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.

Definition12.1.2.Vector-Valued Functions.

A vector-valued function is a function of the form

where \(f\text{,}\)\(g\) and \(h\) are real valued functions.

The domain of \(\vec r\) is the set of all values of \(t\) for which \(\vec r(t)\) is defined. The range of \(\vec r\) is the set of all possible output vectors \(\vec r(t)\text{.}\)

Subsection12.1.1Evaluating 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 \(t\text{.}\) For instance, if \(\vec r(t) = \la t^2,t^2+t-1\ra\text{,}\) then \(\vec r(-2) = \la 4,1\ra\text{.}\) We can sketch this vector, as is done in Figure 12.1.3.(a). 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 \(\vec r(t)\text{,}\) where the initial point of each vector is always the origin. In Figure 12.1.3.(b) we sketch the graph of \(\vec r\) ; we can indicate individual points on the graph with their respective vector, as shown.

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.

Example12.1.4.Graphing vector-valued functions.

Graph \(\ds \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra\text{,}\) for \(-2\leq t\leq 2\text{.}\) Sketch \(\vec r(-1)\) and \(\vec r(2)\text{.}\)

Solution1.Video solution

Solution2.

We start by making a table of \(t\text{,}\)\(x\) and \(y\) values as shown in Figure 12.1.5.(a). Plotting these points gives an indication of what the graph looks like. In Figure 12.1.5.(b), we indicate these points and sketch the full graph. We also highlight \(\vec r(-1)\) and \(\vec r(2)\) on the graph.

\(t\)

\(t^3-t\)

\(\ds \frac{1}{t^2+1}\)

\(-2\)

\(-6\)

1/5

\(-1\)

\(0\)

1/2

\(0\)

\(0\)

\(1\)

\(1\)

\(0\)

1/2

\(2\)

\(6\)

1/5

(a)

The image contains the plot of the vector valued function \(\vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra\) for \(-2\leq t\leq 2\text{.}\) The image also contains the vectors \(\vec r(-1) =\la 0,\frac12 \ra\) and \(\vec r(2) =\la 6,\frac15 \ra\text{,}\) which both end at a point on the function \(\vec r(t)\text{.}\) The function \(\vec r(t)\) begins at the point \((-6,\frac15)\) corresponding to the lower bound of \(t=-2\text{,}\) from which it slowly slopes upwards until crossing the \(y\)-axis at \(y=\frac12\text{.}\) From here, the function slopes upwards and slightly outwards away from the \(y\)-axis until curving back and crossing the \(y\)-axis once again at \(y=1\text{.}\) The function then slopes downwards and slightly away from the \(y\)-axis until curving back and crossing the \(y\)-axis again at \(y=\frac12\text{,}\) completing a loop. After crossing the \(y\)-axis, the function continues slowly sloping downwards until reaching the point \((6,\frac15)\) which corresponds to the upper bound of \(t=2\text{.}\) The function is also symmetric about the \(y\)-axis.

(b)

Figure12.1.5.Sketching the vector-valued function of Example 12.1.4

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 \(t\text{,}\) 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 12.1.7. In the graph \(\vec r(7\pi/4)\approx (0.707,-0.707,5.498)\) is highlighted to help us understand the graph.

Figure12.1.7.The graph of \(\vec r(t)\) in Example 12.1.6

Subsection12.1.2Algebra of Vector-Valued Functions

Definition12.1.8.Operations on Vector-Valued Functions.

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

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

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).

Example12.1.10.Adding and scaling vector-valued functions.

Let \(\vec r_1(t) = \la\,0.2t,0.3t\,\ra\text{,}\)\(\vec r_2(t) = \la\,\cos(t) ,\sin(t) \,\ra\) and \(\vec r(t) = \vec r_1(t)+\vec r_2(t)\text{.}\) Graph \(\vec r_1(t)\text{,}\)\(\vec r_2(t)\text{,}\)\(\vec r(t)\) and \(5\vec r(t)\) on \(-10\leq t\leq10\text{.}\)

Solution1.Video solution

Solution2.

We can graph \(\vec r_1\) and \(\vec r_2\) 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 \(\vec r_1(t) = \la\, 0.2t,0.3t\,\ra\) as \(\vec r_1(t) = t\la 0.2,0.3\ra\text{.}\) That is, the function \(\vec r_1\) scales the vector \(\la 0.2,0.3\ra\) by \(t\text{.}\) This scaling of a vector produces a line in the direction of \(\la 0.2,0.3\ra\text{.}\)

We are familiar with \(\vec r_2(t) = \la\, \cos(t) ,\sin(t) \,\ra\text{;}\) it traces out a circle, centered at the origin, of radius 1. Figure 12.1.11.(a) graphs \(\vec r_1(t)\) and \(\vec r_2(t)\text{.}\)

Adding \(\vec r_1(t)\) to \(\vec r_2(t)\) produces \(\vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra\text{,}\) graphed in Figure 12.1.11.(b). The linear movement of the line combines with the circle to create loops that move in the direction of \(\la 0.2,0.3\ra\text{.}\) (We encourage the reader to experiment by changing \(\vec r_1(t)\) to \(\la 2t,3t\ra\text{,}\) etc., and observe the effects on the loops.)

The image contains the plot of the vector valued functions \(\vec r_1(t) = \la 0.2t,0.3t \ra\text{,}\)\(\vec r_2(t) = \la \cos(t) ,\sin(t) \ra\) on \(-10\leq t\leq10\text{.}\) The function \(\vec r_1(t) = \la 0.2t,0.3t \ra\) is a line which begins at the point \((-2,-3)\) corresponding to \(t=-10\) and ends when \(t=10\) at the point \((2,3)\text{.}\) The second function \(\vec r_2(t) = \la \cos(t) ,\sin(t) \ra\) looks like a circle of radius \(1\text{,}\) but as \(-10\leq t\leq10\) the function completes \(\frac{20}{2\pi}\) full circular rotations, which cannot be seen in the graph.

(a)

The image contains the plot of the vector valued function \(\vec r(t) = \vec r_1(t)+\vec r_2(t)\text{.}\) The function is given by \(\vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra\) on \(-10\leq t\leq10\text{.}\) The function begins in the third quadrant, near the point \((-3,-2)\text{.}\) Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards. The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased \(x\) and \(y\) coordinate. The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. The function nearly completes a third loop, before ending near the point \((3,1.5)\text{.}\)

(b)

The image contains the plot of the vector valued function \(5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t)\text{.}\) The function is now given by \(\vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra\) on \(-10\leq t\leq10\text{.}\) Compared to the unscaled function \(\vec r(t)\text{,}\) the function \(5\vec r(t)\) is exactly 5 times larger than the original function. The function begins in the third quadrant, near the point \((-15,-10)\text{.}\) Like the unscaled function, near this point \(5\vec r(t)\) is concave up, starting with a downwards slope and later beginning to slope upwards. The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased \(x\) and \(y\) coordinate. The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. The function nearly completes a third loop, before ending near the point \((15,7.5)\text{.}\) The entirety of the original function \(\vec r(t)\) can be seen to fit between the two loops of the scaled function \(5\vec r(t)\text{.}\)

Multiplying \(\vec r(t)\) by 5 scales the function by 5, producing \(5\vec r(t) = \langle 5\cos(t) +t,5\sin(t) +1.5t\rangle\text{,}\) which is graphed in Figure 12.1.11.(c) along with \(\vec r(t)\text{.}\) The new function is “5 times bigger” than \(\vec r(t)\text{.}\) Note how the graph of \(5\vec r(t)\) in Figure 12.1.11.(c) looks identical to the graph of \(\vec r(t)\) in Figure 12.1.11.(b). This is due to the fact that the \(x\) and \(y\) bounds of the plot in Figure 12.1.11.(c) are exactly 5 times larger than the bounds in Figure 12.1.11.(b).

Example12.1.12.Adding and scaling vector-valued functions.

A cycloid is a graph traced by a point \(p\) on a rolling circle, as shown in Figure 12.1.13. Find an equation describing the cycloid, where the circle has radius 1.

The image shows the cycloid which comes from tracking a point \(p\) on a rolling circle of radius \(1\) on a flat surface. The point \(p\) is initially at the top of the circle of radius \(1\text{.}\) Once the circle rolls, the point \(p\) is tracked to create a graph of a cycloid. The graph coming from tracking the point \(p\) first begins to decrease, until the point \(p\) is at the bottom of the circle, at which point the point \(p\) touches the surface the ball is rolling on. After this point, the graph begins to increase. The circle continues to roll, with the point \(p\) once again becoming the top of the circle, after which the graph begins to decrease. The circle continues rolling, with the point \(p\) becoming the lowest point on the circle two more times, after which the graph stops. Between the starting point and the point at which \(p\) is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle. The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles.

Figure12.1.13.Tracing a cycloid

Solution1.Video solution

Solution2.

This problem is not very difficult if we approach it in a clever way. We start by letting \(\vec p(t)\) 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; \(\vec p(t) = \la \cos(t) , -\sin(t) \ra\text{.}\)

We now want the circle to roll. We represent this by letting \(\vec c(t)\) represent the location of the center of the circle. It should be clear that the \(y\) component of \(\vec c(t)\) 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 \(\vec c(t)\) is a linear function of \(t\text{:}\)\(f(t) = mt\) for some scalar \(m\text{.}\) When \(t=0\text{,}\)\(f(t) = 0\) (the circle starts centered on the \(y\)-axis). When \(t=2\pi\text{,}\) the circle has made one complete revolution, traveling a distance equal to its circumference, which is also \(2\pi\text{.}\) This gives us a point on our line \(f(t) = mt\text{,}\) the point \((2\pi, 2\pi)\text{.}\) It should be clear that \(m=1\) and \(f(t) = t\text{.}\) So \(\vec c(t) = \la t, 1\ra\text{.}\)

We now combine \(\vec p\) and \(\vec c\) together to form the equation of the cycloid: \(\vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra\text{,}\) which is graphed in Figure 12.1.14.

The graph shows the function \(\vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra\text{,}\) which is the graph of a cycloid which comes from tracking a point \(p\) on a rolling a circle of radius \(1\) on a flat surface. The curve begins near the point \((0,2)\text{,}\) after which it begins to decrease until it reaches its lowest point near the point \((2,0)\text{.}\) After this point, the curve begins increasing, until it reaches its highest point when \(p\) is at the top of the circle, after which it decreases until reaching another minimum near the point \((8,0)\text{.}\) The curve continues in the same fashion, reaching another minimum near the point \((14,0)\text{,}\) after which it continues for a slight duration in the same fashion as before.

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

Definition12.1.15.Displacement.

Let \(\vec r(t)\) be a vector-valued function and let \(t_0\lt t_1\) be values in the domain. The displacement \(\vec d\) of \(\vec r\text{,}\) from \(t=t_0\) to \(t=t_1\text{,}\) is

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

Example12.1.17.Finding and graphing displacement vectors.

Let \(\vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra\text{.}\) Graph \(\vec r(t)\) on \(-1\leq t\leq 1\text{,}\) and find the displacement of \(\vec r(t)\) on this interval.

Solution1.Video solution

Solution2.

The function \(\vec r(t)\) traces out the unit circle, though at a different rate than the “usual” \(\la \cos(t) ,\sin(t) \ra\) parametrization. At \(t_0=-1\text{,}\) we have \(\vec r(t_0) = \la 0,-1\ra\text{;}\) at \(t_1=1\text{,}\) we have \(\vec r(t_1) = \la 0,1\ra\text{.}\) The displacement of \(\vec r(t)\) on \([-1,1]\) is thus \(\vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra\text{.}\)

Graph of the function \(\vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra\) on \(-1\leq t\leq 1\text{.}\) The function \(\vec r(t) \) is the right half of the unit circle, beginning at the point \((0,-1)\text{,}\) crossing the \(x\)-axis at the point \((1,0)\text{,}\) and ending at the point \((0,-1)\text{.}\) The graph also contains the displacement vector, which begins at the start of the curve at the point \((0,-1)\) and heads directly upwards until reaching the endpoint of the curve at the point \((0,1)\text{.}\)

Figure12.1.18.Graphing the displacement of a position function in Example 12.1.17

A graph of \(\vec r(t)\) on \([-1,1]\) is given in Figure 12.1.18, along with the displacement vector \(\vec d\) on this interval.

Measuring displacement makes us contemplate related, yet very different, concepts. Considering the semi-circular path the object in Example 12.1.17 took, we can quickly verify that the object ended up a distance of 2 units from its initial location. That is, we can compute \(\vnorm{d} = 2\text{.}\) 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.

Definition12.1.19.Average Rate of Change.

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

\begin{equation*}
\text{ average rate of change } = \frac{\vec r(t_1) - \vec r(t_0)}{t_1-t_0}\text{.}
\end{equation*}

Example12.1.20.Average rate of change.

Let \(\vec r(t) = \la \cos(\frac{\pi}2t),\sin(\frac{\pi}2t)\ra\) as in Example 12.1.17. Find the average rate of change of \(\vec r(t)\) on \([-1,1]\) and on \([-1,5]\text{.}\)

Solution1.Video solution

Solution2.

We computed in Example 12.1.17 that the displacement of \(\vec r(t)\) on \([-1,1]\) was \(\vec d = \la 0,2\ra\text{.}\) Thus the average rate of change of \(\vec r(t)\) on \([-1,1]\) is:

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 \(\la 0,1\ra\) per unit of time.

We can quickly see that the displacement on \([-1,5]\) is the same as on \([-1,1]\text{,}\) so \(\vec d = \la 0,2\ra\text{.}\) The average rate of change is different, though:

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 \([-1,1]\text{.}\)

We considered average rates of change in Sections 1.1 and Section 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.

Exercises12.1.4Exercises

Terms and Concepts

1.

Vector-valued functions are closely related to of graphs.

2.

When sketching vector-valued functions, technically one isn’t graphing points, but rather .

3.

It can be useful to think of as a vector that points from a starting position to an ending position.

4.

In the context of vector-valued functions, average rate of change is divided by time.

Problems

Exercise Group.

In the following exercises, sketch the vector-valued function on the given interval.

5.

\(\vec r(t) = \la t^2,t^2-1\ra\text{,}\) for \(-2\leq t\leq 2\text{.}\)

6.

\(\vec r(t) = \la t^2,t^3\ra\text{,}\) for \(-2\leq t\leq 2\text{.}\)

7.

\(\vec r(t) = \la 1/t,1/t^2\ra\text{,}\) for \(-2\leq t\leq 2\text{.}\)

8.

\(\vec r(t) = \la \frac1{10}t^2,\sin(t) \ra\text{,}\) for \(-2\pi\leq t\leq 2\pi\text{.}\)

9.

\(\vec r(t) = \la \frac1{10}t^2,\sin(t) \ra\text{,}\) for \(-2\pi\leq t\leq 2\pi\text{.}\)

10.

\(\vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra\text{,}\) on \([0,2]\text{.}\)

11.

\(\vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra\text{,}\) on \([0,2\pi]\text{.}\)

12.

\(\vec r(t) = \la 2\sec(t) ,\tan(t) \ra\text{,}\) on \([-\pi,\pi]\text{.}\)

Exercise Group.

In the following exercises, sketch the vector-valued function on the given interval in \(\mathbb{R}^3\text{.}\) Technology may be useful in creating the sketch.

If \(\vec r(t) = \la 2\cos(t) ,2\sin(t) ,t\ra\text{,}\) then \(\norm{\vec r(t)}=\).

20.

\(\vec r(t) = \la \cos(t) ,t,t^2\ra\text{.}\)

Exercise Group.

Create a vector-valued function whose graph matches the given description.

21.

A circle of radius \(2\text{,}\) centered at \((1,2)\text{,}\) traced counter-clockwise once at constant speed on \([0,2\pi)\text{.}\)

22.

A circle of radius 3, centered at \((5,5)\text{,}\) traced clockwise once on \([0,2\pi]\text{.}\)

23.

An ellipse, centered at \((0,0)\) with vertical major axis of length 10 and minor axis of length 3, traced once counter-clockwise on \([0,2\pi]\text{.}\)

24.

An ellipse, centered at \((3,-2)\) with horizontal major axis of length 6 and minor axis of length 4, traced once clockwise on \([0,2\pi]\text{.}\)

25.

A line through \((2,3)\) with a slope of \(5\text{.}\)

26.

A line through \((1,5)\) with a slope of \(-1/2\text{.}\)

27.

The line through points \((1,2,3)\) and \((4,5,6)\text{,}\) where