Functions of Several Variables, taught from zero

1 · A function of two variables: feed in a point, get out a height §14.1

Until now a function took one number and returned one number. A function of two variables takes a pair — a point (x, y) in the plane — and returns a single number:

The picture that makes everything in §14.1 concrete: a landscape. Every location (x, y) on the map has an altitude f(x, y). Temperature over a room, pressure over a weather map, the depth of a lake at each spot — all functions of two variables. f(x, y) = x² + y² is a bowl whose floor is at the origin; f(x, y) = x + 2y is a tilted flat plane.

The domain is now a region of the plane

In one variable, "domain" was a set of allowed numbers — an interval or two. Now it's a set of allowed points: a shape drawn in the plane. The rules for what's forbidden haven't changed at all: no dividing by zero, no square root of a negative, no log of a non-positive number. What's new is that the answer is a picture. Two examples, done the way an exam wants them:

f(x, y) = √(y − x²). Need y − x² ≥ 0, i.e. y ≥ x². That's every point on or above the parabola y = x². Sketch the parabola, shade the inside. The range — the set of outputs — is [0, ∞): a square root is never negative, it is 0 on the parabola itself, and it grows without bound as you go up.

f(x, y) = ln(x² + y² − 1). Need x² + y² − 1 > 0, i.e. x² + y² > 1: everything strictly outside the unit circle (the circle itself excluded — ln 0 doesn't exist). Range: the argument of ln runs over (0, ∞), and ln of that is every real number, so the range is (−∞, ∞).

2 · Seeing the function: surfaces and level curves

The graph of z = f(x, y) is a surface floating over the domain — the landscape itself. Surfaces are hard to draw on paper, so §14.1 uses the trick every hiking map uses: contour lines. Fix a height c, and mark every point where the landscape is at exactly that height:

Draw these for several values of c and you have a contour map: closely spaced curves mean steep, widely spaced mean flat. Three examples — sketch each yourself before reading the answer, that's the skill being tested:

• f = x² + y²: the level curve x² + y² = c is a circle of radius √c for c > 0, just the origin for c = 0, and empty for c < 0. Concentric circles: the contour map of a bowl.
• f = y/x² (domain x ≠ 0): y/x² = c means y = cx² — a parabola through the origin for each c ≠ 0 (opening up for positive c, down for negative), with the origin itself removed since it isn't in the domain. For c = 0: the x-axis minus the origin.
• f = xy: xy = c is a hyperbola with the axes as asymptotes (first/third quadrants for c > 0, second/fourth for c < 0); for c = 0 it's the two axes themselves.

Slide the level c below and watch the level curves of f = x² − y² — a saddle surface — change family as c passes through zero. This is what "describe the level curves" means: name the curve family and say how it depends on c.

3 · The vocabulary of regions: interior, boundary, open, closed

Because domains are now shapes, the course needs words to describe shapes precisely — and tutorial/exam questions ask for them by name ("find the boundary; is the domain open, closed, or neither?"). There are only five words, and they're all built from one picture: a small disk drawn around a point.

The practical recipe: the boundary is almost always the curve you get by turning the domain's inequality into an equation. Then: strict inequality (<, >, ≠) ⟹ boundary excluded ⟹ open; non-strict (≤, ≥) ⟹ boundary included ⟹ closed. A region can be neither (it includes part of its boundary) — and, a genuine oddity worth knowing, the whole plane is both open and closed: it has no boundary at all.

Back to section 1's examples: the domain of √(y − x²) is y ≥ x²; boundary is the parabola y = x², included, so closed; it stretches upward forever, so unbounded. The domain of ln(x² + y² − 1) is x² + y² > 1; boundary is the unit circle, excluded, so open; unbounded.

4 · Three variables: level surfaces

Everything above extends to w = f(x, y, z), with one twist: the graph would need four dimensions, so we only have the contour idea. Setting f(x, y, z) = c gives a level surface in space. For f = x² + y² + z² the level surfaces are spheres of radius √c; for f = x + 2y + 3z they're parallel planes. Domains become regions of space, "disk" becomes "ball" in the definitions of section 3, and the words interior/boundary/open/closed mean exactly the same things. That's all of §14.1's three-variable content.

5 · Limits in two variables: the same idea, infinitely many roads in §14.2

lim_{(x,y)→(a,b)} f(x, y) = L means precisely what it meant before: the outputs get as close to L as anyone demands, provided the input point is close enough to (a, b) — the same ε–δ game, with "within δ of a" now meaning "inside a disk of radius δ around (a, b)". The limit still ignores what happens at the point itself.

Here is the new part. In one variable, "the outputs settle" had to be checked from two directions — left and right — and if they disagreed, no limit. In two variables you can approach (a, b) along the x-axis, the y-axis, any line, any parabola, any spiral. For the limit to exist, every one of those routes must give the same answer. That's an impossibly long list to check — but it makes disproving a limit easy:

Watch it on the standard example, f(x, y) = (x² − y²)/(x² + y²) as (x, y) → (0, 0). Plugging in gives 0/0, so dig:

• Along the x-axis (set y = 0, let x → 0): f(x, 0) = x²/x² = 1 for every x ≠ 0. So along this road the values are stuck at 1.
• Along the y-axis (set x = 0): f(0, y) = −y²/y² = −1. Stuck at −1.

Two roads, two answers: the limit does not exist. Three lines of working, full marks. Geometrically the surface has a cliff at the origin that runs at different heights in different directions.

When the axes agree: try the family y = mx

Sometimes both axes give the same value and you need to look harder. The standard second move is to approach along every line at once by substituting y = mx. If the result depends on m, different lines give different values — done. For f = xy/(x² + y²), both axes give 0, but y = mx gives mx²/(x² + m²x²) = m/(1 + m²) — ½ along y = x, 0 along y = 0. No limit. The notes page writes this one out as a model answer.

When even all lines agree: try curves

And sometimes every straight line gives the same answer and the limit still doesn't exist — the trap the examiners love. Take f = x²y/(x⁴ + y²). Along y = mx: mx³/(x⁴ + m²x²) = mx/(x² + m²) → 0 for every m. Looks like the limit is 0. But the x⁴ downstairs is a hint: try the parabolas y = kx², which make y² and x⁴ the same size:

— ½ along y = x², 0 along y = 0. No limit after all. The lesson: choose the path that balances the powers in the denominator. If the denominator has x⁴ and y², try y = kx²; if x² and y², lines suffice.

6 · Showing a limit does exist — the four tools

If you suspect the limit exists (all your test paths agree), you need one argument that covers all paths at once. There are four, and the first two are the old §2.4 toolbox verbatim.

Tool 1: substitute. If the formula is built from polynomials, roots, exponentials, logs and trig by sums, products, quotients and compositions, it is continuous wherever it's defined (section 7), so the limit is the value: lim_{(x,y)→(1,2)} (x²y + 3) = 1·2 + 3 = 5. The 0/0 cases below are the only real work.

Tool 2: simplify algebraically — factor and cancel, or multiply by a conjugate, exactly as in one variable. For example,

(valid for the approach along any path with x ≠ y — the points where the function is defined). After cancelling, the leftover is continuous, so substitution finishes it — for every road at once.

Tool 3: switch to polar coordinates. This is the tool that's new, and it's the most powerful for limits at the origin. Put x = r cos θ, y = r sin θ. Then x² + y² = r², and "(x, y) → (0, 0) along any path" becomes simply "r → 0, with θ doing whatever it likes". If you can show the expression → L as r → 0 no matter what θ does, you have covered every path. Example:

because |r cos³θ| ≤ r — whatever θ is, the size is at most r, and r → 0. That last inequality is the whole proof; write it. (Numerically: along y = x the values are 0.05, 0.005, 0.0005 at x = 0.1, 0.01, 0.001 — collapsing to 0 as promised.)

Tool 4: the squeeze (sandwich). Same theorem as §2.4: trap the function between two things that both go to L. In two variables the trap is usually an inequality like x²/(x² + y²) ≤ 1 (a part is at most the whole). For 3x²y/(x² + y²): |3x²y/(x² + y²)| = 3|y| · x²/(x² + y²) ≤ 3|y| → 0. The notes page does this one both ways (polar and squeeze) as a model answer.

7 · Continuity — same three-part check

f is continuous at (a, b) if (1) f(a, b) is defined, (2) the limit exists, and (3) they're equal — the identical checklist from the limits lesson. And the same theorems carry over: sums, products, quotients (nonzero denominator) and compositions of continuous functions are continuous, so every formula you'll meet is continuous on its domain. The only places to inspect are points outside the natural domain that the problem has patched by hand.

The exam pattern: "f(x, y) = (x² − y²)/(x² + y²) for (x, y) ≠ (0, 0), and f(0, 0) = 0. Is f continuous at the origin?" Away from the origin it's a quotient of polynomials with nonzero denominator — continuous. At the origin, condition (1) holds by the patch, but section 5 showed the limit doesn't exist, so (2) fails: not continuous there, and no choice of f(0, 0) could fix it. Compare x³/(x² + y²) patched with 0: the limit is 0 (section 6) and equals the patch, so that one is continuous everywhere.

8 · You're ready — what to do next

• Practise the two workouts this module examines: (i) the full domain / range / level curves / boundary / open-closed workout on a given f (Tutorial 4, Q5 is exactly this), and (ii) "does the limit exist?" — two-path to disprove, polar/squeeze/algebra to prove. The notes page's practice table gives Thomas exercise ranges.
• Revise from the notes page, not this one, before the quiz.