11.1: Vector–Valued Functions

We are very familiar with real valued functions, that is, functions whose output is a real number. This section introduces vector-valued functions - functions whose output is a vector.

Definition $\PageIndex{1}$: Vector-Valued Functions

A vector-valued function is a function of the form

$$\vecs r(t) = \langle\, f(t),g(t)\,\rangle$$

or

$$\vecs r(t) = \langle \,f(t),g(t),h(t)\,\rangle,$$

where $f$, $g$ and $h$ are real valued functions.

The domain of $\vecs r$ is the set of all values of $t$ for which $\vecs r(t)$ is defined. The range of $\vecs r$ is the set of all possible output vectors $\vecs r(t)$.

Evaluating and Graphing Vector-Valued Functions

Evaluating a vector-valued function at a specific value of $t$ is straightforward; simply evaluate each component function at that value of $t$. For instance, if $\vecs r(t) = \langle t^2,t^2+t-1\rangle$, then $\vecs r(-2) = \langle 4,1\rangle$. We can sketch this vector, as is done in Figure $\PageIndex{1a}$. Plotting lots of vectors is cumbersome, though, so generally we do not sketch the whole vector but just the terminal point. The graph of a vector-valued function is the set of all terminal points of $\vecs r(t)$, where the initial point of each vector is always the origin. In Figure $\PageIndex{1b}$ we sketch the graph of $\vecs r$; we can indicate individual points on the graph with their respective vector, as shown.

Figure $\PageIndex{1}$: Sketching the graph of a vector-valued function.

Vector-valued functions are closely related to parametric equations of graphs. While in both methods we plot points $\big(x(t), y(t)\big)$ or $\big(x(t),y(t),z(t)\big)$ to produce a graph, in the context of vector-valued functions each such point represents a vector. The implications of this will be more fully realized in the next section as we apply calculus ideas to these functions.

Example $\PageIndex{1}$: Graphing vector-valued functions

Graph $\vecs r(t) = \langle t^3-t, \dfrac{1}{t^2+1}\rangle$, for $-2\leq t\leq 2$. Sketch $\vecs r(-1)$ and $\vecs r(2)$.

Solution

We start by making a table of $t$, $x$ and $y$ values as shown in Figure $\PageIndex{1a}$. Plotting these points gives an indication of what the graph looks like. In Figure $\PageIndex{1b}$, we indicate these points and sketch the full graph. We also highlight $\vecs r(-1)$ and $\vecs r(2)$ on the graph.

Figure $\PageIndex{2}$: Sketching the vector-valued function of Example 11.1.1

Example $\PageIndex{2}$: Graphing vector-valued functions.

Graph $\vecs r(t) = \langle \cos t,\sin t,t\rangle$ for $0\leq t\leq 4\pi$.

Solution

We can again plot points, but careful consideration of this function is very revealing. Momentarily ignoring the third component, we see the $x$ and $y$ components trace out a circle of radius 1 centered at the origin. Noticing that the $z$ component is $t$, we see that as the graph winds around the $z$-axis, it is also increasing at a constant rate in the positive $z$ direction, forming a spiral. This is graphed in Figure $\PageIndex{3}$. In the graph $\vecs r(7\pi/4)\approx (0.707,-0.707,5.498)$ is highlighted to help us understand the graph.

Figure $\PageIndex{3}$: Viewing a vector-valued function, and its derivative at one point.

Algebra of Vector-Valued Functions

Definition $\PageIndex{2}$: Operations on Vector-Valued Functions

Let $\vecs r_1(t)=\langle f_1(t),g_1(t)\rangle$ and $\vecs r_2(t)=\langle f_2(t),g_2(t)\rangle$ be vector-valued functions in $\mathbb{R}^2$ and let $c$ be a scalar. Then:

1. $\vecs r_1(t) \pm \vecs r_2(t) = \langle\, f_1(t)\pm f_2(t),g_1(t)\pm g_2(t)\,\rangle$.
2. $c\vecs r_1(t) = \langle\, cf_1(t),cg_1(t)\,\rangle$.

A similar definition holds for vector-valued functions in $\mathbb{R}^3$.

This definition states that we add, subtract and scale vector-valued functions component-wise. Combining vector-valued functions in this way can be very useful (as well as create interesting graphs).

Example $\PageIndex{3}$: Adding and scaling vector-valued functions.

Let $\vecs r_1(t) = \langle\,0.2t,0.3t\,\rangle$, $\vecs r_2(t) = \langle\,\cos t,\sin t\,\rangle$ and $\vecs r(t) = \vecs r_1(t)+\vecs r_2(t)$. Graph $\vecs r_1(t)$, $\vecs r_2(t)$, $\vecs r(t)$ and $5\vecs r(t)$ on $-10\leq t\leq10$.

Solution

We can graph $\vecs r_1$ and $\vecs r_2$ easily by plotting points (or just using technology). Let's think about each for a moment to better understand how vector-valued functions work.

We can rewrite $\vecs r_1(t) = \langle\, 0.2t,0.3t\,\rangle$ as $\vecs r_1(t) = t\langle 0.2,0.3\rangle$. That is, the function $\vecs r_1$ scales the vector $\langle 0.2,0.3\rangle$ by $t$. This scaling of a vector produces a line in the direction of $\langle 0.2,0.3\rangle$.

We are familiar with $\vecs r_2(t) = \langle\, \cos t,\sin t\,\rangle$; it traces out a circle, centered at the origin, of radius 1. Figure $\PageIndex{4a}$ graphs $\vecs r_1(t)$ and $\vecs r_2(t)$.

Adding $\vecs r_1(t)$ to $\vecs r_2(t)$ produces $\vecs r(t) = \langle\,\cos t + 0.2t,\sin t+0.3t\,\rangle$, graphed in Figure $\PageIndex{4b}$. The linear movement of the line combines with the circle to create loops that move in the direction of $\langle 0.2,0.3\rangle$. (We encourage the reader to experiment by changing $\vecs r_1(t)$ to $\langle 2t,3t\rangle$, etc., and observe the effects on the loops.)

Figure $\PageIndex{4}$: Graphing the functions in Example $\PageIndex{3}$

Multiplying $\vecs r(t)$ by 5 scales the function by 5, producing $5\vecs r(t) = \langle 5\cos t+1,5\sin t+1.5\rangle$, which is graphed in Figure $\PageIndex{4c}$ along with $\vecs r(t)$. The new function is "5 times bigger'' than $\vecs r(t)$. Note how the graph of $5\vecs r(t)$ in (c) looks identical to the graph of $\vecs r(t)$ in $(b)$. This is due to the fact that the $x$ and $y$ bounds of the plot in $(c)$ are exactly 5 times larger than the bounds in (b).

Example $\PageIndex{4}$: Adding and scaling vector-valued functions.

A cycloid is a graph traced by a point $p$ on a rolling circle, as shown in Figure $\PageIndex{5}$. Find an equation describing the cycloid, where the circle has radius 1.

Figure $\PageIndex{5}$: Tracing a cycloid.

Solution

This problem is not very difficult if we approach it in a clever way. We start by letting $\vecs p(t)$ describe the position of the point $p$ on the circle, where the circle is centered at the origin and only rotates clockwise (i.e., it does not roll). This is relatively simple given our previous experiences with parametric equations; $\vecs p(t) = \langle \cos t, -\sin t\rangle$.

We now want the circle to roll. We represent this by letting $\vecs c(t)$ represent the location of the center of the circle. It should be clear that the $y$ component of $\vecs c(t)$ should be 1; the center of the circle is always going to be 1 if it rolls on a horizontal surface.

The $x$ component of $\vecs c(t)$ is a linear function of $t$: $f(t) = mt$ for some scalar $m$. When $t=0$, $f(t) = 0$ (the circle starts centered on the $y$-axis). When $t=2\pi$, the circle has made one complete revolution, traveling a distance equal to its circumference, which is also $2\pi$. This gives us a point on our line $f(t) = mt$, the point $(2\pi, 2\pi)$. It should be clear that $m=1$ and $f(t) = t$. So $\vecs c(t) = \langle t, 1\rangle$.

We now combine $\vecs p$ and $\vecs c$ together to form the equation of the cycloid:

$$\vecs r(t) = \vecs p(t) + \vecs c(t) = \langle \cos t+ t,-\sin t+1\rangle, \nonumber$$

which is graphed in Figure $\PageIndex{6}$.

Figure $\PageIndex{6}$: The cycloid in Example $\PageIndex{4}$.

Displacement

A vector-valued function $\vecs r(t)$ is often used to describe the position of a moving object at time $t$. At $t=t_0$, the object is at $\vecs r(t_0)$; at $t=t_1$, the object is at $\vecs r(t_1)$. Knowing the locations $\vecs r(t_0)$ and $\vecs r(t_1)$ give no indication of the path taken between them, but often we only care about the difference of the locations, $\vecs r(t_1)-\vecs r(t_0)$, the displacement.

Definition $\PageIndex{3}$: Displacement

Let $\vecs r(t)$ be a vector-valued function and let $t_0<t_1$ be values in the domain. The displacement $\vecs d$ of $\vecs r$, from $t=t_0$ to $t=t_1$, is $$\vecs d=\vecs r(t_1)-\vecs r(t_0).$$

When the displacement vector is drawn with initial point at $\vecs r(t_0)$, its terminal point is $\vecs r(t_1)$. We think of it as the vector which points from a starting position to an ending position.

Example $\PageIndex{5}$: Finding and graphing displacement vectors

Let $\vecs r(t) = \langle \cos (\dfrac{\pi}{2}t),\sin (\dfrac{\pi}2 t)\rangle$. Graph $\vecs r(t)$ on $-1\leq t\leq 1$, and find the displacement of $\vecs r(t)$ on this interval.

Solution

The function $\vecs r(t)$ traces out the unit circle, though at a different rate than the "usual'' $\langle \cos t,\sin t\rangle$ parametrization. At $t_0=-1$, we have $\vecs r(t_0) = \langle 0,-1\rangle$; at $t_1=1$, we have $\vecs r(t_1) = \langle 0,1\rangle$. The displacement of $\vecs r(t)$ on $[-1,1]$ is thus

$$\vecs d = \langle 0,1\rangle - \langle 0,-1\rangle = \langle 0,2\rangle. \nonumber$$

Figure $\PageIndex{7}$: Graphing the displacement of a position function in Example $\PageIndex{5}$.

A graph of $\vecs r(t)$ on $[-1,1]$ is given in Figure $\PageIndex{7}$, along with the displacement vector $\vecs d$ on this interval.

Measuring displacement makes us contemplate related, yet very different, concepts. Considering the semi-circular path the object in Example $\PageIndex{5}$ took, we can quickly verify that the object ended up a distance of 2 units from its initial location. That is, we can compute $\norm{d} = 2$. However, measuring distance from the starting point is different from measuring distance traveled. Being a semi-circle, we can measure the distance traveled by this object as $\pi\approx 3.14$ units. Knowing distance from the starting point allows us to compute average rate of change.

Definition $\PageIndex{4}$: Average Rate of Change

Let $\vecs r(t)$ be a vector-valued function, where each of its component functions is continuous on its domain, and let $t_0<t_1$. The average rate of change of $\vecs r(t)$ on $[t_0,t_1]$ is

$$\text{average rate of change} = \dfrac{\vecs r(t_1) - \vecs r(t_0)}{t_1-t_0}.$$

Example $\PageIndex{6}$: Average rate of change

Let $\vecs r(t) = \langle \cos(\dfrac{\pi}2t),\sin(\dfrac{\pi}2t)\rangle$ as in Example 11.1.5. Find the average rate of change of $\vecs r(t)$ on $[-1,1]$ and on $[-1,5]$.

Solution

We computed in Example $\PageIndex{5}$ that the displacement of $\vecs r(t)$ on $[-1,1]$ was $\vecs d = \langle 0,2\rangle$. Thus the average rate of change of $\vecs r(t)$ on $[-1,1]$ is:

$$\dfrac{\vecs r(1) -\vecs r(-1)}{1-(-1)} = \dfrac{\langle 0,2\rangle}{2} = \langle 0,1\rangle. \nonumber$$

We interpret this as follows: the object followed a semi-circular path, meaning it moved towards the right then moved back to the left, while climbing slowly, then quickly, then slowly again. On average, however, it progressed straight up at a constant rate of $\langle 0,1\rangle$ per unit of time.

We can quickly see that the displacement on $[-1,5]$ is the same as on $[-1,1]$, so $\vecs d = \langle 0,2\rangle$. The average rate of change is different, though:

$$\dfrac{\vecs r(5)-\vecs r(-1)}{5-(-1)} = \dfrac{\langle 0,2\rangle}{6} = \langle 0,1/3\rangle. \nonumber$$

As it took "3 times as long'' to arrive at the same place, this average rate of change on $[-1,5]$ is $1/3$ the average rate of change on $[-1,1]$.

We considered average rates of change in Sections 1.1 and 2.1 as we studied limits and derivatives. The same is true here; in the following section we apply calculus concepts to vector-valued functions as we find limits, derivatives, and integrals. Understanding the average rate of change will give us an understanding of the derivative; displacement gives us one application of integration.

Contributors and Attributions

• Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/