Introduction to Vector-Valued Functions: Learn It 2

Giselle Miller

Introduction to Vector-Valued Functions: Learn It 2

Graphing Vector-Valued Functions

A plane vector has two components: direction and magnitude. Starting from any initial point and moving in a specific direction for a specific distance brings you to the terminal point. The vector’s components are found by subtracting the initial point coordinates from the terminal point coordinates.

A vector is in standard position when its initial point is at the origin. When graphing vector-valued functions, we place vectors in standard position to guarantee a unique graph. This applies to both two-dimensional and three-dimensional vector-valued functions.

The graph of a vector-valued function traces out a curve in space. The form of the function determines what type of curve you get:

Plane curve: For [latex]{\bf{r}},(t)=f,(t),{\bf{i}}+g,(t),{\bf{j}}[/latex], the graph consists of all points [latex](t,\ {\bf{r}},(t))[/latex] and traces a curve in the [latex]xy[/latex]-plane
Space curve: For [latex]{\bf{r}},(t)=f,(t),{\bf{i}}+g,(t),{\bf{j}}+h,(t),{\bf{k}}[/latex], the graph consists of all points [latex](t,\ {\bf{r}},(t))[/latex] and traces a curve in three-dimensional space

Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

Vector-valued functions and parameterized curves are essentially the same thing viewed differently. Given [latex]{\bf{r}},(t)=f,(t),{\bf{i}}+g,(t),{\bf{j}}[/latex], you can write [latex]x=f,(t)[/latex] and [latex]y=g,(t)[/latex] to get the parameterized form. The graphs match exactly, except vector-valued functions show vectors while parameterized curves show points.

How to: Graph a Vector-Valued Function

Create a table of values: Choose several values of [latex]t[/latex] in the domain and calculate [latex]{\bf{r}},(t)[/latex] for each
Plot vectors in standard position: Draw each vector [latex]{\bf{r}},(t)[/latex] with its initial point at the origin
Mark terminal points: Identify where each vector ends
Connect the points: Draw a smooth curve through the terminal points in order of increasing [latex]t[/latex]
Indicate orientation: Add arrows to show the direction of the curve as [latex]t[/latex] increases

Create a graph of each of the following vector-valued functions:

[latex]{\bf{r}}\,(t)=4\cos{t{\bf{i}}}+3\sin{t{\bf{j}}},\ 0\,\leq\,t\,\leq\,2\pi[/latex]
[latex]{\bf{r}}\,(t)=4\cos{t^{3}{\bf{i}}}+3\sin{t^{3}\,{\bf{j}}},\ 0\,\leq\,t\,\leq\,2\pi[/latex]
[latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}+t\,{\bf{k}},\ 0\,\leq\,t\,\leq\,4\pi[/latex]

As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form a curve. This curve turns out to be an ellipse centered at the origin.

[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]	[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]
[latex]0[/latex]	[latex]4\,{\bf{i}}[/latex]	[latex]\pi[/latex]	[latex]-4\,{\bf{i}}[/latex]
[latex]\frac{\pi}{4}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}+\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]	[latex]\frac{5\pi}{4}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}-\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]
[latex]\frac{\pi}{2}[/latex]	[latex]3\,{\bf{j}}[/latex]	[latex]\frac{3\pi}{2}[/latex]	[latex]-3\,{\bf{j}}[/latex]
[latex]\frac{3\pi}{4}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}+\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]	[latex]\frac{7\pi}{4}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}-\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]
[latex]2\pi[/latex]	[latex]4\,{\bf{i}}[/latex]

This figure is a graph of an ellipse centered at the origin. The graph is the vector-valued function r(t)=4cost i + 3sint j. The ellipse has arrows on the curve representing counter-clockwise orientation. There are also line segments inside of the ellipse to the curve at different increments of t. The increments are t=0, t=pi/4, t=pi/2, t=3pi/4. — Figure 1. The graph of the first vector-valued function is an ellipse.

The table of values for [latex]{\bf{r}}\,(t)=4\cos{t\,{\bf{i}}}+3\sin{t\,{\bf{j}}},\ 0\ \leq\ t\ \leq\ 2\pi[/latex] is as follows:

[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]	[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]
[latex]0[/latex]	[latex]4\,{\bf{i}}[/latex]	[latex]\frac{\pi}{2}[/latex]	[latex]-4\,{\bf{i}}[/latex]
[latex]\frac{\pi}{8}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}+\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]	[latex]\frac{5\pi}{8}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}-\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]
[latex]\frac{\pi}{4}[/latex]	[latex]3\,{\bf{j}}[/latex]	[latex]\frac{3\pi}{4}[/latex]	[latex]-3\,{\bf{j}}[/latex]
[latex]\frac{3\pi}{8}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}+\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]	[latex]\frac{7\pi}{8}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}-\frac{3\sqrt{2}}{2}\,{\bf{j}}[/latex]
[latex]\pi[/latex]	[latex]4\,{\bf{i}}[/latex]

The graph of this curve is also an ellipse centered at the origin.

This figure is a graph of an ellipse centered at the origin. The graph is the vector-valued function r(t)=4cost^3 i + 3sint^3 j. — Figure 2. The graph of the second vector-valued function is also an ellipse.

We go through the same procedure for a three-dimensional vector function.

[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]	[latex]t[/latex]	[latex]{\bf{r}}\,(t)[/latex]
[latex]0[/latex]	[latex]4\,{\bf{i}}[/latex]	[latex]\pi[/latex]	[latex]-4\,{\bf{j}}+\pi\,{\bf{k}}[/latex]
[latex]\frac{\pi}{4}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}+2\sqrt{2}\,{\bf{j}}+\frac{\pi}{4}\,{\bf{k}}[/latex]	[latex]\frac{5\pi}{4}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}-2\sqrt{2}\,{\bf{j}}+\frac{5\pi}{4}\,{\bf{k}}[/latex]
[latex]\frac{\pi}{2}[/latex]	[latex]4\,{\bf{j}}+\frac{\pi}{2}{\bf{k}}[/latex]	[latex]\frac{3\pi}{2}[/latex]	[latex]-4\,{\bf{j}}+\frac{3\pi}{2}\,{\bf{k}}[/latex]
[latex]\frac{3\pi}{4}[/latex]	[latex]-2\sqrt{2}\,{\bf{i}}+2\sqrt{2}\,{\bf{j}}+\frac{3\pi}{4}\,{\bf{k}}[/latex]	[latex]\frac{7\pi}{4}[/latex]	[latex]2\sqrt{2}\,{\bf{i}}-2\sqrt{2}\,{\bf{j}}+\frac{7\pi}{4}\,{\bf{k}}[/latex]
[latex]2\pi[/latex]	[latex]4\,{\bf{i}}+2\pi\,{\bf{k}}[/latex]

The values then repeat themselves, except for the fact that the coefficient of k is always increasing (Figure 3.4). This curve is called a helix. Notice that if the k component is eliminated, then the function becomes [latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}[/latex], which is a unit circle centered at the origin.

This figure is the graph of a helix in the 3 dimensional coordinate system. The curve represents the function r(t) = cost i + sint j + tk. The curve spirals in a circular path around the vertical z-axis and has the look of a spring. The arrows on the curve represent orientation. — Figure 3. The graph of the third vector-valued function is a helix.

Reparameterization

Notice that the graphs in parts (a) and (b) are identical. This happens because the function in part (b) is a reparameterization of the function in part (a). Any curve has infinitely many reparameterizations. For example, replacing [latex]t[/latex] with [latex]2t[/latex] in any of the three curves above doesn’t change the shape—only the interval over which [latex]t[/latex] is defined may change. We’ll explore arc-length parameterization later in this chapter.

Helix

The curve in part (c) is a helix with a circular cross-section. Helices can also have elliptical cross-sections—for instance, [latex]{\bf{r}},(t)=4\cos{t,{\bf{i}}}+3\sin{t,{\bf{j}}}+t,{\bf{k}}[/latex] describes an elliptical helix whose projection into the [latex]xy[/latex]-plane is an ellipse. The arrows in the graph indicate the orientation of the curve as [latex]t[/latex] progresses from [latex]0[/latex] to [latex]4\pi[/latex].

helix

A helix is a three-dimensional curve that resembles a spring. When viewed along the [latex]z[/latex]-axis, it has a circular (or elliptical) cross-section. The curve spirals upward or downward as the parameter increases.