Math Core

A point moving through space, such as a drone, an electron in a magnetic field or a strand of DNA traced from end to end, can't be described by a single function $y = f(x)$ . What you need is a rule that hands you a whole position vector for each moment in time. That rule is a vector-valued function, and the path its tip traces is a space curve.

Functions whose outputs are vectors

Every function you've studied so far returns a number. A vector-valued function takes a real number $t$ as input and returns a vector as output.

Definition

Vector-valued function

A vector-valued function in space has the form

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

where $f$ , $g$ and $h$ are ordinary real-valued functions called the component functions. The domain of $\mathbf{r}$ is the set of $t$ for which all three components are defined.

Picture $\mathbf{r}(t)$ as an arrow drawn from the origin. As $t$ changes, the arrow swings and stretches, and its tip sweeps out a curve $C$ . The points on $C$ are exactly the points $(f(t), g(t), h(t))$ , so the equations

x = f(t), \qquad y = g(t), \qquad z = h(t)

are parametric equations for $C$ with parameter $t$ . Vector notation and parametric notation describe the same object; the vector form is simply more compact and makes calculus easier later.

A parametrization carries more information than the curve alone. It has an orientation (the direction of travel as $t$ increases) and a pace. The functions $\langle \cos t, \sin t, 0 \rangle$ and $\langle \cos 2t, \sin 2t, 0 \rangle$ trace the same circle, but the second one goes around twice as fast.

Domain, limits and continuity

Because a vector is just its list of components, limits are taken one component at a time.

Limits are componentwise

If $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ , then

\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 limits exist. The function $\mathbf{r}$ is continuous at $a$ when $\displaystyle\lim_{t \to a} \mathbf{r}(t) = \mathbf{r}(a)$ , which happens exactly when every component is continuous at $a$ .

A continuous vector function traces a curve without jumps, which is what you'd expect from a moving object.

Worked example: Domain and a limit

Let $\mathbf{r}(t) = \left\langle \sqrt{9 - t^2},\; \ln t,\; \dfrac{\sin t}{t} \right\rangle$ . Find the domain of $\mathbf{r}$ and $\displaystyle\lim_{t \to 0^+} \mathbf{r}(t)$ , if it exists.

Domain. The square root needs $9 - t^2 \ge 0$ , so $-3 \le t \le 3$ . The logarithm needs $t > 0$ . The third component needs $t \ne 0$ . All three hold when $0 < t \le 3$ , so the domain is $(0, 3]$ .

Limit. As $t \to 0^+$ , the first component tends to $3$ and the third tends to $1$ , but $\ln t \to -\infty$ . Since one component has no finite limit, the vector limit does not exist.

A gallery of space curves

Lines. You met these already: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ is the line through the tip of $\mathbf{r}_0$ in the direction $\mathbf{v}$ . Every component is linear in $t$ .

Helices. The curve $\mathbf{r}(t) = \langle a\cos t,\; a\sin t,\; bt \rangle$ with $a > 0$ and $b \ne 0$ satisfies $x^2 + y^2 = a^2$ , so it lies on a circular cylinder of radius $a$ around the $z$ -axis. While the $x$ and $y$ coordinates circle around, $z$ climbs steadily, so the curve spirals upward like a spring. Each full turn raises it by $2\pi b$ .

Twisted cubic. The curve $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ looks like a parabola from above and like the cubic $z = x^3$ from the side.

Seeing a space curve through its shadows

Drawing a 3D curve on paper is hard, so a good strategy is to look at its projections onto the coordinate planes. To project onto the $xy$ -plane, drop the $z$ -component: the shadow of $\langle f(t), g(t), h(t) \rangle$ is the plane curve $\langle f(t), g(t) \rangle$ . Projecting onto the $xz$ -plane or $yz$ -plane works the same way.

For the helix $\langle \cos t, \sin t, t \rangle$ , the shadow on the $xy$ -plane is the unit circle, and the shadow on the $xz$ -plane is the wave $x = \cos t$ , $z = t$ .

Two shadows of the helix ⟨cos t, sin t, t⟩ for 0 ≤ t ≤ 2π: the unit circle in the xy-plane, and the wave x = cos t, z = t in the xz-plane (vertical axis is z).Open in grapher →

Combining the two pictures tells you what the helix does: it loops around the circle while its height increases.

Tip

To find a surface a curve lies on, look for a relation among the components that doesn't involve $t$ . For $\langle 2\cos t, \sin t, t \rangle$ , you get $\dfrac{x^2}{4} + y^2 = 1$ , so the curve lies on an elliptic cylinder and its shadow on the $xy$ -plane is that ellipse.

Curves of intersection

Two surfaces usually meet in a curve, and a common task is to parametrize it. The idea is to parametrize one surface's constraint first, then use the other equation to solve for the remaining coordinate.

Worked example: A cylinder meets a plane

Find a vector function for the curve where the cylinder $x^2 + y^2 = 4$ meets the plane $z = x + 3$ , and find the highest point on that curve.

The cylinder is handled by $x = 2\cos t$ , $y = 2\sin t$ , since then $x^2 + y^2 = 4$ automatically. The plane then forces $z = x + 3 = 2\cos t + 3$ . So

\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 2\cos t + 3 \rangle, \qquad 0 \le t \le 2\pi.

The height $z = 2\cos t + 3$ is largest when $\cos t = 1$ , that is, at $t = 0$ . The highest point is $\mathbf{r}(0) = (2, 0, 5)$ . The curve is an ellipse, tilted because the plane is tilted.

Worked example: Do two paths meet?

Two particles move along $\mathbf{r}_1(t) = \langle t, t^2, 2t \rangle$ and $\mathbf{r}_2(s) = \langle 4 - s,\; 4s - 4,\; s + 2 \rangle$ . Do their paths intersect? Do the particles collide?

Paths intersect if some $t$ and some $s$ give the same point:

t = 4 - s, \qquad t^2 = 4s - 4, \qquad 2t = s + 2.

Adding the first and third equations gives $3t = 6$ , so $t = 2$ and $s = 2$ . Check the middle one: $t^2 = 4$ and $4s - 4 = 4$ . It works, so the paths cross at $(2, 4, 4)$ .

A collision requires the particles to be there at the same time, meaning the same parameter value in both. Here the crossing happens at $t = 2$ and $s = 2$ , so if both parameters measure the same clock, the particles do collide.

Common mistake

When you check whether two curves intersect, use different parameter names for the two curves. If you set $\mathbf{r}_1(t) = \mathbf{r}_2(t)$ , you're only looking for collisions, and you can miss crossing points that the particles reach at different times.

Practice

Practice 1

Let $\mathbf{r}(t) = \langle t^2 - 1,\; 4 - t,\; 2t^3 \rangle$ . Find $\mathbf{r}(2)$ . Enter it as a point like $(a, b, c)$ .

Enter a point like (2, -3)

Practice 2

Find the domain of $\mathbf{r}(t) = \langle \sqrt{t + 1},\; \ln(5 - t),\; t^2 \rangle$ . Write it as an inequality in $t$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 3

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

Enter a point like (2, -3)

Practice 4

Which best describes the curve $\mathbf{r}(t) = \langle 2\cos t,\; \sin t,\; t \rangle$ ?

Practice 5

The cylinder $x^2 + y^2 = 1$ meets the plane $y + z = 2$ in a curve. Parametrize the curve and find its highest point.

Enter a point like (2, -3)

Practice 6

For which values of $t$ does the helix $\mathbf{r}(t) = \langle \cos t,\; \sin t,\; t \rangle$ meet the sphere $x^2 + y^2 + z^2 = 5$ ?

Separate answers with commas, e.g. 2, -5

Practice 7

Find the point where the paths $\mathbf{r}_1(t) = \langle t,\; 1 - t,\; 3 + t^2 \rangle$ and $\mathbf{r}_2(s) = \langle 3 - s,\; s - 2,\; s^2 \rangle$ intersect.

Enter a point like (2, -3)

Practice 8

The curve $\mathbf{r}(t) = \langle t,\; t^2,\; t^3 \rangle$ is projected onto the $xz$ -plane. What is the projection?