 + 2 = 3$ ✓, and $f (1, 3) = 3 - 1 = 2$ ✓. Why verify? Checking the same point two ways (from the solved form and from the original $f$ ) confirms our algebra of solving for $y$ was reversible.

Ex 4 — Domain restriction bites (cell F)

Worked example

$f (x, y) = ln (x^{2} + y^{2})$ . Find where $f$ is defined, then the level curve $f = 0$ . Forecast: the domain will have a hole. And $f = 0$ should be a circle of a specific radius — which one?

Step 1. Domain: $ln$ needs a positive input, so require $x^{2} + y^{2} > 0$ . Why strict $>$ ? $ln$ of $0$ is undefined (and negatives too, but a sum of squares is never negative). So the only excluded point is $x^{2} + y^{2} = 0$ , i.e. the origin. Domain $= R^{2} ∖ {(0, 0)}$ — the plane with a single puncture. (cell F)

Step 2. Level curve $f = 0$ : $ln (x^{2} + y^{2}) = 0$ . Why is a log easy here? Apply $e^{(\cdot)}$ to undo $ln$ : $x^{2} + y^{2} = e^{0} = 1$ . Why $e^{0}$ ? $ln$ and $e^{x}$ are inverses; $e^{0} = 1$ .

Step 3. So $f = 0$ is the circle of radius $1$ . And $f = c$ generally gives $x^{2} + y^{2} = e^{c}$ , a circle of radius $e^{c /2}$ — real for every $c$ (since $e^{c} > 0$ always). No empty cell, but the origin is never on any level curve.

Verify: $(1, 0)$ : $ln (1^{2} + 0) = ln 1 = 0$ ✓. For $c = 2$ , radius $= e^{1} \approx 2.718$ ; test point $(e, 0)$ : $ln (e^{2}) = 2$ ✓. Why verify? Because a log/exp step is easy to invert wrongly; substituting the point confirms the circle radius really produces the intended level value.

Ex 5 — A level surface in 3D (cell G)

Worked example

$f (x, y, z) = x^{2} + y^{2} - z$ . Describe the level surface $f = 0$ and $f = 1$ . Forecast: with three inputs, one equation should give a surface (dimension $3 - 1 = 2$ ), not a curve. Which quadric?

Step 1. $f = 0$ : $x^{2} + y^{2} - z = 0 \Rightarrow z = x^{2} + y^{2}$ . Why solve for $z$ ? Then it reads as "height $z$ over the point $(x, y)$ " — this is a paraboloid (a bowl opening upward). (cell G)

Step 2. $f = 1$ : $x^{2} + y^{2} - z = 1 \Rightarrow z = x^{2} + y^{2} - 1$ . Why? Same bowl, shifted down by $1$ (vertex at $z = - 1$ ). Different $c$ just slides the bowl vertically.

Step 3. Cross-check the dimension rule: 3 variables, 1 equation ⇒ level set is $3 - 1 = 2$ dimensional ⇒ a surface. ✓ (This is the "level-set dimension = variables − 1" fact stated at the top of this page.)

Figure 4 — Two stacked paraboloid level surfaces of $f (x, y, z) = x^{2} + y^{2} - z$ in 3D: the blue bowl is $f = 0$ ( $z = x^{2} + y^{2}$ ) with vertex at the origin, the orange bowl is $f = 1$ ( $z = x^{2} + y^{2} - 1$ ) shifted down so its vertex sits at $z = - 1$ . Both are 2D sheets floating in 3D space.

Look at the figure: two stacked bowls in 3D — the blue one is $f = 0$ ( $z = x^{2} + y^{2}$ ) with its lowest point at the origin, and the orange one is $f = 1$ , the identical bowl slid down so its vertex sits at $z = - 1$ . Both are clearly 2D sheets (surfaces) floating in 3D space, confirming the dimension count.

These paraboloids are examples of Quadric surfaces.

Verify: on $f = 0$ , point $(1, 1, 2)$ : $1 + 1 - 2 = 0$ ✓. On $f = 1$ , point $(1, 1, 1)$ : $1 + 1 - 1 = 1$ ✓. Why verify? A 3D point is easy to misplace; substituting confirms it lies on the correct shell rather than between the two bowls.

Ex 6 — Real-world contour map (cells A, C, H)

Worked example A hill's height is

$H (x, y) = 100 - x^{2} - 4 y^{2}$ , where $H$ is in metres. To keep the formula dimensionally clean we write it with an implicit length scale $L = 1 m$ so that $x, y$ are pure numbers (multiples of $L$ ): $H = (100 - x^{2} - 4 y^{2}) m$ . Draw the " $H =$ const" trails at $H = 100, 64, 0$ metres and interpret. Is $H = 120$ m possible? Forecast: the $x^{2}$ and $4 y^{2}$ have different weights — guess whether the contours are circles or squished ellipses, and which way squished.

Step 0 (units check). Inside the bracket, $100$ , $x^{2}$ and $4 y^{2}$ are all pure numbers (because $x, y$ are measured as multiples of the unit length $L = 1 m$ ), and the whole bracket is multiplied by $1 m$ to give a height in metres. So the addition $100 - x^{2} - 4 y^{2}$ is dimensionally consistent — we never add a length to an area. Why this step? You can only add or subtract quantities with matching units; making $x, y$ dimensionless (numbers of metres) keeps every term in the bracket a pure number and prevents the classic "length + area" blunder.

Step 1. Set $H = c$ (in metres): $100 - x^{2} - 4 y^{2} = c \Rightarrow x^{2} + 4 y^{2} = 100 - c$ . Why rearrange? To move all the square terms to one side, giving a sum of positive squares on the left — a recognisable ellipse form whose size is set by the single number $100 - c$ .

Step 2. $H = 100$ (the summit): $x^{2} + 4 y^{2} = 100 - 100 = 0 \Rightarrow x = y = 0$ — a single point, the top of the hill. (cell B/A boundary) Why does it collapse to a point? A sum of positive squares equals $0$ only when each square is $0$ , forcing $x = 0$ and $y = 0$ .

Step 3. $H = 64$ : $x^{2} + 4 y^{2} = 100 - 64 = 36$ , i.e. $\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1$ . Why divide by 36? Dividing through by the right-hand number puts it in the standard ellipse form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with $a = 6$ , $b = 3$ : it reaches $6$ along $x$ but only $3$ along $y$ — squished in the $y$ -direction because the $4 y^{2}$ term grows four times faster than $x^{2}$ . (cell A, cell H)

Step 4. $H = 0$ (sea level): $x^{2} + 4 y^{2} = 100 - 0 = 100$ , so $\frac{x ^{2}}{100} + \frac{y ^{2}}{25} = 1$ , an ellipse with $a = 10, b = 5$ — the widest contour. Why is it the widest? Smaller $c$ leaves a bigger number $100 - c$ on the right, and larger right-hand value means larger semi-axes, so lower trails are bigger loops.

Step 5. $H = 120$ : $x^{2} + 4 y^{2} = 100 - 120 = - 20 < 0$ — impossible, empty set. The hill never reaches $120$ m; its max height is $100$ m at the summit. (cell C) Why does negativity mean empty? The left side $x^{2} + 4 y^{2}$ is a sum of squares, so it can never equal a negative number — no point satisfies the equation.

Figure 5 — Contour map of the hill $H = 100 - x^{2} - 4 y^{2}$ : the blue ellipse ( $H = 64$ , semi-axes $6$ and $3$ ) nested inside the wider orange ellipse ( $H = 0$ sea level, semi-axes $10$ and $5$ ), both stretched along $x$ and pinched along $y$ . The red centre dot is the summit ( $H = 100$ ) and a corner note states $H = 120$ draws nothing (max height is $100$ ).

Look at the figure: the blue ellipse ( $H = 64$ ) is nested inside the wider orange ellipse ( $H = 0$ , sea level), both stretched along $x$ and pinched along $y$ — that pinch is the $4 y^{2}$ term at work. The red dot at the centre is the summit ( $H = 100$ ), and the corner note reminds you $H = 120$ draws nothing at all.

Verify: $(6, 0)$ gives $H = 100 - 36 - 0 = 64$ ✓ (on the $H = 64$ trail). $(10, 0)$ : $H = 100 - 100 = 0$ ✓ (sea level). Summit $(0, 0)$ : $H = 100$ ✓, and no $(x, y)$ makes $H = 120$ ✓. Why verify? Substituting corner points of each ellipse confirms both the height labels and that the semi-axes ( $6$ and $10$ along $x$ ) are correct.

Ex 7 — Exam twist: recognise the shape without a clean template (cells F, I)

Worked example

$f (x, y) = \frac{x}{y}$ (with $y \neq = 0$ ). What are the level curves $f = 2$ , $f = 0$ , and $f = - 1$ ? Where is $f$ undefined? Forecast: a ratio being constant — that should force a straight-line relationship. Guess through which point every level curve passes.

Step 1. Domain: $y \neq = 0$ (no division by zero). So the entire $x$ -axis is excluded — this is a domain-restriction case. (cell F) Why the whole $x$ -axis? The $x$ -axis is exactly the set of points with $y = 0$ , and every such point makes the denominator zero, so all of them are thrown out.

Step 2. $f = 2$ : $\frac{x}{y} = 2 \Rightarrow x = 2 y$ . Why multiply up? Clearing the fraction (multiplying both sides by $y$ ) turns the ratio into a linear equation. But note $y \neq = 0$ , so the origin is punctured out — it's the line $x = 2 y$ minus the point $(0, 0)$ . Why does the line pass through where the origin would be? Because a constant ratio means $x$ and $y$ scale together — all such points lie on a line through the origin. (cell I)

Step 3. $f = 0$ : $\frac{x}{y} = 0 \Rightarrow x = 0$ (with $y \neq = 0$ ) — the $y$ -axis, minus the origin. Why set $x = 0$ and not $y = 0$ ? A fraction is zero exactly when its numerator is zero while the denominator stays non-zero; so $x = 0, y \neq = 0$ , which is the $y$ -axis with the origin removed.

Step 4. $f = - 1$ : $\frac{x}{y} = - 1 \Rightarrow x = - y$ , the line $y = - x$ punctured at the origin. Why? Clearing the fraction gives $x = - y$ ; again $y \neq = 0$ removes the origin. The negative constant produces a line with negative slope.

Figure 6 — Level curves of $f (x, y) = x / y$ : three straight rays through (but excluding) the origin — blue line $x = 2 y$ ( $f = 2$ ), the orange vertical $y$ -axis ( $f = 0$ ), and the green line $y = - x$ ( $f = - 1$ ). The origin is drawn as a hollow circle to show it is removed from every level curve.

Look at the figure: three straight lines through the origin — blue ( $x = 2 y$ ), orange (the vertical $y$ -axis for $f = 0$ ), and green ( $y = - x$ ). The origin is drawn as a hollow circle to signal it is removed from every level curve. All lines fan out from that one hole, which is the visual signature of a constant ratio.

So every level curve of $x / y$ is a line through the origin (with the origin removed), and the slope of that line is set by $c$ . The gradient idea (Gradient vector) explains why these are all rays fanning out from one point.

Verify: $(2, 1)$ on $f = 2$ : $2/1 = 2$ ✓. $(0, 5)$ on $f = 0$ : $0/5 = 0$ ✓. $(3, - 3)$ on $f = - 1$ : $3/ (- 3) = - 1$ ✓. Why verify? Substituting confirms each test point respects $y \neq = 0$ and lands on the claimed line, guarding against accidentally including the forbidden origin.

Recall Quick self-test on the matrix

Which cell does "empty set" belong to, and give a function+value that hits it? ::: Cell C — e.g. $x^{2} + y^{2} = - 4$ (sum of squares can't be negative). Sign of $c$ flipping a hyperbola from left-right to up-down is which cell? ::: Cell D (Ex 2). For $f (x, y, z) = c$ , what dimension is the level set and why? ::: Dimension $2$ (a surface); $3$ variables minus $1$ equation. In Ex 6, why is $H = 120$ impossible? ::: It needs $x^{2} + 4 y^{2} = - 20 < 0$ ; the hill maxes out at $100$ m.

Connections

Gradient vector — perpendicular to every level curve above; explains the fanning rays in Ex 7
Partial derivatives — how $f$ changes across the contours
Directional derivatives — rate of change as you walk between trails on the Ex 6 map
Quadric surfaces — the paraboloid of Ex 5, the ellipses of Ex 6
Tangent planes and linear approximation — flat sheet touching the Ex 5 surface
Limits and continuity of multivariable functions — behaviour near the punctures in Ex 4 and Ex 7