Vector functions

A vector function \(\mathbf{r}:\mathbb{R} \to \mathbb{R}^n\):

\[\mathbf{r}(t) = \left\langle r_1(t), r_2(t), \dots\right\rangle\]

There are several ways to interpret this. Two of them are:

\(\mathbf{r}(t)\) is a position vector of an object moving through the \(n\)-dimensional space at time \(t\).
\(\mathbf{r}\) is a “mapping” that takes points from the number line and places them into the \(n\)-dimensional space. The point \(t\) is placed at the location given by the position vector \(\mathbf{r}(t)\).

If the function is continuous, it will look like this:

Plotting vector functions and identifying the plots

As an example, consider the following vector function:

\[\mathbf{r}(t) = \left\langle 1 + \cos 2\pi t, 1 + \sin 2\pi t, t\right\rangle\]

Let’s try to figure out how the curve defined by this function for \(0 \le t \le 2\) looks like.

One useful tool for studying plots of vector functions is projecting the curve onto the coordinate planes.

Projection to the \(xy\)-plane

We have some curve in the 3d space. Each point on this curve can be projected to the \(xy\)-plane, simply by setting its \(z\)-coordinate to 0. If we do that, we get a new vector function

\[\mathbf{r_{xy}}(t) = \left\langle 1 + \cos 2\pi t, 1 + \sin 2\pi t, 0\right\rangle\]

We know that in the \(xy\)-plane, the equations

\[\begin{aligned} x &= \cos(2\pi t)\\ y &= \sin(2\pi t) \end{aligned}\]

form a circle with center at the origin and radius 1, traveled counterclockwise, starting at \((1,0)\) for \(t = 0\), and going once around the whole circle every time \(t\) increases by 1. This follows from the very definition of trigonometric functions \(\cos\) and \(\sin\).

Here we have

\[\begin{aligned} x &= 1 + \cos(2\pi t)\\ y &= 1 + \sin(2\pi t) \end{aligned}\]

which will move the circle so that its center is at the point \((1,1)\).

So the projection of our curve onto the \(xy\)-plane is the circle with radius 1, center at \((1,1)\), traveled counterclokwise twice around, starting at the point \((2,1)\).

Projection to the \(xz\)-plane

To project the points on the curve to the \(xz\)-plane, we set its \(y\)-coordinate to 0. This will give us a vector function

\[\mathbf{r_{xz}}(t) = \left\langle 1 + \cos 2\pi t, 0, t\right\rangle\]

In other words, in the \(xz\)-plane, we have equations

\[\begin{aligned} x &= 1 + \cos(2\pi t)\\ z &= t \end{aligned}\]

The second equation says that \(t = z\). We can plug this into the first equation, eliminating the \(t\), and getting

\[x = 1 + \cos(2\pi z) \text{ for } 0 \le z \le 2\].

Here we have \(x\) as a function of \(z\), a function which we can graph:

Projection to the \(yz\)-plane

In a similar way we can see that the projection onto the \(yz\)-plane will be the graph of the function \(y = 1 + \sin(2\pi z)\) for \(0 \le z \le 2\):

The graph of the vector function with the projections

More examples

Try it yourself

Domain of a vector function

In order for \(\mathbf{r}(t)\) to be defined, all of the components \(r_1(t)\), \(r_2(t)\), … \(r_n(t)\) must be defined. For example:

Consider the vector function \(\mathbf{r}(t) = \left\langle 1/t, \sqrt{t+1}, \ln(5-t)\right\rangle\).

The first component is undefined at \(t = 0\).
The second component is undefined at \(t < -1\).
The third component is undefined at \(t \ge 5\).

The domain of the function is then the union of two intervals: \([-1,0) \cup (0,5)\).

Try it yourself

Limits and continuity

Intuitively, a continuous vector function would correspond to a trajectory of a moving object where the object does not do things like sudden jumps through the space, does not cease to exist for a moment, does not fly off to an infinite distance only to come back from an infinite distance, whether form the same or different direction, and so on.

Examples

The function

\[\mathbf{r}(t) = \left\langle t^2, 2t, \frac{1}{t-1}\right\rangle\]

is discontinuous at \(t = 1\). As \(t\) approaches 1 from the left, the \(x\)-coordinate approaches 1, the \(y\)-coordinate approaches 2, while the \(z\)-coordinate goes off towards \(-\infty\). As \(t\) approaches 1 from the right, the \(x\) and \(y\)-coordinates still approach 1 and 2, respectively, but the \(z\)-coordinate now goes off to \(\infty\).

On the illustration below, the part of the curve for \(t < 1\) is shown in red, while the part for \(t>1\) is blue. The arrows show the direction of increasing \(t\). The view is limited to a finite part of the space, in reality the curve stretches indefinitely down and up.
The function

\[\mathbf{r}(t) = \left\langle 2t, \frac{t+1}{(t-1)^2}, \frac{1}{t-1}\right\rangle\]

is discontinuous at \(t = 1\). As \(t\) approaches 1 from the left, the \(x\)-coordinate approaches 1, and the \(y\)-coordinate goes off towards \(\infty\) and the \(z\)-coordinate goes off towards \(-\infty\). As \(t\) approaches 1 from the right, the \(x\) still approaches 1, the \(y\)-coordinate goes off to \(\infty\), and the \(z\)-coordinate now also goes off to \(\infty\).

On the illustration below, the part of the curve for \(t < 1\) is shown in red, while the part for \(t>1\) is blue. The arrows show the direction of increasing \(t\). The view is limited to a finite part of the space, in reality the curve stretches indefinitely down and right and up and right.
The function

\[\mathbf{r}(t) = \left\langle \frac{t-1}{\left\lvert t-1\right\rvert} + 2t, t^2 - 1, \frac{1}{t^2+1} - \frac{1}{2}\right\rangle\]

is discontinuous at \(t = 1\). As \(t\) approaches 1 from the left, the \(x\)-coordinate approaches 1, the \(y\)-coordinate approached 0 and the \(z\)-coordinate approaches 0. As \(t\) approaches 1 from the right, the \(x\) approaches 3, and the \(y\)-coordinate and the \(z\)-coordinate still both approach 0.

On the illustration below, the part of the curve for \(t < 1\) is shown in red, while the part for \(t>1\) is blue. The arrows show the direction of increasing \(t\).

There are many other things that can happen, these are just some simple examples that are easy to visualize.