Chapter 28 — Vector-Valued Functions and Space Curves

28.1 Bridging to Multivariable Calculus

For most of this book, calculus has handled functions $f: \mathbb{R} \to \mathbb{R}$ — one real number in, one real number out. The graph lives in the plane, and the derivative is the slope of a tangent line. In Chapter 25 you took a first step sideways: a parametric curve in the plane is governed by one input $t$ but produces two coordinates, $x(t)$ and $y(t)$, that move together as $t$ advances. You stopped thinking of $t$ as an axis and started thinking of it as time, with the pair $(x(t), y(t))$ tracing the path of a moving point.

This chapter takes the next step and lets the point move through three-dimensional space. We bundle the coordinates into a single object, a vector-valued function:

$$\mathbf{r}(t) = \langle x(t),\, y(t),\, z(t) \rangle = x(t)\,\mathbf{i} + y(t)\,\mathbf{j} + z(t)\,\mathbf{k},$$

which sends each real number $t$ in some interval to a point — equivalently, a position vector — in $\mathbb{R}^3$. As $t$ runs over its interval, the tip of $\mathbf{r}(t)$ sweeps out a space curve. (Everything works identically in $\mathbb{R}^2$; just drop the $z$-component. A planar parametric curve is the special case $z(t) = 0$.)

This is the first genuinely multivariable object in our journey, and it opens Part VI. It is also, paradoxically, the simplest multivariable object, because the input is still a single number. The new feature is only on the output side. And here is the gift that makes the chapter manageable: a vector-valued function can be differentiated and integrated one component at a time. Each component $x(t)$, $y(t)$, $z(t)$ is an ordinary single-variable function, so every rule you have spent twenty-seven chapters mastering applies, three times over, in parallel.

The Key Insight. A vector-valued function is one function of one variable that happens to return a vector. Calculus on it is done component-wise — limits, derivatives, integrals all act separately on $x(t)$, $y(t)$, $z(t)$. So no new computational machinery is needed. What is new is geometric: learning to read $\mathbf{r}'(t)$ as a velocity vector tangent to the path, its length as speed, and the bending of the path as curvature.

This connects directly to our recurring theme that geometry and algebra are inseparable: every vector formula in this chapter has a picture (an arrow on a curve), and every picture has a formula. Keep both in mind at all times.

Warning — a vector-valued function is not a vector field. This is the single most common confusion in all of vector calculus, so we name it on the very first page. A vector-valued function $\mathbf{r}(t)$ takes one number and returns one vector; its graph is a single curve threading through space. A vector field (Chapter 34) takes a point in space and attaches a vector to every point, filling a region with arrows — like wind velocity at every location in a room. One curve versus a whole field of arrows. We build the first object now; the second waits for Chapter 34. Do not let the shared word "vector" blur them together.

28.2 A Gallery of Space Curves

Before any calculus, build intuition for what these curves look like. In each case, read the components as instructions for a moving point.

Straight line

$$\mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v}$$

traces the line through the point $\mathbf{r}_0$ in the direction $\mathbf{v}$. For example, $\mathbf{r}(t) = \langle 1, 2, 3 \rangle + t\,\langle 4, 5, 6 \rangle$ is the line through $(1,2,3)$ pointing along $\langle 4,5,6\rangle$. As $t$ increases the point slides along the line at constant velocity $\mathbf{v}$.

Circle

$$\mathbf{r}(t) = \langle R\cos t,\; R\sin t,\; 0\rangle$$

is a circle of radius $R$ in the $xy$-plane, traced counterclockwise once as $t$ goes from $0$ to $2\pi$. This is exactly the parametric circle of Chapter 25, now sitting in 3D with a flat $z$-coordinate.

Helix

$$\mathbf{r}(t) = \langle \cos t,\; \sin t,\; t \rangle$$

is the helix — a corkscrew. The pair $(\cos t, \sin t)$ circles the $z$-axis while $z = t$ rises steadily, so the point spirals upward. The helix is the path of anything that turns and climbs at once: the thread of a screw, a coiled spring, the strands of DNA, and — as we will see in §28.11 — a charged particle drifting along a magnetic field line. It will be our running example for the whole chapter, because its curvature and speed are constant and every formula stays clean.

Twisted cubic

$$\mathbf{r}(t) = \langle t,\; t^2,\; t^3 \rangle$$

is the twisted cubic, a celebrated curve from algebraic geometry. Unlike the circle or a parabola, it never lies in a single plane — it genuinely twists through all three dimensions, which makes it a perfect test case for the bending-and-twisting machinery later in the chapter.

Projectile in 3D

$$\mathbf{r}(t) = \big\langle v_x t,\; v_y t,\; h_0 + v_z t - \tfrac{1}{2} g t^2 \big\rangle$$

models a thrown object: horizontal motion at constant velocity in $x$ and $y$, and vertical motion with gravity $g$ pulling down in $z$. We will derive this from $\mathbf{r}''(t) = \langle 0,0,-g\rangle$ by integration in §28.4.

Geometric Intuition. To picture any $\mathbf{r}(t)$, imagine a pen tip at position $\mathbf{r}(t)$ and let $t$ run like a clock. The curve is the trail of ink. Faster-changing components mean the pen sweeps that direction quickly. The helix, for instance, has $z = t$ climbing at unit rate while $(\cos t, \sin t)$ goes once around per $2\pi$ — so one full turn of the spiral raises the pen by exactly $2\pi$.

28.3 Limits, Continuity, Derivatives, and Integrals

All of single-variable calculus lifts to vectors component-wise. Each definition below is simply "do the scalar operation in each slot."

Limit. The limit exists exactly when each component limit exists, and $$\lim_{t \to a} \mathbf{r}(t) = \Big\langle \lim_{t\to a} x(t),\; \lim_{t\to a} y(t),\; \lim_{t\to a} z(t) \Big\rangle.$$

Continuity. $\mathbf{r}$ is continuous at $a$ iff each component is continuous at $a$ — equivalently, iff $\lim_{t\to a}\mathbf{r}(t) = \mathbf{r}(a)$.

Derivative. Defined exactly as in Chapter 6, but with vector subtraction in the difference quotient: $$\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} = \langle x'(t),\; y'(t),\; z'(t) \rangle.$$ The middle expression is the honest limit definition; the right expression is the theorem that it reduces to component-wise differentiation, because the limit of a vector is the vector of limits.

Integral. Indefinite and definite integrals also act component-wise: $$\int \mathbf{r}(t)\,dt = \Big\langle \int x(t)\,dt,\; \int y(t)\,dt,\; \int z(t)\,dt \Big\rangle + \mathbf{C},$$ $$\int_a^b \mathbf{r}(t)\,dt = \Big\langle \int_a^b x(t)\,dt,\; \int_a^b y(t)\,dt,\; \int_a^b z(t)\,dt \Big\rangle.$$

Two things deserve a flag. First, the constant of integration $\mathbf{C}$ is now a constant vector, not a constant scalar — three arbitrary constants bundled together. Second, because the definite integral is built from FTC (Chapter 14) in each component, the Fundamental Theorem of Calculus holds component-wise: if $\mathbf{R}'(t) = \mathbf{r}(t)$, then $\int_a^b \mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a)$. FTC, the keystone of the whole book, comes along for free.

Worked example

Let $\mathbf{r}(t) = \langle t^2,\; \sin t,\; e^t \rangle$.

Differentiate each slot: $$\mathbf{r}'(t) = \langle 2t,\; \cos t,\; e^t \rangle.$$

Integrate each slot: $$\int \mathbf{r}(t)\,dt = \Big\langle \tfrac{1}{3}t^3,\; -\cos t,\; e^t \Big\rangle + \mathbf{C}.$$

Nothing here is harder than Chapter 7 and Chapter 12; you are simply doing three of them at once.

The geometry of the derivative

The component formula is the how; the what is geometric and is the heart of the chapter. Look again at the difference quotient. The numerator $\mathbf{r}(t+h) - \mathbf{r}(t)$ is the secant vector — the arrow from the point at time $t$ to the point at time $t+h$, a chord of the curve. Dividing by $h$ scales it. As $h \to 0$, that chord pivots until it lies along the curve, and the limit $\mathbf{r}'(t)$ is a vector tangent to the curve, pointing in the direction of increasing $t$.

Geometric Intuition. The secant vector $\mathbf{r}(t+h) - \mathbf{r}(t)$ is a chord connecting two nearby points on the curve. As the second point slides back toward the first, the chord swings into the tangent direction — exactly as the secant line became the tangent line in Chapter 5. So $\mathbf{r}'(t)$ always points along the curve, never across it. If the curve is a road, $\mathbf{r}'(t)$ is the direction your headlights face.

The four product rules

Vectors can be combined four ways — sum, scalar multiple, dot product, cross product — and each obeys a product rule that mirrors the ordinary one. Let $\mathbf{u}(t)$ and $\mathbf{v}(t)$ be differentiable vector functions and $f(t)$ a differentiable scalar function:

$$\frac{d}{dt}\big[\mathbf{u} + \mathbf{v}\big] = \mathbf{u}' + \mathbf{v}',$$ $$\frac{d}{dt}\big[f\,\mathbf{u}\big] = f'\,\mathbf{u} + f\,\mathbf{u}',$$ $$\frac{d}{dt}\big[\mathbf{u} \cdot \mathbf{v}\big] = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}' \qquad (\text{scalar output}),$$ $$\frac{d}{dt}\big[\mathbf{u} \times \mathbf{v}\big] = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' \qquad (\text{vector output}).$$

There is also a chain rule for a scalar reparametrization $t = f(s)$: $$\frac{d}{ds}\,\mathbf{r}(f(s)) = f'(s)\,\mathbf{r}'(f(s)).$$

Each product rule is proved by writing out components and applying the ordinary product rule in each — the same "component-wise" principle yet again.

Common Pitfall. In the cross-product rule the order of factors must be preserved, because $\mathbf{u} \times \mathbf{v} = -\,\mathbf{v} \times \mathbf{u}$. Writing $\frac{d}{dt}[\mathbf{u}\times\mathbf{v}] = \mathbf{v}'\times\mathbf{u} + \mathbf{u}'\times\mathbf{v}$ flips a sign and gives the wrong vector. The first factor stays first: $\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$. The dot-product rule has no such trap, since $\mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}$.

A small but powerful consequence of the dot-product rule: if $\mathbf{u}(t)$ has constant length, then $\mathbf{u}$ is perpendicular to $\mathbf{u}'$. Indeed, constant length means $\mathbf{u}\cdot\mathbf{u} = \|\mathbf{u}\|^2$ is constant, so differentiating gives $\frac{d}{dt}(\mathbf{u}\cdot\mathbf{u}) = 2\,\mathbf{u}\cdot\mathbf{u}' = 0$, hence $\mathbf{u}\cdot\mathbf{u}' = 0$. We will use this fact three times in this chapter — it is the reason the unit normal is perpendicular to the unit tangent.

Check Your Understanding. Let $\mathbf{r}(t) = \langle t^2,\; t^3,\; t \rangle$. Find $\mathbf{r}'(t)$, then $\mathbf{r}''(t)$, then $\int_0^1 \mathbf{r}(t)\,dt$.

Answer

$\mathbf{r}'(t) = \langle 2t,\; 3t^2,\; 1\rangle$ and $\mathbf{r}''(t) = \langle 2,\; 6t,\; 0\rangle$. For the integral, integrate each slot from $0$ to $1$: $\int_0^1 t^2\,dt = \tfrac13$, $\int_0^1 t^3\,dt = \tfrac14$, $\int_0^1 t\,dt = \tfrac12$, so $\int_0^1 \mathbf{r}(t)\,dt = \big\langle \tfrac13,\; \tfrac14,\; \tfrac12 \big\rangle$. The result is a single point/vector — a definite vector integral returns a vector, not a function.

28.4 Velocity, Speed, and Acceleration

The physical reading of these derivatives is the payoff of the chapter. Interpret $\mathbf{r}(t)$ as the position of a particle at time $t$. Then:

• Velocity is the derivative: $\mathbf{v}(t) = \mathbf{r}'(t)$. It is tangent to the path, and it points in the direction the particle is actually moving.
• Speed is the length of velocity: $|\mathbf{v}(t)| = \|\mathbf{r}'(t)\| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}$. Speed is a non-negative scalar — "how fast," with no direction.
• Acceleration is the second derivative: $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$. It records how the velocity changes — and crucially, velocity can change by speeding up, by slowing down, or by turning. Acceleration captures all three.

The distinction between velocity (a vector) and speed (its length) is the conceptual core. A car rounding a bend at a steady 30 mph has constant speed but changing velocity, because the direction is turning — so it is accelerating even though the speedometer never moves. This is calculus as the mathematics of change: the velocity is changing, and acceleration measures exactly that.

Helix example

Take the helix $\mathbf{r}(t) = \langle \cos t,\; \sin t,\; t \rangle$.

$$\mathbf{v}(t) = \mathbf{r}'(t) = \langle -\sin t,\; \cos t,\; 1 \rangle, \qquad |\mathbf{v}(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}.$$

The speed is the constant $\sqrt 2$ — the particle climbs the helix at a uniform rate. Now acceleration:

$$\mathbf{a}(t) = \mathbf{r}''(t) = \langle -\cos t,\; -\sin t,\; 0 \rangle, \qquad |\mathbf{a}(t)| = 1.$$

The acceleration lies entirely in the $xy$-plane (its $z$-component is zero) and points from the particle straight back toward the $z$-axis — it is centripetal, the inward pull that keeps the motion curving around the axis. Notice that $\mathbf{v}\cdot\mathbf{a} = (-\sin t)(-\cos t) + (\cos t)(-\sin t) + (1)(0) = 0$: velocity and acceleration are perpendicular here, precisely because the speed is constant (recall the constant-length fact from §28.3). When you neither speed up nor slow down, all of your acceleration goes into turning.

Recovering motion by integration

Run the logic backward. If you know a particle's acceleration and its initial velocity and position, you recover the whole trajectory by integrating twice — the vector form of "integrate a rate to get the total change" (the Net Change Theorem, Chapter 14). This is how the 3D projectile of §28.2 arises.

A projectile feels only gravity, so $\mathbf{a}(t) = \langle 0,\, 0,\, -g \rangle$. Integrate once: $$\mathbf{v}(t) = \int \mathbf{a}(t)\,dt = \langle 0,\, 0,\, -gt \rangle + \mathbf{C}_1.$$ The constant vector $\mathbf{C}_1$ is fixed by the initial velocity $\mathbf{v}(0) = \langle v_x,\, v_y,\, v_z\rangle$, giving $\mathbf{v}(t) = \langle v_x,\; v_y,\; v_z - gt \rangle$. Integrate again: $$\mathbf{r}(t) = \langle v_x t,\; v_y t,\; v_z t - \tfrac12 g t^2 \rangle + \mathbf{C}_2,$$ and $\mathbf{C}_2 = \mathbf{r}(0) = \langle 0,\,0,\,h_0\rangle$ supplies the launch height. The result is exactly the projectile curve of §28.2. The horizontal components are linear (no horizontal force), the vertical component is the familiar downward parabola, and the path through space is a parabola tilted in whatever plane the launch velocity selects.

Real-World Application — Ballistics and spacecraft guidance (physics/engineering). Integrating acceleration to recover position is the literal job of an inertial navigation system. A spacecraft or aircraft carries accelerometers measuring $\mathbf{a}(t)$ in three axes; an onboard computer integrates once for velocity and twice for position, updating thousands of times per second. With no GPS and no view of the stars, a submarine or a rocket can know where it is purely by accumulating its own acceleration — the Net Change Theorem of Chapter 14, executed in real time on a vector.

Common Pitfall. Acceleration does not point in the direction of motion, and its magnitude is not the rate of change of speed in general. Students often write $|\mathbf{a}| = \frac{d}{dt}|\mathbf{v}|$. For the helix this would claim $|\mathbf{a}| = \frac{d}{dt}\sqrt{2} = 0$, yet $|\mathbf{a}| = 1$. The resolution is that acceleration also turns the velocity. Only the tangential part of acceleration changes speed; the rest curves the path. We make this split precise in §28.9.

28.5 Arc Length of a Space Curve

How long is a space curve? The answer generalizes the parametric arc length from Chapter 25 by adding a third coordinate under the root. Over $a \le t \le b$,

$$L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt.$$

The idea is the same Riemann-sum argument as always: approximate the curve by short straight chords, each of length $|\mathbf{r}'(t)|\,\Delta t$ to first order, and let the chords shrink. Since $|\mathbf{r}'(t)|$ is the speed, the formula reads "length equals the integral of speed over time" — distance is accumulated speed, the Net Change Theorem once more.

Example: one turn of the helix

For $\mathbf{r}(t) = \langle \cos t,\; \sin t,\; t \rangle$ on $0 \le t \le 2\pi$, the speed is the constant $\sqrt 2$, so

$$L = \int_0^{2\pi} \sqrt{2}\,dt = 2\pi\sqrt{2}.$$

Compare the underlying circle $\langle \cos t, \sin t, 0\rangle$, whose length over the same range is $2\pi$. The helix is longer by the factor $\sqrt 2 \approx 1.414$ — the extra length is the vertical climb, exactly the Pythagorean combination of "around" and "up."

Common Pitfall — arc length depends on the curve, not the parametrization. If you re-trace the helix twice as fast, $\mathbf{r}(2t)$ on $[0,\pi]$, the speed doubles to $2\sqrt2$ but the time interval halves, and the length integral still gives $2\pi\sqrt2$. The geometric length cannot depend on how quickly you drive the curve. A frequent error is to integrate speed over the wrong interval after rescaling $t$ — always integrate over the $t$-range that traces the curve exactly once.

The arc-length function

Freeze the lower limit and let the upper limit move — the same move that built the accumulation function in Chapter 14. Define

$$s(t) = \int_a^t |\mathbf{r}'(u)|\,du,$$

the arc-length function: the distance traveled along the curve from the start up to parameter $t$. By FTC Part 1 (Chapter 14),

$$\frac{ds}{dt} = |\mathbf{r}'(t)| = \text{speed},$$

so $s$ is strictly increasing whenever the speed is positive (the particle never stops or reverses). This single identity, $ds/dt = |\mathbf{r}'(t)|$, is the bridge between "parameter time" and "distance along the curve," and it is the engine behind the next section.

28.6 The Unit Tangent and Unit-Speed Reparametrization

The unit tangent vector

The velocity $\mathbf{r}'(t)$ gives both a direction (along the curve) and a magnitude (the speed). Often we want the direction alone, scrubbed of speed. Divide out the length:

$$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}.$$

This is the unit tangent vector: length $1$, pointing in the direction of motion. As $t$ advances, $\mathbf{T}(t)$ swivels to stay tangent to the curve, and watching it turn is exactly how we will measure bending in §28.8.

For the helix, $\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$ with $|\mathbf{r}'| = \sqrt 2$, so

$$\mathbf{T}(t) = \frac{1}{\sqrt 2}\,\langle -\sin t,\; \cos t,\; 1 \rangle.$$

Its $z$-component is the constant $1/\sqrt 2$: the tangent always tips upward at the same angle, while its horizontal part rotates around the axis. That is the geometric signature of a helix climbing at a fixed pitch.

Reparametrizing by arc length

The cleanest possible way to drive a curve is at unit speed — covering exactly one unit of length per unit of parameter. To arrange this, use the arc length $s$ itself as the parameter. Since $s(t)$ is increasing, it can be inverted to give $t(s)$; substitute to obtain $\mathbf{r}(s) := \mathbf{r}(t(s))$. By the chain rule and $ds/dt = |\mathbf{r}'(t)|$,

$$\frac{d\mathbf{r}}{ds} = \frac{d\mathbf{r}}{dt}\cdot\frac{dt}{ds} = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \mathbf{T},$$

a unit vector. So the arc-length parametrization always has speed $1$, and its velocity is the unit tangent. This is the arc-length (or unit-speed) parametrization, also called the natural parametrization. In it, geometric quantities like curvature take their simplest possible form, because the parameter is the geometry — distance itself.

Example: the helix at unit speed

Because the helix has constant speed $\sqrt 2$, the arc length from $t=0$ is simply $s = \sqrt 2\,t$, so $t = s/\sqrt 2$. Substituting:

$$\mathbf{r}(s) = \Big\langle \cos\tfrac{s}{\sqrt 2},\; \sin\tfrac{s}{\sqrt 2},\; \tfrac{s}{\sqrt 2} \Big\rangle.$$

Differentiate and check: $\frac{d\mathbf{r}}{ds} = \frac{1}{\sqrt2}\langle -\sin\frac{s}{\sqrt2}, \cos\frac{s}{\sqrt2}, 1\rangle$, which has length $\frac{1}{\sqrt2}\sqrt{1+1} = 1$. Unit speed, confirmed.

Computational Note. Most curves cannot be reparametrized by arc length in closed form, because the arc-length integral $s(t) = \int_a^t |\mathbf{r}'(u)|\,du$ often has no elementary antiderivative (recall Chapter 17's improper and non-elementary integrals — most arc lengths land there). The ellipse is the classic offender: its arc length defines the elliptic integrals, which provably escape elementary functions. The helix is special precisely because its speed is constant, making $s(t) = \sqrt2\,t$ trivially invertible. In practice we compute arc length and reparametrize numerically, as the Python in §28.10 does.

28.7 A Geometric Detour: What Curvature Should Measure

Before the formula, fix the intuition. Curvature should answer: how sharply is the curve bending right here? A straight line does not bend at all — its curvature should be $0$. A tight little circle bends hard — large curvature. A vast, gentle circle barely bends — small curvature. And curvature should be a property of the curve itself, independent of how fast you drive along it: speeding up does not make a road bendier.

The cleanest way to capture "bending" is to watch the unit tangent $\mathbf{T}$ turn. On a straight line $\mathbf{T}$ never changes — zero bending. On a sharp curve $\mathbf{T}$ whips around quickly. So curvature is the rate at which the direction of motion turns, measured per unit distance traveled (not per unit time, so that speed drops out). That is the definition we make precise next.

28.8 Curvature

Definition. Using the arc-length parameter $s$ so that speed is factored out, the curvature is

$$\kappa = \left|\frac{d\mathbf{T}}{ds}\right|.$$

It is a non-negative scalar with units of (1/length). Because we usually have the curve in terms of $t$, not $s$, convert with the chain rule $\frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}/dt}{ds/dt} = \frac{\mathbf{T}'(t)}{|\mathbf{r}'(t)|}$, giving the working formula

$$\kappa(t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}.$$

A more practical formula

Computing $\mathbf{T}'$ by hand is painful (it needs the quotient rule on a square root). A cleaner equivalent, derived from the definition, uses only $\mathbf{r}'$ and $\mathbf{r}''$:

$$\boxed{\;\kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.\;}$$

This is almost always the formula to reach for. For a plane curve $y = f(x)$ (parametrize as $\langle x, f(x), 0\rangle$), it collapses to the tidy single-variable version

$$\kappa(x) = \frac{|f''(x)|}{\big(1 + f'(x)^2\big)^{3/2}}.$$

Notice the second derivative $f''$ in the numerator: curvature is fundamentally about concavity, the bending captured by $f''$ back in Chapter 9. Where the graph is straight ($f''=0$), curvature vanishes.

Geometric meaning: the osculating circle

For a circle of radius $R$, curvature is constant and equals $\kappa = 1/R$ (check below). A small circle has large curvature, a large circle small curvature — matching the intuition of §28.7. This inspires a definition: the radius of curvature is

$$\rho = \frac{1}{\kappa},$$

the radius of the circle that best hugs the curve at a point — the osculating circle (Latin osculum, "kiss"). It is the circle that matches the curve's position, tangent direction, and curvature there: the circle the curve is momentarily "trying to be."

Geometric Intuition. Drive along a winding road in the dark. At each instant your headlights sweep out the arc of a circle — the osculating circle — whose radius is how gently or sharply you are turning right now. On a near-straight stretch that circle is enormous (radius $\to\infty$, curvature $\to 0$); in a hairpin it shrinks to a tight loop (small radius, large curvature). Curvature is just $1/(\text{turning radius})$ at each point.

Verifying the circle

For $\mathbf{r}(t) = \langle R\cos t, R\sin t, 0\rangle$: $\mathbf{r}' = \langle -R\sin t, R\cos t, 0\rangle$ with $|\mathbf{r}'| = R$, and $\mathbf{r}'' = \langle -R\cos t, -R\sin t, 0\rangle$. The cross product $\mathbf{r}'\times\mathbf{r}''$ has only a $z$-component, $(-R\sin t)(-R\sin t) - (R\cos t)(-R\cos t) = R^2$, so $|\mathbf{r}'\times\mathbf{r}''| = R^2$. Then $$\kappa = \frac{R^2}{R^3} = \frac{1}{R}. \checkmark$$ A circle of radius $R$ has curvature $1/R$ everywhere, exactly as promised.

Helix example, carefully

For $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$:

$$\mathbf{r}'(t) = \langle -\sin t,\; \cos t,\; 1\rangle, \qquad \mathbf{r}''(t) = \langle -\cos t,\; -\sin t,\; 0\rangle.$$

Compute the cross product component by component: $$\mathbf{r}'\times\mathbf{r}'' = \big\langle (\cos t)(0)-(1)(-\sin t),\;\; (1)(-\cos t)-(-\sin t)(0),\;\; (-\sin t)(-\sin t)-(\cos t)(-\cos t)\big\rangle$$ $$= \langle \sin t,\; -\cos t,\; \sin^2 t + \cos^2 t\rangle = \langle \sin t,\; -\cos t,\; 1\rangle.$$

Its length is $|\mathbf{r}'\times\mathbf{r}''| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt 2$, and $|\mathbf{r}'|^3 = (\sqrt2)^3 = 2\sqrt2$. Therefore

$$\kappa = \frac{\sqrt 2}{2\sqrt 2} = \frac{1}{2}.$$

The helix has constant curvature $\tfrac12$ — it bends at exactly the same rate everywhere, which is why a helix looks so uniform. (Symbolic computation in §28.10 confirms this in one line.)

Computational Note. The general helix $\mathbf{r}(t) = \langle a\cos t,\; a\sin t,\; b\,t\rangle$ — radius $a$, rising $2\pi b$ per turn — has constant curvature $\kappa = \dfrac{a}{a^2+b^2}$. For our example $a=b=1$, this gives $\kappa = \dfrac{1}{1+1} = \dfrac12$, matching the hand computation. Two checks worth noting: as $b\to 0$ the helix flattens to a circle of radius $a$ and $\kappa \to 1/a$, the circle value; and as $b\to\infty$ the helix stretches toward a straight vertical line and $\kappa\to 0$. Both limits are exactly what the geometry demands.

Check Your Understanding. Find the curvature of the parabola $y = x^2$ at the origin $x=0$ and at $x=1$.

Answer

Here $f'(x) = 2x$ and $f''(x) = 2$, so $\kappa(x) = \dfrac{2}{(1+4x^2)^{3/2}}$. At $x=0$: $\kappa = 2$ (the parabola bends most sharply at its vertex — radius of curvature $\tfrac12$). At $x=1$: $\kappa = \dfrac{2}{5^{3/2}} = \dfrac{2}{5\sqrt5}\approx 0.179$ — much gentler, because the parabola straightens out as it opens. Curvature is largest at the vertex and decays to $0$ far out on the arms.

28.9 The Unit Normal and the Components of Acceleration

The principal unit normal

The unit tangent $\mathbf{T}$ tells you which way the curve heads. To describe which way it bends, look at how $\mathbf{T}$ itself changes. Define the principal unit normal:

$$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}.$$

Because $\mathbf{T}$ has constant length $1$, the constant-length fact from §28.3 guarantees $\mathbf{T}\cdot\mathbf{T}' = 0$ — so $\mathbf{N}$ is perpendicular to $\mathbf{T}$. And $\mathbf{N}$ points toward the concave side of the curve, into the center of the osculating circle — the direction the curve is turning. Together $\{\mathbf{T}, \mathbf{N}\}$ are two perpendicular unit vectors riding along the curve: one along the motion, one across it into the bend.

Math Major Sidebar — the binormal and the Frenet frame (optional). A third unit vector completes an orthonormal trio: the binormal $\mathbf{B} = \mathbf{T}\times\mathbf{N}$, perpendicular to both. The right-handed frame $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ moving along the curve is the Frenet–Serret frame, and the rate at which $\mathbf{B}$ rotates defines a second invariant, the torsion $\tau$, measuring how the curve twists out of its instantaneous plane (a plane curve has $\tau = 0$; the helix $\langle\cos t,\sin t,t\rangle$ has constant $\tau = \tfrac12$). The deep result — the Fundamental Theorem of Space Curves — is that curvature and torsion as functions of arc length determine a curve completely, up to where you place it and how you rotate it. That is the exact space-curve analog of "an antiderivative is determined up to a constant" (Chapter 14): the geometry is recovered from its rates of bending and twisting by integration. We keep the main text focused on $\kappa$, $\mathbf{T}$, and $\mathbf{N}$, which suffice for all the motion problems below; torsion and the full Frenet apparatus belong to a differential-geometry course.

Splitting acceleration

Here is where $\mathbf{T}$ and $\mathbf{N}$ earn their keep. The §28.4 pitfall warned that acceleration is not simply the rate of change of speed. The clean statement is that acceleration always lies in the plane of $\mathbf{T}$ and $\mathbf{N}$, and splits into exactly two pieces:

$$\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N},$$

with

$$a_T = \frac{d}{dt}|\mathbf{v}| = \frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|}, \qquad a_N = \kappa\,|\mathbf{v}|^2 = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}.$$

The tangential component $a_T$ is the rate of change of speed — it speeds you up or slows you down along the path. The normal component $a_N$ is the turning part — it bends the velocity without changing its length, and it grows with both the sharpness of the curve ($\kappa$) and the square of the speed. There is no third component: acceleration never points out of the $\mathbf{T}$–$\mathbf{N}$ plane.

The Key Insight. Every acceleration is a blend of two jobs: changing how fast you go ($a_T\mathbf{T}$) and changing which way you go ($a_N\mathbf{N}$). The turning part $a_N = \kappa v^2$ depends on the square of speed — which is why doubling your speed through a bend quadruples the sideways force you feel.

Helix, decomposed

For the helix, speed is the constant $\sqrt 2$, so $a_T = \frac{d}{dt}\sqrt 2 = 0$ — no tangential acceleration, consistent with constant speed. The normal component is $a_N = \kappa\,|\mathbf{v}|^2 = \tfrac12\cdot(\sqrt2)^2 = \tfrac12\cdot 2 = 1$, which matches $|\mathbf{a}| = 1$ found directly in §28.4. All of the helix's acceleration is turning; none of it is speeding up. The picture is fully consistent.

Real-World Application — highway curve design (civil engineering). The lateral force a driver feels rounding a curve is $a_N = \kappa v^2$ per unit mass. Designers cap the lateral acceleration at a comfortable, safe fraction of $g$ (often around $0.1$–$0.15\,g$ for high-speed roads), which forces the curvature down: $\kappa \le a_{N,\max}/v^2$. Because $a_N$ scales with $v^2$, a highway rated for high speeds needs very large radii — long, sweeping bends — while a slow city street can turn tightly. Real on-ramps go further and use a clothoid (a spiral whose curvature grows linearly with distance) so that $\kappa$ ramps up smoothly from $0$, sparing the driver a sudden jerk in lateral force. The same $\kappa v^2$ governs how fast a roller coaster can take a loop before the g-forces exceed human tolerance.

Check Your Understanding. A particle moves with velocity $\mathbf{v} = \langle 1, 2, 2\rangle$ and acceleration $\mathbf{a} = \langle 3, 0, 0\rangle$ at some instant. Find $a_T$, the tangential component of acceleration.

Answer

$|\mathbf{v}| = \sqrt{1+4+4} = 3$ and $\mathbf{v}\cdot\mathbf{a} = (1)(3)+(2)(0)+(2)(0) = 3$. So $a_T = \dfrac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|} = \dfrac{3}{3} = 1$. The particle is speeding up at $1$ unit/s². (For the curious, $a_N = \sqrt{|\mathbf{a}|^2 - a_T^2} = \sqrt{9-1} = 2\sqrt2$, since $\mathbf{T}$ and $\mathbf{N}$ are perpendicular and the two components are the legs of a right triangle with hypotenuse $|\mathbf{a}|=3$.)

28.10 Computation: Plotting and Measuring a Space Curve

Hand computation built the understanding; now let the machine give us power — plotting curves we cannot picture and computing curvature where the algebra is hopeless. We follow the book's three-tier pattern: state it, do it by hand, then confirm with Python.

First, plot the helix in 3D and overlay its underlying circle, so the climb is visible:

# Plot a helix as a 3D space curve r(t) = <cos t, sin t, t/(2*pi)>
# (z scaled so one full turn rises by 1 unit).
import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0, 4 * np.pi, 400)
x, y, z = np.cos(t), np.sin(t), t / (2 * np.pi)   # two full turns

fig = plt.figure(figsize=(7, 6))
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z, lw=2, label='helix')
ax.plot(np.cos(t), np.sin(t), 0 * t, lw=1, ls='--', label='base circle')
ax.set_xlabel('x'); ax.set_ylabel('y'); ax.set_zlabel('z')
ax.set_title('Figure 28.1 — A helix climbing above its base circle')
ax.legend()
plt.show()
# Figure 28.1: the helix spirals upward; its shadow on the xy-plane is the dashed circle.

Now verify the curvature symbolically — no approximation, the exact value:

# Verify kappa = 1/2 for the helix r(t) = <cos t, sin t, t> using sympy.
import sympy as sp

t = sp.symbols('t', real=True)
r = sp.Matrix([sp.cos(t), sp.sin(t), t])
r1 = r.diff(t)                 # r'(t)
r2 = r1.diff(t)                # r''(t)
cross = r1.cross(r2)
kappa = sp.simplify(cross.norm() / r1.norm()**3)
print("speed =", sp.simplify(r1.norm()))   # speed = sqrt(2)
print("kappa =", kappa)                     # kappa = 1/2

Finally, compute curvature numerically for a curve with no clean formula — the twisted cubic $\langle t, t^2, t^3\rangle$ — using finite-difference derivatives. This is how curvature is found in practice for data-driven curves:

# Numerical curvature of the twisted cubic r(t) = <t, t^2, t^3>.
import numpy as np

t = np.linspace(-2, 2, 2001)
r = np.stack([t, t**2, t**3], axis=1)          # shape (N, 3)
dt = t[1] - t[0]
r1 = np.gradient(r, dt, axis=0)                 # r'(t), component-wise
r2 = np.gradient(r1, dt, axis=0)                # r''(t)
cross = np.cross(r1, r2)
kappa = np.linalg.norm(cross, axis=1) / np.linalg.norm(r1, axis=1)**3
print("max curvature ~", round(kappa.max(), 3), "near t =", round(t[kappa.argmax()], 3))
# max curvature ~ 2.0 near t = 0.0  -> the twisted cubic bends most sharply at the origin

Computational Note. np.gradient differentiates each column (each component) independently — it is the numerical embodiment of "differentiate component-wise" from §28.3. The curvature formula $|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$ is applied to the whole array at once, returning $\kappa$ at every sample point. Finite differences degrade near the ends of the interval, so trust the interior values; the peak at $t=0$ is robust. This is the workflow for any curve known only as sampled points — a GPS track, a motion-capture trajectory, a fitted spline.

28.11 Motion in Space: Two Application Fields

Vector-valued functions are the natural language of anything that moves. We close with two physical settings, then point the way into the rest of multivariable calculus.

Planetary and satellite orbits (physics)

A planet of position $\mathbf{r}(t)$ orbiting the Sun obeys Newton's law of gravitation, which in vector form is a second-order vector ODE:

$$\mathbf{r}''(t) = -\,\frac{GM}{|\mathbf{r}|^3}\,\mathbf{r},$$

— acceleration points back toward the Sun (the $-\mathbf{r}$ direction) with magnitude $GM/|\mathbf{r}|^2$, the inverse-square law. Two of the chapter's tools reveal the orbit's structure immediately.

First, angular momentum is conserved. Define $\mathbf{L} = \mathbf{r}\times\mathbf{r}'$ (angular momentum per unit mass). Differentiate using the cross-product rule from §28.3: $$\mathbf{L}' = \mathbf{r}'\times\mathbf{r}' + \mathbf{r}\times\mathbf{r}'' = \mathbf{0} + \mathbf{r}\times\Big(\!-\tfrac{GM}{|\mathbf{r}|^3}\mathbf{r}\Big) = \mathbf{0},$$ because $\mathbf{r}'\times\mathbf{r}' = \mathbf{0}$ and $\mathbf{r}\times\mathbf{r} = \mathbf{0}$. So $\mathbf{L}$ is a constant vector — which means $\mathbf{r}$ and $\mathbf{r}'$ always lie in the fixed plane perpendicular to $\mathbf{L}$. Orbits are planar, and we have just proved it from the product rule. Moreover, the constancy of $|\mathbf{L}|$ is precisely Kepler's Second Law (equal areas swept in equal times), since the area swept rate is $\tfrac12|\mathbf{r}\times\mathbf{r}'|$. The conic-section orbits of Chapter 27 — ellipses, parabolas, hyperbolas — emerge from solving this very ODE, tying Part VI back to Part V.

Real-World Application — satellite station-keeping (aerospace). A geostationary satellite must hold a near-circular orbit, but the Moon, the Sun, and Earth's equatorial bulge perturb $\mathbf{r}''(t)$ away from the ideal inverse-square law. Mission controllers integrate the perturbed vector ODE forward (numerically, as in §28.10) to predict drift, then fire thrusters to correct it — adding a small $\Delta\mathbf{v}$ to the velocity vector. Every GPS fix, weather image, and satellite-TV signal depends on someone integrating a vector-valued acceleration accurately.

Charged particles and the aurora (physics)

A particle of charge $q$ and mass $m$ moving with velocity $\mathbf{v}$ through a uniform magnetic field $\mathbf{B}$ feels the Lorentz force $\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}$, so $m\,\mathbf{r}'' = q\,\mathbf{r}'\times\mathbf{B}$. Because the force is always perpendicular to the velocity (a cross product), it does no work and never changes the speed — it only turns the particle. The motion splits cleanly: the velocity component along $\mathbf{B}$ is unaffected (constant drift), while the component perpendicular to $\mathbf{B}$ is forced into a circle. A constant drift plus a circle is exactly a helix — our running example, now produced by physics. The radius of the circular part is the gyroradius

$$r = \frac{m\,v_\perp}{|q|\,B},$$

set by the perpendicular speed $v_\perp$, the charge, and the field strength.

This is not an abstraction. Charged particles from the solar wind spiral along Earth's magnetic field lines in just such helices, funneling toward the poles where the field lines converge. When they strike the upper atmosphere they make it glow — the aurora borealis and australis. The shimmering curtains of the northern lights are vast numbers of particles each tracing the space curve we computed the curvature of in §28.8.

Curves in biology

The DNA double helix is geometrically a helix: two strands winding around a common axis. Modeling each strand as $\langle a\cos t, a\sin t, b\,t\rangle$ with radius $a\approx 1$ nm and a rise of about $3.4$ nm per ten base pairs (so $b = 3.4/(2\pi) \approx 0.54$ nm per radian) gives, from the general-helix formula of §28.8, a curvature of roughly $\kappa = a/(a^2+b^2) \approx 0.77$ nm$^{-1}$.

Warning — model versus measurement. That $\kappa \approx 0.77$ nm$^{-1}$ is the curvature of an idealized smooth-helix model of one strand, and it is sensitive to which radius and rise you plug in; biophysics papers report a range of effective curvature and twist values depending on the model (rigid rod, worm-like chain, base-pair-level geometry) and on sequence and solvent conditions. Treat the number as an order-of-magnitude illustration of how vector calculus applies to DNA, not as a fixed physical constant. The genuine, model-independent point stands: DNA's mechanical behavior — how it bends to pack into a nucleus and how proteins recognize its shape — is described in the language of curvature and torsion built in this chapter.

Add to Your Modeling Portfolio. Add a vector-valued trajectory to your model — a quantity that moves through space (or state space) over time, described by $\mathbf{r}(t)$, with velocity $\mathbf{r}'(t)$ and acceleration $\mathbf{r}''(t)$. Biology: model the path of a swimming microorganism or a strand of DNA as a space curve; compute its arc length and curvature. Economics: track a two-variable system (say price and inventory) as a curve $\mathbf{r}(t) = \langle p(t), q(t)\rangle$ in state space; the velocity vector shows the direction the economy is heading at each instant. Physics: model a projectile or orbiting body, integrating $\mathbf{r}''(t) = \mathbf{a}(t)$ twice from initial conditions to recover the full trajectory. Data Science: treat a time series of feature vectors as a curve in $\mathbb{R}^n$; arc length measures total "distance traveled" through feature space and curvature flags moments of sharp change.

28.12 Connecting to What's Coming

This chapter is a deliberate bridge. It has the formal structure of single-variable calculus — one input $t$, one derivative, component-wise everything — but the geometric content of multivariable calculus: vectors living in space, tangents, normals, and curvature. That is why it opens Part VI.

What changes next is the input side, not the output side. Chapter 29 introduces functions of several variables, $f(x,y)$, which take two numbers in and return one out — the graph is now a surface, a landscape of hills and valleys, rather than a curve. Chapter 30 differentiates such functions, producing the gradient $\nabla f$, and you will find the unit-tangent and perpendicularity ideas from this chapter resurfacing: the gradient is perpendicular to level curves just as $\mathbf{N}$ is perpendicular to $\mathbf{T}$. Chapter 31 optimizes functions of several variables, Chapters 32–33 integrate over regions, and in Chapter 34 the other kind of vector object finally arrives — the vector field, a vector at every point of space (the object we so carefully distinguished from $\mathbf{r}(t)$ on page one). The space curves of this chapter then return as the flow lines of those fields and as the paths of line integrals in Chapter 35, where the slogan of the entire book — the integral of a derivative over a region equals the values on the boundary (Chapter 14, generalized) — reaches its vector-calculus form.

You arrived at this chapter able to do calculus on numbers. You leave it able to do calculus on motion through space. The toolkit — velocity, speed, acceleration, arc length, the unit tangent and normal, curvature — is the vocabulary of every moving thing, from a tossed ball to a planet to a strand of DNA. Turn the page, and we let the inputs multiply.