Vector-Valued Functions and Space Curves

Learning Outcomes

Write the general equation of a vector-valued function in component form and unit-vector form.
Recognize parametric equations for a space curve.
Describe the shape of a helix and write its equation.

Definition of a Vector-Valued Function

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

Definition

A vector-valued function is a function of the form

[latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}} + g\,(t)\,{\bf{j}}\;\text{ or }\;{\bf{r}}\,(t)=f\,(t)\,{\bf{i}} + g\,(t)\,{\bf{j}} + h\,(t)\,{\bf{k}}[/latex],

where the component functions [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex], are real-valued functions of the parameter [latex]t[/latex]. Vector-valued functions are also written in the form

[latex]{\bf{r}}\,(t)={\langle}f\,(t),\ g\,(t){\rangle}\;\text{ or }\;{\bf{r}}\,(t)={\langle}f\,(t),\ g\,(t),\ h\,(t){\rangle}[/latex].

In both cases, the first form of the function defines a two-dimensional vector-valued function; the second form describes a three-dimensional vector-valued function.

The parameter [latex]t[/latex] can lie between two real numbers: [latex]a \leq t \leq b[/latex]. Another possibility is that the value of [latex]t[/latex] might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of [latex]t[/latex]. We often use [latex]t[/latex] as a parameter because [latex]t[/latex] can represent time.

Example: Evaluating Vector-Valued Functions and Determining Domains

For each of the following vector-valued functions, evaluate [latex]{\bf{r}}(0), {\bf{r}}(\frac{\pi}{2})[/latex], and [latex]{\bf{r}}(\frac{2{\pi}}{3})[/latex]. Do any of these functions have domain restrictions?

[latex]{\bf{r}}\,(t)=4\cos{t}\,{\bf{i}}+3\sin{t}{\bf{j}}[/latex]
[latex]{\bf{r}}\,(t)=3\tan{t}\,{\bf{i}}+4\sec{t}{\bf{j}}+5t{\bf{k}}[/latex]

Show Solution

To calculate each of the function values, substitute the appropriate value of t into the function:
[latex]\begin{array}{ccc}\hfill {\bf{r}}(0) & =\hfill & 4\cos{(0)}{\bf{i}}+3\sin{(0)}{\bf{j}} \hfill \\ \hfill & =\hfill & 4{\bf{i}}+0{\bf{j}}=4{\bf{i}}\hfill\\\hfill{\bf{r}}\left(\frac{\pi}{2}\right) & =\hfill & 4\cos{\left(\frac{\pi}{2}\right)}{\bf{i}}+3\sin{\left(\frac{\pi}{2}\right)}{\bf{j}}\hfill\\\hfill&=\hfill&0{\bf{i}}+3{\bf{j}}=3{\bf{j}}\hfill\\\hfill{\bf{r}}\left(\frac{2\pi}{3}\right)& =\hfill & 4\cos{\left(\frac{2\pi}{3}\right)}{\bf{i}}+3\sin{\left(\frac{2\pi}{3}\right)}{\bf{j}}\hfill \\ \hfill & =\hfill & 4\left(-\frac{1}{2}\right){\bf{i}}+3\left(\frac{\sqrt{3}}{2}\right){\bf{j}}=-2{\bf{i}}+\frac{3\sqrt{3}}{2}{\bf{j}}\end{array}[/latex].

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is [latex]f(t)=4\cos{t}[/latex] and the second component function is [latex]g(t)=3\sin{t}[/latex] Neither of these functions has a domain restriction, so the domain of [latex]{\bf{r}}(t)=4\cos{t{\bf{i}}}+3\sin{t\,{\bf{j}}}[/latex] is all real numbers.
To calculate each of the function values, substitute the appropriate value of [latex]t[/latex] into the function:
[latex]\begin{array}{ccc}\hfill {\bf{r}}(0) & =\hfill & 3\tan{(0){\bf{i}}}+4\sec{(0){\bf{j}}}+5\,(0)\,{\bf{k}} \hfill \\ \hfill & =\hfill & 0{\bf{i}}+4{\bf{j}}+0{\bf{k}}=4{\bf{j}}\hfill\\\hfill{\bf{r}}\left(\frac{\pi}{2}\right) & =\hfill & 3\tan{\left(\frac{\pi}{2}\right){\bf{i}}}+4\sec{\left(\frac{\pi}{2}\right){\bf{j}}}+5\,\left(\frac{\pi}{2}\right)\,{\bf{k}},\text{ which does not exist}\hfill\\\hfill{\bf{r}}\left(\frac{2\pi}{3}\right) & =\hfill & 3\tan{\left(\frac{2\pi}{2}\right){\bf{i}}}+4\sec{\left(\frac{2\pi}{2}\right){\bf{j}}}+5\,\left(\frac{2\pi}{2}\right)\,{\bf{k}} \hfill \\ \hfill & =\hfill & 3\,(-\sqrt{3})\,{\bf{i}}+4(-{2})\,{\bf{j}}+\frac{10\pi}{3}\,{\bf{k}} \hfill \\ \hfill & =\hfill & -3\sqrt{3}\,{\bf{i}}-8{\bf{j}}+\frac{10\pi}{3}\,{\bf{k}}.\hfill \\ \hfill \end{array}[/latex]

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is [latex]f\,(t)=3\tan{t}[/latex], the second component function is [latex]g\,(t)=4\sec{t}[/latex], and the third component function is [latex]h\,(t)=5t[/latex]. The first two functions are not defined for odd multiples of [latex]\frac{\pi}{2}[/latex], so the function is not defined for odd multiples of [latex]\frac{\pi}{2}[/latex]. Therefore, [latex]\text{dom}({\bf{r}}\,(t))=\bigg\{t\,\bigg|\,t\neq\frac{(2n+1){\pi}}{2}\bigg\}[/latex], where [latex]n[/latex] is any integer.

Try It

For the vector-valued function [latex]{\bf{r}}\,(t)=(t^{2}-3t)\,{\bf{i}}+(4t+1)\,{\bf{j}}[/latex], evaluate [latex]{\bf{r}}\,(0),\ {\bf{r}}\,(1)[/latex], and [latex]{\bf{r}}\,(-4)[/latex]. Does this function have any domain restrictions?

Show Solution

The above example illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.

Graphing Vector-Valued Functions

Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive at a second point. This represents the terminal point of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal point.

A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}[/latex] consists of the set of all [latex](t,\ {\bf{r}}\,(t))[/latex], and the path it traces is called a plane curve. The graph of a vector-valued function of the form [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}[/latex] consists of the set of all [latex](t,\ {\bf{r}}\,(t))[/latex] and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

Example: Combining Vectors

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]

Show Solution

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.

You may notice that the graphs in parts 1 and 2 are identical. This happens because the function describing curve 2 is a so-called reparameterization of the function describing curve 1. In fact, any curve has an infinite number of reparameterizations; for example, we can replace [latex]t[/latex] with [latex]2t[/latex] in any of the three previous curves without changing the shape of the curve. The interval over which t is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization.

As mentioned, the name of the shape of the curve of the graph in part 3 of the previous example is a helix (Figure 3). The curve resembles a spring, with a circular cross-section looking down along the [latex]z[/latex]-axis. It is possible for a helix to be elliptical in cross-section as well. For example, the vector-valued function [latex]{\bf{r}}\,(t)=4\cos{t\,{\bf{i}}}+3\sin{t\,{\bf{j}}}+t\,{\bf{k}}[/latex] describes an elliptical helix. The projection of this helix into the [latex]x,y\text{-plane}[/latex] is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as [latex]t[/latex] progresses from [latex]0[/latex] to [latex]4\pi[/latex].

Try It

Create a graph of the vector valued function [latex]{\bf{r}}\,(t)=(t^{2}-1)\,{\bf{i}}+(2t-3)\,{\bf{j}},\ 0\ \leq\ t\ \leq\ 3.[/latex]

Show Solution

This figure is a graph of the function r(t) = (t^2-1)i + (2t-3)j, for the values of t from 0 to 3. The curve begins in the 3rd quadrant at the ordered pair (-1,-3) and increases up through the 1st quadrant. It is increasing and has arrows on the curve representing orientation to the right.

Figure 4. Graph of [latex]{\bf{r}}\,(t)=(t^{2}-1)\,{\bf{i}}+(2t-3)\,{\bf{j}},\ 0\ \leq\ t\ \leq\ 3[/latex].

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.2” here (opens in new window).At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}[/latex], we can define [latex]x=f\,(t)[/latex] and [latex]y=g\,(t)[/latex]. If a restriction exists on the values of [latex]t[/latex] (for example, [latex]t[/latex] is restricted to the interval [latex][a,\ b][/latex] for some constants [latex]a\ \le\ b[/latex]). then this restriction is enforced on the parameter The graph of the parameterized function would then agree with the graph of the vector-valued function, except that the vector-valued graph would represent vectors rather than points. Since we can parameterize a curve defined by a function [latex]y=f\,(x)[/latex], it is also possible to represent an arbitrary plane curve by a vector-valued function.

Module 3: Vector-Valued Functions