3.1 Vector-Valued Functions and Space Curves

Big idea

Every function so far has taken a number and returned a number. This chapter changes the output. A vector-valued function takes a single real number and returns a vector, so its value is a point in space rather than a height above an axis. Nothing about the input changes: there is still one variable, usually called $t$, and still one thing happening at a time.

What such a function produces is a curve. As $t$ runs through an interval, the tip of the vector $\mathbf{r}(t)$ moves, and the set of positions it visits is a space curve. That picture is worth holding onto, because it also supplies the vocabulary: $t$ behaves like time, $\mathbf{r}(t)$ is a position, and the curve is a trajectory. A graph of an ordinary function is a static object you read left to right; a space curve is traced, with a direction of travel built in.

The machinery for handling these functions is lighter than you might expect, because a vector-valued function is really three ordinary functions bundled together. Those three are the component functions, and nearly every question about $\mathbf{r}$ is answered by asking the same question of each component separately. The domain is the intersection of the three domains. A limit exists when each component limit exists. Continuity is continuity of all three at once.

This section is about the objects themselves: what they are, where they are defined, what curves they trace, and when they behave without jumps. The next section differentiates and integrates them, and the two after that use them to measure arc length and describe motion. Getting comfortable with the curve first makes all of that easier, because the derivative will turn out to have a geometric meaning you can see only if you can already picture the curve.

Decoder

A vector-valued function of one variable assigns a vector to each number in its domain, and it is completely determined by its component functions, each of which is an ordinary real-valued function of that same variable.

The second half is the useful half. Writing $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ says that the new object is three old objects in a container. So every technique for real-valued functions is still available; you apply it three times and reassemble. There is nothing genuinely new to learn about limits or continuity, only a new place to put the results.

The first half is where the geometry hides. Because the output is a vector, it can be drawn as an arrow from the origin, and the tip of that arrow is a point. As $t$ varies the point moves and sweeps out the curve. Two different functions can sweep out the same set of points at different speeds or in opposite directions, so the function carries more information than the curve does. When a problem asks for a parametrization of a given curve, it is asking you to choose one of many.

The classic mistake is to confuse the curve with a graph. The graph of $y = f(x)$ lives in the plane of inputs and outputs; a space curve lives entirely in the space of outputs, with the input $t$ nowhere visible on the picture. A curve can therefore cross itself, loop, or return to its starting point, none of which a function graph is allowed to do.

Definitions and results

Vector-valued function. A function whose domain is a set of real numbers and whose range is a set of vectors. In space,

$$ \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} = \langle f(t), g(t), h(t) \rangle $$

and $f$, $g$, $h$ are the component functions. With two components it is a plane curve.

Domain. Unless stated otherwise, the domain is the set of $t$ for which all component functions are defined, that is, the intersection of their individual domains.

Space curve and orientation. The graph of $\mathbf{r}$ is the set of points traced by the tip of $\mathbf{r}(t)$ as $t$ runs over the domain, with $\mathbf{r}$ regarded as a position vector. Increasing $t$ gives the curve a direction of travel, called its orientation, which is part of the data.

Parametrizations are not unique. Replacing $t$ by another increasing function of a new parameter traces the same curve with the same orientation at a different rate, and replacing $t$ by a decreasing function reverses the orientation. So a curve has infinitely many parametrizations.

Lines as vector-valued functions. The line through $\mathbf{r}_0$ with direction $\mathbf{v}$ is $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$, which is already in this form. Every result of this chapter applies to lines as the simplest case.

Circles and helices. In the plane, $\mathbf{r}(t) = \langle a\cos t, a\sin t \rangle$ traces the circle of radius $a$ counterclockwise once as $t$ goes from $0$ to $2\pi$. Adding a linear third component gives a circular helix

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

which winds around the cylinder $x^2 + y^2 = a^2$, rising $2\pi |c|$ per turn.

Curves as intersections of surfaces. A space curve often lies on two surfaces at once, and the pair of surface equations is another way to present it. Eliminating $t$ from the component functions is how you find those equations.

Limits. The limit of a vector-valued function is taken componentwise:

$$ \lim_{t \to a} \mathbf{r}(t) = \left\langle \lim_{t \to a} f(t), \; \lim_{t \to a} g(t), \; \lim_{t \to a} h(t) \right\rangle $$

provided all three component limits exist. Equivalently, $\mathbf{r}(t)$ approaches $\mathbf{L}$ when $\|\mathbf{r}(t) - \mathbf{L}\|$ approaches $0$.

Continuity. A vector-valued function is continuous at $t = a$ when $\mathbf{r}(a)$ is defined, the limit exists, and the two agree. This happens exactly when all three component functions are continuous at $a$. A curve traced by a continuous function has no breaks.

Worked examples

Finding a domain

Find the domain of

$$ \mathbf{r}(t) = \left\langle \sqrt{t-1}, \; \ln(5 - t), \; \frac{1}{t} \right\rangle $$

Take each component in turn. The square root requires $t - 1 \ge 0$, so $t \ge 1$. The logarithm requires $5 - t > 0$, so $t < 5$. The reciprocal requires $t \ne 0$, which is already implied by $t \ge 1$. Intersecting the three conditions gives the domain

$$ [1, 5) $$

At the left endpoint the function is defined, with $\mathbf{r}(1) = \langle 0, \ln 4, 1 \rangle$, while as $t$ approaches $5$ from below the second component goes to negative infinity and the curve runs off without end.

A helix and the cylinder it lies on

Describe the curve traced by $\mathbf{r}(t) = \langle 3\cos t, \; 3\sin t, \; 4t \rangle$.

The first two components satisfy $x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9$ for every $t$, so every point of the curve lies on the cylinder of radius $3$ about the $z$-axis. Within that cylinder, the horizontal position goes counterclockwise once as $t$ increases by $2\pi$, while the height increases steadily at a rate of $4$ units per unit of $t$. So the curve is a circular helix, and each complete turn raises it by $8\pi \approx 25.1$ units.

The orientation matters. Replacing $t$ by $-t$ gives $\langle 3\cos t, -3\sin t, -4t \rangle$, which traces the same cylinder but spirals the other way and descends. Replacing $t$ by $2t$ instead traces the identical curve in the identical direction, only twice as fast.

A curve on two surfaces

Identify the curve $\mathbf{r}(t) = \langle \cos t, \; \sin t, \; \cos 2t \rangle$.

The first two components again give $x^2 + y^2 = 1$, so the curve lies on the unit cylinder. For the third, use the double-angle identity $\cos 2t = 1 - 2\sin^2 t$. Since $y = \sin t$,

$$ z = 1 - 2y^2 $$

which is a parabolic cylinder running along the $x$-axis. The curve is exactly the intersection of those two surfaces. It is closed, since all three components have period $2\pi$, and it rides up and down between $z = -1$, reached when $y = \pm 1$, and $z = 1$, reached when $y = 0$. Check a value: at $t = \pi/2$ the formula gives the point $(0, 1, -1)$, and indeed $0 + 1 = 1$ and $1 - 2(1)^2 = -1$.

A limit taken componentwise

Evaluate

$$ \lim_{t \to 0} \left\langle \frac{\sin 3t}{t}, \; \frac{e^t - 1}{t}, \; \frac{1 - \cos t}{t^2} \right\rangle $$

Each component is an indeterminate form on its own, and each is a standard limit. Writing the first as $3 \cdot \frac{\sin 3t}{3t}$ gives $3$. The second is the definition of the derivative of $e^t$ at $t = 0$, which is $1$. The third is the familiar half. So the limit is

$$ \left\langle 3, \; 1, \; \tfrac{1}{2} \right\rangle $$

The original function is undefined at $t = 0$, so it is not continuous there as written. Defining $\mathbf{r}(0)$ to be that limit vector removes the discontinuity, and the resulting function is continuous on all of $\mathbf{R}$ since each repaired component is.

Practice

These problems cover the objects of this section: identifying component functions, finding domains, matching a vector-valued function to the curve it traces, recognizing helices and curves lying on given surfaces, and evaluating limits and testing continuity componentwise.

Quiz

Five items on component functions and domains, the curves traced by standard parametrizations, orientation, and limits and continuity of vector-valued functions.