Vector Functions and Space Curves, multivariable and vector calculus lesson

A drone’s flight log records three numbers at every moment, its $x$ , $y$ and $z$ position. Put them together and you get one arrow from the origin that moves as time passes, and the tip of that arrow draws the flight path. A vector function is exactly this package of three ordinary functions of one variable, and almost everything about it can be checked one component at a time.

Key terms

Component functions

Definition A vector function, or vector-valued function, with values in $\mathbb{R}^3$ is a rule $\mathbf{r}$ that assigns to each number $t$ in a set $D \subseteq \mathbb{R}$ a vector $\mathbf{r}(t)$ in $\mathbb{R}^3$ . Writing

\begin{aligned} \mathbf{r}(t) &= \langle f(t), g(t), h(t) \rangle \\[0.4em] &= f(t)\,\mathbf{i} + g(t)\,\mathbf{j} + h(t)\,\mathbf{k} \end{aligned}

defines three real-valued functions $f$ , $g$ and $h$ on $D$ , the component functions of $\mathbf{r}$ . Vector functions with values in $\mathbb{R}^2$ are defined the same way with two components.

In plain words A vector function is three ordinary functions of the same variable, read off together as the coordinates of one point.

Example For $\mathbf{r}(t) = \langle t^2, \, 3t - 1, \, \cos(\pi t) \rangle$ the component functions are $f(t) = t^2$ , $g(t) = 3t - 1$ and $h(t) = \cos(\pi t)$ . At $t = 2$ the vector is $\mathbf{r}(2) = \langle 4, 5, 1 \rangle$ , the position vector of the point $(4, 5, 1)$ .

Parameter domain

Definition The parameter domain of $\mathbf{r}$ is the set $D$ of values of $t$ where $\mathbf{r}(t)$ is defined. When $\mathbf{r}$ is given only by a formula, its natural domain is the intersection of the natural domains of its component functions. A problem may also restrict $t$ to a smaller set, such as an interval $[a, b]$ , and then only that part of the curve is meant.

In plain words The vector exists only when all three components exist at once.

Example Take

\mathbf{r}(t) = \Big\langle \sqrt{4 - t}, \; \ln t, \; \frac{1}{t - 3} \Big\rangle.

The first component needs $t \le 4$ , the second needs $t > 0$ and the third needs $t \ne 3$ . The domain is $(0, 3) \cup (3, 4]$ .

Limits of vector functions

Definition Let $\mathbf{r}$ be defined on an open interval containing $a$ , except possibly at $a$ itself, and let $\mathbf{L}$ be a vector. Then $\lim_{t \to a} \mathbf{r}(t) = \mathbf{L}$ means

\lim_{t \to a} \lvert \mathbf{r}(t) - \mathbf{L} \rvert = 0.

If $\mathbf{r} = \langle f, g, h \rangle$ and $\mathbf{L} = \langle L_1, L_2, L_3 \rangle$ , this holds if and only if $\lim_{t \to a} f(t) = L_1$ , $\lim_{t \to a} g(t) = L_2$ and $\lim_{t \to a} h(t) = L_3$ . The vector limit fails to exist as soon as one component limit fails to exist.

In plain words The arrow settles on $\mathbf{L}$ exactly when each coordinate settles on the matching coordinate of $\mathbf{L}$ .

Example The vector function

\mathbf{r}(t) = \Big\langle \frac{\sin t}{t}, \; t^2 + 1, \; \frac{e^t - 1}{t} \Big\rangle

is undefined at $t = 0$ , because its first and third components are, but the component limits are $1$ , $1$ and $1$ , the first and third being standard limits. So $\lim_{t \to 0} \mathbf{r}(t) = \langle 1, 1, 1 \rangle$ . By contrast $\langle t, \frac{1}{t}, 0 \rangle$ has no limit at $0$ , because its second component does not.

Continuity

Definition A vector function $\mathbf{r}$ is continuous at a point $a$ of its domain if

\lim_{t \to a} \mathbf{r}(t) = \mathbf{r}(a).

This holds if and only if each component function is continuous at $a$ . At an endpoint of an interval domain the limit is one-sided. The function is continuous on an interval $I$ if it is continuous at every point of $I$ .

In plain words No jumps and no holes, in any of the three coordinates.

Example Define $\mathbf{r}(t) = \langle t, \frac{\sin t}{t}, t^2 \rangle$ for $t \ne 0$ and $\mathbf{r}(0) = \langle 0, 1, 0 \rangle$ . Every component is continuous at $0$ , so $\mathbf{r}$ is continuous on all of $\mathbb{R}$ . If we had set $\mathbf{r}(0) = \langle 0, 0, 0 \rangle$ instead, the second component would jump and $\mathbf{r}$ would be discontinuous at $0$ .

Space curve

Definition Let $\mathbf{r} = \langle f, g, h \rangle$ be continuous on an interval $I$ . The set

C = \{ (f(t), g(t), h(t)) : t \in I \}

is a space curve, and the equations $x = f(t)$ , $y = g(t)$ , $z = h(t)$ for $t \in I$ are parametric equations of $C$ . The vector function $\mathbf{r}$ is called a parametrization of $C$ , and $\mathbf{r}(t)$ is the position vector of the point of $C$ at parameter value $t$ .

In plain words The curve is the trail left by the tip of the arrow $\mathbf{r}(t)$ as $t$ runs through the interval.

Example The twisted cubic $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ , $t \in \mathbb{R}$ , passes through $(2, 4, 8)$ at $t = 2$ . The point $(2, 4, 7)$ is not on it, since $x = 2$ forces $t = 2$ and then $z$ must be $8$ .

$t$	$x$	$y$	$z$
$-1$	$-1$	$1$	$-1$
$0$	$0$	$0$	$0$
$1$	$1$	$1$	$1$
$2$	$2$	$4$	$8$

Helix

Definition For constants $a > 0$ and $b \ne 0$ , the curve

\begin{gathered} \mathbf{r}(t) = \langle a\cos t, \; a\sin t, \; bt \rangle, \\[0.4em] t \in \mathbb{R}, \end{gathered}

is a circular helix. It lies on the cylinder $x^2 + y^2 = a^2$ , and each time $t$ increases by $2\pi$ the point goes once around the $z$ -axis and its height changes by $2\pi b$ , so the pitch of the helix is $2\pi\lvert b\rvert$ . With right-handed axes and $b > 0$ , the point turns counterclockwise as seen from above while it rises, a right-handed helix.

In plain words A spring or a spiral staircase. Its shadow on the $xy$ -plane goes around a circle while its height climbs at a steady rate.

Example For $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, \frac{t}{2} \rangle$ , every point satisfies $x^2 + y^2 = 4$ . At $t = 0$ the point is $(2, 0, 0)$ , at $t = \frac{\pi}{2}$ it is $(0, 2, \frac{\pi}{4})$ , and at $t = 2\pi$ it is $(2, 0, \pi)$ , directly above the start. The pitch is $\pi$ .

Orientation

Definition A parametrization $\mathbf{r}$ on an interval $I$ orients its curve in the direction of increasing $t$ . The same set of points can be traced in the opposite direction, for example by $\mathbf{r}(-t)$ for $t \in -I$ , or by $\mathbf{r}(a + b - t)$ for $t \in [a, b]$ when $I = [a, b]$ . A curve together with a chosen direction of travel is an oriented curve. Orientation matters later for line integrals of vector fields.

In plain words The formula tells you not just where the curve is but which way you walk along it as time moves forward.

Example The circle $\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle$ , $0 \le t \le 2\pi$ , starts at $(1, 0, 0)$ and reaches $(0, 1, 0)$ at $t = \frac{\pi}{2}$ , so it runs counterclockwise as seen from the positive $z$ -axis. The parametrization $\langle \cos t, -\sin t, 0 \rangle$ draws the same circle but reaches $(0, -1, 0)$ at $t = \frac{\pi}{2}$ , so it runs clockwise.

Intersection curves

Definition Two surfaces often meet in a curve. To parametrize it, find functions $x(t)$ , $y(t)$ and $z(t)$ on an interval such that the point $(x(t), y(t), z(t))$ satisfies both surface equations for every $t$ , and such that every point of the intersection arises for some $t$ . A common route is to parametrize one equation with a known curve, such as $x = \cos t$ , $y = \sin t$ for $x^2 + y^2 = 1$ , and then solve the other equation for the remaining coordinate.

In plain words Let one surface tell you how two coordinates move, then let the other surface tell you the third.

Example The cylinder $x^2 + y^2 = 1$ meets the plane $y + z = 2$ in a closed curve. Setting $x = \cos t$ and $y = \sin t$ and solving the plane equation for $z$ gives

\begin{gathered} \mathbf{r}(t) = \langle \cos t, \; \sin t, \; 2 - \sin t \rangle, \\[0.4em] 0 \le t \le 2\pi. \end{gathered}

Both equations hold, since $\cos^2 t + \sin^2 t = 1$ and $\sin t + (2 - \sin t) = 2$ . The curve is an ellipse, a slanted slice of the cylinder.

Check yourself

Find the domain of $\mathbf{r}(t) = \big\langle \sqrt{t + 1}, \, \frac{1}{t^2 - 4}, \, \ln(5 - t) \big\rangle$ .

Show the answer

The components need $t \ge -1$ , then $t \ne \pm 2$ , then $t < 5$ . Together that is $[-1, 2) \cup (2, 5)$ .

Evaluate $\lim_{t \to 1} \big\langle \frac{t^2 - 1}{t - 1}, \, \cos(\pi t), \, \frac{\sqrt{t} - 1}{t - 1} \big\rangle$ .

Show the answer

Work component by component. The first quotient simplifies to $t + 1$ , the second component is continuous, and the third simplifies to $\frac{1}{\sqrt{t} + 1}$ for $t \ne 1$ . The limit is $\langle 2, -1, \frac{1}{2} \rangle$ .

Parametrize the curve where the cone $z = \sqrt{x^2 + y^2}$ meets the plane $z = 1 + y$ .

Show the answer

On the curve $\sqrt{x^2 + y^2} = 1 + y$ , and squaring gives $x^2 = 1 + 2y$ . Squaring is reversible here because $1 + y = \frac{x^2 + 1}{2} > 0$ . With $x = t$ this gives

\begin{gathered} \mathbf{r}(t) = \Big\langle t, \; \frac{t^2 - 1}{2}, \; \frac{t^2 + 1}{2} \Big\rangle, \\[0.4em] t \in \mathbb{R}, \end{gathered}

a parabola.

Vector Functions and Space Curves

Key terms

Component functions

Parameter domain

Limits of vector functions

Continuity

Space curve

Helix

Orientation

Intersection curves

Check yourself

How this lesson connects