
# 11.1: Vector–Valued Functions

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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

• 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/