1. Vector-valued functions
A vector-valued function is a function whose output is a vector. The input is a scalar — usually time $t$:
$$ \mathbf{r}(t) = \langle x(t),\ y(t),\ z(t)\rangle. $$You can read this two ways. As an algebra: three scalar functions packaged together. As a geometry: at each moment $t$, the tip of the vector $\mathbf{r}(t)$ sits at a point in space, and as $t$ varies the tip traces a curve.
The image $\{\mathbf{r}(t) : t \in I\}$ of a continuous vector-valued function on an interval $I$ — a one-dimensional trail through three-dimensional space.
This is the natural language of trajectories. The orbit of a planet, the path of a charged particle in a magnetic field, the helix of a spring — all are described not by $y = f(x)$ but by a single vector function $\mathbf{r}(t)$ that tells you where the thing is as a function of when.
Boldface ($\mathbf{r}$) or an over-arrow ($\vec r$) marks a vector. Components are written in angle brackets $\langle\, ,\, ,\, \rangle$ or as a column. The arithmetic — addition, scalar multiplication, dot and cross products — works componentwise.
2. Derivatives: velocity and acceleration
The derivative of a vector-valued function is taken componentwise:
$$ \mathbf{r}'(t) = \langle x'(t),\ y'(t),\ z'(t)\rangle. $$That's the cleanest possible definition, and it's also the right one — it follows from the limit $\mathbf{r}'(t) = \lim_{h\to 0}\tfrac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h}$, with the subtraction and division done componentwise.
Two interpretations stack on top of each other:
- Velocity: $\mathbf{v}(t) = \mathbf{r}'(t)$ is the instantaneous velocity. Its direction is tangent to the curve; its magnitude $|\mathbf{v}(t)|$ is the speed.
- Acceleration: $\mathbf{a}(t) = \mathbf{r}''(t) = \mathbf{v}'(t)$ measures how velocity is changing. Crucially, acceleration can change direction as well as magnitude — and that distinction will drive the entire rest of the page.
Speeding up and turning are different things, even though both involve "changing velocity." For a vector function, the derivative captures both: change in $|\mathbf{v}|$ and change in direction. Pulling those apart is the engine behind the TNB frame.
3. Unit tangent vector
$\mathbf{r}'(t)$ points along the curve, but its length depends on how fast the parametrization sweeps. Strip the length out and keep only the direction:
$\mathbf{T}(t)$ is a unit vector ($|\mathbf{T}| = 1$) tangent to the curve at the point $\mathbf{r}(t)$, pointing in the direction of motion.
Why bother normalizing? Because the shape of the curve is the same regardless of how fast you traverse it — but $\mathbf{r}'$ rescales when you reparametrize. The unit tangent $\mathbf{T}$ depends on the curve, not on the timing. It's an intrinsic property.
4. Arc length
The same logic as the parametric case in 2-D, extended to 3-D. Between $t$ and $t + dt$, the displacement is $\mathbf{r}'(t)\,dt$, whose length is $|\mathbf{r}'(t)|\,dt$. Sum over $[a, b]$:
$$ L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt. $$The integrand is the speed, and the integral is the total distance traveled. The arc-length function measures distance traveled from a fixed starting time $a$ up to time $t$:
$$ s(t) = \int_a^t |\mathbf{r}'(u)|\,du. $$By the Fundamental Theorem of Calculus, $\dfrac{ds}{dt} = |\mathbf{r}'(t)|$ — the rate at which arc length grows is precisely the speed. That little identity is the bridge to reparameterization.
5. Arc-length reparameterization
The parameter $t$ is convenient (it's clock time), but it's arbitrary. The same curve admits infinitely many parametrizations — slow, fast, accelerating, even backwards. Among them, one is canonical:
The curve is parametrized by arc length when the parameter is the distance traveled. Equivalently, the speed is constantly $1$: $|\mathbf{r}'(s)| = 1$ for all $s$.
To reparametrize by arc length, compute $s(t) = \int_a^t |\mathbf{r}'(u)|\,du$, invert to get $t(s)$, and substitute back: $\mathbf{r}(s) = \mathbf{r}(t(s))$. The result is a parametrization where the parameter is geometrically meaningful — it's the distance you've traveled along the curve, not a stopwatch reading.
This isn't always practical (inverting $s(t)$ often has no closed form), but it's the right theoretical move. With $s$ as parameter, derivatives like $d\mathbf{T}/ds$ become intrinsic to the curve — they don't depend on how fast you traversed it. That sets up the cleanest possible definition of curvature.
Two cars driving the same loop at different speeds traverse the same curve, but their $\mathbf{r}'(t)$ vectors are different. Reparametrizing by arc length erases the speed difference and leaves only the curve's geometry. Curvature, the TNB frame, and the Frenet–Serret formulas are all about that geometry.
6. Curvature
Curvature measures how sharply a curve bends — independently of how fast you travel along it. A straight line has curvature $0$. A circle of radius $R$ has curvature $1/R$ everywhere (small circles bend a lot; large ones bend a little).
The intrinsic definition: $\kappa = \!\left|\dfrac{d\mathbf{T}}{ds}\right|$. With $s$ being arc length, $\mathbf{T}$ is unit-length, so this rate of change measures only how fast the direction is rotating.
That definition is clean but inconvenient — you'd need to reparametrize by arc length first. Two equivalent formulas in terms of the original parameter $t$ are much more usable:
$$ \kappa(t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}. $$The second form is the one to reach for in practice. Compute $\mathbf{r}'$ and $\mathbf{r}''$, take the cross product, take magnitudes — done.
7. The TNB frame
At each point on a smooth space curve, three mutually perpendicular unit vectors live together — a local coordinate system that travels along with the curve.
- $\mathbf{T}$ — tangent. Points in the direction of motion. Defined: $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$.
- $\mathbf{N}$ — principal normal. Points in the direction the tangent is turning. Defined: $\mathbf{N} = \mathbf{T}'/|\mathbf{T}'|$. Lies in the plane of the curve's bend and points toward the center of the osculating circle.
- $\mathbf{B}$ — binormal. Perpendicular to both: $\mathbf{B} = \mathbf{T} \times \mathbf{N}$. Points out of the plane of the bend.
Together, $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ form an orthonormal basis — the Frenet–Serret frame. The plane spanned by $\mathbf{T}$ and $\mathbf{N}$ is called the osculating plane: it's the plane that best contains the curve at the point of interest. For a planar curve, the osculating plane is just the plane the curve lives in, and $\mathbf{B}$ points perpendicular to that plane.
The binormal $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ depends on the right-hand rule, which depends on the orientation of your coordinate system. Reverse the direction of motion along the curve and $\mathbf{T}$ flips, $\mathbf{N}$ may flip (it's always toward the center of the bend), so $\mathbf{B}$ may or may not flip — work it out from the definitions, don't guess.
8. Tangential and normal acceleration
Acceleration $\mathbf{a} = \mathbf{r}''$ does two distinct things: it changes the magnitude of velocity (speeds you up or slows you down), and it changes the direction of velocity (turns you). The TNB frame lets you decompose acceleration cleanly into those two parts:
$$ \mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, $$where
$$ a_T = \frac{d|\mathbf{v}|}{dt} = \frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|}, \qquad a_N = \kappa\,|\mathbf{v}|^2 = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}. $$$a_T$ is exactly the rate of change of speed — that's the "speeding up" part. $a_N$ is what turns you, and the formula $a_N = \kappa\,|\mathbf{v}|^2$ encodes a familiar physical fact: driving a curve of radius $R$ at speed $v$ requires centripetal acceleration $v^2/R$.
Importantly, the acceleration vector has no binormal component — $\mathbf{a}$ always lies in the osculating plane. That's not an accident; it's a theorem, and it's the cleanest place to feel the geometry of the TNB frame at work.
9. Common pitfalls
$\mathbf{T}$ requires dividing $\mathbf{r}'$ by its magnitude. Same for $\mathbf{N}$. Skipping the normalization leaves you with vectors that point in the right direction but have wrong lengths, and every downstream computation (curvature, the binormal, the acceleration decomposition) breaks.
Curvature is not the magnitude of acceleration. A particle moving in a straight line at increasing speed has $|\mathbf{r}''| > 0$ but $\kappa = 0$ — the curve isn't bending, even though the speed is changing. Curvature measures bending; the magnitude of acceleration mixes bending with speeding up.
$\mathbf{B} = \mathbf{T} \times \mathbf{N}$ obeys the right-hand rule. If you reverse the parametrization direction, $\mathbf{T}$ flips sign and so does $\mathbf{B}$ — the osculating plane is the same, but its "above" and "below" swap. Don't memorize a sign; rederive from the cross product when you need it.
$\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$ is only defined where $|\mathbf{r}'| \neq 0$. A curve with a momentary stop (e.g. $\mathbf{r}(t) = \langle t^3, t^2, 0\rangle$ at $t = 0$) has no well-defined tangent direction there. Such points often correspond to cusps, and curvature blows up.
The cross-product formula for $\kappa$ assumes $\mathbf{r}$ lives in $\mathbb{R}^3$. To use it for a planar curve $\mathbf{r}(t) = \langle x(t), y(t)\rangle$, embed in 3-D by setting $z(t) = 0$. Then $\mathbf{r}' \times \mathbf{r}''$ has only a $z$-component, and its magnitude is the familiar $|x'y'' - y'x''|$.
10. Worked examples
Example 1 · Arc length of the helix $\mathbf{r}(t) = \langle\cos t, \sin t, t\rangle$
One full turn corresponds to $t \in [0, 2\pi]$.
Step 1. Derivative:
$$ \mathbf{r}'(t) = \langle -\sin t,\ \cos t,\ 1\rangle. $$Step 2. Speed:
$$ |\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}. $$The speed is constant — the helix is already very close to being arc-length parameterized (just off by a factor of $\sqrt 2$).
Step 3. Integrate:
$$ L = \int_0^{2\pi} \sqrt{2}\,dt = 2\pi\sqrt{2}. $$Example 2 · Curvature of the same helix
Using the cross-product formula $\kappa = |\mathbf{r}' \times \mathbf{r}''|/|\mathbf{r}'|^3$.
Step 1. Compute $\mathbf{r}''$:
$$ \mathbf{r}''(t) = \langle -\cos t,\ -\sin t,\ 0\rangle. $$Step 2. Cross product:
$$ \mathbf{r}' \times \mathbf{r}'' = \det\!\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin t & \cos t & 1 \\ -\cos t & -\sin t & 0\end{pmatrix} = \langle \sin t,\ -\cos t,\ \sin^2 t + \cos^2 t\rangle = \langle \sin t,\ -\cos t,\ 1\rangle. $$Step 3. Magnitudes:
$$ |\mathbf{r}' \times \mathbf{r}''| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}, \qquad |\mathbf{r}'|^3 = (\sqrt{2})^3 = 2\sqrt{2}. $$Step 4. Combine:
$$ \kappa = \frac{\sqrt{2}}{2\sqrt{2}} = \tfrac{1}{2}. $$Curvature is constant along the helix — it bends the same amount at every point, which matches the symmetry of the shape.
Example 3 · Unit tangent for $\mathbf{r}(t) = \langle t, t^2, t^3\rangle$ at $t = 1$
Step 1. Differentiate:
$$ \mathbf{r}'(t) = \langle 1,\ 2t,\ 3t^2\rangle. $$Step 2. Evaluate at $t = 1$:
$$ \mathbf{r}'(1) = \langle 1,\ 2,\ 3\rangle, \qquad |\mathbf{r}'(1)| = \sqrt{1 + 4 + 9} = \sqrt{14}. $$Step 3. Normalize:
$$ \mathbf{T}(1) = \frac{1}{\sqrt{14}}\langle 1,\ 2,\ 3\rangle. $$Example 4 · Tangential and normal acceleration for circular motion
Consider uniform circular motion: $\mathbf{r}(t) = \langle R\cos(\omega t),\ R\sin(\omega t),\ 0\rangle$, with $R, \omega$ constants.
Step 1. Velocity and acceleration:
$$ \mathbf{v}(t) = \langle -R\omega \sin(\omega t),\ R\omega \cos(\omega t),\ 0\rangle, $$ $$ \mathbf{a}(t) = \langle -R\omega^2 \cos(\omega t),\ -R\omega^2 \sin(\omega t),\ 0\rangle. $$Step 2. Speed is constant: $|\mathbf{v}| = R\omega$.
Step 3. Tangential component: $a_T = d|\mathbf{v}|/dt = 0$ — no speeding up or slowing down.
Step 4. Normal component: $a_N = |\mathbf{a}| = R\omega^2$. Alternatively, $\kappa = 1/R$ (circle of radius $R$) and $|\mathbf{v}|^2 = R^2\omega^2$, so
$$ a_N = \kappa\,|\mathbf{v}|^2 = \tfrac{1}{R}\cdot R^2\omega^2 = R\omega^2. \quad\checkmark $$All of the acceleration is normal — pointing toward the center. That's centripetal acceleration, and the TNB decomposition has handed it back automatically.
Example 5 · Reparametrizing the helix by arc length
From Example 1, $|\mathbf{r}'(t)| = \sqrt 2$ for the helix $\mathbf{r}(t) = \langle\cos t, \sin t, t\rangle$.
Step 1. Compute $s(t)$:
$$ s(t) = \int_0^t \sqrt{2}\,du = \sqrt{2}\,t. $$Step 2. Invert: $t = s/\sqrt{2}$.
Step 3. Substitute back:
$$ \mathbf{r}(s) = \!\left\langle\cos\!\left(\tfrac{s}{\sqrt 2}\right),\ \sin\!\left(\tfrac{s}{\sqrt 2}\right),\ \tfrac{s}{\sqrt 2}\right\rangle. $$Step 4. Verify: $\mathbf{r}'(s) = \tfrac{1}{\sqrt 2}\langle -\sin(s/\sqrt 2),\ \cos(s/\sqrt 2),\ 1\rangle$, and $|\mathbf{r}'(s)| = \tfrac{1}{\sqrt 2}\sqrt{1 + 1} = 1$. ✓
The parameter $s$ now measures distance along the curve, and the speed is constantly $1$.