# Vector-Valued Functions

We have briefly looked over a few topics regarding vectors in the Calculus section of this site. We are now ready to look at a new type of function known as a vector-valued function which we will now loosely define.

Definition: A Vector-Valued Function (also known as Vector Function) denoted $\vec{r}(t)$ on an interval $I$ takes every $t \in I$ and maps it to a position vector or point $\vec{r}(t)$. In $\mathbb{R}^2$, we can denote a vector-valued function $\vec{r}(t) = x(t) \vec{i} + y(t) \vec{j}$ by the parametric equations $\left\{\begin{matrix} x = x(t)\\ y = y(t) \end{matrix}\right.$ that respectively define the $x$ and $y$ position of $\vec{r}(t)$. Similarly, in $\mathbb{R}^3$ we can denote a vector-valued function by $\vec{r}(t) = x(t) \vec{i} + y(t) \vec{j} + z(t) \vec{k} = (x(t), y(t), z(t))$ by the set of parametric equations $\left\{\begin{matrix} x = x(t)\\ y = y(t)\\ z = z(t) \end{matrix}\right.$ that respectively definite the $x$, $y$, and $z$ position of $\vec{r}(t)$. |

*Of course, we can define vector-valued functions in $\mathbb{R}^n$.*

For example, $\vec{r}(t) = 2t \vec{i} + 3t^2 \vec{j} + t \vec{k} = (2t, 3t^2, t)$ for $-\infty < t < \infty$ is a vector valued function. At $t = 0$, $\vec{r}(0) = (0, 0, 0)$ is the zero vector, and at $t = 1$, $\vec{r}(t) = (2, 3, 1)$. Of course, we could plug in any value of $t$ and get a resulting vector $\vec{r}(t)$. The following image represents an arbitrary curve $C$ defined from a vector-valued function.

Of course, we do not necessarily need to work with $\mathbb{R}^3$ exclusively. We can also look at vector-valued functions in $\mathbb{R}^2$, or even $\mathbb{R}^n$ though vector-valued functions in $\mathbb{R}^n$ are trickier to visualize.

For example, consider the curve $C$ traced by the vector-valued function $\vec{r}(t) = t \vec{i} + t^2 \vec{j} = (t, \sin t)$, $0 ≤ t ≤ 2 \pi$. This vector-valued function traces out a one period of the sine function. This can easily be seen since $\vec{r}(0) = (0, 0)$, $\vec{r} \left ( \frac{\pi}{2} \right ) = \left ( \frac{\pi}{2}, 1 \right )$, $\vec{r} (\pi) = (\pi, 0)$, … The following image depicts the curve $C$ as well as some position vectors.

We will now look at another important definition analogous to that of real-valued functions.

Definition: Let $\vec{r}(t)$ be a vector-valued function. Then the Domain of $\vec{r}$ denoted $D(\vec{r})$ is the set of all values $t$ for which $\vec{r}(t)$ is defined. |

For example, consider the vector-valued function $\vec{r}(t) (\sqrt{t}, t, t^2)$. We notice that $t ≥ 0$ otherwise $\vec{r}(t)$ is undefined since $\sqrt{t}$ would be undefined. So $D(\vec{r}) = [0, \infty)$. Of course, sometimes we purposely restrict the domain as some of the examples above and below state.

## Example 1

**Describe the space curve $C$ traced out by the vector-valued function $\vec{r}(t) = (1 + t, 2 + t, 3 + t)$ for $-\infty < t < \infty$.**

The curve $C$ is traced by the set of parametric equations $\left\{\begin{matrix} x(t) = 1 + t\\ y(t) = 2 + t\\ z(t) = 3 + t \end{matrix}\right.$. We've seen already from the Equations of Lines in Three-Dimensional Space that this set of parametric equations form a line in $\mathbb{R}^3$ that passes through the point $(1, 2, 3)$ at $t = 0$, and that is parallel to the vector $(1, 1, 1)$.

## Example 2

**Describe the space curve $C$ traced out by the vector-valued function $\vec{r}(t) = (1, \cos t, 2 \sin t)$ for $-\infty < t < \infty$.**

We first notice that since $x(t) = 1$ then all points on the curve $C$ will be located on the plane $x = 1$. Furthermore, we note that the parametric equations $y(t) = \cos t$ and $z(t) = 2 \sin t$ represent an ellipse on this plane with semi-minor axis $1$ and semi=major axis $2$ since $\left [y(t) \right]^2 + \left [\frac{z(t)}{2} \right]^2 = \cos^2 t + \sin^2 t = 1$.

We also note that this ellipse is traced out infinitely many times since $\cos t$ and $2\sin t$ are periodic functions, both with periods of $2\pi$. The following animation depicts this curve.

## Example 3

**Describe the space curve $C$ traced out by the vector valued function $\vec{r}(t) = (4 \sin t, 4 \cos t, t)$.**

Notice that $[x(t)]^2 + [y(t)]^2 = 16 \sin ^2 t + 16 \cos ^2 t = 16 (\sin ^2 t + \cos ^2 t) = 16$, and therefore $C$ must lie on the circular cylinder $x^2 + y^2 = 4^2$ with radius $r$.

At $t = 0$, $\vec{r}(0) = (0, 4, 0)$, and at $t = 2\pi$, $\vec{r}(0) = (0, 4, 2\pi)$, so our curve is traced upwards. The following animations depicts this curve.