# Derivatives and Integrals of Vector Functions • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

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 ${c}.$