We study differentiation and integration of vector functions of a single variable. Tangent vectors and unit normal vectors are also considered. Several exercises are given at the end.

## 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**. The difference quotient of a vector function ${F}$ is the vector function \begin{equation} \frac{\Delta {F}}{\Delta t}=\frac{{F}(t+\Delta t)-{F}(t)}{\Delta t} \end{equation} where $\Delta t$ is an increment of the variable $t.$

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

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

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

**Theorem**. (** Differentiable Vector Functions**) Any vector function $${F}(t)=f_1(t) {i}+f_2(t) {j}+f_3(t) {k} $$ is differentiable whenever the component functions $f_1$, $f_2$, and $f_3$ are each differentiable and in this case $$ {F}'(t)=f_1′(t) {i} +f_2′(t) {j}+f_3′(t){k}. $$

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

**Example**. Find the derivative of the vector function \begin{equation} {F}(t)=(\ln t){i}+\frac{1}{2}t^3{j}-t {k}. \end{equation}

**Solution**. We find that\begin{equation}{F}'(t)=\frac{1}{t}{i}+\frac{3}{2}t^2{j}-{k}\end{equation} is the derivative is the vector function ${F}.$

**Theorem**. (** Derivative Rules for Vector Functions**) If the vector functions ${F},$ ${G}$ and the scalar function $h$ are differentiable at $t$, and if $a$ and $b$ are constants, then $a {F}+b {G},{F}\cdot {G},$ and ${F}\times {G} $ are differentiable at $t$ and,

$(1) \quad (a{F}+b \, {G})'(t)=a {F}'(t)+b {G}'(t)$

$(2) \quad (h{F})'(t)=h'(t){F}(t)+h(t){F}'(t)$

$(3) \quad ({F}\cdot {G})'(t)={F}'(t)\cdot {G}(t)+{F}(t)\cdot {G}'(t)$

$(4) \quad ({F}\times {G})'(t)={F}'(t)\times {G}(t)+{F}(t)\times {G}'(t)$

$(5) \quad {F}(h(t))’=h'(t){F}'(h(t)).$

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

**Solution**. We find \begin{align} & [{F}(t)\times {G}(t)]’={F}'(t)\times {G}(t)+{F}(t)\times {G}'(t) \\ & \qquad =\left[(2t {i}+{j})\times \left({i} + t {j} + t^2 {k} \right) \right]+\left[\left(t^2{i}+t {j}+{k}\right)\times ( {j}+2t {k})\right] \\ & \qquad = \left| \begin{array}{ccc} {i} & {j} & {k} \\ 2 t & 1 & 0 \\ 1 & t & t^2 \\ \end{array} \right| + \left| \begin{array}{ccc} {i} & {j} & {k} \\ t^2 & t & 1 \\ 0 & 1 & 2t \end{array} \right| \\ & \qquad =\left(3t^2-1\right){i}-4t^3{j}+\left(3t^2-1\right){k}. \end{align} as needed.

## Tangent Vectors

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

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

**Solution**. We find a tangent vector \begin{equation} {F}'(2)=\left.(2t+1){i}-e^t{j}+\frac{1}{2\sqrt{t}}{k}\right|_{t=2}=5{i}-e^2{j}+\frac{1}{2\sqrt{2}}{k}; \end{equation} and the tangent line to the graph of ${F}(t)$ 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 \begin{equation} 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 \end{equation} because this line passes through ${F}(2)=6{i}-e^2{j}+\sqrt{2}{k}$ and is parallel to the tangent vector at $t=2.$

**Example**. Find parametric equations for the tangent line to the graph of $${R}(t) = t e^{-2t}{i}+t^2{j}+t e^{-2t}{k} $$ 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, \begin{equation} {R}'(t)=\left(e^{-2t}-2t e^{-2t}\right){i}+2t {j}+\left(e^{-2t}-2t e^{-2t}\right){k} \end{equation} and so

\begin{equation} {R}’\left(\frac{1}{2}\right)=\left(e^{-1}- e^{-1} \right) \vec{i}+{j}+\left(e^{-1}- e^{-1} \right)\vec{k}=0\vec{i}+\vec{j}+0\vec{k}.

\end{equation} Also, $$ {R}\left(\frac{1}{2}\right)=\frac{1}{2e^{1/2}} {i}+\frac{1}{4}{j}+\frac{1}{2e}{k} $$ 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.

## Unit Tangent and Unit Normal Vector

**Definition**. If the graph of the vector function ${R}(t)$ is smooth, then at each point $t$ a unit tangent vector is defined by \begin{equation} {T}(t) = \frac{{R}'(t)}{\left|\left| {R}'(t)\right|\right| } \end{equation} and the principal unit normal vector function is defined by \begin{equation} {N}(t)=\frac{{T}'(t)}{\left|\left| {T}'(t)\right|\right| }.\end{equation}

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

**Theorem**. Suppose that ${R}$ is a vector such that $\left|\left| R(t)\right|\right| =c$ for all t. Then ${R}(t)$ is orthogonal to ${R}'(t).$

**Proof**. Notice ${R}(t)\cdot {R}(t)=\left|\left| {R}(t)\right|\right| ^2=c^2$, for all $t.$ We differentiate with respect to $t$ to find, \begin{equation} {R}'(t)\cdot {R}(t)+{R}'(t)\cdot {R}(t)=0 \end{equation} which yields ${R}'(t)\cdot {R}(t)=0$ and so ${R}'(t)$ is orthogonal to ${R}(t).$

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

**Solution**. We find \begin{align} & {T}(t) = \frac{{R}'(t)}{\left|\left| {R}'(t)\right|\right| } \\ & \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 $$ {N}(t) =\frac{{T}'(t)}{\left|\left| {T}'(t)\right|\right| } =\frac{\left\langle 0,\frac{-2}{\sqrt{5}}\sin 2t,\frac{2}{\sqrt{5}}\cos 2t\right\rangle }{\sqrt{\frac{4}{5}}} =\langle 0,-\sin 2t,\cos 2t\rangle. $$ as the unit normal vector.

## 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**. Let ${F}(t)=f_1(t){i}+f_2(t){j}+f_3(t){k}$, 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 ${F}(t)$ is \begin{equation} \int {F}(t)\, dt=\left[\int f_1(t)\, dt\right]{i}+\left[\int f_2(t)\, dt\right]{j}+\left[\int f_3(t)\, dt\right]{k}. \end{equation}

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

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

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

**Definition**. Let ${F}(t)=f_1(t){i}+f_2(t){j}+f_3(t){k}$,

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 ${F}(t)$ is the vector \begin{equation} \int _b^a{F}(t)\, dt=\left[\int _b^af_1(t)\, dt\right]{i}+\left[\int _b^af_2(t)\, dt\right]{j}+\left[\int _b^af_3(t)\, dt\right]{k}. \end{equation}

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

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

## Exercises on Derivatives and Integrals of Vector Functions

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

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

**Exercise**. Find the vector derivative ${F}’$ for each of the following functions.

$(1) \quad {F}(s)=\left(s {i}+s^2{j}+s^2{k}\right)+\left(2s^2 {i}-s {j}+3{k}\right).$

$(2) \quad {F}(s)=\left(1-2s^2\right){i}+(s \cos s) {j}-s {k}.$

$(3) \quad {F}(\theta )=\left(\sin ^2\theta \right){i}+(\cos 2\theta ) {j}+\theta ^2 {k}.$

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

$(1) \quad \int \langle\cos t,\sin t,-2t\rangle \, dt.$

$(2) \quad \int \langle 3 e ,t^2, t \sin t\rangle \,dt.$

$(3) \quad \int e^{-t} \langle 3,t,\sin t\rangle \,dt.$

$(4) \quad \int \langle\sinh t, -3, \cosh t\rangle \, dt.$

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

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

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

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

$(1) \quad {F}(t)=(\ln t)\left[t {i}+5 {j}-e^t{k}\right]$

$(2) \quad {F}(t)=(\sin t){i}+(\cos t) {j}+t^2{k}$

$(3) \quad f(x)=\left|\left(x {i}+x^2{j}-20{k}\right)+\left(x^3 {i}+ x {j}-x {k}\right)\right|.$

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

f'(x)=-a x^2-\frac{18}{a}x \end{equation} given \begin{equation}

f(x)=[x {i}+(x+1){j}]\cdot \left[2x {i}-3x^2{j}\right]. \end{equation}

**Exercise**. Determine all real numbers $a$ such that $$ F”(t)=a {i}+a t^{-3}{j}+2a e^{a t}{k} $$ given ${F}(t)=t^2{i}+t^{-1}{j}+e^{2 t}{k}.$

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

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

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

**Exercise**. Prove that if ${F}, {G},$ and ${H}$ are differentiable vector functions, then \begin{equation} [{F}\times ({G}\times {H})] ‘ =[({H}\cdot {F}){G}] ‘-[({G}\cdot {F}){H}] ‘. \end{equation}

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

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

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

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

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

**Exercise**. 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**. Prove that \begin{equation} \frac{d}{dt}\left[{R}(t)\times {R}\,'(t)\right] ={R}(t)\times {R}\,”(t). \end{equation}

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