Vector-valued functions are functions that assign a vector to every point in a given space. In this book, we will explore the properties of these functions and how they can be used to model real-world situations. We’ll start with the basics and work our way up to more complex concepts. Along the way, you’ll learn about space curves and how to calculate their derivatives and integrals. So whether you’re a beginner or a pro, this book has something for you!

A vector function (or a, vector-valued function) is a function that takes one or more inputs and outputs a vector. Vector functions have a wide range of applications in science and technology.

Vector functions are important in mathematics, physics, and engineering. In physics, vector functions are used to describe the motion of particles and the behavior of waves. In engineering, vector functions are used to design bridges, buildings, and other structures. And in mathematics, vector functions are used to study geometry and topology.

Vector functions are quite interesting once you wrap your head around them. So let’s take a few minutes to do just that.

A vector function is simply a function that takes one or more scalars as input and outputs a vector. In other words, it’s a function that gives you a vector when you plug in a number(s). For example, the function $$f(x) = (2x, 3x-1)$$ is a vector function. You can think of it as a machine that takes in a number (x) and spits out a vector (2x, 3x-1). Pretty simple, right?

Now, vector functions can be used to model all sorts of things in the real world, from the trajectory of a baseball to the movement of particles in a fluid. So the next time you’re struggling with a vector function problem, just remember that you’re really just trying to understand the movement of something in the world around us. And that’s not so tough after all.

Derivatives and integrals of vector functions may sound like a mouthful, but they’re really just mathematical tools that allow us to use vector functions. In other words, they let us find out how vector functions change over time or space. And that can be really useful information, whether we’re trying to figure out the trajectory of a projectile or the motion of a planet in orbit.

So what exactly are derivatives and integrals of vector functions? Well, derivatives tell us how vector functions change with respect to changes in their input variables, while integrals give us information about how vector functions change over time or space. Basically, they’re two sides of the same coin: one tells us how a vector function changes in response to changes in its input variables, while the other tells us how a vector function changes over time or space. But together, they give us a pretty complete picture of how vector functions behave.

Vector functions are used in a variety of fields, such as physics, engineering, and economics. One example of their use is in the analysis of electrical circuits. Vector-valued functions can be used to model the voltage and current in an electrical circuit. This information can then be used to design products that are more efficient and have fewer problems.

In addition, vector-valued functions can be used to model the movement of objects in space. This information can be used to optimize the trajectory of a spacecraft or to predict the path of a hurricane. As you can see, vector-valued functions have a wide range of applications in the real world.

There’s more to a curve than meets the eye. In fact, one of the most important properties of a curve is its arc length. Arc length is simply the length of the curve between two points, and it can be calculated using vector functions.

Of course, this assumes that you have a vector function for your curve. If you don’t, you could always try finding one. In any case, arc length is a vital property of curves, and it’s something that everyone should know about.

It’s a well-known fact that the shortest distance between two points is a straight line. But what if those points are moving? This is where the concept of arc length comes in. Arc length is simply the distance traveled by a point moving along a curve. In other terms, it’s the length of a vector function. And just like with regular old straight-line distances, we can use arc length to calculate things like speed and acceleration.

So the next time you find yourself on a winding road, take comfort in knowing that you’re traveling the mathematically shortest distance possible!

Curvature is a measure of how vector functions change as you move along them. In other words, it’s a way to quantify how “curved” a function is. For example, the function $$f(x) = x^2$$ is more curved than the function $$g(x) = x$$. In this chapter, you’ll learn how to find the curvature vector”. This vector tells you how quickly the direction of the vector function is changing at any given point. It also tells you which way the vector function is curving - whether it’s curving to the left or right. Curvature can be positive or negative, depending on which way the vector function is curving.

And finally, the curvature can be calculated for any (smooth) vector function - not just those that represent curves in two-dimensional space.

This book is for anyone who wants to learn about vector-valued functions and their applications. Whether you’re a student, a professional, or just someone curious about the topic, this book will give you a thorough understanding of vector-valued functions. We’ll start with the basics and work our way up to more advanced material:

• What are vector-valued functions and what do they do?
• How are vector-valued functions used in the real world?
• What is arc length and how can it be calculated?
• What is curvature and how can it be calculated?

In this book, you’ll find short, concise chapters that get straight to the point. I’ll also be using a lot of pictures and diagrams to help you understand the material. And finally, I’ll be giving you plenty of opportunities to practice what you’ve learned with lots of exercises.

In general, the best way I teach vector functions is to start with simple examples and then gradually increase the complexity of the examples. This gives you a chance to learn the basics and solidify your understanding of the material before being overwhelmed by too much information at once.

So if you’re ready to learn about vector-valued functions, then let’s get started!

## 1.1 Space Curves

Before we prove any theorems and develop any definitions let’s see how natural it is to use vector functions to model the world around us.

In the professor’s classroom, there is little to no air movement so we assume the only force acting on the arrow is gravity.

Example 1.1 To start a calculus semester a professor wants to light a torch with a flaming arrow that sits on top of a podium on the other side of a large classroom.

Solution. Suppose the professor lit the arrow and shot it at a height of 6 ft above the ground level 90 ft from the 30 ft high podium and he wanted the arrow to reach a maximum height exactly 4 ft above the center of the podium. What will the initial firing angle of the arrow be?

Assume that the arrow is fired from the initial point $$(x_0,y_0)=(0,6)$$ corresponding to $$t=0$$. Using the definition of sine and cosine the initial velocity is found to be \begin{align*} \vec{v}_0 & =\left(x_0 +\norm{\vec{v}_0} \cos \theta \right) \vec{i}+\left(y_0+\norm{\vec{v}_0} \sin \theta \right)\vec{j} \\ & =\norm{\vec{v}_0} \cos \theta \, \vec{i}+\left(6+\norm{\vec{v}_0} \sin \theta \right)\vec{j} \end{align*} where $$\theta$$ is the angle that $$\vec{v}_0$$ makes with the horizontal. Recall Newton’s second law of motion which says the the force acting on the projectile is equal to the projectile’s mass $$m$$ times its acceleration, or $\vec{F}=m \frac{d^2 \, \vec{r}}{d t^2}$ when $$\vec{r}$$ is the projectile’s position vector and $$t$$ is time. In the
We assume the only force acting on the flaming arrow is the gravitational force $$-m g \, \vec{j}$$, then $\begin{equation} \label{fa1} \frac{d^2 \, \vec{r}}{d t^2} = - g \, \vec{j}, \quad \vec{r_0}=0\, \vec{i}+6\, \vec{j}, \quad \text{and} \quad \left.\frac{d\vec{r}}{dt}\right|_{t=0}=\vec{v}_0. \end{equation}$ The first integration gives $$\frac{d\vec{r}}{dt}=-g t \, \vec{j}+\vec{v}_0$$. The second integration yields $\begin{equation}\label{fa2} \vec{r}(t)=\frac{-g t^2}{2}\, \vec{j} +\vec{v}_0 t+\vec{r}_0. \end{equation}$ Now using substitution of the initial conditions in $$\eqref{fa1}$$ into equation $$\eqref{fa2}$$ we find an expression for the position function of the arrow $\begin{equation}\label{fa3} \vec{r}(t)=\left(\norm{\vec{v}_0} \cos\theta\right)t \, \vec{i}+\left(6+(\norm{\vec{v}_0}\sin \theta)t -\frac{1}{2} g t^2\right) \, \vec{j}. \end{equation}$ The arrow reaches its highest point when $$\frac{dy}{dt}=0$$, and solving for $$t$$ we obtain $t=\frac{\norm{\vec{v}_0} \sin \theta}{g}.$ For this value of $$t$$, the value of $$y$$ is $6+\frac{ (\norm{\vec{v}_0} \sin \theta)^2 }{2g}.$ Using $$y_{\text{max}}=34$$ and $$g=32$$, we see that $34=6+\frac{ (\norm{\vec{v}_0} \sin \theta)^2 }{2(32)}$ or $$\norm{\vec{v}_0} \sin \theta =\sqrt{(28)(64)}$$.

In order to find $$\theta$$ we wish to find a similar expression for $$\norm{\vec{v}_0} \cos \theta$$. When the arrow reaches $$y_{\text{max}}=34$$, the horizontal distance is $$x=90$$ ft and by substitution into equation $$\eqref{fa3}$$ we obtain $90=(\norm{\vec{v}_0} \cos \theta) \left(\frac{\norm{\vec{v}_0} \sin \theta}{g}\right).$ Therefore we find $\tan \theta =\frac{\norm{\vec{v}_0} \sin \theta}{\norm{\vec{v}_0} \cos \theta} = \frac{\left(\sqrt{(28)(64)}\right)^2}{(90)(32)}=\frac{28}{45}.$ So the approximate firing angle will be $\begin{equation} \theta = \tan^{-1}(28/45) \approx 31.8908^\circ. \end{equation}$

## 1.2 Definition of a Vector-Valued Function

Vector functions can be used to model the motion of an object. For example, suppose a fly takes off from the top of a coffee cup at the front of a classroom. The path the fly traces as it travels through the classroom is a one-dimensional path which can be described as a vector-valued function. You may want to study the path (geometry only), but you may also want to know its speed, direction of motion, and acceleration at each point in time, in that case you will want more than its path – you will want a vector-valued function that describes this fly’s motion in three dimensions. Given a vector-valued function defined at each point in time, you will not only have position but also its speed, direction of motion, and acceleration at each point in time.

A vector-valued function can be defined with more than one variable and with more components. In general, a vector-valued function is a function that takes an $$n$$-tuple of variables and outputs a unique vector with $$m$$ components. We start of by defining a vector function of one variable and then give several examples.

Definition 1.1 A vector-valued function $$\vec{F}$$ of a real variable $$t$$ with $$D$$ assigns to each number $$t$$ in the set $$D$$ a unique vector $$\vec{F}(t)$$. The set of all vectors $$\vec v$$ of the form $$\vec v=\vec{F}(t)$$ for $$t$$ in $$D$$ is the range of $$\vec F$$.

In three dimensions vector functions can be expressed in the form
$\vec{F}(t)=f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k}$ where $$f_1,f_2,$$ and $$f_3$$ are real-valued functions of the real variable $$t$$ defined on the domain set $$D$$. A vector function may also be denoted by $\vec{F}(t)= \langle f_1(t),f_2(t),f_3(t)\rangle .$ Unless stated otherwise, the domain of a vector function $$\vec{F}$$ is the intersection of the domains of the scalar component functions $$f_1, f_2,$$ and $$f_3.$$

Next we take a given vector-valued function and explicitly say what the scalar component functions are and ascertain the domain.

Example 1.2 Given the vector function $\vec{F}(t)= t^3\, \vec{i} +\ln (3-t)\, \vec{j}+\sqrt{t}\, \vec{k}$ find the component functions and the domain of $$\vec{F}$$, and then evaluate $$\vec{F}$$ at $$t=1$$.

Solution. The component functions are $$f(t)=t^3,$$ $$g(t)=\ln (3-t),$$ and $$h(t)=\sqrt{t}$$
where $$\vec{F}(t)=f(t)\, \vec{i}+g(t)\, \vec{j}+h(t)\, \vec{k}.$$ The domain of $$\vec{F}(t)$$ is the interval $$[0,3)$$ since $$\sqrt{t}$$ requires $$t\geq 0$$ and $$\ln (3-t)$$ requires $$t<3.$$ We can evaluate the vector function $$\vec{F}$$ at $$t=1$$ as follows $\vec{F}(1)=f(1)\, \vec{i}+g(1)\, \vec{j}+h(1)\, \vec{k} =\vec{i}+\ln 2\, \vec{j}+\vec{k}.$

## 1.3 Graphs of Vector Functions

Vector functions are useful for tracing out graphs of curves and for describing motion along a path. Often the variable $$t$$ represents time and since each $$\vec{F}(t)$$ represents a vector, we have a position $$(x,y,z)$$ at time $$t.$$ That is to say, given a time value of $$t$$ we have a vector $\vec{F}(t)= \langle f_1(t),f_2(t),f_3(t)\rangle$ which represents a point $$(x,y,z)$$ where $$x=f_1(t),$$ $$y=f_2(t)$$, and $$z=f_3(t)$$. In this manner, we use arrowheads on the curve to indicate the curve’s orientation by pointing in the direction of increasing values of $$t.$$

Sketching the graph a given vector function can be time-consuming, especially if we are given an unfamiliar function. In the next example we notice that each scalar component function is linear and so claim that the vector function is linear in three dimensions. Thus graphing this vector function is easy; just pick any two points in the domain you wish, plot them and then draw a straight line.

Example 1.3 Sketch the graph of the vector-valued function defined by $\vec{F}(t)=-3 \sin t \vec{i}+3\cos t\vec{j}+0.1t \vec{k}.$

Solution. The graph is the set of all points $$(x,y,z)$$ with $x=-3\sin t, \quad y=3\cos t, \quad \text{ and } \quad z=0.1t.$ The graph is a circular helix that lies on the surface of the cylinder with equation $x^2+y^2=(-3\sin t)^2+(3\cos t)^2=9.$ The cylinder is centered at (0,0) in the $$x y$$-plane as shown in $$\ref{fig:vf2}$$.

Example 1.4 Sketch the graph of the vector-valued function $\vec{F}(t)=(5-2t)\vec{i}+(3+2t)\vec{j}+ 2t \vec{k}.$

Solution. The graph is the set of all points $$(x,y,z)$$ with $$x=5-2t,$$ $$y=3+2t,$$ and $$z=5t.$$ The graph is a line that passes through the point $$(5,3,0)$$ (when $$t=0$$) and the point $$(3,5,5)$$ (when $$t=1$$) as shown in $$\ref{fig:vf1}$$.

## 1.4 Spaces Curves and Parameterizations

It is important to realize that there is not a one-to-one correspondence between a one-dimensional graph in three dimensions and a vector function. That is to say a vector-valued function has a graph, and only one set of points in three dimensions is its graph. But a graph (set of points in three dimensions) can be represented by more than one functional rule. The classic two dimensional example is that of the unit circle, which can be parametrized as a vector-valued function by $$\vec{F}(t)=\sin t\vec{i}+\cos t\vec{j}$$ with $0t<2$; and can also be parametrized as a vector-valued function by $$\vec{F}(t)=\sin (2t)\vec{i}+\cos (2t)\vec{j}$$ with $0t<$. With both of these vector-valued functions we have the same graph: the unit circle. It is this relationship between algebra and geometry that make vector-valued functions extremely useful and important to study.

Finding a functional rule for a vector function, from given geometrical information, can also be a challenging endeavor. In fact, given a geometrical object in three dimensions, there can be quite a few ways to represent that object using algebra. In short, parametrizing a geometrical object, even one as simple as the unit circle, can lead to many parametric representations and thus different vector functions.

In the next few examples we illustrate two ways in which this process might be carried out. We are given a geometric object and asked to find a vector function, (find a parametrization), that represents the given object.

There are many other ways to accomplish the task of introducing the variable $$t$$, e.g. try $$x=2t$$. What’s interesting is that no matter how you introduce $$t$$ the vector function obtained will have the same graph.

Example 1.5 Find a vector-valued function $$\vec{F}$$ whose graph is the curve of intersection of the hemisphere $$z=\sqrt{1-x^2-y^2}$$ and the parabolic cylinder $$y=x^2$$.

Solution. One way to accomplish the task is by letting $$x=t.$$ Then $$y=t^2$$ and $z=\sqrt{1-x^2-y^2}=\sqrt{1-t^2-t^4}.$ Therefore $\vec{F}(t)=t \vec{i}+t^2\vec{j}+\sqrt{1-t^2-t^4}\vec{k}.$ is a value-valued function for this intersection.

Try reworking this example with $$x=2t$$. and graph the resulting vector function.

Example 1.6 Find a vector-valued function $$\vec{F}$$ whose graph is the curve of intersection of the plane $$2 x+y+3z=6$$ and the plane $$x-y-z=1.$$

Solution. One way to accomplish the task is by letting $$x=t.$$ Then to find relations for $$y$$ and $$z$$ we will solve the system
$\begin{cases} t-y-z=1 \\ 2t+y+3z=6. \end{cases}$
Eliminating $$y$$ we have, $$3t+2z=7$$ and so $$z=(7-3t)/2$$. Solving the first for $$y$$ we find
$y=t-1-z=t-1-\left(\frac{7-3t}{2}\right)=\frac{2t-2-7+3t}{2}=\frac{5t-9}{2}$ Therefore $\begin{equation} \label{vfplanes} \vec{F}(t)=t \, \vec{i}+\frac{5t-9}{2}\vec{j}+\frac{7-3t}{2}\vec{k}. \end{equation}$ is a vector-valued function for this intersection.

## 1.5 Operations with Vector Functions

Next we show how to perform basic operations like addition, subtraction, dot product and the cross product with vector functions.

Definition 1.2 Let $$\vec{F}$$ and $$\vec{G}$$ be vector-valued functions of the real variable $$t$$, and let $$f(t)$$ be a real-valued function. Then $$\vec{F}+\vec{G},$$ $$\vec{F}-\vec{G},$$ $$f\vec{F},$$ $$\vec{F}\times \vec{G},$$ and $$\vec{F}\cdot \vec{G}$$ are vector functions defined as follows

• $$(\vec{F}+\vec{G})(t)=\vec{F}(t)+\vec{G}(t)$$
• $$(f\vec{F})(t)=f(t)\vec{F}(t)$$
• $$(\vec{F}-\vec{G})(t)=\vec{F}(t)-\vec{G}(t)$$
• $$(\vec{F}\cdot \vec{G})(t)=\vec{F}(t)\cdot \vec{G}(t)$$
• $$(\vec{F}\times \vec{G})(t)=\vec{F}(t)\times \vec{G}(t)$$ These operations are defined on the intersection of the domain of the vector-valued and real-valued functions that occur in the definitions, respectively.

Example 1.7 Let $$\vec F$$, $$\vec G$$ , and $$\vec H$$ be the vector functions defined by \begin{align*} \vec{F}(t)=(2t)\vec{i}-5\vec{j}+t^2\vec{k}, \quad \vec{G}(t)=(1-t)\vec{i}+\frac{1}{t}\vec{k}, \quad \vec{H}(t)=(\sin t)\vec{i}+e^t\vec{j}. \end{align*} Find the vector functions

• $$\vec{F}(t)\times \vec{G}(t)$$
• $$2e^t\vec{F}(t)+t\vec{G}(t)+10\vec{H}(t)$$
• $$\vec{G}(t)\cdot [\vec{H}(t)\times \vec{F}(t)]$$
• Solution.
• Using $$\ref{limdf}$$, we find \begin{align*} & 2e^t\vec{F}(t)+t\vec{G}(t)+10\vec{H}(t) \\ & \qquad =\left(4t e^t+t-t^2+10 \sin t\right)\vec{i}+\left(2t^2e^t+1\right)\vec{k}. \end{align*}
• Using the determinant formula for a cross product of vectors, we find \begin{align*} \vec{F}(t)\times \vec{G}(t) & = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 t & -5 & t^2 \\ 1-t & 0 & \frac{1}{t} \end{array} \right| \\ & =\left(\frac{-5}{t}\right)\vec{i}+\left(-t^3+t^2-2\right)\vec{j}+(-5t+5)\vec{k}. \end{align*}
• We find $\vec{G}(t)\cdot [\vec{H}(t)\times \vec{F}(t)] = (1-t)t^2e^t-\frac{2 t e^t+5\sin t}{t}$ by using
\begin{align*} & \vec{G}(t)\cdot [\vec{H}(t)\times \vec{F}(t)] \\ & \quad =\left[(1-t)\vec{i}+\frac{1}{t}\vec{k}\right]\cdot \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \sin t & e^t & 0 \\ 2 t & -5 & t^2 \end{array} \right|. \end{align*}

## 1.6 Limits of Vector Functions

The idea is now to extend the notion of a limit of a one-variable function and continuity of a one-variable function to vector-valued functions by capitalizing on the properties of vector functions.

Definition 1.3 Suppose the components function $$f_1$$, $$f_2$$, and $$f_3$$ of the vector-valued function $\begin{equation} \vec{F}(t)=f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k} \end{equation}$ all have finite limits as $$t\rightarrow t_0$$, where $$t_0$$ is any real number or . Then the limit of $$\vec{F}(t)$$ as $$t\rightarrow t_0$$ is defined as the vector $\lim _{t\to t_0}\vec{F}(t)=\left( \lim _{t\to t_0}f_1(t)\right)\vec{i}+\left(\lim _{t\to t_0}f_2(t)\right)\vec{j}+\left(\lim _{t\to t_0}f_3(t)\right)\vec{k}.$

::: {#lem- } [Limits of Vector-Functions] If the vector-valued functions $$\vec{F}$$ and $$\vec{G}$$ are functions of a real variable $$t$$ and $$h(t)$$ is a real-valued function such that all three functions have finite limits as $$t\to t_0$$, then

• $$\lim _{t\to t_0}[\vec{F}(t)+\vec{G}(t)] =\lim _{t\to t_0}\vec{F}(t)+\lim _{t\to t_0}\vec{G}(t)$$
• $$\lim _{t\to t_0}[\vec{F}(t)-\vec{G}(t)] =\lim _{t\to t_0}\vec{F}(t)-\lim _{t\to t_0}\vec{G}(t)$$
• $$\lim _{t\to t_0}[h(t) \vec{F}(t)] =\left[\lim _{t\to t_0}h(t)\right] \left[\lim _{t\to t_0}\vec{F}(t)\right]$$
• $$\lim _{t\to t_0}[\vec{F}(t)\cdot \vec{G}(t)] =\left[\lim _{t\to t_0}\vec{F}(t)\right]\cdot \left[\lim _{t\to t_0}\vec{G}(t)\right]$$
• $$\lim _{t\to t_0}[\vec{F}(t)\times \vec{G}(t)] =\left[\lim _{t\to t_0}\vec{F}(t)\right]\times \left[\lim _{t\to t_0}\vec{G}(t)\right]$$

These limit formulas are also valid when $$t\to \pm \infty$$ provided all limits are finite. :::

Proof. Let $$\vec{F}$$ and $$\vec{G}$$ have (respectively) standard forms $\begin{equation} \label{stform} \vec{F}(t)=f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k} \qquad \vec{G}(t)=g_1(t)\vec{i}+g_2(t)\vec{j}+g_3(t)\vec{k}. \end{equation}$ (i) Using $$\eqref{stform}$$ we find, \begin{align*} & \lim _{t\to t_0}[\vec{F}(t)+\vec{G}(t)] \\ & = \left(\lim _{t\to t_0}(f_1+g_1)(t)\right) \vec{i} +\left(\lim _{t\to t_0}(f_2+g_2)(t)\right) \vec{j}+\left(\lim _{t\to t_0}(f_3+g_3)(t)\right) \vec{k} \\ &= \lim _{t\to t_0}[f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k} ]+\lim _{t\to t_0}[g_1(t)\vec{i}+g_2(t)\vec{j}+g_3(t)\vec{k}] \\ & = \lim _{t\to t_0}\vec{F}(t)+\lim _{t\to t_0}\vec{G}(t) \end{align*} (iv) Again using $$\eqref{stform}$$ we find, \begin{align*} \lim _{t\to t_0}[\vec{F}(t)\cdot \vec{G}(t)] & = \lim _{t\to t_0}[f_1(t)g_1(t)+f_2(t)g_2(t)+f_3(t)g_3(t)] \\ & = [f_1(t_0)g_1(t_0)+f_2(t_0)g_2(t_0)+f_3(t_0)g_3(t_0)] \\ & = \vec{F}(t_0)\cdot \vec{G}(t_0) \\ & = \left[\lim _{t\to t_0}\vec{F}(t)\right]\cdot \left[\lim _{t\to t_0}\vec{G}(t)\right] \end{align*} The remainder of the proof is left as Exercise $$\ref{ex:vecfunclimthm}$$.

Example 1.8 Find $$\lim _{t\to 0}h(t) \vec{F}(t)$$ given $\vec{F}(t)=(\sin t)\vec{i}-t \vec{k}$ and $$h(t)=1/(t^2+t-1).$$

Solution. By $$\ref{vecfunclimthm}$$, we find
\begin{align*} & \lim _{t\to 0}\left(\frac{\sin t}{t^2+t-1}\right)\vec{i}-\frac{t}{t^2+t-1} \vec{k} \\ & \ =\lim _{t\to 0}\left(\frac{\sin t}{t^2+t-1}\right)\vec{i}-\lim _{t\to 0}\left(\frac{t}{t^2+t-1}\right) \vec{k} = (0)\vec{i}-(0)\vec{k} =\vec{0}. \end{align*}

Example 1.9 Given $$\vec{F}(t)=2\vec{i}- t \vec{j}+e^t\vec{k}$$ and $$\vec{G}(t)=t^2\vec{i}+4 \sin t \vec{j}$$ find $\lim _{t\to 2}\vec{F}\times \vec{G}.$

Solution. By $$\ref{vecfunclimthm}$$, we find \begin{align*} \lim _{t\to 2}\vec{F}\times \vec{G} & =\left(2\vec{i}-2\vec{j}+e^2\vec{k}\right)\times (4\vec{i}+4 \sin 2 \vec{j}) \\ & = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 & -2 & e^2 \\ 4 & 4\sin 2 & 0 \end{array} \right| \\ & =\left(-4e^2 \sin 2\right)\vec{i} -\left(-4e^2\right)\vec{j}+(8 \sin 2+8)\vec{k}. \tag*{} \end{align*}

## 1.7 Continuous Vector Functions

One of the reasons why vector-valued functions are so favorable is that it is easy to extend known results from functions of one-variable and results in two-dimensional space to results concerning vector functions of one-variable, or space curves, that have graphs in 3 (or more) dimensions. Continuity of vector functions is one such example.

Definition 1.4 A vector-valued function $$\vec F$$ is continuous at $$t_0$$ means $$t_0$$ is in the domain of $$\vec{F}$$ and $$\lim _{t\to t_0}\vec{F}(t)=\vec{F}\left(t_0\right)$$. Further, a vector function is continuous on an interval $$I$$ if it is continuous at every point in the interval.

::: {#thm- } Continuous Vector Functions A vector function is continuous at $$a$$ if and only if each of its component functions is continuous at $$a$$. :::

Proof. Suppose that $$\vec{F}$$ is a vector function that is continuous at $$a$$. Let $\vec{F}(t)=f_1(t)\vec{i}+f_1(2)\vec{j}+f_3(t)\vec{k}.$ Then we have \begin{align*} \vec{F}(a)=\lim_{t\to a} \vec{F}(t) & =\lim_{t\to a} \left[ f_1(t)\vec{i}+f_1(2)\vec{j}+f_3(t)\vec{k} \right] \\ & = \left(\lim_{t\to a} f_1(t)\right)\vec{i}+\left(\lim_{t\to a} f_2(t)\right)\vec{j}+\left(\lim_{t\to a} f_3(t)\right)\vec{k} %\\ & =f_1(a)\vec{i}+f_2(a)\vec{j}+f_3(a)\vec{k} \end{align*} Hence $\lim_{t\to a} f_1(t)=f_1(a), \quad \lim_{t\to a} f_2(t)=f_2(a), \quad \lim_{t\to a} f_3(t)=f_3(a).$ Therefore, each of the component functions of $$\vec{F}$$ is continuous at $$a$$. The reminder of the proof is Exercise $$\ref{ex:vecfunccontthm}$$.

Example 1.10 Determine where the vector-valued function $\vec{F}(t)=t e^t \vec{i}+\frac{e^t}{t}\vec{j}+3e^t\vec{k}$ is continuous.

Solution. The component functions $$t e^t$$ and $$3e^t$$ are continuous for all real numbers. The component function $$e^t/t$$ is continuous on its domain. By $$\ref{vecfunccontthm}$$, the function $$\vec{F}$$ is continuous for all real numbers in its domain which is $$\{t\in \mathbb{R}\, | \, t \neq 0\}.$$

Example 1.11 Determine where the vector-valued function $$\vec F$$ defined by
$\vec{F}(t)=\displaystyle \frac{\vec{u}}{\|\vec{u}\|}, \qquad \text{ } \vec{u}=t\vec{i}+\frac{1}{\sqrt{t}}\vec{j}+e^t\vec{k}$ is continuous.

Solution. By $$\ref{vecfunccontthm}$$, the function $$\vec{F}$$ is continuous for all real numbers in its domain which is $$\{t\in \mathbb{R} \, : \, t>0\}$$ because
$\|\vec{u}\|= \sqrt{t^2+\left(\frac{1}{\sqrt{t}}\right)^2+e^{2t}}$ and therefore \begin{align*} \lim _{t\to t_0}\vec{F}(t) & =\lim _{t\to t_0}\left(\frac{t\vec{i}+\frac{1}{\sqrt{t}}\vec{j}+e^t\vec{k}}{\sqrt{t^2+\left(\frac{1}{\sqrt{t}}\right)^2+e^{2t}}}\right) =\frac{t_0\vec{i}+\frac{1}{\sqrt{t_0}}\vec{j}+e^{t_0}\vec{k}}{\sqrt{\left(t_0\right)^2+\left(\frac{1}{\sqrt{t_0}}\right)^2+e^{2t_0}}} \\ & =\left(\frac{t_0}{\sqrt{t_0^2+\frac{1}{t_0}+e^{2t_0}}}\right)\vec{i}+\left(\frac{\frac{1}{\sqrt{t_0}}}{\sqrt{t_0^2+\frac{1}{t_0}+e^{2t_0}}}\right)\vec{j}+\left(\frac{e^{t_0}}{\sqrt{t_0^2+\frac{1}{t_0}+e^{2t_0}}}\right)\vec{k} \end{align*} for all real numbers such that $$t_0>0.$$

## 1.8 Exercises

Exercise 1.1 Find the domain of the vector function $\vec{F}(t)=(1-t)\vec{i}+\left(\sqrt{t}\right)\vec{j}-\frac{1}{t-2}\vec{k}.$

Exercise 1.2 Find the domain of the vector function $\vec{F}(t)=(\cos t)\vec{i}-(\cot t)\vec{j}+(\csc t)\vec{k}.$

Exercise 1.3 Find the domain of the vector function $$\vec{F}(t)+\vec{G}(t)$$ where $$\vec{F}(t)=3t \vec{j}+t^{-1}\vec{k}$$ and $$\vec{G}(t)=5 t \vec{i}+\sqrt{10-t}\vec{j}.$$

Exercise 1.4 Find the domain of the vector function $$\vec{F}(t)\times \vec{G}(t)$$ where $$\vec{F}(t)=t^2\vec{i}-t \vec{j}+2t \vec{k}$$ and $$\vec{G}(t)=\frac{1}{t+2} \vec{i}+(t+4)\vec{j}-\sqrt{-t}\vec{k}.$$

Exercise 1.5 Describe the graph in words and sketch a graph by hand for the vector function $$\vec F$$ defined by $\vec{F}(t)=(a \cos t)\vec{i}+(a \sin t)\vec{j}+t \vec{k}.$

Exercise 1.6 Show that the vector function $$\vec F$$ defined by
$\vec{F}(t)=t \vec{i}+2t \cos t \vec{j}+2t \sin t \vec{k}$ lies on the cone $$4x^2=y^2+z^2.$$ Sketch the curve.

Exercise 1.7 Describe the graph in words and sketch a graph by hand for the vector function $$\vec F$$ defined by $\vec{F}(t)=\left(e^{a t}\right)\vec{i}+\left(e^{a t}\right)\vec{j}+\left(e^{-t}\right) \vec{k}.$

Exercise 1.8 How many revolutions are made by the circular helix $\vec{F}(t)=(4 \sin t)\vec{i}+(4 \cos t) \vec{j}+\frac{7}{12}t \, \vec{k}$ in a vertical distance of $$12$$ units?

Exercise 1.9 Find the domain of the vector function $$\vec F$$ defined by $$(\vec{F}\times \vec{G})\times \vec{H}$$ where

• $$\vec{F}(t)=t^2\vec{i}-t \vec{j}+2t \vec{k},$$
• $$\vec{G}(t)=\frac{1}{t+2}\vec{i}+(t+4)\vec{j}-\sqrt{-t}\vec{k},$$ and
• $$\vec{H}(t)=\frac{1}{t+3}\vec{i}+t^2\vec{j}-\sqrt{t}\vec{k}.$$

Exercise 1.10 Find a vector function whose graph is the curve of intersection of the hemisphere $$z=\sqrt{9-x^2-y^2}$$ and the parabolic cylinder $$x=y^2.$$

Exercise 1.11 Find a vector function whose graph is the line of intersection of the planes $$2x+y+3z=6$$ and $$x-y-z=1.$$

Exercise 1.12 Determine the component functions for the vector function defined by
$$\vec{D}(t)=2 e^t\vec{F}(t)+t \vec{G}(t)+10\vec{H}(t)\times \vec{G}(t)$$
where $$\vec{F}(t)=2 t \vec{i}-5\vec{j}+t^2\vec{k},$$ $$\vec{G}(t)=(1-t)\vec{i}+\frac{1}{t}\vec{k},$$ and $$\vec{H}(t)=(\sin t)\vec{i} +e^t\vec{j}.$$

Exercise 1.13 Determine the function given by
$\vec{E}(t)=\vec{F}(t)\cdot [\vec{G}(t)\times \vec{H}(t)]$ where $$\vec{F}(t)=2 t \vec{i}-5 \vec{j}+t^2\vec{k},$$ $$\vec{G}(t)=(t+1)\vec{i}+\frac{1}{t-1}\vec{k},$$ and $$\vec{H}(t)=(\cos t)\vec{i}+e^{-t}\vec{j}.$$

• Determine a function $$A$$ that satisfies
$A(t)e^t+\frac{5}{t}\sin t=\vec{H}(t)\cdot [\vec{G}(t)\times \vec{F}(t)]$
where $$\vec{F}(t)=2 t \vec{i}-5 \vec{j}+t^2\vec{k},$$ $$\vec{G}(t)=(1-t)\vec{i}+\frac{1}{t}\vec{k},$$ and $$\vec{H}(t)=(\sin t)\vec{i}+e^t\vec{j}.$$

Exercise 1.14 Find the limit of each of the following vector-valued function.

• $$\lim _{t\to 1}\left[\frac{t^3-1}{t-1}\vec{i}+\frac{t^2-3t+2}{t^2+t-2}\vec{j}+\left(t^2+1\right)e^{t-1}\vec{k}\right]$$
• $$\lim _{t\to \infty }\left[\left(e^{-t}\right) \vec{i}+\left(\frac{t-1}{t+1}\right)\vec{j}+\left(\tan ^{-1}t \right)\vec{k}\right]$$
• $$\lim _{t\to 0^+}\left[\frac{\sin 3t}{\sin 2t} \vec{i}+\frac{\ln (\sin t)}{\ln (\tan t)}\vec{j}+(t \ln t)\vec{k}\right]$$

Exercise 1.15 Determine all real numbers $$a$$ that satisfies
$\lim _{t\to 0}\left[\frac{ t}{\sin a t} \vec{i}+\frac{a}{a-\cos t}\vec{j}+\left(e^{a-t}\right)\vec{k}\right]=\frac{3}{2}\vec{i}-2\vec{j}+e^{2/3}\vec{k}.$

Exercise 1.16 Find a vector function $$\vec{F}(t)= \langle f_1(t),f_2(t),f_3(t)\rangle$$ such that $$\|\vec{F}(t)\|$$ is continuous at $$t=0$$ but $$\vec{F}(t)$$ is not continuous at $$t=0.$$

Exercise 1.17 Determine the intervals for which both vector functions $$\vec{F}(t)= \langle e^{-t},t^2,\tan t\rangle$$ and $$\vec{G}(t)= \langle 8,\sqrt{t},\sqrt{t}\rangle$$ are continuous.

Exercise 1.18 Show that if $$\vec{F}$$ is a vector function that is continuous at $$c,$$ then $$\|\vec{F}\|$$ is continuous at $$c.$$

Exercise 1.19 Find the interval(s) on which the vector function $$\vec{r}(t)=e^{-2t}\vec{i}+\cos\sqrt{9-t} \, \vec{j}+\frac{1}{t^2-1}\vec{k}$$ is continuous.

## 1.9 Derivatives of Vector Functions

As with functions of one variable we define the derivative as the limit of a difference quotient; and then we develop a theorem which allows us to compute derivatives based on previously known differentiation rules.
Recall the definition of difference quotient from single variable calculus, namely, the difference quotient of $$y=f(x)$$ with respect to a change $$\Delta x$$ in $$x$$ is defined by $\frac{f(x+\Delta x)-f(x)}{\Delta x}.$ We now generalize this idea to vector functions.

Definition 1.5 The difference quotient of a vector function $$\vec{F}$$ is the vector function $\frac{\Delta \vec{F}}{\Delta t}=\frac{\vec{F}(t+\Delta t)-\vec{F}(t)}{\Delta t}$ where $$\Delta t$$ is an increment of the variable $$t.$$

Notice that if the component functions of a vector function $$\vec{F}$$ are $$f_1$$, $$f_2$$, and $$f_3$$, then $\begin{equation} \label{diffvec} \frac{\Delta \vec{F}}{\Delta x} =\frac{\Delta f_1}{\Delta x} \vec{i} +\frac{\Delta f_2}{\Delta x} \vec{j}+\frac{\Delta f_3}{\Delta x} \vec{k} \end{equation}$

In light of $$\ref{diffvec}$$, the definition of the derivative of vector function seems natural.

Definition 1.6 The derivative of a vector function $$\vec{F}$$ is the vector function $$\vec{F}\,'$$ defined as the limit
$\vec{F}\,'(t)=\lim_{\Delta t\to 0} \frac{\Delta \vec{F}}{\Delta t}=\lim_{\Delta t\to 0} \frac{\vec{F}(t+\Delta t)-\vec{F}(t)}{\Delta t}$ provided this limit exists. If $$\vec{F}\,'(t)$$ exists for a given value of $$t$$, then we say $$\vec{F}$$ is differentiable at $$t$$.

::: {#lem- } [Differentiable Vector Functions] Any vector function
$\vec{F}(t)=f_1(t) \vec{i}+f_2(t) \vec{j}+f_3(t) \vec{k}$ is differentiable whenever the component functions $$f_1$$, $$f_2$$, and $$f_3$$ are each differentiable and in this case $\vec{F}\, '(t)=f_1'(t) \vec{i} +f_2'(t) \vec{j}+f_3'(t)\vec{k}.$ :::

Proof. By the definition of the derivative of a vector function and $$\ref{diffvec}$$ we find that \begin{align*} \vec{F}\, '(t) & =\lim_{\Delta t\to 0} \frac{\vec{F}(t+\Delta t)-\vec{F}(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0} \left[ \frac{\Delta f_1}{\Delta x} \vec{i} +\frac{\Delta f_2}{\Delta x} \vec{j}+\frac{\Delta f_3}{\Delta x} \vec{k} \right] \\ & = \left(\lim_{\Delta t\to 0} \frac{\Delta f_1}{\Delta x} \right) \vec{i} + \left(\lim_{\Delta t\to 0} \frac{\Delta f_2}{\Delta x} \right) \vec{j}+ \left( \lim_{\Delta t\to 0} \frac{\Delta f_3}{\Delta x} \right) \vec{k} \end{align*} By the hypothesis that the component functions $$f_1$$, $$f_2$$, and $$f_3$$ are each differentiable we have $\vec{F}\, '(t) = f_1'(t) \vec{i} +f_2'(t) \vec{j}+f_3'(t)\vec{k}.$ as needed.

Example 1.12 Find the derivative of the vector function $\vec{F}(t)=(\ln t)\vec{i}+\frac{1}{2}t^3\vec{j}-t \vec{k}.$

Solution. By $$\ref{vfder}$$, we find that $\vec{F}\, '(t)=\frac{1}{t}\vec{i}+\frac{3}{2}t^2\vec{j}-\vec{k}$
is the derivative is the vector function $$\vec{F}$$.

::: {#lem- } [Derivative Rules for Vector Functions] If the vector functions $$\vec{F},$$ $$\vec{G}$$ and the scalar function $$h$$ are differentiable at $$t$$, and if $$a$$ and $$b$$ are constants, then $$a \, \vec{F}+b\, \vec{G},\vec{F}\cdot \vec{G}, \text{and}\text{ }\vec{F}\times \vec{G}$$ are differentiable at $$t$$ and,

• $$(a \, \vec{F}+b \, \vec{G})'(t)=a \, \vec{F}\, '(t)+b \, \vec{G}'(t)$$
• $$(h \, \vec{F})'(t)=h'(t)\vec{F}(t)+h(t)\vec{F}\, '(t)$$
• $$(\vec{F}\cdot \vec{G})'(t)=\vec{F}\, '(t)\cdot \vec{G}(t)+\vec{F}(t)\cdot \vec{G}'(t)$$
• $$(\vec{F}\times \vec{G})'(t)=\vec{F}\, '(t)\times \vec{G}(t)+\vec{F}(t)\times \vec{G}'(t)$$
• $$\vec{F}(h(t))'=h'(t)\vec{F}\, '(h(t)).$$

:::

Proof. The proof is left for the reader.

Example 1.13 Compute the derivative of the vector function given by $$\vec{F}(t)\times \vec{G}(t)$$ where $$\vec{F}(t)=t^2\vec{i}+t \vec{j}+\vec{k}$$ and $$\vec{G}(t)=\vec{i}+t \vec{j}+t^2\vec{k}.$$

Solution. By $$\ref{VFD}$$, we find \begin{align*} & [\vec{F}(t)\times \vec{G}(t)]'=\vec{F}\, '(t)\times \vec{G}(t)+\vec{F}(t)\times \vec{G}'(t) \\ & \qquad =\left[(2t \vec{i}+\vec{j})\times \left(\vec{i}+t \vec{j}+t^2\vec{k}\right)\right]+\left[\left(t^2\vec{i}+t \vec{j}+\vec{k}\right)\times ( \vec{j}+2t \vec{k})\right] \\ & \qquad = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 t & 1 & 0 \\ 1 & t & t^2 \\ \end{array} \right| + \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ t^2 & t & 1 \\ 0 & 1 & 2t \end{array} \right| \\ & \qquad =\left(3t^2-1\right)\vec{i}-4t^3\vec{j}+\left(3t^2-1\right)\vec{k}. \end{align*}

## 1.10 Tangent Vectors

::: {#thm- } Tangent Vectors Suppose $$\vec{F}(t)$$ is differentiable at $$t_0$$ and that $$\vec{F}'\left(t_0\right)\neq 0.$$ Then $$\vec{F}'\left(t_0\right)$$ is a tangent vector to the graph of $$\vec{F}(t)$$ at the point where $$t=t_0$$ and points in the direction of increasing $$t$$.
:::

Proof. Let $$t_0$$ be a number in the domain of the vector function $$\vec{F}$$, and let $$P$$ be the point on the graph of $$\vec{F}$$ that corresponds to $$t_0$$. Then for any positive number $$\Delta t$$, the difference quotient $\frac{\Delta \vec{F}}{\Delta t}=\frac{\vec{F}(t+\Delta t)-\vec{F}(t)}{\Delta t}$ is a vector that points in the same direction as the secant vector $P Q=\vec{F}\left(t_0+\Delta t\right)-\vec{F}\left(t_0\right)$ where $$Q$$ is the point on the graph of $$\vec{F}$$ that corresponds to $$t=t_0+\Delta t$$. Suppose the difference quotient $$\Delta \vec{F}/\Delta t$$ has a limit as $$\Delta t\to 0$$ and that $\lim_{\Delta \to 0} \frac{\Delta \vec{F}}{\Delta t}\neq 0.$ Then, as $$\Delta t\to 0$$, the direction of the secant vector, $$P Q$$, and hence that of the difference quotient $$\Delta \vec{F}\Delta t$$, will approach the direction of the tangent vector of $$P$$. Thus we expect the tangent vector at $$P$$ to be the limit vector $\lim_{\Delta \to 0} \frac{\Delta \vec{F}}{\Delta t}$ which is the vector derivative $$\vec{F}'\left(t_0\right).$$

Example 1.14 Find a tangent vector at the point where $$t=2$$ for
$\vec{F}(t)=\left(t^2+t\right)\vec{i}-e^t\vec{j}+\sqrt{t}\vec{k}.$ Also find parametric equations for the tangent line to the graph of $$\vec{F}$$ that passes through the point corresponding to $$t=2$$.

Solution. Using $$\ref{vfder}$$, we find a tangent vector $\vec{F}\, '(2)=\left.(2t+1)\vec{i}-e^t\vec{j}+\frac{1}{2\sqrt{t}}\vec{k}\right|_{t=2}=5\vec{i}-e^2\vec{j}+\frac{1}{2\sqrt{2}}\vec{k};$ and the tangent line to the graph of $$\vec{F}(t)$$ for $$t=2$$ is the line that passes through the point $$\left(6,-e^2,\sqrt{2}\right)$$ and is determined by the parametric equations $x(t)=6+5t, \quad y(t)=-e^2-e^2t, \quad \text{ and } \quad z(t)=\sqrt{2}+\frac{1}{2\sqrt{2}}t$ because this line passes through $$\vec{F}(2)=6\vec{i}-e^2\vec{j}+\sqrt{2}\vec{k}$$ and is parallel to the tangent vector at $$t=2.$$

Example 1.15 Find parametric equations for the tangent line to the graph of
$\vec{R}(t)=t e^{-2t}\vec{i}+t^2\vec{j}+t e^{-2t}\vec{k}$ at the highest point on the graph.

Solution. We want $$\frac{d z}{d t}=0$$ where $$z=t e^{-2t}$$ and since
$$\frac{dz}{dt}=e^{-2t}-2t e^{-2t}$$
we find that $$t=1/2.$$ Further,
$\vec{R}'(t)=\left(e^{-2t}-2t e^{-2t}\right)\vec{i}+2t \vec{j}+\left(e^{-2t}-2t e^{-2t}\right)\vec{k}$ and so
$\vec{R}'\left(\frac{1}{2}\right)=\left(e^{-1}- e^{-1}\right)\vec{i}+\vec{j}+\left(e^{-1}- e^{-1}\right)\vec{k}=0\vec{i}+\vec{j}+0\vec{k}.$ Also, $\vec{R}\left(\frac{1}{2}\right)=\frac{1}{2e^{1/2}}\vec{i}+\frac{1}{4}\vec{j}+\frac{1}{2e}\vec{k}$ and therefore, $x(s)=\frac{1}{2e}, \quad y(s)=s+\frac{1}{4}, \quad z(s)=\frac{1}{2e}.$ are the parametric equations for the tangent line at the highest point.

## 1.11 Unit Tangent and Unit Normal Vectors

Definition 1.7 If the graph of the vector function $$\vec{R}(t)$$ is smooth, then at each point $$t$$ a unit tangent vector is defined by $\vec{T}(t)=\frac{\vec{R}'(t)}{\norm{\vec{R}'(t)}}$ and the principal unit normal vector function is defined by $\vec{N}(t)=\frac{\vec{T}'(t)}{\norm{\vec{T}'(t)}}.$

The unit normal vector is orthogonal (or normal, or perpendicular) to the unit tangent vector and hence to the curve as well.

Lemma 1.1 Suppose that $$\vec{R}$$ is a vector such that $$\norm{R(t)}=c$$ for all t. Then $$\vec{R}(t)$$ is orthogonal to $$\vec{R}'(t)$$.

Proof. Notice $$\vec{R}(t)\cdot \vec{R}(t)=\norm{\vec{R}(t)}^2=c^2$$, for all $$t$$. We differentiate with respect to $$t$$ to find, $\vec{R}'(t)\cdot \vec{R}(t)+\vec{R}'(t)\cdot \vec{R}(t)=0$ which yields $$\vec{R}'(t)\cdot \vec{R}(t)=0$$ and so $$\vec{R}'(t)$$ is orthogonal to $$\vec{R}(t)$$.

Example 1.16 Given the vector function defined by $\vec{R}(t)=\langle -4t,\sin 2t,-\cos 2t\rangle$ find the unit tangent vector $$\vec{T}(t)$$ and unit normal vector $$\vec{N}(t).$$

Solution. We find \begin{align*} & \vec{T}(t) = \displaystyle \frac{\vec{R}'(t)}{\norm{\vec{R}'(t)}} \\ & \qquad =\frac{\left\langle -4,2\cos 2t,2\sin 2 t\right\rangle }{\sqrt{(-4)^2+4\cos ^22t+4\sin ^22t}} \\ &\qquad =\frac{\langle -4, 2\cos 2t, 2\sin2t \rangle}{2\sqrt{5}} \\ & \qquad =\left\langle -\frac{4}{2\sqrt{5}},\frac{2}{2\sqrt{5}}\cos 2t,\frac{2}{2\sqrt{5}}\sin 2 t \right\rangle \\ & \qquad =\left\langle -\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\cos 2t,\frac{1}{\sqrt{5}}\sin 2 t\right\rangle\end{align*} and
$\vec{N}(t) =\frac{\vec{T}'(t)}{\norm{\vec{T}'(t)}} =\frac{\left\langle 0,\frac{-2}{\sqrt{5}}\sin 2t,\frac{2}{\sqrt{5}}\cos 2t\right\rangle }{\sqrt{\frac{4}{5}}} =\langle 0,-\sin 2t,\cos 2t\rangle.$ as the unit normal vector.

## 1.12 Integrals of Vector Functions

Next we study vector integration . Since integration is a linear process, studying vector functions and integration together is natural. Indeed recall that with functions of one variable, definite integration is the process of taking a limit of a Riemann sum. Basically, since taking limits and Riemann sums both are linear processes, so is integration.

Definition 1.8 Let
$$\vec{F}(t)=f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k}$$,
where $$f_1(t)$$, $$f_2(t)$$, and $$f_3(t)$$ are continuous on the closed interval $$a\leq t\leq b.$$ Then the indefinite integral of $$\vec{F}(t)$$ is $\int \vec{F}(t)\, dt=\left[\int f_1(t)\, dt\right]\vec{i}+\left[\int f_2(t)\, dt\right]\vec{j}+\left[\int f_3(t)\, dt\right]\vec{k}.$

Recall integration by parts: $\int u \, dv =u v -\int v du$ and so use $$u=\ln t$$ and $$dv=t \, dt.$$

Example 1.17 Evaluate $$\int \langle t \ln t,-\sin (1-t),t\rangle \, dt.$$

Solution. We find that the given integral is equal to \begin{align*} & =\left(\int t \ln t \, dt\right)\vec{i}+\left(\int -\sin(1-t)\, dt\right)\vec{j}+\left(\int t\, dt\right)\vec{k} \\ & =\left(\frac{-t^2}{4}+\frac{t^2}{2}\ln t\right)\vec{i}-\cos (1-t)\vec{j}+\left(\frac{t^2}{2}\right)\vec{k}+\vec{C} \end{align*} where $$\vec{C}$$ is a constant vector.

Definition 1.9 Let $$\vec{F}(t)=f_1(t)\vec{i}+f_2(t)\vec{j}+f_3(t)\vec{k}$$,
where $$f_1(t)$$, $$f_2(t)$$, and $$f_3(t)$$ are continuous on the closed interval $$a\leq t\leq b.$$ Then the definite integral of $$\vec{F}(t)$$ is the vector $\int _b^a\vec{F}(t)\, dt=\left[\int _b^af_1(t)\, dt\right]\vec{i}+\left[\int _b^af_2(t)\, dt\right]\vec{j}+\left[\int _b^af_3(t)\, dt\right]\vec{k}.$

Example 1.18 Given the vector function
$\vec{F}(t)=\left(t\sqrt{1+t^2}\right)\vec{i}+\left(\frac{1}{1+t^2}\right)\vec{j}.$ Find a value of $$a$$ for which $$\int_0^a \vec{F}(t) \, dt=\frac{2\sqrt{2}}{3}\vec{i}+\frac{\pi }{4}\vec{j}.$$

Solution. By definition, we find \begin{align*} \int_0^a \vec{F}(t) \, dt & =\left(\int _0^at\sqrt{1+t^2}\, dt\right)\vec{i}+\left(\int _0^a\frac{1}{1+t^2}\, dt\right)\vec{j} \\ & =\left(\frac{1}{3}\sqrt{\left(1+a^2\right)^3}\right)\vec{i}+\left(\tan ^{-1} a\right)\vec{j}. \end{align*} Thus we require $$\left(1+a^2\right)^3=8$$ and $$\tan ^{-1}a=\frac{\pi }{4}.$$ Therefore we find $$a=1.$$

vMotion of an Object

If the graph of a vector function $$\vec{R}(t)$$ is a smooth curve $$C$$ then the nonzero derivative $$\vec{R}'(t)$$ is tangent to $$C$$ at the point $$P$$ that corresponds to $$t$$; and in this case we can make the following definition.

Definition 1.10 If an object moves in such a way that its position at time $$t$$ is given by the vector function $$\vec{R}(t)$$ whose graph is a smooth curve $$C,$$ then

• $$\vec{V(t)}=\frac{d \vec{R}}{d t}$$ is the object’s velocity vector at time $$t$$,
• $$\|\vec{V(t)}\|$$ is the speed of the object at time $$t$$,
• $$\frac{\vec{V(t)}}{\|\vec{V(t)}\|}$$ is the direction of the object’s motion at time $$t$$, and
• $$\frac{d \vec{V(t)}}{d t}$$ is the object’s acceleration at time $$t$$.

Example 1.19 Suppose the position vector for a particle in space at time $$t$$ is given by $\vec{R}(t)=e^t\vec{i}+e^{-t}\vec{j}+e^{2t}\vec{k}.$ Find the particle’s velocity vector, acceleration vector, speed, and direction of motion vector at time $$t=\ln 2.$$

Solution. At time $$t=\ln 2$$, the particle’s velocity vector is
$\vec{R}'(t)|_{t=\ln 2}=e^t\vec{i}-e^{-t}\vec{j}+2e^{2t}\vec{k}|_{t=\ln 2}=2\vec{i}-\frac{1}{2}\vec{j}+8\vec{k},$ the acceleration vector is $\vec{R}''(t)|_{t=\ln 2}=e^t\vec{i}+e^{-t}\vec{j}+4e^{2t}\vec{k}|_{t=\ln 2}=2\vec{i}+\frac{1}{2}\vec{j}+16\vec{k},$ the speed is
$\vec{R}'(t)|_{t=\ln 2}=\sqrt{2^2+\left(\frac{-1}{2}\right)^2+8^2}=\sqrt{68.25},$ and the direction of motion vector is
$\begin{equation} \frac{2}{\sqrt{68.25}}\vec{i}-\frac{1}{2\sqrt{68.25}}\vec{j}+\frac{8}{\sqrt{68.25}}\vec{k}. \tag*{} \end{equation}$

Example 1.20 Find the position vector $$\vec{R}(t)$$ and velocity vector $$\vec{V}(t),$$ given the acceleration vector function $\vec{A}(t)=t^2 \vec{i}-2\sqrt{t } \vec{j}+e^{3t} \vec{k},$ initial position vector $$\vec{R}(0)=2 \vec{i}+\vec{j}-\vec{k},$$ and initial velocity vector $$\vec{V}(0)=\vec{i}-\vec{j}-2\vec{k}.$$

Solution. Given $$\vec{A}(t)=t^2 \vec{i}-2\sqrt{t } \vec{j}+e^{3t} \vec{k}$$ the velocity vector function is
$\vec{V}(t)=\left(\int t^2 dt\right)\vec{i}-\left(\int 2\sqrt{t }dt\right) \vec{j}+\left(\int e^{3t} dt\right) \vec{k}$ $\vec{V}(t)=\left(\frac{t^3}{3}+C_1\right) \vec{i}-\left(\frac{4 t^{3/2}}{3}+C_2\right)\vec{j}+\left(\frac{e^{3 t}}{3}+C_3\right) \vec{k}$ where $$C_1, C_2,$$ and $$C_3$$ are constants to be determined. By using
$\vec{V}(0)=\vec{i}-\vec{j}-2\vec{k} =\left(0+C_1\right)\vec{i}+\left(-0+C_2\right)\vec{j}+\left(\frac{1}{3}+C_3\right)\vec{k}$ we find $$C_1=1,$$ $$C_2=1,$$ $$C_3=-7/3.$$ Therefore,
$\vec{V}(t)=\left(\frac{t^3}{3}+1\right)\vec{i}+\left(-\frac{4 t^{3/2}}{3}-1\right)\vec{j}+\left(\frac{e^{3 t}-7}{3}\right) \vec{k}.$ So the position vector function is
$\vec{R}(t)=\left(\int \left(\frac{t^3}{3}+1\right) \, dt\right) \vec{i}+\left(\int \left(-\frac{4 t^{3/2}}{3}-1\right) \, dt\right)\vec{j}+\left(\int \frac{e^{3 t}-7}{3} \, dt\right) \vec{k}$ which is $\vec{R}(t)=\left(\frac{t^4}{12}+t+K_1\right) \vec{i}+\left(-\frac{8 t^{5/2}}{15}-t+K_2\right)\vec{j}+\left(\frac{1}{3} \left(\frac{e^{3 t}}{3 }-7 t\right)+k_3\right) \vec{k}$ where $$K_1, K_2,$$ and $$K_3$$ are constants to be determined. By using
$\vec{R}(0)=2 \vec{i}+\vec{j}-\vec{k} =\left(0+K_1\right)\vec{i}+\left(-0+K_2\right)\vec{j}+\left(\frac{1}{9}+K_3\right)\vec{k}$ we find $$K_1=2,$$ $$K_2=1,$$ $$K_3=-10/9.$$ Therefore,
$\vec{R}(t)=\left(\frac{t^4}{12}+t+2\right) \vec{i}+\left(-\frac{8 t^{5/2}}{15}-t+1\right)\vec{j}+\left(\frac{1}{3} \left(\frac{e^{3 t}}{3 }-7 t\right)-\frac{10}{9}\right) \vec{k}$ is the required vector function.

Theorem 1.1 Neglecting air resistance, the path of a projectile launched from an initial height $$h$$ with initial speed $$v_0$$ and an angle of elevation $$\theta$$ is described by the vector function $\begin{equation} \label{prolaueq} \vec{R}(t) =(v_0\cos \theta)t\vec{i}+\left[h+(v_0\sin \theta)t-\frac{1}{2}g t^2)\right]\vec{j} \end{equation}$ where $$g$$ is he gravitional constant.

Proof. Suppose a projectile of mass $$m$$ is launched from an initial position $$\vec{R}_0$$ with an initial velocity $$\vec{V}_0$$. First we will find its position vector as a function of time. We begin with acceleration $$\vec{A}(t)=-g \vec{j}$$ and we integrate twice, namely, \begin{align*} & \vec{V}(t)=\int \vec{A}(t)\, dt=\int -g \vec{j} \, dt =-g t \vec{j}+\vec{C}_1 \\ & \vec{R}(t)= \int \vec{V}(t) \, dt =\int (g t \vec{j}+\vec{C}_1) \, dt =-\frac{1}{2} g t^2 \vec{i}+\vec{C}_1 t+\vec{C}_2 \end{align*} where $$\vec{C}_1$$ and $$\vec{C}_2$$ can be determined from the initial conditions. Using $$\vec{V}(0)=\vec{V}_0$$, $$\vec{R}(0)=\vec{R}_0$$, produces $$\vec{C}_1=\vec{V}_0$$ and $$\vec{C}_2=\vec{R}_0$$. Therefore the position vector function is $\begin{equation} \label{proeq} \vec{R}(t)=-\frac{1}{2} g t^2 \vec{j} +t \vec{V}_0 +\vec{R}_0. \end{equation}$ Now from the given height $$h$$ we realize that $\vec{R}_0=h \vec{j}$ and because speed is the magnitude of the velocity, that is $$V_0=||\vec{V}_0||$$, we have $\vec{V}_0=x\vec{i}+y \vec{j} =\left(||\vec{V}_0|| \cos \theta \right) \vec{i}+\left(||\vec{V}_0|| \sin \theta \right) \vec{j} =V_0\cos \theta\vec{i}+V_0\sin \theta\vec{j}.$ Now we can derive $$\ref{prolaueq}$$ as follows \begin{align*} R(t) & =-\frac{1}{2} g t^2 \vec{j} +t \vec{V}_0+\vec{R}_0 \\ & = -\frac{1}{2} g t^2 \vec{j} +tV_0 \cos \theta \vec{i} + t (V_0\cos \theta\vec{i}+V_0\sin \theta\vec{j}) +h \vec{j} \\ & =(v_0\cos \theta)t\vec{i}+\left[h+(v_0\sin \theta)t-\frac{1}{2}g t^2)\right]\vec{j} \end{align*} as desired.

## 1.13 Exercises

Exercise 1.20 Evaluate $$\int_0^6 \langle t^3-t,t^2-2\rangle \, dt$$.

Exercise 1.21 Evaluate $$\int \left( 2t \vec{i}+9t^2 \vec{j} +7\vec{k}\right) \, dt$$.

Exercise 1.22 Find the vector derivative $$\vec{F}\, '$$ for each of the following functions.

• $$\vec{F}(s)=\left(s \vec{i}+s^2\vec{j}+s^2\vec{k}\right)+\left(2s^2 \vec{i}-s \vec{j}+3\vec{k}\right).$$
• $$\vec{F}(s)=\left(1-2s^2\right)\vec{i}+(s \cos s) \vec{j}-s \vec{k}.$$
• $$\vec{F}(\theta )=\left(\sin ^2\theta \right)\vec{i}+(\cos 2\theta ) \vec{j}+\theta ^2 \vec{k}.$$

Exercise 1.23 Find the indefinite vector integral for each of the following.

• $$\int \langle\cos t,\sin t,-2t\rangle \, dt.$$
• $$\int \langle 3 e ,t^2, t \sin t\rangle \,dt.$$
• $$\int e^{-t} \langle 3,t,\sin t\rangle \,dt.$$
• $$\int \langle\sinh t, -3, \cosh t\rangle \, dt.$$

Exercise 1.24 Given $$\vec{F}(t)=t^2\vec{i}+2t \vec{j}+\left(t^3+t^2\right)\vec{k}$$ determine all real numbers $$a$$ such that $$\vec{F}\, '(0)+\vec{F}\, '(1)+\vec{F}\, '(-1)=a \vec{j}+a \vec{k}.$$

Exercise 1.25 Given $$\vec{F}(t)=t^2\vec{i}+\cos t \vec{j}+t^2\cos t \vec{k}$$ determine all real numbers $$a$$ such that $$\vec{F}\, '(0)+\vec{F}\, '(a)=\pi \vec{i} -\vec{j}-a^2 \vec{k}.$$

Exercise 1.26 Determine all real numbers $$a$$ such that the parametric equations for the tangent line to the graph of the vector function $$\vec{F}(t)=t^{-3}\vec{i}+t^{-2}\vec{j}+t^{-1}\vec{k}$$ at the point corresponding to $$t=-1$$ are
$x=\frac{-3}{a}-a t, \quad y=\frac{a}{3}+\frac{6}{a}t, \quad z=\frac{-3}{a}-\frac{3}{a}t.$

Exercise 1.27 Find the first and second derivatives for each of the following functions.

• $$\vec{F}(t)=(\ln t)\left[t \vec{i}+5 \vec{j}-e^t\vec{k}\right]$$
• $$\vec{F}(t)=(\sin t) \vec{i}+(\cos t) \vec{j}+t^2\vec{k}$$
• $$f(x)=\left\|\left(x \vec{i}+x^2\vec{j}-20\vec{k}\right)+\left(x^3 \vec{i}+ x \vec{j}-x \vec{k}\right)\right\|$$.

Exercise 1.28 Determine all real numbers $$a$$ such that $f'(x)=-a x^2-\frac{18}{a}x$ given $f(x)=[x \vec{i}+(x+1)\vec{j}]\cdot \left[2x \vec{i}-3x^2\vec{j}\right].$

Exercise 1.29 Determine all real numbers $$a$$ such that $F''(t)=a \vec{i}+a t^{-3}\vec{j}+2a e^{a t}\vec{k}$ given $$\vec{F}(t)=t^2\vec{i}+t^{-1}\vec{j}+e^{2 t}\vec{k}.$$

Exercise 1.30 Determine all real numbers $$a$$ such that $$\vec{F}\, '(t)$$ and $$\vec{F}\, '''(t)$$ are parallel for all $$t$$ given $$\vec{F}(t)=e^{a t}\vec{i}+e^{- a t}\vec{j}.$$

Exercise 1.31 Prove that if $$\vec{F}$$ is a differentiable vector function such that $$\vec{F}(t)\neq 0,$$ then
$\frac{d}{d t}\left(\frac{\vec{F}(t)}{\|\vec{F}(t)\|}\right)=\frac{\vec{F}\, '(t)}{\|\vec{F}(t)\|}-\frac{[\vec{F}(t)\cdot \vec{F} '(t)]\vec{F}(t)}{\left\|\vec{F}(t)\|^3\right.}.$

Exercise 1.32 Prove that if $$\vec{F}, \vec{G},$$ and $$\vec{H}$$ are differentiable vector functions, then
\begin{align*} %& [\vec{F}\cdot (\vec{G}\times \vec{H})] ' %\\ & \qquad =\vec{F} \, '\cdot (\vec{G}\times \vec{H})+\vec{F}\cdot (\vec{G} \, ' \times \vec{H})+\vec{F}\cdot (\vec{G}\times \vec{H} \,' ). \end{align*}

Exercise 1.33 Prove that if $$\vec{F}, \vec{G},$$ and $$\vec{H}$$ are differentiable vector functions, then $[\vec{F}\times (\vec{G}\times \vec{H})] ' =[(\vec{H}\cdot \vec{F})\vec{G}] '-[(\vec{G}\cdot \vec{F})\vec{H}] '.$

Exercise 1.34 Find a value of $$a$$ and $$b$$ such that
$\int_0^a \left[t\sqrt{1+t^2}\vec{i}+\left(\frac{1}{1+t^2}\right)\vec{j}\right] \, dt=\frac{2\sqrt{2}-1}{3}\vec{i}+\frac{\pi }{4}\vec{j}$ and
$\int_0^b [\cos t \vec{i}+\sin t \vec{j}+\sin t \cos t \vec{k}] \, dt=\vec{i}+\vec{j}+\frac{1}{2}\vec{k}.$

Exercise 1.35 Show that if $$\vec{R}(t)= f(t)\, \vec{i}+g(t)\, \vec{j}+h(t)\, \vec{k}$$ is differentiable at $$t=t_0$$, then it is continuous at $$t_0$$ as well.

Exercise 1.36 Suppose that the scalar function $$u(t)$$ and the vector function $$\vec{R}(t)$$ are both defined for $$a \leq t \leq b$$. Show that $$u \, \vec{R}$$ is continuous on $$[a,b]$$ if $$u$$ and $$\vec{R}$$ are continuous on $$[a,b]$$. If $$u$$ and $$\vec{R}$$ are both differentiable on $$[a,b]$$, show that $$u \, \vec{R}$$ is differentiable on $$[a,b]$$ and that $\frac{d}{dt}(u\, \vec{R} )=u\, \frac{d \vec{R}}{dt}+\vec{R} \frac{du}{dt}.$

Exercise 1.37 Use the Mean Value Theorem to show that if $$\vec{R}_1(t)$$ and $$\vec{R}_2(t)$$ have identical derivatives on an interval $$I$$, then the vector functions differ by a constant vector value throughout $$I$$.

Exercise 1.38 Suppose $$\vec{r}$$ is a continuous vector function on $$[a,b]$$. Show that if $$\vec{R}$$ is any antiderivative of a $$\vec{r}$$ on $$[a,b]$$ then if $\vec{r}(t)=\frac{d}{dt} \int_a^t \vec{r}(s) \, ds \quad \text{and} \quad \int_a^b \vec{r}(t)\, dt = \vec{R}(a)-\vec{R}(b).$

Exercise 1.39 Find a vector function describing the curve of intersection of the plane $$x+y+2z=2$$ and the paraboloid $$z=x^2+y^2$$. Also find the point(s) on the curve that are closest to and farthest from the origin.

Exercise 1.40 Prove that $\frac{d}{dt}\left[\vec{R}(t)\times \vec{R}\,'(t)\right] =\vec{R}(t)\times \vec{R}\,''(t).$

Exercise 1.41 Let $$\vec{R}(t)=\sin t \, \vec{i}+\cos t \, \vec{j} +t \, \vec{k}$$ and $$c=2\, \vec{i}+3\, \vec{j}-\vec{k}$$ show that $\int_a^b \vec{c}\cdot \vec{R}(t)= \vec{c} \cdot \int_a^b \vec{R}(t);$ also show that this formula holds true for any vector function $$\vec{R}(t)$$ that is integrable on $$[a,b]$$ and for any vector constant $$\vec{c}$$.

Exercise 1.42 Find the unit tangent vector $$\vec{T}(t)$$ for $$\vec{r}(t)=8t\vec{i}+8t\vec{j}+4t\vec{k}$$ at $$t=2$$.

Exercise 1.43 Find the unit tangent vector $$\vec{T}(t)$$ and the unit normal vector $$\vec{N}(t)$$.

• $$\vec{R}(t)=t^2\vec{i}+\sqrt{t}\vec{j},$$ with $$t>0$$
• $$\vec{R}(t)=(t \cos t)\vec{i}+(t \sin t)\vec{j}$$
• $$\vec{R}(t)=t\vec{i}+(\ln \cos t)\vec{j}, \ -\pi/2<t<\pi/2$$
• $$\vec{R}(t)=(2t+3)\vec{i}+(4-t^2)\vec{j}$$
• $$\vec{R}(t)=(\ln \sec t)\vec{i}+t\vec{j} \ -\pi/2<t<\pi/2$$
• $$\vec{R}(t)=\sin t \vec{i}-\cos t \vec{j}+t \vec{k}$$
• $$\vec{R}(t)=(3\sin t)\vec{i}+(3\cos t)\vec{j}+4t\vec{k}$$
• $$\vec{R}(t)=(e^t\cos t)\vec{i}+(e^t \sin t)\vec{j}+4\vec{k}$$
• $$\vec{R}(t)=(t^3/3)\vec{i}+\vec{j}+(t^2/2)\vec{k}, \ t>0$$
• $$\vec{R}(t)=(\cos^3 t)\vec{i}+(\sin^3 t)\vec{j}+\vec{k}, \ 0<t<\pi/2$$

Exercise 1.44 Given $$\vec{R}(t)=(\ln t)\vec{i}+t^2\vec{j}$$ find $$a$$ so that
$\vec{N}(t)=\frac{-2}{\sqrt{1+a t^a}}\vec{i}+\frac{1}{\sqrt{1+a t^a}}\vec{j}.$

Exercise 1.45 Given $$\vec{R}(t)=(\cos t)\vec{i}+(\sin t)\vec{j}+t \vec{k}$$ find $$a$$ so that $$\vec{N}(t)=(- \cos a t) \vec{i}+(-\sin a t)\vec{j}.$$

Exercise 1.46 Determine values for $$a$$ and $$b$$ so that the following vector functions are smooth over $$[a,b].$$

• $$\vec{F}(t)=a t^3\vec{i}+b t^2\vec{j}+ a b t \vec{k}$$
• $$\vec{G}(t)=(a \sin b t)\vec{i}+(a \cos b t)\vec{j}+ a b \vec{k}$$
• $$\vec{H}(t)=\left(e^{a t}-b t\right)\vec{i}+t^{a/b}\vec{j}+ (b \cos a t) \vec{k}$$

Exercise 1.47 The velocity of a particle moving in space is $$\vec{V}(t)=e^t\vec{i}+t^2\vec{j}.$$ Find the vector $$\vec{R}(0)$$ so that the particle’s position as a function of $$t$$ is
$\vec{R}(t)=e^t\vec{i}+\left(\frac{1}{3}t^3-1\right)\vec{j}.$

Exercise 1.48 The acceleration of a moving particle is $$\vec{A}(t)=24t^2\vec{i}+4 \vec{j}.$$ Find the vectors $$\vec{R}(0)$$ and $$\vec{V}(0)=0$$ so that the particle’s position as a function of $$t$$ is $\vec{R}(t)=\left(2t^4+1\right)\vec{i}+\left(2t^2+2\right)\vec{j}.$

Exercise 1.49 Show that $$\vec{N}(t)=-g'(t)\vec{i}+f'(t)\vec{j}$$ and $$-\vec{N}(t)=g'(t)\vec{i}-f'(t)\vec{j}$$ are both normal to the curve $$\vec{R}(t)+f(t)\vec{i}+g(t)\vec{j}$$ at the point $$(f(t),g(t))$$.

Exercise 1.50 A baseball is hit above ground at 100 feet per second at an angle of $$\pi/4$$ with respect to the ground. Find the maximum height reached by the baseball. Will it clear a 10-foot high fence located 300 feet from home plate?

Exercise 1.51 Find the vector function for the path pf a projectile launched at a height of 10 feet above the ground with an initial velocity of 88 feet per second and at an angle of $$30^\circ$$ above the horizontal. Sketch the path of the projectile.

Exercise 1.52 Find the angle at which an object must be thrown to obtain (a) the maximum range, and (b) he maximum m height.

Exercise 1.53 A projectile is fired from ground level at an angle of $$10^\circ$$ with the horizontal. Find the minimum initial velocity necessary of the projectile is to have a range of 100 feet.

Exercise 1.54 Use linear approximation by finding a set of parametric equations for the tangent line to the graph at $$t=t_0$$ and use the equations for the line to approximate $$\vec{R}(t_0+.01)$$.

• $$\vec{R}(t)=t \vec{i}-t^2 \vec{j} +\frac{t^3}{4}$$, $$t_0=1$$
• $$\vec{R}(t)=t \vec{i}+\sqrt{25-t^2} \vec{j} +\sqrt{25-t}$$, $$t_0=3$$

## 1.14 Smooth Curves

The one variable function $$f(x)=|x|$$ is not differentiable at $$x=0$$ because it has a corner" at $x=0.$ In three dimensions we have the concept of a vector function representing a smooth curve; that is, where there are no so calledcorners”. With an extra degree of freedom we require the derivative to not only exist but also to be continuous and nonzero.

Definition 1.11 The graph of the vector function defined by $$\vec{F}(t)$$ is smooth on any interval of $$t$$ where $$\vec{F}\,'$$ is continuous and $$\vec{F}'(t)\neq \vec{0}$$.

Example 1.21 Find the intervals on which the epicycloid $$C$$ given by $\vec{R}(t)=(5\cos t -\cos 5t) \vec{i} + (5\sin t - \sin 5t )\vec{j}, \qquad 0\leq t \leq 2\pi$ is smooth.

Solution. The derivative of $$\vec{R}(t)$$ is $\vec{R}(t)=(-5\sin t+5\sin 5t)\vec{i}+(5\cos t-5\cos 5t)\vec{j}.$ Notice that $$\vec{R}'$$ is continuous on $$[0,2\pi)$$. In the interval $$[0,2\pi)$$, the only values of $$t$$ for which $$\vec{R}'(t)=0\vec{i}+0\vec{j}$$ are found by solving the trigonometric equations: $-5\sin t+5\sin 5t =0 \qquad 5\cos t-5\cos 5t=0, \quad \text{ on } [0,2\pi).$ The first equation yields: $t=0, \quad t=\frac{7 \pi }{6}, \quad t=\frac{3 \pi }{2}, \quad t=\frac{11 \pi }{6}, \quad t=\frac{\pi }{6}, \quad t=\frac{\pi }{2}, \quad t=\frac{5 \pi }{6}$ The second equation yields: $t=0, \quad t=\pi, \quad t=\frac{4 \pi }{3}, \quad t=\frac{3 \pi }{2}, \quad t=\frac{5 \pi }{3}, \quad t=\frac{\pi }{3}, \quad t=\frac{\pi }{2}, \quad t=\frac{2 \pi }{3}$ Therefore, we can conclude that $$C$$ is smooth on the intervals $\left(0,\frac{\pi}{2}\right), \qquad \left(\frac{\pi}{2},\pi\right), \qquad \left(\pi, \frac{3\pi}{2}\right), \qquad \left(\frac{3\pi}{2}, 2\pi\right), \qquad$ as shown in $$\ref{fig:epismooth}$$

The graph of a vector function $$\vec{F}$$ is piecewise smooth on any interval that can be subdivided into a finite number of subintervals on which $$\vec{F}$$ is smooth. For instances, the graph in $$\ref{fig:epismooth}$$ is piecewise smooth on $$[0,2\pi)$$ since the interval $$[0,2\pi)$$ can be subdivided into a finite number of subintervals on which $$\vec{F}$$ is smooth.

Example 1.22 Determine where the graph of the vector function $\vec{F}(t)=\left(2t^3+\left(4-\frac{3\pi }{2}\right)t^2-4\pi t+2\pi \right)\vec{i}+\left(-\frac{1}{2}\cos 2 t \right)\vec{j}+(\sin t)\vec{k}.$ is piecewise smooth.

Solution. The graph of the vector function $$\vec{F}$$ is smooth over any interval not containing $$t=\pi/2$$ because
$\vec{F}'(t)=\left(6t^2+(8-3\pi )t-4\pi \right)\vec{i}+(\sin 2t)\vec{j}+(\cos t)\vec{k},$ $$\vec{F}'(t)\neq \vec{0}$$ for any $$t$$ except $$t=\frac{\pi }{2}$$, and $$\vec{F}'$$ is continuous everywhere.

Example 1.23 Determine where the graph of the vector function $\vec{F}(t) =\left(2t^2-\pi t+2\pi \right)\vec{i}+\left(\frac{1}{2}\sin 2t\right)\vec{j}+\left(\frac{-1}{4}\cos 4t\right)\vec{k}.$ is piecewise smooth.

Solution. The graph of the vector function $$\vec{F}$$ is piecewise smooth everywhere because $\vec{F}'(t)=(4t-\pi )\vec{i}+(\cos 2t)\vec{j}+(\sin 4t)\vec{k},$
$$\vec{F}'(t)\neq \vec{0}$$ except for $$t=\frac{\pi }{4}$$, and $$\vec{F}'$$ is continuous everywhere.

## 1.15 Arc Length

In this section, we first generalize the arc length formula from two to three dimensions. Recall that the arc length of the graph of a differentiable function $$y=f(x)$$ on the interval $$[a,b]$$ is given by
$s=\int _a^b\sqrt{1+\left(\frac{d y}{d x}\right)^2}dx.$ Now if the graph is parametrized by equations $$x=x(t)$$ and $$y=y(t)$$ then by the chain rule $\frac{d y}{d x} =\frac{d y}{d t} \frac{d t}{d x}$ whenever $$d x/d t \neq 0$$. After algebraic manipulation we find that $s=\int _{t_1}^{t_2}\sqrt{\frac{ \left( \frac{d x}{d t} \right)^2 +\left(\frac{d y}{d t}\right)^2}{\left(\frac{d x}{d t}\right)^2}}\left(\frac{dx}{dt}\right)dt =\int_{t_1}^{t_2}\sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2}\, dt$ represents the arc length of the curve $$y=f(x)$$ from the points corresponding to $$x=a$$ to $$x=b$$.

Suppose that $$C$$ is a plane curve described by the vector function $\vec{R}(t)=x(t)\vec{i}+y(t)\vec{j}$ instead. Then $$\vec{R}'(t)=x'(t)\vec{i}+y'(t)\vec{j}$$ where $$y=f(x)$$ and $|\vec{R}'(t)|=\sqrt{R'(t)\cdot R'(t)}=\sqrt{[x'(t)]^2+[y'(t)]^2}$ form which we see that $$s$$ can also be written in the form $s=\int_{t_1}^{t_2}\sqrt{ [x'(t)]^2+[y'(t)]^2 }\, dt = \int_{t_1}^{t_2}|R'(t)| \, dt$ In $$\mathbb{R}^3$$ we have $s=\int _{t_1}^{t_2}\sqrt\left(\frac{d x}{d t}\right)^2 +\left(\frac{d y}{d t}\right)^2+\left(\frac{d z}{d t}\right)^2}\, dt.$ But in fact the arc length of a graph is independent of its parametrization and thus,
$s(t)=\int_{t_0}^{ t} \sqrt{\left(\frac{d x}{d u}\right)^2 +\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2} \, du$ where $$P_0$$ is the base point corresponding to $$t_0.$$
This discussion motivates the following theorem.

::: {#thm- } [Arc Length Function] Let $$C$$ be a piecewise smooth curve that is the graph of the vector function described parametrically by $\vec{R}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}, \qquad a \leq t \leq b$ and let $$P_0=P\left(a\right)$$ be a particular point on $$C$$ ( base point ). If $$C$$ is transversed exactly once as $$t$$ increases from $$a$$ to $$t$$, then the length of $$C$$ from the base point $$P_0$$ to the variable $$P(t)$$ is given by the arc length function
$\begin{equation*} s(t)=\int _{a}^t\sqrt{\left(\frac{d x}{d u}\right)^2+\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du =\int_{a}^{t}||\vec{R}'(t)|| \, dt \end{equation*}$ :::

Example 1.24 Find the arc length of the curve $\vec{R}(t)=(1-2\cos t)\vec{i}+2\sin t \vec{j}+ 0.3t \vec{k}$ from $$t=-5$$ to $$t=5$$.

Solution. We find the arc length to be $$\sqrt{409}$$ units because $\frac{d x}{d t}=2\sin t, \quad \frac{d y}{d t}=2 \cos t, \quad \frac{d z}{d t}=0.3$ and so the arc length is $\int_{-5}^{5} \sqrt{4\sin ^2t+4\cos ^2t+(0.3)^2} \, dt=\int_{-5}^5 \frac{\sqrt{409}}{10} \, dt =\sqrt{409}.$ as claimed.

::: {#cor- } [Speed Along an Arc] Suppose an object moves along a smooth curve $$C$$ that is the graph of the position function $\vec{R}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t) \vec{k},$ where $$\vec{R}'(t)$$ is continuous on the interval $$\left[t_1,t_2\right].$$ Then the object has speed $$d s/d t$$ for $$t_1\leq t\leq t_2$$ where $s(t)=\int _{t_1}^t\sqrt{\left(\frac{d x}{d u}\right)^2+\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du.$ :::

Proof. Given that $$\vec{R}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}$$ is the position vector function for an object which moves along the graph of $$\vec{R}$$ and given that $$\vec{R}'(t)$$ is continuous on $$\left[t_1,t_2\right]$$ we can apply the Fundamental Theorem of Calculus to
$s(t)=\int _{t_1}^t\sqrt{\left(\frac{d x}{d u}\right)^2+\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du$ and obtain
$\frac{d s}{d t}=\sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2+\left(\frac{d z}{d t}\right)^2}=\norm{\vec{R}'(t)}=\norm{\vec{V}(t)}.\,$

Example 1.25 If a moving object has a position vector function of
$\vec{R}(t)=\langle e^{3t},\sqrt{18}t-5,-e^{-3t}\rangle$ then find the speed of the object at time $$t$$ and the distance traveled by the object between times $$t=0$$ and $$t=1.$$

Solution. The speed of the object at time $$t$$ is $$3e^{3t}+3e^{-3t}$$ because
$\vec{V}(t)=\langle 3e^{3t},\sqrt{18},3e^{-3t}\rangle,$ $\|\vec{V}(t)\|=\sqrt{9e^{6t}+18+9e^{-6t}}=3\sqrt{\left(e^{3t}+e^{-3t}\right)^2}=3\left(e^{3t}+e^{-3t}\right)$ and the distance traveled by the object between times $$t=0$$ and $$t=1$$ is $s=\int _0^1\left(3e^{3t}+3e^{-3t}\right)\, dt=e^3-\frac{1}{e^3}.$

Example 1.26 Express the vector function $$\vec{R}(t)=\langle e^{-t},1-e^{-t}\rangle$$ in terms of arc length measured from the point corresponding to $$t=0$$, in the direction of increasing $$t$$.

Solution. We have $s=s(t)=\int _0^t\sqrt{e^{-2u}+e^{-2u}}\, du=\sqrt{2}\int _0^te^{-u}\, du=\sqrt{2}-\sqrt{2}e^{-t}.$ Solving for $$e^{-t}$$ we have $$e^{-t}=\frac{\sqrt{2}-s}{\sqrt{2}}.$$ Thus $\vec{R}(s)=\left\langle\frac{\sqrt{2}-s}{\sqrt{2}},\frac{s}{\sqrt{2}}\right\rangle.$ which is expressed in terms of arc length $$s$$.

::: {#thm- } [Normal Vectors]

If $$\vec{R}(t)$$ has a piecewise smooth graph and is represented as $$\vec{R}(s)$$ in terms of the arc length parameter $$s$$, then the unit tangent vector $$\vec{T}$$ and the principal unit normal vector satisfies
$\vec{T}=\frac{d \vec{R}}{d s} \qquad \text{and} \qquad \vec{N}=\frac{1}{\kappa } \frac{d \vec{T}}{d s}$ where $$\kappa =\norm{d \vec{T} / d s}$$ is a scalar function of $$s.$$ :::

Geometrically, the curvature $$\kappa$$ measures how fast the unit tangent vector to the curve rotates. If a curve keeps close to the same direction, the unit tangent vector changes very little and the curvature is small; where the curve undergoes a tight turn, the curvature is large.

Proof. Given a piecewise smooth graph represented by $$\vec{R}(t)$$ and in terms of arc length by $$\vec{R}(s)$$, then by the chain rule,
$\frac{d\vec{R}}{d s}=\frac{\frac{d \vec{R}}{d t}}{\frac{d s}{d t}}=\frac{\vec{R}'(t)}{\norm{\vec{R}'(t)}}=\vec{T}.$ Also
$\vec{N}=\frac{\vec{T}'(t)}{\norm{\vec{T}'(t)}}=\frac{d \vec{T}/d t}{\norm{\vec{T}'(t)}}=\frac{d \vec{T}}{d s}\cdot \frac{d s/d t}{\norm{\vec{T}'(t)}}$ and since $$d s/ d t >0$$ and $$\norm{\vec{T}'(t)}>0$$, $$\vec{N}$$ points in the same direction as $$d \vec{T}/d s$$ and since $$\vec{N}$$ is a unit vector $\vec{N}=\frac{\frac{d \vec{T}}{d s}}{\norm{d s/d t}}=\frac{1}{\kappa }\frac{d \vec{T}}{d s}$ where $$\kappa =\norm{d s/ d t}.$$

## 1.16 Curvature

Suppose the smooth curve $$C$$ is the graph of the vector function $$\vec{R}(s)$$, parametrized in terms of the arc length $$s$$. Then the curvature of $$C$$ is the function $\kappa (s)=\norm{\frac{d \vec{T}}{d s}}$ where $$\vec{T}(s)$$ is the unit tangent vector.

Example 1.27 Find the curvature of a circle.

Solution. A circle can be parametrized by $$\vec{R}(t)=\langle r \cos t,r \sin t\rangle$$ where $$r$$ is the radius. We determine the arc length function as
$s=\int _0^t\sqrt{r^2\cos ^2u+r^2\sin ^2u}\, du=r t.$ Solving for $$t$$ we find the component functions to be $x(s)=r \cos \left(\frac{s}{r}\right) \qquad \text{and} \qquad y(s)=r \sin \left(\frac{s}{r}\right).$ Thus the unit tangent and unit normal vectors are $\vec{T}(s) =\left\langle -\sin \left(\frac{s}{r}\right),\cos \left(\frac{s}{r}\right)\right\rangle$ and
$\vec{N}(s) =\left\langle-\frac{1}{r}\cos \left(\frac{s}{r}\right),-\frac{1}{r}\sin \left(\frac{s}{r}\right)\right\rangle\left(\frac{1}{\kappa }\right)$ where
$\kappa =\sqrt{\frac{1}{r^2}\cos ^2\left(\frac{s}{r}\right)+\frac{1}{r^2}\sin ^2\left(\frac{s}{r}\right)}=\frac{1}{r}.$
Therefore the curvature of a circle is $$\frac{1}{r}.$$

Example 1.28 Let $$C$$ be the curve given as the graph of the vector function
$\vec{R}(t)=(t-\sin t)\vec{i}+(1-\cos t)\vec{j}+\left(4\sin \frac{t}{2}\right)\vec{k}.$ Find the unit tangent vector $$\vec{T}(t)$$ to $$C,$$ $$\frac{d \vec{T}}{d s},$$ and the curvature $$\kappa (t).$$

Solution. The derivative is
$\vec{R}'(t)=(1-\cos t)\vec{i}+(\sin t) \vec{j}+\left(2 \cos \frac{t}{2}\right)\vec{k}$ and the magnitude of the derivative is \begin{align*} \norm{\vec{R}'(t)} & =\sqrt{(1-\cos t)^2+\sin ^2t+4 \cos ^2 \frac{t}{2}} \\ & =\sqrt{2-2\cos t+4 \cos ^2\frac{t}{2}} =\sqrt{4\sin ^2\frac{t}{2}+4\cos ^2\frac{t}{2}} =2 \end{align*} thus the unit tangent vector is $\vec{T}(t)=\frac{\vec{R}'(t)}{\norm{\vec{R}'(t)}}=\frac{1}{2}\left[(1-\cos t)\vec{i}+(\sin t)\vec{j}+\left(2 \cos \frac{t}{2}\right)\vec{k}\right].$ Then,
$\vec{T}'(t)=\frac{1}{2}\left[(\sin t)\vec{i}+(\cos t)\vec{j}-\left(\sin \frac{t}{2}\right)\vec{k}\right]$ because $$\frac{d \vec{T}}{d s}=\frac{\vec{T}(t)}{d s / d t}$$ we have $\frac{d \vec{T}}{d s}=\frac{1}{4}\left[(\sin t)\vec{i}+(\cos t)\vec{j}-\left(\sin ^2\frac{t}{2}\right)\vec{k}\right]$ Therefore the curvature is $\kappa (t)=\norm{\frac{d \vec{T}}{d s}}=\frac{1}{4}\sqrt{\sin ^2t+\cos ^2t+\sin ^2\frac{t}{2}}=\frac{1}{4}\sqrt{1+\sin ^2\frac{t}{2}}.$ as a function of $$t$$.

::: {#cor- } [Curvature in Vector Form] Suppose the smooth curve $$C$$ is the graph of the vector function $$\vec{R}(t).$$ Then the curvature is given by
$\begin{equation} \label{curform} \kappa(t) =\frac{\norm{\vec{R}'(t) \times \vec{R}''(t)}}{\norm{\vec{R}'(t)}^3}. \end{equation}$ :::

Proof. Given two vector functions $$\vec{R}(t)$$ and $$\vec{R}(s)$$ with the same smooth graph $$C$$ and where $$\vec{R}(s)$$ is a parametrization vector function in terms of the arc length of $$C$$, then according to the rule
$\frac{d \vec{T}}{d t}=\frac{d \vec{T}}{d s}\frac{d s}{d t}$ we have
$\kappa(t) =\norm{\frac{d \vec{T}}{d s}}=\frac{\norm{\frac{d \vec{T}}{d t}}}{\norm{\vec{R}'(t)}}.$ Therefore,
\begin{align*} \kappa (t) &=\frac{\norm{\vec{T}'(t)}}{\norm{\vec{R}'(t)}} =\frac{\norm{\vec{T}'(t)}\norm{\vec{R}'(t)}^2}{\norm{\vec{R}'(t)}^3} =\frac{\norm{\vec{T}'(t)} \left(\frac{d s}{d t}\right)^2}{\norm{\vec{R}'(t)}^3} \\ & =\frac{\norm{\vec{T}'(t)}\norm{\vec{T} (t)}\left(\frac{d s}{d t}\right)^2}{\norm{\vec{R}'(t)}^3} =\frac{\norm{\vec{T}'(t)}\norm{\vec{T} (t)} \sin \left(\frac{\pi }{2}\right) \left(\frac{d s}{d t}\right)^2}{\norm{\vec{R}'(t)}^3} \\ & =\frac{\norm{\vec{T} (t)\times \vec{T}' (t)}\ \left(\frac{d s}{d t}\right)^2}{\norm{\vec{R}'(t)}^3}.\end{align*} Since
$\vec{R}' (t)=\norm{\vec{R}' (t)}\vec{T} (t)=\left(\frac{d s}{d t}\right)\vec{T} (t)$ and
$\vec{R}'' (t)=\frac{d}{d t}\left(\frac{d s}{d t}\right)\vec{T} (t)+\frac{d s}{d t}\vec{T}' (t)=\frac{d^2s}{d t^2}\vec{T} (t)+\frac{d s}{d t}\vec{T}' (t)$ we have,
\begin{align*} \vec{R}'(t)\times \vec{R}''(t)&=\left(\frac{d s}{d t}\right)\vec{T}(t)\times \left(\frac{d^2s}{d t^2}\vec{T}(t)+\frac{d s}{d t}\vec{T}'(t)\right) \\ & =\left(\frac{d s}{d t}\right)\left(\frac{d^2 s}{d t^2}\right)(\vec{T}(t)\times \vec{T}(t))+\left(\frac{d s}{d t}\right)^2(\vec{T}(t)\times \vec{T}'(t)) \\ & =\left(\frac{d s}{d t}\right)^2(\vec{T}(t)\times \vec{T}'(t)). \end{align*} Therefore,
$\norm{\vec{R}'(t)\times \vec{R}''(t)}=\left(\frac{d s}{d t}\right)^2\norm{\vec{T}(t)\times \vec{T}'(t)}.$ Finally $\kappa (t)=\frac{\norm{\vec{R}' (t) \times \vec{R}'' (t)}}{\norm{\vec{R}'(t)}^3}$ as desired.

Example 1.29 Given the curve defined by $\vec{R}(t)=\sin t \vec{i}+\cos t \vec{j}+t \vec{k}$ find a unit tangent vector $$\vec{T}$$ at the point on the curve where $$t=\pi$$, the curvature at $$t=\pi$$, and find the length of the curve from $$t=0$$ to $$t=\pi$$.

Solution. We have $$\vec{R}'(t)=\cos t \vec{i}-\sin t \vec{j}+\vec{k},$$ $$\vec{R}''(t)=-\sin t \vec{i} -\cos t \vec{j},$$ and
\begin{align*} \norm{\vec{R}'(t)\times \vec{R}''(t)}& =\norm{ \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \cos t & -\sin t & 1 \\ -\sin t & -\cos t & 0 \end{array} } \\ & =\norm{(\cos t)\vec{i}+(-\text{sint}) \vec{j}+\left(-\cos ^2t-\sin ^2t\right)\vec{k}} \\ & =\sqrt{2}. \end{align*} Therefore the curvature is $$\kappa =\frac{\sqrt{2}}{\left(\sqrt{2}\right)^3}=\frac{1}{2},$$ and also
$\vec{T}=\left(\frac{1}{\sqrt{2}}\cos t\right) \vec{i}+\left(-\frac{1}{\sqrt{2}}\sin t\right)\vec{j}+\left(\frac{1}{\sqrt{2}}\right)\vec{k}$ and
$\int _0^{\pi }\norm{\vec{R}'}\, dt=\int _0^{\pi }\sqrt{2}\, dt=\sqrt{2}\pi .$

## 1.17 Maximum Curvature

Now that we know that the curvature formula in $$\eqref{curform}$$ is a one variable function we can apply one variable calculus to find maximum curvature. We will derive two more curvature formulas, one for planar functional form $$\ref{pfc}$$ and another for parametric form $$\ref{pafc}$$.
There is a fourth formula (for polar form) which we leave for the reader to explore as Exercise $$\ref{ex:pfc}$$. For example, consider the cardioid given by equation $$r=2(1+\cos \theta)$$ with derivative $$dr/d\theta=-2\sin\theta$$. We can use this information to determine where the curvature is maximum. From the graph ($$\ref{cargraph}$$) can you tell at which points will the curvature be maximum?

::: {#cor- } [Curvature in Planar Form] The graph $$C$$ of the function $$y=f(x)$$ has curvature
$\begin{equation} \label{pfc} \kappa(x) =\frac{|f''(x)|}{\left(1+[f'(x)]^2\right)^{3/2}} \end{equation}$ where $$f(x),$$ $$f'(x),$$ and $$f''(x)$$ all exist. :::

Proof. Given the vector function $$x(t)=t$$ and $$y(t)=f(t)$$ defined by $$y=f(x)$$ we have $$\vec{R}(t)=t \vec{i}+f(t) \vec{j}.$$ Using the formula $\kappa (t) =\frac{\norm{\vec{R}'(t) \times \vec{R}''(t) }}{\norm{\vec{R}'(t) }^3},$ we have
$\kappa (t) =\frac{\norm{\ \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & f'(t) & 0 \\ 0 & f''(t) & 0 \end{array}\right|\ }}{\norm{\vec{R}'(t) }^3}=\frac{|f''(t)|}{\left(\sqrt{1+[f''(t)]^2}\right)^3}.$

Example 1.30 Find the maximum curvature for the graph of
$$y=\ln x$$.

Solution. We have $$y'=\frac{1}{x},$$ $$y''=\frac{-1}{x^2},$$ and
$\kappa (x) =\frac{\frac{1}{x^2}}{\left(1+\frac{1}{x^2}\right)^{3/2}}=\frac{1}{\left(1+\frac{1}{x^2}\right)^{3/2} x^2}$ To maximize the curvature we find the first derivative $\frac{d \kappa }{d x}=\frac{3}{\left(1+\frac{1}{x^2}\right)^{5/2} x^5}-\frac{2}{\left(1+\frac{1}{x^2}\right)^{3/2} x^3}=\frac{1-2 x^2}{\sqrt{1+\frac{1}{x^2}} x \left(x^2+1\right)^2}$ Applying the first derivative test we have the maximum curvature at $$x=\frac{1}{\sqrt{2}}$$ with curvature of $$\kappa \left(\frac{1}{\sqrt{2}}\right)=\frac{2}{3 \sqrt{3}}.$$

Example 1.31 Find the maximum curvature for the graph of
$$y=e^{2 x}$$.

Solution. We have $$y'=2e^{2x},$$ $$y\text{''}=4e^{2x},$$ and
$\kappa =\frac{|y''|}{\left[1+(y')^2\right]^{3/2}}=\frac{4e^{2x}}{\left(1+4e^{4x}\right)^{3/2}}$ To maximize the curvature we find the first derivative $\frac{d \kappa }{d x}=-\frac{96 e^{6 x}}{\left(1+4 e^{4 x}\right)^{5/2}}+\frac{8 e^{2 x}}{\left(1+4 e^{4 x}\right)^{3/2}}=-\frac{8 e^{2 x} \left(-1+8 e^{4 x}\right)}{\left(1+4 e^{4 x}\right)^{5/2}}$ Applying the first derivative test we have the maximum curvature at $$x=\frac{-3}{4}\ln2$$ with curvature of
$\begin{equation} \kappa \left(\frac{-3}{4}\ln2\right)=\frac{4\sqrt{3}}{9}. \end{equation}$

::: {#cor- } [Curvature in Parametric Form] If $$C$$ is a smooth curve in $$\mathbb{R}^2$$ described by the parametric equations $$x(t)$$ and $$y(t),$$ then the curvature is given by
$\begin{equation} \label{pafc} \kappa =\frac{|x'y''-y'x''|}{\left[(x')^2+(y')^2\right]^{3/2}}. \end{equation}$ :::

Example 1.32 Find the point(s) where the ellipse $$9x^2+4y^2=36$$ has maximum curvature.

Solution. Write the ellipse as $\frac{x^2}{2^2}+\frac{y^2}{3^2}=1$ and so we parametrize using $$x(t)=2 \cos t$$ and $$y(t)=3 \sin t.$$ Then we find $x'(t)=-2 \sin t, \quad y'(t)=3 \cos t, \quad x''(t)=-2 \cos t , \quad y''(t)=-3 \sin t.$ Now using $$\ref{thm:Curvature in Parametric Form}$$, we find

\begin{align*} \kappa & =\frac{|x'y''-y'x''|}{\left[(x')^2+(y')^2\right]^{3/2}} \\ & =\frac{|(-2 \sin t )(-3 \sin t)-(3 \cos t)(-2 \cos t)|}{\left[(-2 \sin t)^2+(3 \cos t)^2\right]^{3/2}} \\ & =\frac{6}{\left(4 \sin ^2t+9\cos ^2t\right)^{3/2}} \\ & = 6\left(4+5\cos ^2t\right)^{-3/2} \end{align*} See $$\ref{fig:ellipse}$$ for a sketch of the curvature function. Then we apply the first derivative test, using $\frac{d\kappa }{dt}=-9\left(4+5 \cos ^2t\right)^{-5/2}(-10\cos t \sin t)=0$ when $$t=0, \pi/2 ,\pi , 3\pi /2, 2\pi$$. The point(s) where the ellipse $$9x^2+4y^2=36$$ has maximum curvature are $$(0,\pm 3)$$ from when $$t=\pi/2$$ and $$t=3\pi/2$$ with curvature $$\kappa =3/4$$.

## 1.18 Exercises

Exercise 1.55 Find the length of the given curve over the given interval.

• $$\vec{R}(t)=t \vec{i}+3 t \vec{j}$$ over the interval $$[0,4]$$
• $$\vec{R}(t)=t \vec{i}+2t \vec{j}+3 t \vec{k}$$ over the interval $$[0,2]$$
• $$\vec{R}(t)=\cos ^3t \vec{i}+\cos ^2t \vec{k}$$ over the interval $$\left[0,\frac{\pi }{2}\right]$$

Exercise 1.56 Express the following vector functions in terms of the arc length parameter $$s$$ measured from the point where $$t=0$$ in the direction of increasing $$t.$$

• $$\vec{R}(t)=\langle \sin t,\cos t \rangle$$ at the point where $$t=0$$
• $$\vec{R}(t)=\langle 3 \cos t+3 t \sin t ,2t^2,3 \sin t-3 t \cos t\rangle$$ at the point where $$t=0$$

Exercise 1.57 $$\vec{R}(t)=[\ln (\sin t)]\vec{i}+[\ln (\cos t)]\vec{j}$$ at the point where $$t=\pi /3$$

Exercise 1.58 Find the length of the given curve described by the vector function $$\vec{R}(t)=3 t \vec{i}+(3 \cos t)\vec{j} +(3 \sin t)\vec{k}$$ over the interval $$\left[0,\frac{\pi }{2}\right].$$

Exercise 1.59 Determine $$a$$ so that the length of the given curve described by the vector function $$\vec{R}(t)=a \cos t \vec{i}+(a \sin t)\vec{j} +(5 t)\vec{k}$$ has length $$\sqrt{41}\pi$$ over the interval $$[0,\pi ].$$

Exercise 1.60 Express the vector function $$\vec{R}(t)=\langle 2+3t,1-t,-4t-9\rangle$$ in terms of the arc length parameter $$s$$ measured from the point where $$t=0$$ in the direction of increasing $$t.$$

Exercise 1.61 Determine $$a$$ so that the vector function $$\vec{R}(t)=\langle 2-3t,1+t,-4t\rangle$$ written in terms of the arc length parameter $$s$$ measured from the point where $$t=0$$ in the direction of increasing $$t$$ is $\vec{R}(s)=\left\langle 2-\frac{3s}{a},1+\frac{s}{a},\frac{-4s}{a}\right\rangle.$

Exercise 1.62 Find the curvature at the given point.

• $$\vec{R}(t)=(2t+3)\vec{i}+(5-t^2)\vec j$$
• $$\vec{R}(t)=t\vec{i} +\cos t\vec{j}$$
• $$\vec{R}(t)=t\vec{i} + e^t\vec{j}$$
• $$\vec{R}(t)=3\cos t \vec{i} + 2\sin t\vec{j}$$
• $$\vec{R}(t)=(2t-1)\vec{i}+(t-3)\vec{j}+(2-3t)\vec{k}$$, $$(1,-2,-1)$$
• $$\vec{R}(t)=t\vec{i}+\ln t\vec{j}+0.5 t\vec{k}$$, $$(2,\ln 2,1)$$
• $$\vec{R}(t)=\sin t\vec{i}+\cos t\vec{j}+2t\vec{k}$$, $$(1,0,\pi)$$
• $$\vec{R}(t)=(\cos t+t\sin t)\vec{i}+(\sin t-t\cos t)\vec{j}+3\vec{k}$$
• $$\vec{R}(t)=(e^t\sin t)\vec{i}+(e^t \cos t)\vec{j}+2\vec{k}$$
• $$\vec{R}(t)=6 \sin 2t \vec{i}+6 \cos 2t \vec{j}+5t\vec{k}$$

Exercise 1.63 Find the curvature of the plane curve at the given location.

• $$y=x-\frac{1}{9}x^2$$, $$x=3$$
• $$y=\sin x$$, $$x=\frac{\pi }{2}$$
• $$y=\ln x$$, $$x=1$$
• $$y=1/\ln x$$, $$x=2$$
• $$y=x^3-3x^2+2$$, $$x=2$$
• $$y=e^{x^2}$$, $$x=0$$
• $$y=\sec x$$, $$\pi/3$$
• $$1/x^2$$, $$x=2$$
• $$y=\cos 2x$$, $$0$$
• $$y=x^4-x^2+3$$, $$x=-1$$

Exercise 1.64 Find the curvature of the plane curve at the given location.

• $$y=\ln (\cos x)$$, $$-\pi/2 < x <\pi /2$$
• $$x=t-1, y=\sqrt{t}$$, $$(x,y)=(3,2)$$
• $$x=t-t^2, y=1-t^3$$, $$(x,y)=(0,1)$$
• $$x=1-\sin t, y=2+\cos t$$, $$(x,y)=(1,3)$$
• $$x=\cos^3 t, y=\sin^3 t$$, $$(x,y)=(\sqrt{2}/4,\sqrt{2}/4)$$

Exercise 1.65 Given the vector function $\vec{R}(t)=\sin t \vec{i}+ \cos t \vec{j}+t \vec{k}$ and $$t_0=\pi ,$$ find the unit tangent vector and the curvature at $$t_0.$$

Exercise 1.66 Find all points on the curve at which the curvature is zero.

• $$y=\cot x$$
• $$y=e^{-x^2}$$
• $$y=x^4-12x^2$$
• $$\vec{R}(t)=t^2\vec{i}-t^2\vec{j}$$
• $$\vec{R}(t)=e^x\vec{i}+\ln x\vec{j}$$
• $$\vec{R}(t)=e^{-x}\vec{i}+e^x\vec{j}$$

Exercise 1.67 Find the point(s) where the ellipse $$36x^2+9y^2=121$$ has maximum curvature.

Exercise 1.68 Find the points on the curve in the $$xy$$-plane described by $$y=3x-x^3$$ at which the curvature is maximum.

Exercise 1.69 Show that the curvature of the parabola $$y=x^2+3$$ approaches 0 as $$x\to \infty$$.

Exercise 1.70 Show that the curvature at every point on a line is 0.

Exercise 1.71 Show that the maximum curvature of a parabola occurs at the vertex.

Exercise 1.72 Show that for an ellipse, the maximum and minim curvature occurs at ends of the major and minor axes, respectively.

• Show that for a hyperbola, the maximum curvature occurs at the ends of the traverse axis.

Exercise 1.73 Show that the curvature of the helix $\vec{R}(t)= a\cos t \vec{i}+a \sin t \vec{j}+b t$ is $$\kappa=a/(a^2+b^2)$$. What is the largest value that $$\kappa$$ can have for a given value of $$b$$?

Exercise 1.74 Suppose a curve $$C$$ in the $$xy$$-plane described by a polar equation $$r=f(\theta)$$. Show that the curvature of $$C$$ is $\begin{equation} \kappa(\theta)=\frac{| \, \, [f(\theta)]^2+2[f'(\theta)]^2-f(\theta)f''(\theta)\, \, |}{\, \{[f(\theta)]^2+[f'(\theta)]^2\, \}^{3/2}}. \end{equation}$

Exercise 1.75 Find the curvature of the polar curve at $$P(r,\theta)$$.

• the cardoid $$r=a(1-\cos \theta)$$, where $$0< \theta<2\pi$$.
• the four-leafed rose $$r=\sin 2\theta$$, where $$0<\theta < \pi/2$$
• the spiral $$r=e^{a\theta}$$
• $$r=1-\cos \theta$$, $$\theta=3\pi/2$$
• $$r=1-\sin\theta -\cos\theta$$, $$\theta=2\pi$$

Exercise 1.76 Show that lines and circles are the only planes curves that have constant curvature. Is this true for space curves?