Vector functions can be used to model the motion of an object. You may want the object’s 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 the object’s motion.

## 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**. 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} {v}0 & =\left(x_0 +\left|\left| {v}_0\right|\right| \cos \theta \right) {i}+\left(y_0+\left|\left| {v}_0\right|\right| \sin \theta \right){j} \\ & =\left|\left| {v}_0\right|\right| \cos \theta \, {i}+\left(6+\left|\left| {v}_0\right|\right| \sin \theta \right){j} \end{align} where $\theta$ is the angle that ${v}_0$ 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 \begin{equation} {F}=m \frac{d^2 \, {r}}{d t^2} \end{equation} when ${r}$ 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 $-m g \, {j}$, then \begin{equation} \label{fa1} \frac{d^2 \, {r}}{d t^2} = – g \, {j}, \quad {r_0}=0\, {i}+6\, {j}, \quad \text{and} \quad \left.\frac{d{r}}{dt}\right|{t=0}={v}0. \end{equation} The first integration gives $\frac{d{r}}{dt}=-g t \, {j}+{v}_0.$ The second integration yields \begin{equation}\label{fa2} {r}(t)=\frac{-g t^2}{2}\, {j} +{v}_0 t+{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} {r}(t)=\left(\left|\left| {v}_0\right|\right| \cos\theta\right)t \, {i}+\left(6+(\left|\left| {v}_0\right|\right|\sin \theta)t -\frac{1}{2} g t^2\right) \, {j}. \end{equation} The arrow reaches its highest point when $\frac{dy}{dt}=0$, and solving for $t$ we obtain $$ t=\frac{\left|\left| {v}_0\right|\right| \sin \theta}{g}. $$ For this value of $t$, the value of $y$ is $$ 6+\frac{ (\left|\left| {v}_0\right|\right| \sin \theta)^2 }{2g}. $$ Using $y{\text{max}}=34$ and $g=32$, we see that \begin{equation}34=6+\frac{ (\left|\left| {v}_0\right|\right| \sin \theta)^2 }{2(32)}\end{equation} or $\left|\left| {v}_0\right|\right| \sin \theta =\sqrt{(28)(64)}.$

In order to find $\theta$ we wish to find a similar expression for $\left|\left| {v}0\right|\right| \cos \theta.$ When the arrow reaches $y{\text{max}}=34$, the horizontal distance is $x=90$ ft and by substitution into equation \eqref{fa3} we obtain \begin{equation}90=(\left|\left| {v}_0\right|\right| \cos \theta) \left(\frac{\left|\left| {v}_0\right|\right| \sin \theta}{g}\right).\end{equation} Therefore we find \begin{equation} \tan \theta =\frac{\left|\left| {v}_0\right|\right| \sin \theta}{\left|\left| {v}_0\right|\right| \cos \theta} = \frac{\left(\sqrt{(28)(64)}\right)^2}{(90)(32)}=\frac{28}{45}. \end{equation} So the approximate firing angle will be \begin{equation} \theta = \tan^{-1}(28/45) \approx 31.8908^\circ \end{equation} as desired.

## Vector Functions

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

In three dimensions vector functions can be expressed in the form \begin{equation} {F}(t)=f_1(t){i}+f_2(t){j}+f_3(t){k} \end{equation} 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 \begin{equation} {F}(t) = \langle f_1(t),f_2(t),f_3(t)\rangle .\end{equation} Unless stated otherwise, the domain of a vector function ${F}$ 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**. Given the vector function $$ {F}(t)= t^3\, {i} +\ln (3-t)\, {j}+\sqrt{t}\, {k} $$ find the component functions and the domain of ${F}$, and then evaluate ${F}$ at $t=1.$

**Solution**. The component functions are $f(t)=t^3,$ $g(t)=\ln (3-t),$ and $h(t) = \sqrt{t}$ where ${F}(t)=f(t)\, {i}+g(t)\, {j}+h(t)\, {k}.$ The domain of ${F}(t)$ 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 ${F}$ at $t=1$ as follows $$ {F}(1)=f(1)\, {i}+g(1)\, {j}+h(1)\, {k} ={i}+\ln 2\, {j}+{k} $$ as desired.

## 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 ${F}(t)$ 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 $$ {F}(t)= \langle f_1(t),f_2(t),f_3(t)\rangle $$ 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**. Sketch the graph of the vector-valued function defined by \begin{equation} {F}(t)=-3 \sin t {i}+3\cos t{j}+0.1t {k}. \end{equation}

**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 below.

**Example**. Sketch the graph of the vector-valued function \begin{equation} {F}(t)=(5-2t){i}+(3+2t){j}+ 2t {k}. \end{equation}

**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.

## 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 $$ {F}(t)=\sin t{i}+\cos t{j} $$ with $0\leq t<2\pi $; and can also be parametrized as a vector-valued function by $${F}(t)=\sin (2t){i}+\cos (2t){j} $$ with $0\leq t<\pi .$ 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**. Find a vector-valued function ${F}$ 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 \begin{equation} {F}(t)=t {i}+t^2{j}+\sqrt{1-t^2-t^4}{k}. \end{equation} is a value-valued function for this intersection. The intersection of the hemisphere \mbox{$z=\sqrt{1-x^2-y^2}$} and the parabolic cylinder $y=x^2.$ A sketch of the graph of the vector function ${F}(t)=t {i}+t^2{j}+\sqrt{1-t^2-t^4}{k}.$ Try reworking this example with $x=2t.$ and graph the resulting vector function.

**Example**. Find a vector-valued function ${F}$ 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{equation} \begin{cases} t-y-z=1 \ 2t+y+3z=6. \end{cases}\end{equation} Eliminating $y$ we have, $3t+2z=7$ and so $z=(7-3t)/2.$ Solving the first for $y$ we find \begin{equation}y=t-1-z=t-1-\left(\frac{7-3t}{2}\right)=\frac{2t-2-7+3t}{2}=\frac{5t-9}{2}\end{equation} Therefore \begin{equation}\label{vfplanes} {F}(t)=t \, {i}+\frac{5t-9}{2}{j}+\frac{7-3t}{2}{k}.\end{equation}is a vector-valued function for this intersection.

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

$(1) \quad ({F}+{G})(t)={F}(t)+{G}(t)$

$(2) \quad (f{F})(t)=f(t){F}(t)$

$(3) \quad ({F}-{G})(t)={F}(t)-{G}(t)$

$(4) \quad ({F}\cdot {G})(t)={F}(t)\cdot {G}(t)$

$(5) \quad ({F}\times {G})(t)={F}(t)\times {G}(t)$

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**. Let $\vec F$, $\vec G$ , and $\vec H$ be the vector functions defined by

\begin{align} {F}(t)=(2t){i}-5{j}+t^2{k}, \quad {G}(t)=(1-t){i}+\frac{1}{t}{k}, \quad {H}(t)=(\sin t){i}+e^t{j}. \end{align} Find the vector functions

$(a) \quad {F}(t)\times {G}(t)$

$(b) \quad 2e^t{F}(t)+t{G}(t)+10{H}(t)$

$(c) \quad {G}(t)\cdot [{H}(t)\times {F}(t)]$

**Solution**. We find \begin{align} & 2e^t{F}(t)+t{G}(t)+10{H}(t) \\ & \qquad =\left(4t e^t+t-t^2+10 \sin t\right){i}+\left(2t^2e^t+1\right){k}. \end{align} Using the determinant formula for a cross product of vectors, we find

\begin{align} {F}(t)\times {G}(t) & = \left| \begin{array}{ccc} {i} & {j} & {k} \\ 2 t & -5 & t^2 \\ 1-t & 0 & \frac{1}{t} \end{array} \right| \\ & =\left(\frac{-5}{t}\right){i}+\left(-t^3+t^2-2\right){j}+(-5t+5){k}. \end{align} For the third, we find $$ {G}(t)\cdot [{H}(t)\times {F}(t)] = (1-t)t^2e^t-\frac{2 t e^t+5\sin t}{t} $$ by using \begin{align} & {G}(t)\cdot [{H}(t)\times {F}(t)] \\ & \quad =\left[(1-t){i}+\frac{1}{t}{k}\right]\cdot \left| \begin{array}{ccc} {i} & {j} & {k} \\ \sin t & e^t & 0 \\ 2 t & -5 & t^2 \end{array} \right|. \end{align} as desired.

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

**Theorem**. (** Limits of Vector-Functions**) If the vector-valued functions ${F}$ and ${G}$ 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

$(1) \quad \displaystyle \lim_{t\to t_0}[{F}(t)+{G}(t)] =\lim_{t\to t_0}{F}(t)+\lim_{t\to t_0}{G}(t)$

$(2) \quad \displaystyle \lim_{t\to t_0}[{F}(t)-{G}(t)] = \lim_{t\to t_0}{F}(t)-\lim_{t\to t_0}{G}(t)$

$(3) \quad \displaystyle \lim_{t\to t_0}[h(t) {F}(t)] =\left[\lim_{t\to t_0}h(t)\right] \left[\lim_{t\to t_0}{F}(t)\right] $

$(4) \quad \displaystyle \lim_{t\to t_0}[{F}(t)\cdot {G}(t)] = \left[\lim_{t\to t_0}{F}(t)\right]\cdot \left[\lim_{t\to t_0}{G}(t)\right]$

$(5) \quad \displaystyle \lim_{t\to t_0}[{F}(t)\times {G}(t)] =\left[\lim_{t\to t_0}{F}(t) \right]\times \left[\lim_{t\to t_0}{G}(t)\right]$

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

**Proof**. Let ${F}$ and ${G}$ have (respectively) standard forms \begin{equation} \label{stform} {F}(t) = f_1(t) {i} + f_2(t) {j} + f_3(t) {k} \qquad {G}(t)=g_1(t){i}+g_2(t){j}+g_3(t){k}. \end{equation} We find, \begin{align} & \lim_{t\to t_0}[{F}(t)+{G}(t)] \\ & = \left(\lim_{t\to t_0}(f_1+g_1)(t)\right) {i} +\left(\lim_{t\to t_0}(f_2+g_2)(t)\right) {j}+\left(\lim_{t\to t_0}(f_3+g_3)(t)\right) {k} \\ &= \lim_{t\to t_0}[f_1(t){i}+f_2(t){j}+f_3(t){k} ]+\lim_{t\to t_0}[g_1(t){i}+g_2(t){j}+g_3(t){k}] \\ & = \lim_{t\to t_0}{F}(t)+\lim_{t\to t_0}{G}(t) \end{align} Again we find, \begin{align} \lim_{t\to t_0}[{F}(t)\cdot {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)] \\ & = {F}(t_0)\cdot {G}(t_0) \\ & = \left[\lim_{t\to t_0}{F}(t)\right]\cdot \left[\lim_{t\to t_0}{G}(t)\right] \end{align} The remainder of the proof is left as an exercise.

**Example**. Find $\lim_{t\to 0}h(t) {F}(t)$ given $$ {F}(t)=(\sin t){i}-t {k} \qquad \qquad h(t)=1/(t^2+t-1). $$

**Solution**. We find \begin{align} & \lim_{t\to 0}\left(\frac{\sin t}{t^2+t – 1 } \right){i}-\frac{t}{t^2+t-1} {k} \\ & =\lim_{t\to 0}\left(\frac{\sin t}{t^2+t-1}\right){i}-\lim_{t\to 0}\left(\frac{t}{t^2+t-1}\right) {k} = (0){i}-(0){k} ={0}. \end{align} as desired.

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

**Solution**. We find \begin{align} \lim_{t\to 2}{F}\times {G} & = \left(2{i}-2{j}+e^2{k}\right)\times (4{i}+4 \sin 2{j}) \\ & = \left| \begin{array}{ccc} {i} & {j} & {k} \\ 2 & -2 & e^2 \\ 4 & 4\sin 2 & 0 \end{array} \right| \\ & =\left(-4e^2 \sin 2\right){i} -\left(-4e^2\right){j}+(8 \sin 2+8){k}. \end{align} as desired.

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

**Theorem**. A vector function is continuous at $a$ if and only if each of its component functions is continuous at $a.$

**Proof**. Suppose that ${F}$ is a vector function that is continuous at $a.$ Let

$$ {F}(t)=f_1(t){i}+f_1(2){j}+f_3(t){k}. $$ Then we have

\begin{align} {F}(a)=\lim_{t\to a} {F}(t) & =\lim_{t\to a} \left[ f_1(t){i}+f_1(2){j}+f_3(t){k} \right] \\ & = \left(\lim_{t\to a} f_1(t)\right){i}+\left(\lim_{t\to a} f_2(t)\right){j}+\left(\lim_{t\to a} f_3(t)\right){k} \\ & =f_1(a){i}+f_2(a){j}+f_3(a){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 ${F}$ is continuous at $a.$ The reminder of the proof is left as exercise.

**Example**. Determine where the vector-valued function \begin{equation} {F}(t) = t e^t {i}+\frac{e^t}{t}{j}+3e^t{k} \end{equation} 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. The function ${F}$ is continuous for all real numbers in its domain which is ${t\in \mathbb{R}\, | \, t \neq 0}.$

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

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

## Exercises on Vector Functions and Space Curves

**Exercise**. Find the domain of the vector function \begin{equation}{F}(t)=(1-t){i}+\left(\sqrt{t}\right){j}-\frac{1}{t-2}{k}.\end{equation}

**Exercise**. Find the domain of the vector function \begin{equation}{F}(t)=(\cos t){i}-(\cot t){j}+(\csc t){k}.\end{equation}

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

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

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

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

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

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

**Exercise**. Find the domain of the vector function $({F}\times {G})\times {H}$ where

$(1) \quad \displaystyle {F}(t)=t^2{i}-t {j}+2t {k},$

$(2) \quad \displaystyle {G}(t)=\frac{1}{t+2}{i}+(t+4){j}-\sqrt{-t}{k},$

$(3) \quad \displaystyle {H}(t)=\frac{1}{t+3}{i}+t^2{j}-\sqrt{t}{k}.$

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

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

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

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

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

$(1) \quad \displaystyle \lim_{t\to 1}\left[\frac{t^3-1}{t-1}{i}+\frac{t^2-3t+2}{t^2+t-2}{j}+\left(t^2+1\right)e^{t-1}{k}\right]$

$(2) \quad \displaystyle \lim_{t\to \infty }\left[\left(e^{-t}\right) {i}+\left(\frac{t-1}{t+1}\right){j}+\left(\tan ^{-1}t \right){k}\right]$

$(3) \quad \displaystyle \lim_{t\to 0^+}\left[\frac{\sin 3t}{\sin 2t} {i}+\frac{\ln (\sin t)}{\ln (\tan t)}{j}+(t \ln t){k}\right]$

**Exercise**. Determine all real numbers $a$ that satisfies \begin{equation}

\lim_{t\to 0}\left[\frac{ t}{\sin a t} {i}+\frac{a}{a-\cos t}{j}+\left(e^{a-t}\right){k}\right]=\frac{3}{2} {i}-2{j}+e^{2/3}{k}. \end{equation}

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

**Exercise**. Determine the intervals for which both vector functions ${F}(t)= \langle e^{-t},t^2,\tan t\rangle$ and ${G}(t)= \langle 8,\sqrt{t},\sqrt[3]{t}\rangle$ are continuous.

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

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