
# 1.8: Vector-Valued Functions

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Now that we are familiar with vectors and their operations, we can begin discussing functions whose values are vectors.

Definition 1.10

A vector-valued function of a real variable is a rule that associates a vector $$\textbf{f}(t)$$ with a real number $$t$$, where $$t$$ is in some subset $$D$$ of $$\mathbb{R}^1$$ (called the domain of $$f$$). We write f: $$D →$$$$\mathbb{R}^ 3$$ to denote that f is a mapping of $$D$$ into $$\mathbb{R}^ 3$$ .

For example, $$\textbf{f}(t) = t\textbf{i}+ t^ 2 \textbf{j}+ t^ 3\textbf{k}$$ is a vector-valued function in $$\mathbb{R}^ 3$$ , defined for all real numbers $$t$$. We would write f : $$\mathbb{R} → \mathbb{R}^ 3$$ . At $$t = 1$$ the value of the function is the vector i + j + k, which in Cartesian coordinates has the terminal point $$(1,1,1)$$.

A vector-valued function of a real variable can be written in component form as

$\nonumber \textbf{f}(t) = f_1(t)\textbf{i}+ f_2(t)\textbf{j}+ f_3(t)\textbf{k}$

or in the form

$\nonumber \textbf{f}(t) = (f_1(t), f_2(t), f_3(t))$

for some real-valued functions $$f_1(t), f_2(t), f_3(t)$$, called the component functions of f. The first form is often used when emphasizing that $$\textbf{f}(t)$$ is a vector, and the second form is useful when considering just the terminal points of the vectors. By identifying vectors with their terminal points, a curve in space can be written as a vector-valued function.

Example 1.35

Define f: $$\mathbb{R} → \mathbb{R}^ 3$$ by $$\textbf{f}(t) = (\cos t,\sin t,t)$$. This is the equation of a helix (see Figure 1.8.1). As the value of $$t$$ increases, the terminal points of $$\textbf{f}(t)$$ trace out a curve spiraling upward. For each $$t$$, the $$x- \text{ and }y-$$coordinates of $$\textbf{f}(t) \text{ are }x = \cos t \text{ and }y = \sin t$$, so

$\nonumber x^2 + y^ 2 = \cos^2 t+\sin^2 t = 1$

Thus, the curve lies on the surface of the right circular cylinder $$x^ 2 + y^ 2 = 1$$.

Figure 1.8.1

It may help to think of vector-valued functions of a real variable in $$\mathbb{R}^ 3$$ as a generalization of the parametric functions in $$\mathbb{R}^ 2$$ which you learned about in single-variable calculus. Much of the theory of real-valued functions of a single real variable can be applied to vector-valued functions of a real variable. Since each of the three component functions are real-valued, it will sometimes be the case that results from single-variable calculus can simply be applied to each of the component functions to yield a similar result for the vector-valued function. However, there are times when such generalizations do not hold (see Exercise 13). The concept of a limit, though, can be extended naturally to vector-valued functions, as in the following definition.

Definition 1.11

Let $$\textbf{f}(t)$$ be a vector-valued function, let $$a$$ be a real number and let c be a vector. Then we say that the limit of $$\textbf{f}(t)$$ as $$t$$ approaches $$a$$ equals c, written as $$\lim\limits_{t \to a} \textbf{f}(t) = \textbf{c}$$, if $$\lim\limits_{t \to a} \norm{\textbf{f}(t)−\textbf{c}} = 0$$. If $$\textbf{f}(t) = (f_1(t), f_2(t), f_3(t))$$, then

$\nonumber \lim\limits_{t \to a}\textbf{f}(t) = \left ( \lim\limits_{t \to a}f_1(t),\,\lim\limits_{t \to a}f_2(t),\,\lim\limits_{t \to a}f_3(t)\right )$

provided that all three limits on the right side exist.

The above definition shows that continuity and the derivative of vector-valued functions can also be defined in terms of its component functions.

Definition 1.12

Let $$\textbf{f}(t) = (f_1(t), f_2(t), f_3(t))$$ be a vector-valued function, and let $$a$$ be a real number in its domain. Then $$\textbf{f}(t)$$ is continuous at $$a$$ if $$\lim\limits_{t \to a} \textbf{f}(t) = \textbf{f}(a)$$. Equivalently, $$\textbf{f}(t)$$ is continuous at $$a$$ if and only if $$f_1(t), f_2(t), \text{ and }f_3(t)$$ are continuous at $$a$$.

The derivative of $$\textbf{f}(t)$$ at $$a$$, denoted by $$\textbf{f}′(a)$$ or $$\dfrac{d\textbf{f}}{ dt} (a)$$, is the limit

$\nonumber \textbf{f} ′ (a) = \lim\limits_{h \to 0} \dfrac{\textbf{f}(a+ h)−\textbf{f}(a)}{ h}$

if that limit exists. Equivalently, $$\textbf{f} ′ (a) = (f_1 ′ (a), f_2 ′ (a), f_3 ′ (a))$$, if the component derivatives exist. We say that $$\textbf{f}(t)$$ is differentiable at $$a$$ if $$\textbf{f} ′ (a)$$ exists.

Recall that the derivative of a real-valued function of a single variable is a real number, representing the slope of the tangent line to the graph of the function at a point. Similarly, the derivative of a vector-valued function is a tangent vector to the curve in space which the function represents, and it lies on the tangent line to the curve (see Figure 1.8.2).

Figure 1.8.2 Tangent vector $$\textbf{f} ′ (a)$$ and tangent line $$L = \textbf{f}(a)+ s\textbf{f} ′ (a)$$

Example 1.36

Let $$\textbf{f}(t) = (\cos t,\sin t,t)$$. Then $$\textbf{f} ′ (t) = (−\sin t,\cos t,1) \text{ for all }t$$. The tangent line $$L$$ to the curve at $$\textbf{f}(2π) = (1,0,2π) \text{ is }L = \textbf{f}(2π)+ s \textbf{f} ′ (2π) = (1,0,2π)+ s(0,1,1)$$, or in parametric form: $$x = 1, y = s, z = 2π+ s \text{ for }−∞ < s < ∞$$.

A scalar function is a real-valued function. Note that if $$u(t)$$ is a scalar function and $$\textbf{f}(t)$$ is a vector-valued function, then their product, defined by ($$u\textbf{f})(t) = u(t)\textbf{f}(t)$$ for all $$t$$, is a vector-valued function (since the product of a scalar with a vector is a vector).

The basic properties of derivatives of vector-valued functions are summarized in the following theorem.

Theorem 1.20

Let $$\textbf{f}(t) \text{ and }\textbf{g}(t)$$ be differentiable vector-valued functions, let $$u(t)$$ be a differentiable scalar function, let $$k$$ be a scalar, and let c be a constant vector. Then

1. $$\dfrac{d}{ dt} (\textbf{c}) = \textbf{0}$$
2. $$\dfrac{d}{ dt} (k\textbf{f}) = k \dfrac{d\textbf{f}}{ dt}$$
3. $$\dfrac{d}{ dt} (\textbf{f}+\textbf{g}) = \dfrac{d\textbf{f}}{ dt} + \dfrac{d\textbf{g}}{ dt}$$
4. $$\dfrac{d}{ dt} (\textbf{f−g}) = \dfrac{d\textbf{f}}{ dt} − \dfrac{d\textbf{g}}{ dt}$$
5. $$\dfrac{d}{ dt} (u\textbf{f}) = \dfrac{du}{ dt} \textbf{f} + u \dfrac{d\textbf{f}}{ dt}$$
6. $$\dfrac{d}{ dt} (\textbf{f}· \textbf{g}) = \dfrac{d\textbf{f}}{ dt} · \textbf{g} + \textbf{f}· \dfrac{d\textbf{g}}{ dt}$$
7. $$\dfrac{d}{ dt} (\textbf{f} × \textbf{g}) = \dfrac{d\textbf{f}}{ dt} × \textbf{g} + \textbf{f} × \dfrac{d\textbf{g}}{ dt}$$

Proof

The proofs of parts (a)-(e) follow easily by differentiating the component functions and using the rules for derivatives from single-variable calculus. We will prove part (f), and leave the proof of part (g) as an exercise for the reader.

(f) Write $$\textbf{f}(t) = (f_1(t), f_2(t), f_3(t)) \text{ and }\textbf{g}(t) = (g_1(t), g_2(t), g_3(t))$$, where the component functions $$f_1(t), f_2(t), f_3(t), g_1(t), g_2(t), g_3(t)$$ are all differentiable real-valued functions. Then

\nonumber \begin{align}\dfrac{d}{dt} (\textbf{f}(t)· \textbf{g}(t))&=\dfrac{d}{ dt} (f_1(t) g_1(t)+ f_2(t) g_2(t)+ f_3(t) g_3(t)) \\ \nonumber &= \dfrac{d}{ dt} (f_1(t) g_1(t))+ \dfrac{d}{ dt} (f_2(t) g_2(t))+ \dfrac{d}{ dt} (f_3(t) g_3(t)) \\ \nonumber &=\dfrac{d f_1}{ dt} (t) g_1(t)+ f_1(t)\dfrac{ d g_1}{ dt} (t)+ \dfrac{d f_2}{ dt} (t) g_2(t)+ f_2(t) \dfrac{d g_2}{ dt} (t)+ \dfrac{d f_3}{ dt} (t) g_3(t)+ f_3(t) \dfrac{d g_3}{ dt} (t) \\ \nonumber &=\left ( \dfrac{d f_1}{ dt} (t),\dfrac{d f_2}{ dt} (t), \dfrac{d f_3}{ dt} (t) \right ) · (g_1(t), g_2(t), g_3(t)) \\ \nonumber &\quad +(f_1(t), f_2(t), f_3(t))· \left ( \dfrac{d g_1}{ dt} (t), \dfrac{d g_2}{ dt} (t), \dfrac{d g_3}{ dt} (t)\right ) \\ \nonumber &=\dfrac{d\textbf{f}}{ dt} (t)· \textbf{g}(t) + \textbf{f}(t)· \dfrac{d\textbf{g}}{ dt} (t) \text{ for all }t.\tag{$$\textbf{QED}$$} \\ \end{align}

$$\square$$

Example 1.37

Suppose $$\textbf{f}(t)$$ is differentiable. Find the derivative of $$\norm{\textbf{f}(t)}$$.

Solution

Since $$\norm{\textbf{f}(t)}$$ is a real-valued function of $$t$$, then by the Chain Rule for real-valued functions, we know that $$\dfrac{d}{ dt} \norm{\textbf{f}(t)}^2 = 2\norm{\textbf{f}(t)} \dfrac{d}{ dt} \norm{\textbf{f}(t)}$$.

But $$\norm{\textbf{f}(t)}^ 2 = \textbf{f}(t)·\textbf{f}(t)$$, so $$\dfrac{d}{ dt} \norm{\textbf{f}(t)}^2 = \dfrac{d}{ dt} (\textbf{f}(t)·\textbf{f}(t))$$. Hence, we have

\nonumber \begin{align} 2\norm{\textbf{f}(t)} \dfrac{d}{ dt} \norm{\textbf{f}(t)} &= \dfrac{d}{ dt} (\textbf{f}(t)·\textbf{f}(t)) = \textbf{f} ′ (t)·\textbf{f}(t) + \textbf{f}(t)·\textbf{f} ′ (t)\text{ by Theorem 1.20(f), so} \\ \nonumber &= 2\textbf{f} ′ (t)·\textbf{f}(t) , \text{ so if }\norm{\textbf{f}(t)} \neq 0 \text{ then} \\ \nonumber \dfrac{d}{ dt} \norm{\textbf{f}(t)} &= \dfrac{\textbf{f} ′ (t)·\textbf{f}(t)}{ \norm{\textbf{f}(t)}} \\ \end{align}

We know that $$\norm{\textbf{f}(t)}$$ is constant if and only if $$\dfrac{d}{ dt} \norm{\textbf{f}(t)} = 0$$ for all $$t$$. Also, $$\textbf{f}(t) ⊥ \textbf{f} ′ (t) \text{ if and only if }\textbf{f} ′ (t)·\textbf{f}(t) = 0$$. Thus, the above example shows this important fact:

If $$\norm{\textbf{f}(t)} \neq 0, \text{ then }\norm{\textbf{f}(t)} \text{ is constant if and only if }\textbf{f}(t) ⊥ \textbf{f} ′ (t)\text{ for all }t$$.

This means that if a curve lies completely on a sphere (or circle) centered at the origin, then the tangent vector $$\textbf{f} ′ (t)$$ is always perpendicular to the position vector $$\textbf{f}(t)$$.

Example 1.38

The spherical spiral $$\textbf{f}(t) = \left ( \dfrac{\cos t}{ \sqrt{ 1+ a^ 2t^ 2}} , \dfrac{\sin t}{ \sqrt{ 1+ a^ 2t^ 2}} , \dfrac{−at}{ \sqrt{ 1+ a^ 2t^ 2}} \right )$$ , for $$a \neq 0$$.

Figure 1.8.3 shows the graph of the curve when $$a = 0.2$$. In the exercises, the reader will be asked to show that this curve lies on the sphere $$x^ 2 + y^ 2 + z^ 2 = 1$$ and to verify directly that $$\textbf{f} ′ (t)·\textbf{f}(t) = 0 \text{ for all }t$$.

Figure 1.8.3: Spherical spiral with $$a = 0.2$$

Just as in single-variable calculus, higher-order derivatives of vector-valued functions are obtained by repeatedly differentiating the (first) derivative of the function:

$\nonumber \textbf{f} ′′(t) = \dfrac{d}{ dt} \textbf{f} ′ (t) ,\qquad \textbf{f} ′′′(t) = \dfrac{d}{ dt} \textbf{f} ′′(t) , ... , \dfrac{d^ n \textbf{f}}{ dt^n} = \dfrac{d}{ dt} \left ( \dfrac{d^{n−1}\textbf{f}}{ dt^{n−1}} \right ) \text{ (for $$n = 2,3,4,...$$)}$

We can use vector-valued functions to represent physical quantities, such as velocity, acceleration, force, momentum, etc. For example, let the real variable $$t$$ represent time elapsed from some initial time ($$t = 0$$), and suppose that an object of constant mass $$m$$ is subjected to some force so that it moves in space, with its position ($$x, y, z$$) at time $$t$$ a function of $$t$$. That is, $$x = x(t), y = y(t), z = z(t)$$ for some real-valued functions $$x(t), y(t), z(t)$$. Call $$\textbf{r}(t) = (x(t), y(t), z(t))$$ the position vector of the object. We can define various physical quantities associated with the object as follows:

\nonumber \begin{align}\textit{position}:\, \textbf{r}(t) &= (x(t), y(t), z(t)) \\ \nonumber\textit{velocity}:\, \textbf{v}(t) &= \dot{\textbf{r}}(t) = \textbf{r} ′ (t) = \dfrac{d\textbf{r}}{ dt} \\ \nonumber &= (x ′ (t), y ′ (t), z ′ (t)) \\ \nonumber \textit{acceleration}:\, \textbf{a}(t) &= \dot{\textbf{v}}(t) = \textbf{v} ′ (t) = \dfrac{d\textbf{v}}{ dt} \\ \nonumber &=\ddot{\textbf{r}}(t) = \textbf{r} ′′(t) = \dfrac{d^ 2 \textbf{r}}{ dt^2} \\ \nonumber &=(x ′′(t), y ′′(t), z ′′(t)) \\ \nonumber \textit{momentum}:\, \textbf{p}(t) &= m\textbf{v}(t) \\ \nonumber \textit{force}:\, \textbf{F}(t) &= \dot{\textbf{p}} (t) = \textbf{p} ′ (t) = \dfrac{d\textbf{p}}{ dt} \text{ (Newton’s Second Law of Motion)} \\ \end{align}

The magnitude $$\norm{\textbf{v}(t)}$$ of the velocity vector is called the speed of the object. Note that since the mass $$m$$ is a constant, the force equation becomes the familiar $$\textbf{F}(t) = m\textbf{a}(t)$$.

Example 1.39

Let $$\textbf{r}(t) = (5\cos t,3\sin t,4\sin t)$$ be the position vector of an object at time $$t ≥ 0$$. Find its (a) velocity and (b) acceleration vectors.

Solution

1. $$\textbf{v}(t) = \dot{\textbf{r}}(t) = (−5\sin t,3\cos t,4\cos t)$$
2. $$\textbf{a}(t) = \dot{\textbf{v}}(t) = (−5\cos t,−3\sin t,−4\sin t)$$

Note that $$\norm{\textbf{r}(t)} = \sqrt{ 25\cos^2 t+25\sin^2 t} = 5$$ for all $$t$$, so by Example 1.37 we know that $$\textbf{r}(t)· \dot{\textbf{r}}(t) = 0 \text{ for all }t$$ (which we can verify from part (a)). In fact, $$\norm{\textbf{v}(t)} = 5$$ for all $$t$$ also. And not only does $$\textbf{r}(t)$$ lie on the sphere of radius 5 centered at the origin, but perhaps not so obvious is that it lies completely within a circle of radius 5 centered at the origin. Also, note that $$\textbf{a}(t) = −\textbf{r}(t)$$. It turns out (see Exercise 16) that whenever an object moves in a circle with constant speed, the acceleration vector will point in the opposite direction of the position vector (i.e. towards the center of the circle).

Recall from Section 1.5 that if $$\textbf{r}_1 , \textbf{r}_2$$ are position vectors to distinct points then $$\textbf{r}_1+t(\textbf{r}_2−\textbf{r}_1)$$ represents a line through those two points as $$t$$ varies over all real numbers. That vector sum can be written as $$(1 − t)\textbf{r}_1 + t\textbf{r}_2$$ . So the function $$\textbf{l}(t) = (1 − t)\textbf{r}_1 + t\textbf{r}_2$$ is a line through the terminal points of $$\textbf{r}_1 \text{ and }\textbf{r}_2$$ , and when $$t$$ is restricted to the interval $$[0,1]$$ it is the line segment between the points, with $$\textbf{l}(0) = \textbf{r}_1 \text{ and }\textbf{l}(1) = \textbf{r}_2$$ .

In general, a function of the form $$\textbf{f}(t) = (a_1 t+b_1 ,a_2 t+b_2 ,a_3 t+b_3)$$ represents a line in $$\mathbb{R}^ 3$$ . A function of the form $$\textbf{f}(t) = (a_1 t^ 2 + b_1 t + c_1 ,a_2 t^ 2 + b_2 t + c_2 ,a_3 t^ 2 + b_3 t + c_3)$$ represents a (possibly degenerate) parabola in $$\mathbb{R}^ 3$$ .

Example 1.40: Bézier curves

Bézier curves are used in Computer Aided Design (CAD) to approximate the shape of a polygonal path in space (called the Bézier polygon or control polygon). For instance, given three points (or position vectors) $$\textbf{b}_0 , \textbf{b}_1 , \textbf{b}_2 \text{ in }$$$$\mathbb{R}^ 3$$ , define

\nonumber \begin{align} \textbf{b}_0^1 (t) &= (1− t)\textbf{b}_0 + t\textbf{b}_1 \\ \nonumber \textbf{b}_1^1 (t) &= (1− t)\textbf{b}_1 + t\textbf{b}_2 \\ \nonumber \textbf{b}_0^2 (t) &= (1− t)\textbf{b}_0^1 (t)+ t\textbf{b}_1^1 (t) \\ \nonumber &=(1− t)^2 \textbf{b}_0 +2t(1− t)\textbf{b}_1 + t^2\textbf{b}_2 \\ \end{align}

for all real $$t$$. For $$t$$ in the interval $$[0,1]$$, we see that $$\textbf{b}_0^1 (t)$$ is the line segment between $$\textbf{b}_0 \text{ and }\textbf{b}_1$$ , and $$\textbf{b}_1^1 (t)$$ is the line segment between $$\textbf{b}_1 \text{ and }\textbf{b}_2$$ . The function $$\textbf{b}_0^2 (t)$$ is the Bézier curve for the points $$\textbf{b}_0 , \textbf{b}_1 , \textbf{b}_2$$ . Note from the last formula that the curve is a parabola that goes through $$\textbf{b}_0$$ (when $$t = 0$$) and $$\textbf{b}_2$$ (when $$t = 1$$).

As an example, let $$\textbf{b}_0 = (0,0,0),\, \textbf{b}_1 = (1,2,3), \text{ and }\textbf{b}_2 = (4,5,2)$$. Then the explicit formula for the Bézier curve is $$\textbf{b}_0^2 (t) = (2t +2t^ 2 ,4t + t^ 2 ,6t −4t^ 2 )$$, as shown in Figure 1.8.4, where the line segments are $$\textbf{b}_0^1 (t)\text{ and }\textbf{b}_1^1 (t)$$, and the curve is $$\textbf{b}_0^2 (t)$$.

Figure 1.8.4 Bézier curve approximation for three points

In general, the polygonal path determined by $$n ≥ 3$$ noncollinear points in $$\mathbb{R}^ 3$$ can be used to define the Bézier curve recursively by a process called repeated linear interpolation. This curve will be a vector-valued function whose components are polynomials of degree $$n − 1$$, and its formula is given by de Casteljau’s algorithm. In the exercises, the reader will be given the algorithm for the case of $$n = 4$$ points and asked to write the explicit formula for the Bézier curve for the four points shown in Figure 1.8.5.

Figure 1.8.5 Bézier curve approximation for four points