Vector Functions and Motion in Space, taught from zero

1 · A curve as a moving point §13.1

Until now a function took a number and returned a number. A vector-valued function takes a number — think of it as time t — and returns a position:

At each instant t the arrow r(t) points from the origin to where a moving particle is. As t runs, the tip of the arrow traces a curve. The picture to keep for the whole chapter: a fly buzzing along a wire. The wire is geometry; how the fly moves along it is motion. This chapter is about the connection between the two.

Three examples to make it concrete — plug in a few values of t by hand for each, it takes a minute and builds the intuition everything else rests on:

• r(t) = ⟨cos t, sin t⟩ — at t = 0 the particle is at (1, 0); at t = π/2 it's at (0, 1). It's going round the unit circle anticlockwise, one lap per 2π seconds.
• r(t) = ⟨t, t²⟩ — position (t, t²): the parabola y = x², traced left to right.
• r(t) = ⟨cos t, sin t, t⟩ — the circle again, but now the third component climbs steadily. A point circling the z-axis while rising: a helix, the spiral staircase. This is the chapter's headline example; the notes page works it to the end.

Notice the parabola: the same curve y = x² could also be traced by ⟨2t, 4t²⟩ — same wire, fly going twice as fast. The curve and the motion along it are different things. Keep that distinction; section 5 depends on it.

Limits, continuity, derivative — all componentwise

What should "r(t) → L" mean? The natural answer — the position gets as close as you like to the point L — turns out to be the same as saying each coordinate gets close to the corresponding coordinate of L. So:

and r is continuous at a exactly when all three components are. That's the whole reason the limits lesson had to come first: the vector versions inherit everything from the scalar versions.

The derivative is defined the way derivatives always are — a difference quotient — but now the difference is a vector:

The second equality is the workhorse: differentiate each component on its own. But the first expression is the one carrying the meaning, so look at it. r(t+h) − r(t) is the arrow from where the fly is now to where it is a moment later — a chord of the curve. Dividing by the scalar h only rescales it. As h → 0 the two points merge and the chord swings into the tangent line. So r′(t) is a vector tangent to the curve, pointing in the direction of travel. Section 2 gives it its physical name.

2 · Velocity, speed, acceleration — and the unit tangent

Once you read t as time, the derivative already has a name from physics. Everything below is just the vocabulary attached to r′ and r″:

Play with it before reading on. The widget traces an ellipse r(t) = ⟨2 cos t, sin t⟩ and draws the velocity arrow live. Watch two things: the arrow is always tangent, and it changes length — the particle is faster on the flat ends and slower at the pointy ends. (The dashed circle is for section 5; ignore it for now.)

Two facts about velocity you'll use constantly:

Speed is a number; velocity is an arrow. On the unit circle r = ⟨cos t, sin t⟩: v = ⟨−sin t, cos t⟩ and |v| = √(sin²t + cos²t) = 1. Constant speed 1 — yet the velocity vector is different at every instant, because it keeps turning. Which means a = v′ = ⟨−cos t, −sin t⟩ = −r is not zero: it points straight at the centre. Constant speed does not mean zero acceleration; that's the single most common conceptual slip in this chapter.

Constant length ⟹ perpendicular derivative. The circle example generalises. If |r(t)| is constant, then r · r = |r|² is constant, and differentiating with the dot-product rule gives r′ · r + r · r′ = 2 r · r′ = 0. So r ⟂ r′. Anything that moves on a sphere has velocity at right angles to its position. Memorise the two-line argument — it's quotable in exams, and section 5 uses it on T.

3 · Going backwards: integrating vector functions §13.2*

If differentiation is componentwise, so is integration — antiderivatives of each slot, plus a constant vector:

The one problem type this produces: given acceleration and the starting state, recover the path. Integrate once to get v, fix C from v(0); integrate again to get r, fix the new constant from r(0). Here it is once, fully:

Suppose a(t) = ⟨2, 6t⟩, v(0) = ⟨1, 0⟩, r(0) = ⟨0, 1⟩.

1. v(t) = ∫a dt = ⟨2t, 3t²⟩ + C. At t = 0 this is C, which must equal ⟨1, 0⟩. So v(t) = ⟨2t + 1, 3t²⟩.
2. r(t) = ∫v dt = ⟨t² + t, t³⟩ + D, and r(0) = D = ⟨0, 1⟩. So r(t) = ⟨t² + t, t³ + 1⟩.

Check by differentiating twice: r′ = ⟨2t+1, 3t²⟩ ✓, r″ = ⟨2, 6t⟩ ✓. Always do this check; it costs ten seconds.

4 · Arc length — how far did the fly travel? §13.3

You know distance = speed × time when speed is constant. When it isn't, the fix is the one calculus always uses: over a tiny interval dt the speed is effectively constant, the distance covered is |v| dt, and you add up the slivers with an integral:

That's the whole formula: integrate the speed. On the helix, speed is the constant √2, so one full turn (0 ≤ t ≤ 2π) has length ∫₀^2π √2 dt = 2π√2 ≈ 8.89 — a bit longer than the circle of circumference 2π it sits over, because it also climbs. Sensible.

The exam skill is the algebra inside the square root. Nearly every arc-length problem is designed so that sin² + cos² = 1, or a perfect square, collapses the radicand into something integrable. If your radicand isn't simplifying, re-check the derivatives before trying to integrate something ugly.

The arc-length parameter s

Now fix a starting time t₀ and let the upper limit float:

s(t) is the odometer reading: distance travelled along the wire since t₀. The second equation — "the odometer ticks at the speed" — is just the Fundamental Theorem of Calculus applied to the integral, and it is the only fact about s the next section needs.

Why bother? Because s is a property of the wire, not the fly. Two flies tracing the same curve at different speeds have different t's but the same s for the same spot. When we want to describe how the road bends, we need to measure change per metre of road, not per second of driving — and that means differentiating with respect to s.

5 · Curvature: how sharply the road bends §13.4

Drive a motorway: the direction you face barely changes per kilometre. Take a hairpin: your direction swings through 180° in a few metres. Curvature κ (kappa) is exactly that — how fast the direction of travel turns, per unit of distance along the road. Direction of travel is T; distance along the road is s. So the definition writes itself:

Per distance, not per time — deliberately. A fast car and a slow car on the same bend turn at different rates per second but the bend is the same bend. Curvature belongs to the wire.

But our functions are written in t, not s. The chain rule converts: dT/dt = (dT/ds) · (ds/dt), and ds/dt = |v| from section 4. Rearranging gives the formula you'll actually compute with:

Two anchor facts to sanity-check every answer against:

• A straight line has κ = 0. T never changes, so dT/dt = 0.
• A circle of radius a has κ = 1/a. Compute it once, it's the calibration for everything: r = ⟨a cos t, a sin t⟩, v = ⟨−a sin t, a cos t⟩, |v| = a, T = ⟨−sin t, cos t⟩, dT/dt = ⟨−cos t, −sin t⟩ with length 1. So κ = 1/a. Big circle, gentle bend; small circle, sharp bend. Exactly what "curvature" should mean.

That second fact gives curvature a picture. At any point of any curve, the circle that hugs it best — same tangent, same curvature — is the circle of curvature (osculating circle), with radius of curvature ρ = 1/κ. Go back to the widget and look at the dashed green circle: it's tiny at the pointy ends of the ellipse (κ = 2, ρ = ½) and huge on the flat parts (κ = ¼, ρ = 4). Press play and watch it breathe.

The principal unit normal N

Curvature says how much the curve turns. N says which way. The direction in which T is changing is, by definition, dT/dt; normalise it:

And here is the section-2 fact paying off: T has constant length 1, so its derivative is perpendicular to it — N ⟂ T, always. N points toward the inside of the bend, at the centre of the circle of curvature (the widget's rust-coloured arrow). On a circle it points at the centre; on a straight line it's undefined, since nothing is turning. Check T · N = 0 at the end of every problem — it catches arithmetic slips for free.

Two things exam questions ask that the definition alone doesn't hand you

Curvature of a graph y = f(x). A graph is a curve too — parametrise it by x itself: r(x) = ⟨x, f(x)⟩, so v = ⟨1, f′⟩, |v| = √(1 + f′²). Pushing that through the definition gives a formula worth having ready:

Sanity-check it on y = x² at the origin: f′ = 0, f″ = 2, so κ = 2 and ρ = ½ — the parabola's tip hugs a circle of radius ½. At x = 1: f′ = 2, f″ = 2, κ = 2/(1+4)^3/2 = 2/(5√5) ≈ 0.18 — flatter, as the picture suggests.

The circle of curvature itself. Its radius is ρ = 1/κ and its centre sits a distance ρ from the point, in the direction of N:

For y = x² at the origin: N = ⟨0, 1⟩ (the curve bends upward), ρ = ½, centre (0, ½), circle x² + (y − ½)² = ¼. For a graph, a quick way to get N without differentiating T: rotate the unit tangent by 90° toward the side the curve bends to (upward where f″ > 0, downward where f″ < 0).

The osculating-circle recipe, end to end — y = x² at x = 1

The origin example above is deceptively easy because the tangent there is horizontal. At a tilted point the same five steps work; this is the full tutorial-sheet pattern ("find a parametrization of the osculating circle"), with nothing skipped.

1. The point. x = 1 on y = x² is the point (1, 1).
2. Curvature and radius there. f′(1) = 2, f″(1) = 2, so κ = 2/(1 + 4)^3/2 = 2/(5√5) and ρ = 1/κ = 5√52 ≈ 5.59. (Notice how much bigger than the origin's ½ — the parabola is much straighter out here.)
3. Which way is the centre? Along N, the unit normal pointing into the bend. The tangent direction is ⟨1, f′⟩ = ⟨1, 2⟩, so T = ⟨1, 2⟩/√5. Rotate 90°: the two candidates are ⟨−2, 1⟩/√5 and ⟨2, −1⟩/√5. The curve is concave up (f″ > 0), so pick the one with positive y-component: N = ⟨−2, 1⟩/√5.
   Unpack this step: why rotating ⟨p, q⟩ by 90° gives ⟨−q, p⟩

   ⟨p, q⟩ · ⟨−q, p⟩ = −pq + qp = 0, so the rotated vector is perpendicular, and it has the same length. The other perpendicular is its negative. You never need trigonometry for this — swap the components and flip one sign.

4. The centre. centre = point + ρN = (1, 1) + 5√52 · ⟨−2, 1⟩√5 = (1, 1) + ⟨−5, 52⟩ = (−4, 72) — the √5's cancel, which is typical and worth expecting.
5. Write the parametrization (the unpack above):
   x(t) = −4 + 5√52 cos t, y(t) = 72 + 5√52 sin t, 0 ≤ t ≤ 2π
   Audit before moving on: the distance from (−4, 7/2) back to (1, 1) is √(25 + 25/4) = √(125/4) = 5√5/2 = ρ ✓ — the circle really passes through the point. That one-line check catches almost every slip in steps 3–4.

Curvature of a parametric plane curve — the formula the ellipse needs

For a curve given as x(t), y(t) (not as a graph), running the definition through the same machinery gives a third form worth having on the formula sheet — it appears in the §13.4 exercises:

Sanity checks: for a graph (x = t, y = f(t)) it collapses to the boxed graph formula above; for the circle ⟨a cos t, a sin t⟩ the top is a², the bottom a³, so κ = 1/a ✓.

Now the classic "show that" it exists for — where does an ellipse bend most? Take x = a cos t, y = b sin t with a > b > 0:

1. Derivatives: x′ = −a sin t, y′ = b cos t, x″ = −a cos t, y″ = −b sin t.
2. Numerator: x′y″ − y′x″ = ab sin²t + ab cos²t = ab — constant, the identity strikes again.
3. So κ(t) = ab / (a² sin²t + b² cos²t)^3/2. The top is fixed; κ is large exactly when the bracket is small, and vice versa. Write the bracket as b² + (a² − b²) sin²t — since a² − b² > 0, it grows exactly as sin²t grows.
4. Smallest bracket: sin t = 0, i.e. t = 0, π — the points (±a, 0), the major axis — giving κ_max = ab/b³ = a/b². Largest bracket: sin t = ±1, the points (0, ±b), the minor axis — κ_min = ab/a³ = b/a². That is the whole "show that": κ written down, bracket monotone in sin²t, endpoints named.

This is exactly what the widget in section 2 shows: the tiny hugging circle at the ellipse's pointy ends (there a = 2, b = 1: κ = 2 vs ¼). Scroll up and watch it with the formula in mind.

One full workout

Take the helix from section 4, r(t) = ⟨3 cos t, 3 sin t, 4t⟩, and compute everything. This is the exam pattern: five steps, always in this order.

1. Velocity and speed. v = ⟨−3 sin t, 3 cos t, 4⟩, |v| = √(9 + 16) = 5.
2. Unit tangent. T = v/5 = ⟨−35 sin t, 35 cos t, 45⟩.
3. Turn rate. dT/dt = ⟨−35 cos t, −35 sin t, 0⟩, so |dT/dt| = 35√(cos²t + sin²t) = 35.
4. Curvature. κ = (3/5) / 5 = 3/25. Sanity: the circle of radius 3 it shadows has κ = 1/3 ≈ 0.33; stretching it upward into a helix straightens it, and 3/25 = 0.12 is indeed smaller. ✓
5. Normal. N = (dT/dt) / (3/5) = ⟨−cos t, −sin t, 0⟩ — horizontal, pointing straight at the z-axis. Check: T · N = 35 sin t cos t − 35 sin t cos t + 0 = 0 ✓, and |N| = 1 ✓.

The notes page runs the same five steps on the unit helix ⟨cos t, sin t, t⟩ (answer: κ = ½). Do that one yourself, cold, before the quiz.

6 · You're ready — what to do next

You now hold the whole conceptual kit for §13.1–13.4: componentwise everything, v/speed/a/T, integrating back from acceleration, arc length as integrated speed, and κ and N from the turn rate of T. To turn it into marks:

• Drill the five-step workout on three curves: the unit helix, ⟨3 cos t, 3 sin t, 4t⟩, and one of your own. The notes page's practice table gives the Thomas exercise ranges (odd numbers — answers in the back).
• Revise from the notes page, not this one — before the quiz you want the one-screen map.