---
title: Vector Functions and Space Curves
module: Vectors and the Geometry of Space
moduleNumber: 10
lessonNumber: 4
order: 1004
summary: >
  A vector function assigns a vector to each value of a parameter, and as the
  parameter runs its tip traces a space curve. Taking limits, derivatives, and
  integrals component by component carries all of single-variable calculus into
  three dimensions. The derivative of a vector function is the tangent vector to
  its curve, and normalizing it gives the unit tangent that points the way
  along the path.
topics: [Vectors and the Geometry of Space]
sources:
  - book: Stewart
    ref: "Ch. 13 — Vector Functions; §13.1 Vector Functions and Space Curves"
  - book: Stewart
    ref: "§13.2 Derivatives and Integrals of Vector Functions"
draft: false
---

A **vector function** assigns a vector to each value of a parameter; as the
parameter runs, its tip sweeps out a curve through space. Each component is an
ordinary scalar function, so limits, derivatives, and integrals all act
componentwise, and single-variable calculus carries over with no new machinery.
The one genuinely new object is the derivative of a vector function, the
**tangent vector** to the curve it traces.

## Vector functions and space curves

> **Definition (Vector function).** A vector function $\vec r$ assigns to
> each real number $t$ in its domain a vector $\vec r(t)$ in $\mathbb{R}^3$,
>
> $$
> \vec r(t) = \vector{f(t), g(t), h(t)} = f(t)\hat\imath + g(t)\hat\jmath + h(t)\hat k.
> $$
>
> The scalar functions $f, g, h$ are its **component functions**. The parameter
> is written $t$ because it usually stands for time.

The domain is the set of $t$ for which every component is defined. For
$\vec r(t) = \vector{t^3,\ \ln(3 - t),\ \sqrt{t}}$, the logarithm needs
$t < 3$ and the root needs $t \ge 0$, so the domain is $[0, 3)$.

Limits and continuity pass through the components untouched.

> **Definition (Limit and continuity).** If $\vec r(t) = \vector{f(t), g(t), h(t)}$,
> then
>
> $$
> \lim_{t \to a} \vec r(t) = \vector{\lim_{t \to a} f(t),\ \lim_{t \to a} g(t),\ \lim_{t \to a} h(t)}
> $$
>
> provided each component limit exists, and $\vec r$ is **continuous at $a$**
> when $\lim_{t \to a} \vec r(t) = \vec r(a)$ — equivalently, when every
> component is continuous at $a$.

A continuous vector function is the same data as a curve. As $t$ ranges over an
interval, the point $P(f(t), g(t), h(t))$ moves through space, and the set of all
such points is a **space curve**.

> **Definition (Space curve).** For continuous $f, g, h$ on an interval $I$, the
> set of points $(x, y, z)$ with
>
> $$
> x = f(t), \qquad y = g(t), \qquad z = h(t), \qquad t \in I,
> $$
>
> is a space curve $C$ with these **parametric equations**. The curve is traced
> by the tip of the position vector $\vec r(t) = \vector{f(t), g(t), h(t)}$.

A linear vector function reproduces a line: $\vec r(t) = \vector{1+t,\ 2+5t,\ -1+6t}$
is the line through $(1, 2, -1)$ with direction $\vector{1, 5, 6}$. Mixing
trigonometric and linear components produces the archetypal space curve, the
**helix**. For $\vec r(t) = \cos t\,\hat\imath + \sin t\,\hat\jmath + t\,\hat k$,
the first two components satisfy $x^2 + y^2 = 1$, so the curve lies on the
cylinder of radius $1$, while $z = t$ raises it steadily as it circles: a
corkscrew, the shape of a spring and of the DNA backbone.

$$
% caption: The helix r(t) = (cos t, sin t, t) circles the unit cylinder while
% rising steadily, so it winds upward one full turn every 2 pi in t.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % projection: (cos t, sin t, z) -> (cos t + 0.5 sin t, 0.35 sin t + z)
  \draw[->, black] (0,0) -- (2.3,0) node[right] {$x$};
  \draw[->, black] (0,0) -- (0.9,0.63) node[right] {$y$};
  \draw[->, black] (0,0) -- (0,4.4) node[above] {$z$};
  % helix as a 2D plot of the projected coordinates (t in degrees)
  \draw[acc, very thick, domain=0:900, samples=200, variable=\t]
     plot ({cos(\t) + 0.5*sin(\t)}, {0.35*sin(\t) + \t/230});
  \node[acc, anchor=west] at (1.15,3.6) {$\mathbf{r}(t)$};
\end{tikzpicture}
$$

Vector functions also parametrize **curves of intersection** of surfaces.

> **Worked example.** Parametrize the curve where the cylinder $x^2 + y^2 = 1$
> meets the plane $y + z = 2$.
>
> Points on the cylinder satisfy $x^2 + y^2 = 1$, so take $x = \cos t$,
> $y = \sin t$ with $0 \le t \le 2\pi$. The plane then forces
> $z = 2 - y = 2 - \sin t$. Assembling the components,
>
> $$
> \vec r(t) = \vector{\cos t,\ \sin t,\ 2 - \sin t}, \qquad 0 \le t \le 2\pi,
> $$
>
> traces the intersection, an ellipse tilted by the plane.

## Derivatives

The derivative of a vector function copies the single-variable definition, with a
difference quotient of vectors.

> **Definition (Derivative and tangent vector).**
>
> $$
> \vec r'(t) = \lim_{h \to 0} \frac{\vec r(t + h) - \vec r(t)}{h},
> $$
>
> when the limit exists. Where $\vec r'(t) \ne \vec 0$, it is the
> **tangent vector** to the curve at $\vec r(t)$, and the **tangent line** is
> the line through that point in the direction $\vec r'(t)$.

The geometry is in the difference quotient. The vector
$\vec r(t + h) - \vec r(t)$ is a **secant** joining two nearby points on
the curve; dividing by $h$ scales it without changing direction, and as $h \to 0$
the secant swings onto the tangent line.

$$
% caption: The secant vector r(t+h) - r(t) joins two nearby points; scaled by
% 1/h and taken to the limit h -> 0, it becomes the tangent vector r'(t).
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick, domain=-1.4:1.7, samples=60, variable=\x]
     plot ({\x}, {0.55*\x*\x + 0.2*\x + 0.6});
  \coordinate (P) at (-0.4,0.608);
  \coordinate (Q) at (1.1,1.485);
  \fill[black!70] (P) circle (1.4pt) node[anchor=north east] {$P$};
  \fill[black!70] (Q) circle (1.4pt) node[anchor=north west] {$Q$};
  \draw[black, thick, ->] (P) -- (Q) node[midway, above left] {$\mathbf{r}(t+h)-\mathbf{r}(t)$};
  \draw[acc, very thick, ->] (P) -- ($(P)+(1.2,0.3)$) node[right] {$\mathbf{r}'(t)$};
  \node[black, anchor=north] at (0.6,2.35) {$C$};
\end{tikzpicture}
$$

Rather than take the limit directly, differentiate each component.

> **Theorem (Componentwise derivative).** If
> $\vec r(t) = \vector{f(t), g(t), h(t)}$ with $f, g, h$ differentiable, then
>
> $$
> \vec r'(t) = \vector{f'(t),\ g'(t),\ h'(t)}.
> $$

For $\vec r(t) = (1 + t^3)\hat\imath + t e^{-t}\hat\jmath + \sin 2t\,\hat k$,
differentiating each component gives
$\vec r'(t) = 3t^2\hat\imath + (1 - t)e^{-t}\hat\jmath + 2\cos 2t\,\hat k$.

Dividing the tangent vector by its length removes the speed information and keeps
only the direction of travel.

> **Definition (Unit tangent vector).** At a point where $\vec r'(t) \ne \vec 0$,
>
> $$
> \hat T(t) = \frac{\vec r'(t)}{\norm{\vec r'(t)}}.
> $$

> **Worked example.** Find the tangent line to
> $\vec r(t) = \vector{2\cos t,\ \sin t,\ t}$ at $t = \pi/2$.
>
> The tangent vector is $\vec r'(t) = \vector{-2\sin t,\ \cos t,\ 1}$. At
> $t = \pi/2$ the point of tangency is $\vec r(\pi/2) = (0, 1, \pi/2)$ and the
> direction is $\vec r'(\pi/2) = \vector{-2, 0, 1}$. The tangent line through
> that point with that direction is
>
> $$
> x = -2t, \qquad y = 1, \qquad z = \tfrac{\pi}{2} + t.
> $$

$$
% caption: The unit tangent vector T(t) at several points along a curve, each
% pointing in the direction of increasing t and each of length one.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % curve y = 0.9 sin(x) (x in radians via the r suffix), x from 0 to 5
  \draw[black, thick, domain=0:5, samples=120, variable=\x]
     plot ({\x}, {1.3 + 0.9*sin(\x r)});
  % unit tangents (length 0.9) hand-computed at x = 1.0, 2.5, 4.0
  \draw[acc, very thick, ->] (1.0,2.057) -- (1.809,2.450);
  \fill[black!70] (1.0,2.057) circle (1.3pt);
  \draw[acc, very thick, ->] (2.5,1.838) -- (3.230,1.312);
  \fill[black!70] (2.5,1.838) circle (1.3pt);
  \draw[acc, very thick, ->] (4.0,0.619) -- (4.776,0.163);
  \fill[black!70] (4.0,0.619) circle (1.3pt);
  \node[acc] at (1.55,2.7) {$\mathbf{T}$};
\end{tikzpicture}
$$

The **second derivative** $\vec r''(t) = (\vec r'(t))'$ is again taken
componentwise; for the helix above, $\vec r''(t) = \vector{-2\cos t,\ -\sin t,\ 0}$.
Read as the motion of a particle, $\vec r'$ is its velocity and $\vec r''$ its
acceleration.

### Differentiation rules

Every product rule of scalar calculus has a vector counterpart, one for each way
two vector functions can be combined.

> **Theorem (Differentiation rules).** For differentiable vector functions
> $\vec u, \vec v$, a scalar function $f$, and a constant $c$,
>
> $$
> \tfrac{\d}{\d t}[\vec u + \vec v] = \vec u' + \vec v',
> \qquad
> \tfrac{\d}{\d t}[c\vec u] = c\vec u',
> \qquad
> \tfrac{\d}{\d t}[f\vec u] = f'\vec u + f\vec u',
> $$
>
> $$
> \tfrac{\d}{\d t}[\vec u \cdot \vec v] = \vec u' \cdot \vec v + \vec u \cdot \vec v',
> \qquad
> \tfrac{\d}{\d t}[\vec u \times \vec v] = \vec u' \times \vec v + \vec u \times \vec v',
> \qquad
> \tfrac{\d}{\d t}[\vec u(f(t))] = f'(t)\,\vec u'(f(t)).
> $$

Order matters in the cross-product rule, since the cross product is
anticommutative. One consequence: a curve confined to a sphere has its tangent
vector perpendicular to its position vector.

> **Corollary.** If $\norm{\vec r(t)}$ is constant, then $\vec r'(t)$ is
> orthogonal to $\vec r(t)$ for all $t$.

A constant length means
$\vec r(t) \cdot \vec r(t) = \norm{\vec r(t)}^2$ is constant, so
differentiating with the dot-product rule gives
$2\,\vec r'(t) \cdot \vec r(t) = 0$, hence $\vec r' \perp \vec r$.
Geometrically, a curve on a sphere centered at the origin always has its tangent
perpendicular to its radius.

## Integrals

Integration is componentwise too, so the antiderivative of a vector function is
the antiderivative of each component, and the Fundamental Theorem of Calculus
extends verbatim.

> **Definition (Integral of a vector function).**
>
> $$
> \int_a^b \vec r(t)\,\d t = \parens{\int_a^b f(t)\,\d t}\hat\imath + \parens{\int_a^b g(t)\,\d t}\hat\jmath + \parens{\int_a^b h(t)\,\d t}\hat k,
> $$
>
> and if $\vec R'(t) = \vec r(t)$ then
> $\int_a^b \vec r(t)\,\d t = \vec R(b) - \vec R(a)$.

An indefinite integral carries a **vector constant** of integration. For
$\vec r(t) = 2\cos t\,\hat\imath + \sin t\,\hat\jmath + 2t\,\hat k$,

$$
\int \vec r(t)\,\d t = 2\sin t\,\hat\imath - \cos t\,\hat\jmath + t^2\hat k + \vec C.
$$

Integration recovers position from a known velocity, and velocity from a known
acceleration, exactly as in one dimension. Applied to Newton's second law,
$\vec F = m\vec a$, it turns a force field into a trajectory.

Each operation applies the scalar version to every component and reassembles the
result.

| Operation | Vector form | How it is computed |
| --- | --- | --- |
| Limit | $\lim_{t \to a}\vec r(t)$ | limit of each component |
| Continuity | $\lim_{t \to a}\vec r(t) = \vec r(a)$ | every component continuous |
| Derivative | $\vec r'(t)$ | derivative of each component |
| Integral | $\int_a^b \vec r(t)\,\d t$ | integral of each component |

The only genuinely vector-valued subtleties are the product rules, where the dot
and cross products each contribute their own version, and the geometric reading of
$\vec r'$ as a tangent vector — neither of which has a one-dimensional analog.

A curve can also be studied through its **projections** onto the coordinate
planes. Dropping the $z$-coordinate of $\vec r(t)$ leaves its shadow on the
$xy$-plane, and comparing the three shadows often clarifies a curve that is hard
to read in perspective.

$$
% caption: A space curve and its shadow on the xy-plane; the projection drops
% the z-coordinate, turning the curve into a plane curve on the floor.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % projection: (cos t, sin t, z) -> (cos t + 0.5 sin t, 0.35 sin t + z)
  \draw[->, black] (0,0) -- (2.4,0) node[right] {$x$};
  \draw[->, black] (0,0) -- (0.95,0.66) node[right] {$y$};
  \draw[->, black] (0,0) -- (0,3.4) node[above] {$z$};
  % space curve (helix) rising from the floor
  \draw[acc, very thick, domain=0:720, samples=180, variable=\t]
     plot ({cos(\t) + 0.5*sin(\t)}, {0.35*sin(\t) + 0.55 + \t/320});
  % shadow (projection onto xy-plane): the flat ellipse at z = 0
  \draw[black, dashed, domain=0:360, samples=90, variable=\t]
     plot ({cos(\t) + 0.5*sin(\t)}, {0.35*sin(\t)});
  \node[acc, anchor=west] at (1.2,2.5) {curve};
  \node[black, anchor=north] at (0.4,-0.35) {shadow};
\end{tikzpicture}
$$