Vector-Valued Functions and Space Curves

You have already described lines in space as $\mathbf{r}(t) = P_0 + t\mathbf{v}$, and curves in the plane with parametric equations. Put the two ideas together and let each coordinate be any function of $t$: $$\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle.$$ This is a vector-valued function. Think of $t$ as time and $\mathbf{r}(t)$ as the position of a moving particle; the path it traces is a space curve.

Key idea

A vector function is three ordinary functions working together. Its domain is the set of $t$ where all components are defined, and limits and continuity work component by component: $$\lim_{t \to a} \mathbf{r}(t) = \left\langle \lim_{t \to a} f(t),\ \lim_{t \to a} g(t),\ \lim_{t \to a} h(t) \right\rangle.$$

Why it works

Why can we take limits one component at a time? "$\mathbf{r}(t)$ approaches $\mathbf{L}$" means the distance $|\mathbf{r}(t) - \mathbf{L}|$ shrinks to ${}0$. Each component difference is at most that distance (a leg of a right triangle is never longer than the diagonal), and the distance is at most the sum of the three component differences. So the distance goes to ${}0$ exactly when all three component differences do.

The helix $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, \frac{t}{3} \rangle$. Move the slider or press play: the shadow on the floor goes around a circle while the height grows steadily.

A curve and its parametrization are not the same thing. $\langle \cos t, \sin t, 0 \rangle$ and $\langle \cos 2t, \sin 2t, 0 \rangle$ trace the same unit circle, the second twice as fast, and $\langle \cos t, -\sin t, 0 \rangle$ traces it the other way around. Vector functions also describe curves where two surfaces meet. The cylinder $x^2 + y^2 = 1$ and the plane $z = 1 + x$ meet in a curve: use $x = \cos t$, $y = \sin t$ to stay on the cylinder, and the plane then forces $z = 1 + \cos t$. So $\mathbf{r}(t) = \langle \cos t, \sin t, 1 + \cos t \rangle$, ${}0 \le t \le 2\pi$: an ellipse tilted in space.

Worked examples

Worked example: Domain

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

$\sqrt{t - 1}$ needs $t \ge 1$. $\ln(4 - t)$ needs $t < 4$. $\frac{1}{t - 2}$ needs $t \ne 2$.
All three at once: ${}1 \le t < 4$ with $t \ne 2$.

Answer: $[1, 2) \cup (2, 4)$

Worked example: Recognize the curve

Describe the curve $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 2t \rangle$.

$x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9$ for every $t$: the curve lies on the cylinder of radius 3 around the $z$-axis.
Meanwhile $z = 2t$ climbs at a steady rate.
So it is a helix; one full turn ($t$ increasing by ${}2\pi$) raises it by ${}4\pi$.

Answer: a helix on the cylinder $x^2 + y^2 = 9$

Worked example: A limit

Find $\displaystyle\lim_{t \to 0} \left\langle \frac{\sin 2t}{t},\ e^{t},\ \frac{t^2 + t}{t} \right\rangle$.

First component: $\frac{\sin 2t}{t} = 2 \cdot \frac{\sin 2t}{2t} \to 2$.
Second: $e^t$ is continuous, so the limit is $e^0 = 1$.
Third: $\frac{t(t + 1)}{t} = t + 1 \to 1$.

Answer: $\langle 2, 1, 1 \rangle$

To see which surface a curve lives on, look for a relation among the components that does not involve $t$, as in the helix example. To find where a curve meets a surface, substitute the components into the surface's equation and solve for $t$.

Common mistake

Reading $t$ as a coordinate. The parameter is time, not a direction in space; the picture shows only the values $(x, y, z)$. Two different vector functions can trace the same curve at different speeds. Taking the union of the domains. The particle exists only when every component is defined, so the domain is the intersection.

Vector-Valued Functions and Space Curves

Worked examples

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