---
title: Arc Length, Curvature, and Motion in Space
module: Vectors and the Geometry of Space
moduleNumber: 10
lessonNumber: 5
order: 1005
summary: >
  Integrating the speed of a vector function gives the length of its curve and a
  natural parameter, arc length, that depends only on the curve's shape.
  Curvature measures how fast the unit tangent turns, and together with the
  normal and binormal it builds the moving TNB frame. Reading the same vector
  function as a trajectory, its first two derivatives are velocity and
  acceleration, and acceleration splits cleanly into tangential and normal parts.
topics: [Vectors and the Geometry of Space]
sources:
  - book: Stewart
    ref: "Ch. 13; §13.3 Arc Length and Curvature"
  - book: Stewart
    ref: "§13.4 Motion in Space: Velocity and Acceleration"
draft: false
---

A [vector function](/calculus/vectors-and-space-curves/vector-functions-and-space-curves)
carries two kinds of information about its curve. Intrinsic to the shape are its
length and how sharply it bends, captured by **arc length** and **curvature**.
Treating the parameter as time instead reads off **velocity** and
**acceleration**. The two meet in one identity: a moving particle's acceleration
splits into a part that changes its speed and a part that turns it, and the
turning part is curvature times speed squared.

## Arc length

Approximate a curve by a chain of short segments and add their lengths; refining
the chain drives the sum to an integral of speed. Since the speed of
$\vec r(t)$ is $\norm{\vec r'(t)}$, the length is its integral.

> **Definition (Arc length).** For a curve $\vec r(t) = \vector{f(t), g(t), h(t)}$,
> $a \le t \le b$, traversed once with $\vec r'$ continuous,
>
> $$
> L = \int_a^b \norm{\vec r'(t)}\,\d t = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}\,\d t.
> $$

For the helix $\vec r(t) = \cos t\,\hat\imath + \sin t\,\hat\jmath + t\,\hat k$,
$\vec r'(t) = \vector{-\sin t,\ \cos t,\ 1}$ has constant length
$\norm{\vec r'(t)} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$, so one full
turn from $t = 0$ to $t = 2\pi$ has length $\int_0^{2\pi}\sqrt{2}\,\d t = 2\sqrt{2}\,\pi$.
The value is independent of how the curve is parametrized: any parametrization of
the same curve gives the same length.

### The arc length function

Fixing the lower limit and letting the upper limit vary turns length into a
function of position along the curve.

> **Definition (Arc length function).**
>
> $$
> s(t) = \int_a^t \norm{\vec r'(u)}\,\d u,
> \qquad\text{so}\qquad
> \frac{\d s}{\d t} = \norm{\vec r'(t)}
> $$
>
> by the Fundamental Theorem of Calculus.

Because $s$ measures distance traveled along the curve, it is a coordinate-free
parameter fixed by the shape alone. **Reparametrizing by arc length** rewrites
the curve so that moving one unit of parameter means moving one unit of distance.
For the helix, $s(t) = \int_0^t \sqrt{2}\,\d u = \sqrt{2}\,t$, so $t = s/\sqrt{2}$
and

$$
\vec r(t(s)) = \vector{\cos\tfrac{s}{\sqrt{2}},\ \sin\tfrac{s}{\sqrt{2}},\ \tfrac{s}{\sqrt{2}}}.
$$

Now $\vec r(t(3))$ is the point exactly $3$ units of length along the curve
from the start.

## Curvature

The [unit tangent](/calculus/vectors-and-space-curves/vector-functions-and-space-curves)
$\hat T$ points along the curve; how fast its direction turns, measured
against distance rather than parameter, is the curvature. Using arc length keeps
the definition independent of parametrization, and since $\hat T$ has constant
length, only its direction changes.

> **Definition (Curvature).** The curvature of a curve is
>
> $$
> \kappa = \norm{\frac{\d\hat T}{\d s}}.
> $$

The chain rule $\d\hat T/\d s = (\d\hat T/\d t)/(\d s/\d t)$ with
$\d s/\d t = \norm{\vec r'}$ converts this to a form computable in the
original parameter:

$$
\kappa(t) = \frac{\norm{\hat T'(t)}}{\norm{\vec r'(t)}}.
$$

A circle of radius $a$, parametrized $\vec r(t) = a\cos t\,\hat\imath + a\sin t\,\hat\jmath$,
has $\hat T(t) = \vector{-\sin t,\ \cos t}$ and $\norm{\hat T'(t)} = 1$
while $\norm{\vec r'(t)} = a$, so $\kappa = 1/a$: small circles bend hard,
large circles bend gently, and a straight line has $\kappa = 0$. A more
convenient formula avoids computing $\hat T$ altogether.

> **Theorem (Curvature from derivatives).**
>
> $$
> \kappa(t) = \frac{\norm{\vec r'(t) \times \vec r''(t)}}{\norm{\vec r'(t)}^3}.
> $$

For the twisted cubic $\vec r(t) = \vector{t, t^2, t^3}$,
$\vec r'(t) = \vector{1, 2t, 3t^2}$ and $\vec r''(t) = \vector{0, 2, 6t}$
cross to $\vector{6t^2, -6t, 2}$, so

$$
\kappa(t) = \frac{2\sqrt{9t^4 + 9t^2 + 1}}{(1 + 4t^2 + 9t^4)^{3/2}},
$$

which equals $2$ at the origin. For a plane curve $y = f(x)$, taking $x$ as the
parameter reduces the theorem to

$$
\kappa(x) = \frac{\abs{f''(x)}}{\bigl[1 + (f'(x))^2\bigr]^{3/2}}.
$$

The parabola $y = x^2$ then has $\kappa(x) = 2/(1 + 4x^2)^{3/2}$, largest at the
vertex ($\kappa(0) = 2$) and tending to $0$ as the arms flatten out.

$$
% caption: Curvature is large where the curve bends sharply and small where it
% is nearly straight; the osculating circle at each point has radius 1/kappa.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[acc, very thick] (-3.2,0.9) .. controls (-1.6,0.9) and (-0.9,1.6) .. (-0.4,1.6)
     .. controls (0.1,1.6) and (0.35,-0.2) .. (0.9,-0.2)
     .. controls (2.0,-0.2) and (2.2,1.3) .. (3.3,1.3);
  % sharp bend marker — at the tight turn where the descent levels out
  \fill[black!70] (0.79,-0.17) circle (1.4pt);
  \draw[black] (0.79,-0.26) -- (0.79,-0.62) node[anchor=north] {sharp bend: large curvature};
  % gentle arc marker
  \fill[black!70] (-2.5,0.9) circle (1.4pt);
  \draw[black] (-2.5,0.9) -- (-2.5,1.7) node[anchor=south] {gentle arc: small curvature};
\end{tikzpicture}
$$

## The normal and binormal vectors

At each point of the curve, many vectors are perpendicular to $\hat T$. One
is singled out by the fact that $\hat T$ has constant length, which forces
$\hat T' \perp \hat T$. Normalizing $\hat T'$ gives the direction the
curve is turning, and crossing it with $\hat T$ completes an orthonormal
triple.

> **Definition (Normal, binormal, and the TNB frame).** Where $\kappa \ne 0$,
>
> $$
> \hat N(t) = \frac{\hat T'(t)}{\norm{\hat T'(t)}},
> \qquad
> \hat B(t) = \hat T(t) \times \hat N(t).
> $$
>
> The **principal unit normal** $\hat N$ points toward the concave side; the
> **binormal** $\hat B$ is a unit vector perpendicular to both. Together
> $\{\hat T, \hat N, \hat B\}$ form the **TNB frame**, a moving set of
> orthogonal axes carried along the curve.

$$
% caption: The TNB frame at a point: T along the curve, N toward the concave
% side in the osculating plane, and B perpendicular to that plane.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (-2.4,-0.6) .. controls (-1.0,0.9) and (1.0,0.9) .. (2.4,-0.6);
  \coordinate (P) at (0,0.68);
  \fill[black!75] (P) circle (1.5pt);
  % T tangent (horizontal at apex)
  \draw[black, very thick, ->] (P) -- ++(1.5,0) node[right] {$\mathbf{T}$};
  % N pointing down (concave side)
  \draw[black, very thick, ->] (P) -- ++(0,-1.3) node[below] {$\mathbf{N}$};
  % B out of plane, drawn up-left with slight offset
  \draw[black, very thick, ->] (P) -- ++(-0.65,1.15) node[above left] {$\mathbf{B}$};
  \node[black] at (2.05,0.55) {$C$};
\end{tikzpicture}
$$

> **Worked example.** Compute $\hat T$, $\hat N$, $\hat B$, and $\kappa$ for the
> helix $\vec r(t) = \vector{\cos t,\ \sin t,\ t}$.
>
> The tangent vector is $\vec r'(t) = \vector{-\sin t,\ \cos t,\ 1}$ with
> constant length $\norm{\vec r'} = \sqrt{2}$, so
>
> $$
> \hat T(t) = \frac{1}{\sqrt{2}}\vector{-\sin t,\ \cos t,\ 1}.
> $$
>
> Differentiating, $\hat T'(t) = \tfrac{1}{\sqrt{2}}\vector{-\cos t,\ -\sin t,\ 0}$
> has length $\norm{\hat T'} = \tfrac{1}{\sqrt{2}}$, so the principal normal is
>
> $$
> \hat N(t) = \frac{\hat T'(t)}{\norm{\hat T'(t)}} = \vector{-\cos t,\ -\sin t,\ 0},
> $$
>
> which points horizontally inward toward the axis. The binormal is
>
> $$
> \hat B(t) = \hat T \times \hat N = \frac{1}{\sqrt{2}}\vector{\sin t,\ -\cos t,\ 1}.
> $$
>
> Curvature follows from $\kappa = \norm{\hat T'}/\norm{\vec r'} = (1/\sqrt{2})/\sqrt{2} = \tfrac12$:
> the helix bends at a constant rate, the same at every point.

The plane through the point spanned by $\hat T$ and $\hat N$ is the
**osculating plane**, the plane the curve is momentarily bending within. Inside
it sits the **osculating circle**: the circle tangent to the curve at the point,
on the concave side, with the same curvature. Its radius $\rho = 1/\kappa$ is the
**radius of curvature**, and its center lies a distance $\rho$ from the point
along $\hat N$. For $y = x^2$ at the origin, $\kappa = 2$, so $\rho = \tfrac12$
and the osculating circle is $x^2 + (y - \tfrac12)^2 = \tfrac14$ — the circle that
best matches the parabola's bend there.

$$
% caption: The osculating circle of y = x^2 at the origin has radius 1/kappa =
% 1/2 and center (0, 1/2); it hugs the parabola's bend at that point.
\begin{tikzpicture}[scale=1.4, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, ->] (-1.4,0) -- (1.5,0) node[right] {$x$};
  \draw[black, ->] (0,-0.25) -- (0,1.7) node[above] {$y$};
  \draw[acc, very thick, domain=-1.15:1.15, samples=80, variable=\x] plot ({\x},{\x*\x});
  \node[acc] at (1.05,1.35) {$y=x^2$};
  \draw[black, thick] (0,0.5) circle (0.5);
  \fill[black] (0,0.5) circle (0.9pt) node[anchor=west] {center $(0,\ 1/2)$};
  \fill[black!75] (0,0) circle (1.1pt);
  \node[black, anchor=north west] at (0.52,0.55) {radius $1/2$};
\end{tikzpicture}
$$

Each of the five quantities of the moving frame is computable from
$\vec r'$ and $\vec r''$ alone, and each has a geometric job.

| Quantity | Formula | What it captures |
| --- | --- | --- |
| Unit tangent $\hat T$ | $\vec r'/\norm{\vec r'}$ | direction of travel |
| Principal normal $\hat N$ | $\hat T'/\norm{\hat T'}$ | direction the curve turns |
| Binormal $\hat B$ | $\hat T \times \hat N$ | normal to the osculating plane |
| Curvature $\kappa$ | $\norm{\vec r' \times \vec r''}/\norm{\vec r'}^3$ | how sharply it bends |
| Radius of curvature $\rho$ | $1/\kappa$ | osculating-circle radius |

## Motion in space

Now read $\vec r(t)$ as the position of a particle at time $t$. The
derivatives that gave tangent and curvature acquire physical names.

> **Definition (Velocity, speed, acceleration).** For a particle with position
> $\vec r(t)$,
>
> $$
> \vec v(t) = \vec r'(t),
> \qquad
> \text{speed} = \norm{\vec v(t)} = \frac{\d s}{\d t},
> \qquad
> \vec a(t) = \vec v'(t) = \vec r''(t).
> $$

Velocity is the tangent vector, so it points along the path; speed is its length,
the rate of distance covered. For $\vec r(t) = t^3\hat\imath + t^2\hat\jmath$,
$\vec v(t) = 3t^2\hat\imath + 2t\hat\jmath$, $\vec a(t) = 6t\hat\imath + 2\hat\jmath$,
and the speed is $\sqrt{9t^4 + 4t^2}$.

Running the derivatives backward, integration recovers motion from its rates.
Given acceleration and initial data,

$$
\vec v(t) = \vec v(t_0) + \int_{t_0}^t \vec a(u)\,\d u,
\qquad
\vec r(t) = \vec r(t_0) + \int_{t_0}^t \vec v(u)\,\d u.
$$

By **Newton's second law**, $\vec F(t) = m\,\vec a(t)$, a known force determines
the acceleration, and integrating twice yields the trajectory.

> **Worked example.** A projectile is fired from the ground with speed $v_0$ at
> elevation angle $\alpha$, under gravity alone. Find its path and the
> horizontal range.
>
> Gravity gives $\vec a = -g\hat\jmath$. With initial velocity
> $\vec v(0) = \vector{v_0\cos\alpha,\ v_0\sin\alpha}$ and start at the origin,
> integrating twice gives
>
> $$
> \vec r(t) = (v_0\cos\alpha)\,t\,\hat\imath + \Bigl[(v_0\sin\alpha)\,t - \tfrac12 g t^2\Bigr]\hat\jmath,
> $$
>
> a parabola. It returns to the ground when the $\hat\jmath$-component vanishes,
> at $t = (2v_0\sin\alpha)/g$. Substituting into the $\hat\imath$-component and
> using $2\sin\alpha\cos\alpha = \sin 2\alpha$,
>
> $$
> R = (v_0\cos\alpha)\frac{2v_0\sin\alpha}{g} = \frac{v_0^2}{g}\sin 2\alpha.
> $$
>
> The range is largest when $\sin 2\alpha = 1$, that is at $\alpha = 45^\circ$.

### Tangential and normal components of acceleration

Acceleration need not point along the motion. Writing speed as $v = \norm{\vec v}$
so that $\vec v = v\hat T$, and differentiating with the product rule and
$\hat T' = \kappa v\,\hat N$, resolves acceleration into two orthogonal
pieces.

> **Theorem (Tangential and normal acceleration).**
>
> $$
> \vec a = a_T\,\hat T + a_N\,\hat N,
> \qquad
> a_T = v' = \frac{\vec r' \cdot \vec r''}{\norm{\vec r'}},
> \qquad
> a_N = \kappa v^2 = \frac{\norm{\vec r' \times \vec r''}}{\norm{\vec r'}}.
> $$

The tangential part $a_T$ changes the speed; the normal part $a_N = \kappa v^2$
changes the direction and always points toward the concave side. The binormal
$\hat B$ never appears — acceleration lies entirely in the osculating plane.
The $\kappa v^2$ form matches experience: a sharp turn (large $\kappa$) or a high
speed both press a passenger against the car door, and doubling the speed
quadruples the sideways push.

$$
% caption: Acceleration resolves into a tangential part along T that changes
% speed and a normal part along N that changes direction; B plays no role.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (-2.6,-0.5) .. controls (-1.1,0.95) and (1.1,0.95) .. (2.6,-0.5);
  \coordinate (P) at (0,0.6);
  \fill[black!75] (P) circle (1.5pt);
  % T and N directions
  \draw[black, thick, ->] (P) -- ++(1.4,0) node[right] {$\mathbf{T}$};
  \draw[black, thick, ->] (P) -- ++(0,-1.25) node[below] {$\mathbf{N}$};
  % total acceleration and components
  \draw[acc, very thick, ->] (P) -- ++(1.0,-0.9) node[anchor=north west] {$\mathbf{a}$};
  \draw[black, dashed] (P) ++(1.0,0) -- ++(0,-0.9);
  \draw[black, dashed] (P) ++(0,-0.9) -- ++(1.0,0);
  \node[black, anchor=south] at (0.5,0.62) {$a_T$};
  \node[black, anchor=east] at (-0.05,0.1) {$a_N$};
\end{tikzpicture}
$$

> **Worked example.** Resolve the acceleration of
> $\vec r(t) = \vector{t^2, t^2, t^3}$ into tangential and normal components.
>
> Differentiating, $\vec r'(t) = \vector{2t, 2t, 3t^2}$ and
> $\vec r''(t) = \vector{2, 2, 6t}$, so $\norm{\vec r'} = \sqrt{8t^2 + 9t^4}$.
> The tangential component uses $\vec r' \cdot \vec r'' = 4t + 4t + 18t^3 = 8t + 18t^3$:
>
> $$
> a_T = \frac{\vec r' \cdot \vec r''}{\norm{\vec r'}} = \frac{8t + 18t^3}{\sqrt{8t^2 + 9t^4}}.
> $$
>
> The normal component uses $\vec r' \times \vec r'' = \vector{6t^2, -6t^2, 0}$,
> whose length is $6\sqrt{2}\,t^2$:
>
> $$
> a_N = \frac{\norm{\vec r' \times \vec r''}}{\norm{\vec r'}} = \frac{6\sqrt{2}\,t^2}{\sqrt{8t^2 + 9t^4}}.
> $$

The same
decomposition, applied to a planet moving under the sun's gravity, reproduces
Kepler's three laws of planetary motion.