APEX Calculus and Vector-Valued Functions

Section 13.2 Calculus and Vector-Valued Functions

The previous section introduced us to a new mathematical object, the vector-valued function. We now apply calculus concepts to these functions. We start with the limit, then work our way through derivatives to integrals.

Subsection 13.2.1 Limits of Vector-Valued Functions

The initial definition of the limit of a vector-valued function is a bit intimidating, as was the definition of the limit in Definition 1.2.2. The theorem following the definition shows that in practice, taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions.

Definition 13.2.1. Limits of Vector-Valued Functions.

Let $I$ be an open interval containing $c\text{,}$ and let $\vec r(t)$ be a vector-valued function defined on $I\text{,}$ except possibly at $c\text{.}$ The limit of $\vec r(t)\text{,}$ as $t$ approaches $c\text{,}$ is $\vec L$, expressed as

\begin{equation*} \lim_{t\to c} \vec r(t) = \vec L\text{,} \end{equation*}

means that given any $\varepsilon \gt 0\text{,}$ there exists a $\delta \gt 0$ such that for all $t\neq c\text{,}$ if $\abs{t-c} \lt \delta\text{,}$ we have $\norm{\vec r(t) - \vec L} \lt \varepsilon\text{.}$

Note how the measurement of distance between real numbers is the absolute value of their difference; the measure of distance between vectors is the vector norm, or magnitude, of their difference.

Theorem 13.2.2 states that we can compute limits of vector-valued functions component-wise.

Theorem 13.2.2. Limits of Vector-Valued Functions.

Let $\vec r(t) = \la \,f(t),g(t)\,\ra$ be a vector-valued function in $\mathbb{R}^2$ defined on an open interval $I$ containing $c\text{,}$ except possibly at $c\text{.}$ Then

\begin{equation*} \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\ra\text{.} \end{equation*}
Let $\vec r(t) = \la \,f(t),g(t),h(t)\,\ra$ be a vector-valued function in $\mathbb{R}^3$ defined on an open interval $I$ containing $c\text{,}$ except possibly at $c\text{.}$ Then

\begin{equation*} \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\,, \,\lim_{t\to c} h(t)\ra \end{equation*}

Figure 13.2.3. Video presentation of Definition 13.2.1 and Theorem 13.2.2

Example 13.2.4. Finding limits of vector-valued functions.

Let $\ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra\text{.}$ Find $\lim\limits_{t\to 0}\vec r(t)\text{.}$

Solution 1.

We apply the theorem and compute limits component-wise.

\begin{align*} \lim_{t\to0} \vec r(t) \amp = \la \lim_{t\to 0}\frac{\sin(t) }{t}\, , \, \lim_{t\to 0} t^2-3t+3\, , \, \lim_{t\to 0} \cos(t) \ra\\ \amp = \la 1,3,1\ra\text{.} \end{align*}

Solution 2. Video solution

Subsection 13.2.2 Continuity

Definition 13.2.5. Continuity of Vector-Valued Functions.

Let $\vec r(t)$ be a vector-valued function defined on an open interval $I$ containing $c\text{.}$

$\vec r(t)$ is continuous at $c$ if $\lim\limits_{t\to c} \vec r(t) = r(c)\text{.}$
If $\vec r(t)$ is continuous at all $c$ in $I\text{,}$ then $\vec r(t)$ is continuous on $I\text{.}$

We again have a theorem that lets us evaluate continuity component-wise.

Theorem 13.2.6. Continuity of Vector-Valued Functions.

Let $\vec r(t)$ be a vector-valued function defined on an open interval $I$ containing $c\text{.}$ Then $\vec r(t)$ is continuous at $c$ if, and only if, each of its component functions is continuous at $c\text{.}$

Figure 13.2.7. Video presentation of Subsection 13.2.2

Example 13.2.8. Evaluating continuity of vector-valued functions.

Let $\ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra\text{.}$ Determine whether $\vec r$ is continuous at $t=0$ and $t=1\text{.}$

Solution.

While the second and third components of $\vec r(t)$ are defined at $t=0\text{,}$ the first component, $(\sin(t) )/t\text{,}$ is not. Since the first component is not even defined at $t=0\text{,}$ $\vec r(t)$ is not defined at $t=0\text{,}$ and hence it is not continuous at $t=0\text{.}$

At $t=1$ each of the component functions is continuous. Therefore $\vec r(t)$ is continuous at $t=1\text{.}$

Subsection 13.2.3 Derivatives

Consider a vector-valued function $\vec r$ defined on an open interval $I$ containing $t_0$ and $t_1\text{.}$ We can compute the displacement of $\vec r$ on $[t_0,t_1]\text{,}$ as shown in Figure 13.2.9.(a). Recall that dividing the displacement vector by $t_1-t_0$ gives the average rate of change on $[t_0,t_1]\text{,}$ as shown in Figure 13.2.9.(b).

Illustration of a vector-valued function on a given interval. — (a)

Illustration of a vector-valued function on a given interval showcasing a derivative vector. — (a)

The derivative of a vector-valued function is a measure of the instantaneous rate of change, measured by taking the limit as the length of $[t_0,t_1]$ goes to 0. Instead of thinking of an interval as $[t_0,t_1]\text{,}$ we think of it as $[c,c+h]$ for some value of $h$ (hence the interval has length $h$). The average rate of change is

\begin{equation*} \frac{\vec r(c+h)-\vec r(c)}{h} \end{equation*}

for any value of $h\neq0\text{.}$ We take the limit as $h\to0$ to measure the instantaneous rate of change; this is the derivative of $\vec r\text{.}$

Definition 13.2.10. Derivative of a Vector-Valued Function.

Let $\vec r(t)$ be continuous on an open interval $I$ containing $c\text{.}$

The derivative of $\vec r$ at $t=c$ is

\begin{equation*} \vrp (c) = \lim_{h\to 0} \frac{\vec r(c+h) - \vec r(c)}{h}\text{.} \end{equation*}
The derivative of $\vec r$ is

\begin{equation*} \vrp (t) = \lim_{h\to 0} \frac{\vec r(t+h) - \vec r(t)}{h}\text{.} \end{equation*}

If a vector-valued function has a derivative for all $c$ in an open interval $I\text{,}$ we say that $\vec r(t)$ is differentiable on $I\text{.}$

Once again we might view this definition as intimidating, but recall that we can evaluate limits component-wise. The following theorem verifies that this means we can compute derivatives component-wise as well, making the task not too difficult.

Theorem 13.2.11. Derivatives of Vector-Valued Functions.

Let $\vec r(t) = \la \, f(t), g(t)\,\ra\text{.}$ Then

\begin{equation*} \vrp(t) = \la\, \fp(t), g'(t)\, \ra\text{.} \end{equation*}
Let $\vec r(t) = \la \, f(t), g(t), h(t)\,\ra\text{.}$ Then

\begin{equation*} \vrp(t) = \la\, \fp(t), g'(t), h'(t)\, \ra\text{.} \end{equation*}

Figure 13.2.12. Video presentation of Definition 13.2.10 and Theorem 13.2.11

Example 13.2.13. Derivatives of vector-valued functions.

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

Sketch $\vec r(t)$ and $\vrp(t)$ on the same axes.
Compute $\vrp(1)$ and sketch this vector with its initial point at the origin and at $\vec r(1)\text{.}$

Solution 1.

Theorem 13.2.11 allows us to compute derivatives component-wise, so

\begin{equation*} \vrp(t) = \la 2t, 1\ra\text{.} \end{equation*}

$\vec r(t)$ and $\vrp(t)$ are graphed together in Figure 13.2.14.(a). Note how plotting the two of these together, in this way, is not very illuminating. When dealing with real-valued functions, plotting $f(x)$ with $\fp(x)$ gave us useful information as we were able to compare $f$ and $\fp$ at the same $x$-values. When dealing with vector-valued functions, it is hard to tell which points on the graph of $\vrp$ correspond to which points on the graph of $\vec r\text{.}$
We easily compute $\vrp(1) = \la 2,1\ra\text{,}$ which is drawn in Figure 13.2.14 with its initial point at the origin, as well as at $\vec r(1) = \la 1,1\ra\text{.}$ These are sketched in Figure 13.2.14.(b).

$Graph of the vector-valued function from the example and its derivative.$
Graph of the vector-valued function $\vec r(t) = \la t^2,t\ra\text{.}$ The graph of the function $\vec r(t) $ is simply the graph of the parabola $y=x^2\text{,}$ but instead of opening towards the positive $y$-axis, the function $\vec r(t) $ opens towards the positive $x$-axis. The function $\vec r(t) $ is plotted on the interval $[-2,2]\text{,}$ where $\vec r(-2)= \la 4,-2\ra$ and $\vec r(2)= \la 4,2\ra\text{.}$ The graph also includes the derivative function $\vrp(t) = \la 2t, 1\ra $ which takes the path of the horizontal line $y=1$ going from left to right in the standard coordinate axes.
(a)

$Graph of the vector-valued function from the example and its derivative.$
Graph of the vector-valued function $\vec r(t) = \la t^2,t\ra$ described in the previous image. The graph also includes two copies of the vector $\vrp(1) = \la 2,1\ra\text{.}$ The first copy of the vector $\vrp(1) = \la 2,1\ra$ begins at the origin, and ends at the point $(2,1)\text{,}$ which is also a point on the derivative function $\vrp(t) = \la 2t, 1\ra $ from the previous image. The second copy of the vector $\vrp(1) = \la 2,1\ra$ begins at the point $(1,1)\text{,}$ which corresponds to the termination point of $\vec r(1) = \la 1,1 \ra\text{.}$ The second copy of the vector $\vrp(1) = \la 2,1\ra$ is tangent to the function $\vec r(t) = \la t^2,t\ra$ at the point $(1,1)$ corresponding to when $t=1$ in the function $\vec r(t)\text{.}$
(b)

Figure 13.2.14. Graphing the derivative of a vector-valued function in Example 13.2.13

Solution 2. Video solution

Example 13.2.15. Derivatives of vector-valued functions.

Let $\vec r(t) = \la \cos(t) , \sin(t) , t\ra\text{.}$ Compute $\vrp(t)$ and $\vrp(\pi/2)\text{.}$ Sketch $\vrp(\pi/2)$ with its initial point at the origin and at $\vec r(\pi/2)\text{.}$

Solution.

We compute $\vrp$ as $\vrp(t) = \la -\sin(t) , \cos(t) , 1\ra\text{.}$ At $t= \pi/2\text{,}$ we have $\vrp(\pi/2) = \la -1,0,1\ra\text{.}$ Figure 13.2.16 shows a graph of $\vec r(t)\text{,}$ with $\vrp(\pi/2)$ plotted with its initial point at the origin and at $\vec r(\pi/2)\text{.}$

Link to full-sized image

Figure 13.2.16. Viewing a vector-valued function and its derivative at one point

In Examples 13.2.13 and 13.2.15, sketching a particular derivative with its initial point at the origin did not seem to reveal anything significant. However, when we sketched the vector with its initial point on the corresponding point on the graph, we did see something significant: the vector appeared to be tangent to the graph. We have not yet defined what “tangent” means in terms of curves in space; in fact, we use the derivative to define this term.

Definition 13.2.17. Tangent Vector, Tangent Line.

Let $\vec r(t)$ be a differentiable vector-valued function on an open interval $I$ containing $c\text{,}$ where $\vrp(c)\neq \vec 0\text{.}$

A vector $\vec v$ is tangent to the graph of $\vec r(t)$ at $t=c$ if $\vec v$ is parallel to $\vrp(c)\text{.}$
The tangent line to the graph of $\vec r(t)$ at $t=c$ is the line through $\vec r(c)$ with direction parallel to $\vrp(c)\text{.}$ An equation of the tangent line is

\begin{equation*} \vec \ell(t) = \vec r(c) + t\,\vrp(c)\text{.} \end{equation*}

Figure 13.2.18. Video presentation of Definition 13.2.17

Example 13.2.19. Finding tangent lines to curves in space.

Let $\vec r(t) = \la t,t^2,t^3\ra$ on $[-1.5,1.5]\text{.}$ Find the vector equation of the line tangent to the graph of $\vec r$ at $t=-1\text{.}$

Solution 1. Video solution

Solution 2.

To find the equation of a line, we need a point on the line and the line’s direction. The point is given by $\vec r(-1) = \la -1,1,-1\ra\text{.}$ (To be clear, $\la -1,1,-1\ra$ is a vector, not a point, but we use the point “pointed to” by this vector.)

The direction comes from $\vrp(-1)\text{.}$ We compute, component-wise, $\vrp(t) = \la 1,2t, 3t^2\ra\text{.}$ Thus $\vrp(-1) = \la 1,-2,3\ra\text{.}$

Link to full-sized image

Figure 13.2.20. Graphing a curve in space with its tangent line

The vector equation of the line is $\ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra\text{.}$ This line and $\vec r(t)$ are sketched in Figure 13.2.20.

Example 13.2.21. Finding tangent lines to curves.

Find the equations of the lines tangent to $\vec r(t) = \la t^3,t^2\ra$ at $t=-1$ and $t=0\text{.}$

Solution 1.

We find that $\vrp(t) = \la 3t^2,2t\ra\text{.}$ At $t=-1\text{,}$ we have

\begin{equation*} \vec r(-1) = \la -1,1\ra \text{ and } \vrp(-1) = \la 3,-2\ra\text{,} \end{equation*}

so the equation of the line tangent to the graph of $\vec r(t)$ at $t=-1$ is

\begin{equation*} \ell(t) = \la -1,1\ra + t\la 3,-2\ra\text{.} \end{equation*}

This line is graphed with $\vec r(t)$ in Figure 13.2.22.

Graph of the vector-valued function from the example and the tangent line to a point on the curve. — Figure 13.2.22. Graphing $\vec r(t)$ and its tangent line in Example 13.2.21

At $t=0\text{,}$ we have $\vrp(0) = \la 0,0\ra=\vec 0\text{!}$ This implies that the tangent line “has no direction.” We cannot apply Definition 13.2.17, hence cannot find the equation of the tangent line.

Solution 2. Video solution

We were unable to compute the equation of the tangent line to $\vec r(t)= \la t^3,t^2\ra$ at $t=0$ because $\vrp(0) = \vec 0\text{.}$ The graph in Figure 13.2.22 shows that there is a cusp at this point. This leads us to another definition of smooth, previously defined by Definition 10.2.22 in Section 10.2.

Definition 13.2.23. Smooth Vector-Valued Functions.

Let $\vec r(t)$ be a differentiable vector-valued function on an open interval $I$ where $\vrp(t)$ is continuous on $I\text{.}$ $\vec r(t)$ is smooth on $I$ if $\vrp(t)\neq \vec 0$ on $I\text{.}$

Figure 13.2.24. Video presentation of Definition 13.2.23

Having established derivatives of vector-valued functions, we now explore the relationships between the derivative and other vector operations. The following theorem states how the derivative interacts with vector addition and the various vector products.

Theorem 13.2.25. Properties of Derivatives of Vector-Valued Functions.

Let $\vec r$ and $\vec s$ be differentiable vector-valued functions, let $f$ be a differentiable real-valued function, and let $c$ be a real number.

$\displaystyle \ds \frac{d}{dt}\Big(\vec r(t) \pm \vec s(t)\Big) = \vrp(t) \pm \vec s\,'(t)$
$\displaystyle \ds \frac{d}{dt}\Big(c\vec r(t)\Big) = c\vrp(t)$
$\ds \frac{d}{dt}\Big(f(t)\vec r(t)\Big) = \fp(t)\vec r(t) + f(t)\vrp(t)$ Product Rule
$\ds \frac{d}{dt}\Big(\vec r(t)\cdot \vec s(t) \Big) = \vrp(t)\cdot \vec s(t) + \vec r(t)\cdot \vec s\,'(t)$ Product Rule
$\ds \frac{d}{dt}\Big(\vec r(t)\times \vec s(t) \Big) = \vrp(t)\times \vec s(t) + \vec r(t)\times \vec s\,'(t)$ Product Rule
$\ds \frac{d}{dt}\Big(\vec r\big(f(t)\big)\Big) = \vrp\big(f(t)\big)\fp(t)$ Chain Rule

Figure 13.2.26. Video presentation of Theorem 13.2.25

Example 13.2.27. Using derivative properties of vector-valued functions.

Let $\vec r(t) = \la t, t^2-1\ra$ and let $\vec u(t)$ be the unit vector that points in the direction of $\vec r(t)\text{.}$

Graph $\vec r(t)$ and $\vec u(t)$ on the same axes, on $[-2,2]\text{.}$
Find $\vec u\,'(t)$ and sketch $\vec u\,'(-2)\text{,}$ $\vec u\,'(-1)$ and $\vec u\,'(0)\text{.}$ Sketch each with initial point the corresponding point on the graph of $\vec u\text{.}$

Solution 1.

To form the unit vector that points in the direction of $\vec r\text{,}$ we need to divide $\vec r(t)$ by its magnitude.

\begin{equation*} \norm{\vec r(t)} = \sqrt{t^2+(t^2-1)^2} \Rightarrow \vec u(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la t,t^2-1\ra\text{.} \end{equation*}

$\vec r(t)$ and $\vec u(t)$ are graphed in Figure 13.2.28. Note how the graph of $\vec u(t)$ forms part of a circle; this must be the case, as the length of $\vec u(t)$ is 1 for all $t\text{.}$

$Graph of the vector-valued function with the unit vector function which points in the direction of the function.$
Graph of the vector-valued function $\vec r(t) = \la t,t^2 -1\ra$ on $[-2,2]\text{.}$ The function looks like the parabola $y=x^2 -1\text{,}$ which takes the path going towards positive $x\text{.}$ The function begins at the point $(-2,3)\text{,}$ decreases until reaching the point $(0,-1)\text{,}$ and then increases until ending at the point $(2,3)\text{.}$ The graph also contains the function $\vec u(t)$ which is the unit vector that points in the direction of $\vec r(t)\text{.}$ The function $\vec u(t)$ looks like the unit circle which is missing a piece of the top. The circular arc from the graph of $\vec u(t)$ begins at the point where the vector $\vec r(-2)=\la -2,3\ra $ crosses the unit circle. The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector$\vec r(2)=\la 2,3\ra $ crosses the unit circle.

Figure 13.2.28. Graphing $\vec r(t)$ and $\vec u(t)$ in Example 13.2.27
To compute $\vec u\,'(t)\text{,}$ we use Theorem 13.2.25, writing

\begin{equation*} \vec u(t) = f(t)\vec r(t), \text{ where } f(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}=\big(t^2+(t^2-1)^2\big)^{-1/2}\text{.} \end{equation*}

(We could write

\begin{equation*} \vec u(t) = \la \frac{t}{\sqrt{t^2+(t^2-1)^2}}, \frac{t^2-1}{\sqrt{t^2+(t^2-1)^2}}\ra \end{equation*}

and then take the derivative. It is a matter of preference; this latter method requires two applications of the Quotient Rule where our method uses the Product and Chain Rules.) We find $\fp(t)$ using the Chain Rule:

\begin{align*} \fp(t) \amp = -\frac12\big(t^2+(t^2-1)^2\big)^{-3/2}\big(2t+2(t^2-1)(2t)\big)\\ \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3} \end{align*}

We now find $\vec u\,'(t)$ using part 3 of Theorem 13.2.25:

\begin{align*} \vec u\,'(t) \amp = \fp(t)\vec u(t) + f(t)\vec u\,'(t)\\ \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3}\la t,t^2-1\ra + \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la 1,2t\ra\text{.} \end{align*}

This is admittedly very “messy;” such is usually the case when we deal with unit vectors. We can use this formula to compute $\vec u\,'(-2)\text{,}$ $\vec u\,'(-1)$ and $\vec u\,'(0)\text{:}$

\begin{align*} \vec u\,'(-2) \amp = \la-\frac{15}{13 \sqrt{13}},-\frac{10}{13 \sqrt{13}}\ra \approx \la -0.320,-0.213\ra\\ \vec u\,'(-1) \amp = \la 0,-2\ra\\ \vec u\,'(0) \amp = \la 1,0\ra \end{align*}

$Graph of the circular unit vector function along with three derivative vectors of the function.$
Graph of the function $\vec u(t)$ which is the unit vector that points in the direction of $\vec r(t)\text{.}$ The graph also contains three unit vectors, $\vec u\,'(-2)\text{,}$ $\vec u\,'(-1)$ and $\vec u\,'(0)\text{.}$ The vector $\vec u\,'(-2)\approx \la -0.320,-0.213\ra$ begins at the starting point of the function $\vec u(t)$ and points in the direction tangent to the circular arc at $t=-2\text{.}$ The vector $\vec u\,'(-1)= \la 0,-2\ra$ begins at the point $(-1,0)$ corresponding to the termination point of $\vec u(-1)$ From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc. The vector $\vec u\,'(0)= \la 1,0\ra$ begins at the point $(0,-1)$ corresponding to the termination point of $\vec u(0)\text{.}$ From here the vector points to the right, as it is tangent to the bottom point of the circular arc. If the vectors $\vec u\,'(-2)\text{,}$ $\vec u\,'(-1)$ and $\vec u\,'(0)$ were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by $\vec u(t)\text{.}$

Figure 13.2.29. Graphing some of the derivatives of $\vec u(t)$ in Example 13.2.27
Each of these is sketched in Figure 13.2.29. Note how the length of the vector gives an indication of how quickly the circle is being traced at that point. When $t=-2\text{,}$ the circle is being drawn relatively slow; when $t=-1\text{,}$ the circle is being traced much more quickly.

Solution 2. Video solution

It is a basic geometric fact that a line tangent to a circle at a point $P$ is perpendicular to the line passing through the center of the circle and $P\text{.}$ This is illustrated in Figure 13.2.29; each tangent vector is perpendicular to the line that passes through its initial point and the center of the circle. Since the center of the circle is the origin, we can state this another way: $\vec u\,'(t)$ is orthogonal to $\vec u(t)\text{.}$

Recall that the dot product serves as a test for orthogonality: if $\vec u\cdot \vec v = 0\text{,}$ then $\vec u$ is orthogonal to $\vec v\text{.}$ Thus in the above example, $\vec u(t)\cdot \vec u\,'(t)=0\text{.}$

This is true of any vector-valued function that has a constant length, that is, that traces out part of a circle. It has important implications later on, so we state it as a theorem (and leave its formal proof as an Exercise.)

Theorem 13.2.30. Vector-Valued Functions of Constant Length.

Let $\vec r(t)$ be a vector-valued function of constant length that is differentiable on an open interval $I\text{.}$ That is, $\norm{\vec r(t)} = c$ for all $t$ in $I\text{;}$ equivalently, $\vec r(t)\cdot \vec r(t) = c^2$ for all $t$ in $I\text{.}$ Then $\vec r(t)\cdot\vrp(t) = 0$ for all $t$ in $I\text{.}$

Subsection 13.2.4 Integration

Before formally defining integrals of vector-valued functions, consider the following equation that our calculus experience tells us should be true:

\begin{equation*} \int_a^b \vrp(t)\, dt = \vec r(b) - \vec r(a)\text{.} \end{equation*}

That is, the integral of a rate of change function should give total change. In the context of vector-valued functions, this total change is displacement. The above equation is true; we now develop the theory to show why.

We can define antiderivatives and the indefinite integral of vector-valued functions in the same manner we defined indefinite integrals in Definition 5.1.2. However, we cannot define the definite integral of a vector-valued function as we did in Definition 5.2.6. That definition was based on the signed area between a function $y=f(x)$ and the $x$-axis. An area-based definition will not be useful in the context of vector-valued functions. Instead, we define the definite integral of a vector-valued function in a manner similar to that of Theorem 5.3.26, utilizing Riemann sums.

Definition 13.2.31. Antiderivatives, Indefinite and Definite Integrals of Vector-Valued Functions.

Let $\vec r(t)$ be a continuous vector-valued function on $[a,b]\text{.}$ An antiderivative of $\vec r(t)$ is a function $\vec R(t)$ such that $\vec R'(t) = \vec r(t)\text{.}$

The set of all antiderivatives of $\vec r(t)$ is the indefinite integral of $\vec r(t)\text{,}$ denoted by

\begin{equation*} \int \vec r(t)\, dt\text{.} \end{equation*}

The definite integral of $\vec r(t)$ on $[a,b]$ is

\begin{equation*} \int_a^b \vec r(t)\, dt =\lim_{\norm{\Delta t}\to 0} \sum_{i=1}^n\vec r(c_i)\Delta t_i\text{,} \end{equation*}

where $\Delta t_i$ is the length of the $i$th subinterval of a partition of $[a,b]\text{,}$ $\norm{\Delta t}$ is the length of the largest subinterval in the partition, and $c_i$ is any value in the $i$th subinterval of the partition.

It is probably difficult to infer meaning from the definition of the definite integral. The important thing to realize from the definition is that it is built upon limits, which we can evaluate component-wise.

The following theorem simplifies the computation of definite integrals; the rest of this section and the following section will give meaning and application to these integrals.

Theorem 13.2.32. Indefinite and Definite Integrals of Vector-Valued Functions.

Let $\vec r(t) = \la f(t),g(t)\ra$ be a vector-valued function in $\mathbb{R}^2$ that is continuous on $[a,b]\text{.}$

$\displaystyle \ds \int \vec r(t)\, dt = \la \int f(t)\, dt, \int g(t)\, dt\ra$
$\displaystyle \ds \int_a^b \vec r(t)\, dt = \la \int_a^b f(t)\, dt, \int_a^b g(t)\, dt\ra$

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

Figure 13.2.33. Video presentation of Definition 13.2.31 and Theorem 13.2.32

Example 13.2.34. Evaluating a definite integral of a vector-valued function.

Let $\vec r(t) = \la e^{2t},\sin(t) \ra\text{.}$ Evaluate $\ds \int_0^1 \vec r(t) \,dt\text{.}$

Solution 1.

We follow Theorem 13.2.32.

\begin{align*} \int_0^1 \vec r(t) \,dt \amp = \int_0^1 \la e^{2t},\sin(t) \ra \,dt\\ \amp = \la \int_0^1 e^{2t}\,dt\,, \int_0^1 \sin(t) \,dt \ra\\ \amp = \la \frac12e^{2t}\Big|_0^1\,, -\cos(t) \Big|_0^1\ra\\ \amp = \la \frac12(e^2-1)\,, -\cos(1)+1\ra\\ \amp \approx \la 3.19,0.460\ra\text{.} \end{align*}

Solution 2. Video solution

Example 13.2.35. Solving an initial value problem.

Let $\vrp'(t) = \la 2, \cos(t) , 12t\ra\text{.}$ Find $\vec r(t)$ where:

$\vec r(0) = \la-7,-1,2\ra$ and
$\vrp(0) = \la 5,3,0\ra\text{.}$

Solution 1.

Knowing $\vrp'(t) = \la 2,\cos(t) , 12t\ra\text{,}$ we find $\vrp(t)$ by evaluating the indefinite integral.

\begin{align*} \int \vrp'(t)\,dt \amp = \la \int 2\,dt\,, \int \cos(t) \,dt\,, \int 12t\,dt\ra\\ \amp = \la 2t+C_1, \sin(t) + C_2, 6t^2 + C_3\ra\\ \amp = \la 2t,\sin(t) ,6t^2 \ra + \la C_1,C_2,C_3\ra\\ \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C\text{.} \end{align*}

Note how each indefinite integral creates its own constant which we collect as one constant vector $\vec C\text{.}$ Knowing $\vrp(0) = \la 5,3,0\ra$ allows us to solve for $\vec C\text{:}$

\begin{align*} \vrp(t) \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C\\ \vrp(0) \amp = \la 0,0,0 \ra + \vec C\\ \la 5,3,0\ra \amp = \vec C\text{.} \end{align*}

So $\vrp(t) = \la 2t,\sin(t) ,6t^2\ra + \la 5,3,0\ra = \la 2t+5, \sin(t) + 3, 6t^2\ra\text{.}$ To find $\vec r(t)\text{,}$ we integrate once more.

\begin{align*} \int \vrp(t)\,dt \amp = \la \int 2t+5\,dt, \int \sin(t) + 3\,dt, \int 6t^2\,dt \ra\\ \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C\text{.} \end{align*}

With $\vec r(0) = \la -7,-1,2\ra\text{,}$ we solve for $\vec C\text{:}$

\begin{align*} \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C\\ \vec r(0) \amp = \la 0,-1,0\ra + \vec C\\ \la -7,-1,2\ra \amp = \la 0,-1,0\ra + \vec C\\ \la -7,0,2\ra \amp = \vec C\text{.} \end{align*}

\begin{align*} \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \la -7,0,2\ra\\ \amp = \la t^2+5t-7,-\cos(t) +3t,2t^3+2\ra\text{.} \end{align*}

Solution 2. Video solution

What does the integration of a vector-valued function mean? There are many applications, but none as direct as “the area under the curve” that we used in understanding the integral of a real-valued function.

A key understanding for us comes from considering the integral of a derivative:

\begin{equation*} \int_a^b \vrp(t)\, dt = \vec r(t)\Big|_a^b = \vec r(b)-\vec r(a)\text{.} \end{equation*}

Integrating a rate of change function gives displacement.

Noting that vector-valued functions are closely related to parametric equations, we can describe the arc length of the graph of a vector-valued function as an integral. Given parametric equations $x=f(t)\text{,}$ $y=g(t)\text{,}$ the arc length on $[a,b]$ of the graph is

\begin{equation*} \text{ Arc Length } = \int_a^b\sqrt{\fp(t)^2+g'(t)^2}\, dt\text{,} \end{equation*}

as stated in Theorem 10.3.17 in Section 10.3. If $\vrt = \la f(t), g(t)\ra\text{,}$ note that $\sqrt{\fp(t)^2+g'(t)^2} = \norm{\vrp(t)}\text{.}$ Therefore we can express the arc length of the graph of a vector-valued function as an integral of the magnitude of its derivative.

Theorem 13.2.36. Arc Length of a Vector-Valued Function.

Let $\vrt$ be a vector-valued function where $\vrp(t)$ is continuous on $[a,b]\text{.}$ The arc length $L$ of the graph of $\vrt$ is

\begin{equation*} L = \int_a^b \norm{\vrp(t)}\, dt\text{.} \end{equation*}

Note that we are actually integrating a scalar-function here, not a vector-valued function.

The next section takes what we have established thus far and applies it to objects in motion. We will let $\vrt$ describe the path of an object in the plane or in space and will discover the information provided by $\vrp(t)$ and $\vrp'(t)\text{.}$

Exercises 13.2.5 Exercises

Terms and Concepts

1.

Limits, derivatives and integrals of vector-valued functions are all evaluated -wise.

2.

The definite integral of a rate of change function gives .

3.

Why is it generally not useful to graph both $\vec r(t)$ and $\vrp(t)$ on the same axes?

4.

Theorem 13.2.25 contains three product rules. What are the three different types of products used in these rules?

Problems

Exercise Group.

Evaluate the given limit.

5.

$\lim\limits_{t\to 5} \la 2t+1,3t^2-1,\sin(t) \ra$

6.

$\lim\limits_{t\to 3} \la e^t,\frac{t^2-9}{t+3}\ra$

7.

$\lim\limits_{t\to 0} \la \frac{t}{\sin(t) }, (1+t)^{\frac1t}\ra$

8.

$\lim\limits_{h\to 0} \frac{\vec r(t+h)-\vec r(t)}{h}\text{,}$ where $\vec r(t) = \la t^2,t,1\ra\text{.}$

Exercise Group.

Identify the interval or union of intervals on which $\vec r(t)$ is continuous.

9.

$\vec r(t) = \la t^2,1/t\ra$

10.

$\vec r(t) = \la \cos(t) , e^t, \ln(t) \ra$

Exercise Group.

Find the derivative of the given function.

11.

$\vec r(t) = \la \cos(t) , e^t, \ln(t) \ra$

12.

$\ds \vec r(t) = \la \frac 1t, \frac {2t-1}{3t+1}, \tan(t) \ra$

13.

$\vec r(t) = (t^2)\la \sin(t) , 2t+5\ra$

14.

$\vec r(t) = \la t^2+1, t-1\ra\cdot \la \sin(t) , 2t+5\ra$

15.

$\vec r(t) = \la t^2+1, t-1,1\ra\times \la \sin(t) , 2t+5,1\ra$

16.

$\ds \vec r(t) = \la \cosh t, \sinh t\ra$

Exercise Group.

First, find $\vrp (t)\text{.}$ Then sketch $\vec r(t)$ and $\vrp(1)\text{,}$ with the initial point of $\vrp(1)$ at $\vec r(1)\text{.}$

17.

$\ds \vec r(t) = \la t^2+t, t^2-t\ra$

18.

$\ds \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra$

19.

$\ds \vec r(t) = \la t^2+1,t^3-t\ra$

20.

$\ds \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra$

Exercise Group.

Give the equation of the line tangent to the graph of $\vec r(t)$ at the given $t$ value.

21.

$\vec r(t) = \la t^2+t, t^2-t\ra\text{,}$ at $t=1$

22.

$\vec r(t) = \la 3\cos(t) ,\sin(t) \ra\text{,}$ at $t=\pi/4$

23.

$\ds \vec r(t) = \la 3\cos(t) ,3\sin(t) ,t\ra$ at $t=\pi\text{.}$

24.

$\vec r(t) = \la e^t,\tan(t) ,t\ra\text{,}$ at $t=0\text{.}$

Exercise Group.

Find the value(s) of $t$ for which $\vec r(t)$ is not smooth.

25.

$\ds \vec r(t) = \la \cos(t) ,\sin(t) - t\ra$

26.

$\vec r(t) = \la t^2-2t+1,t^3+t^2-5t+3\ra$

27.

$\ds \vec r(t) = \la \cos(t) -\sin(t) , \sin(t) - \cos(t) ,\cos(4t)\ra$

28.

$\vec r(t) = \la t^3-3t+2, -\cos(\pi t),\sin^2(\pi t) \ra$

Exercise Group.

The following exercises ask you to verify parts of Theorem 13.2.25. In each let $f(t) = t^3\text{,}$ $\vec r(t) =\la t^2,t-1,1\ra$ and $\vec s(t) = \la \sin(t) , e^t,t\ra\text{.}$ Compute the various derivatives as indicated.

29.

Simplify $f(t)\vec r(t)\text{,}$ then find its derivative; show this is the same as $\fp(t)\vec r(t) + f(t)\vrp(t)\text{.}$

30.

Simplify $\vec r(t)\cdot\vec s(t)\text{,}$ then find its derivative; show this is the same as $\vrp(t)\cdot\vec s(t) + \vec r(t)\cdot\vec s\,'(t)\text{.}$

31.

Simplify $\vec r(t)\times\vec s(t)\text{,}$ then find its derivative; show this is the same as $\vrp(t)\times\vec s(t) + \vec r(t)\times\vec s\,'(t)\text{.}$

32.

Simplify $\ds \vec r\big(f(t)\big)\text{,}$ then find its derivative; show this is the same as $\ds \vrp\big(f(t)\big)\fp(t)\text{.}$

Exercise Group.

In the following exercises, evaluate the given definite or indefinite integral.

33.

$\ds \int \la t^3,\cos(t) , te^t\ra\,dt$

34.

$\ds \int \la \frac{1}{1+t^2},\sec^2(t) \ra\,dt$

35.

$\displaystyle\int_0^{\pi} \la -\sin(t) ,\cos(t) \ra\,dt=$.

36.

$\ds \int_{-2}^{2} \la 2t+1,2t-1\ra\,dt$

Exercise Group.

Solve the given initial value problems.

37.

Find $\vec r(t)\text{,}$ given that $\vrp(t) = \la t,\sin(t) \ra$ and $\vec r(0) = \la 2,2\ra\text{.}$

$\vec r(t)=$

38.

Find $\vec r(t)\text{,}$ given that $\vrp(t) = \la 1/(t+1),\tan(t) \ra$ and

$\vec r(0) = \la 1,2\ra\text{.}$

39.

Find $\vec r(t)\text{,}$ given that $\vrp'(t) = \la t^2,t,1\ra\text{,}$ $\vrp(0) = \la 1,2,3\ra$ and $\vec r(0) = \la 4,5,6\ra\text{.}$

$\vec r(t)=$

40.

Find $\vec r(t)\text{,}$ given that $\vrp'(t) = \la \cos(t) ,\sin(t) ,e^t\ra\text{,}$

$\vrp(0) = \la 0,0,0\ra$ and $\vec r(0) = \la 0,0,0\ra\text{.}$

Exercise Group.

Find the arc length of $\vec r(t)$ on the indicated interval.

41.

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

42.

$\vec r(t) = \la 5\cos(t) , 3\sin(t) , 4\sin(t) \ra$ on $[0,2\pi]\text{.}$

43.

$\vec r(t) = \la t^3,t^2,t^3 \ra$ on $[0,1]\text{.}$

44.

$\vec r(t) = \la e^{-t}\cos(t) ,e^{-t}\sin(t) \ra$ on $[0,1]\text{.}$

45.

Prove Theorem 13.2.30; that is, show if $\vec r(t)$ has constant length and is differentiable, then $\vec r(t)\cdot \vrp(t)=0\text{.}$ (Hint: use the Product Rule to compute $\frac{d}{dt}\big(\vec r(t)\cdot\vec r(t)\big)\text{.}$)

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