# Vector Fields and Gradient Fields • 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.

## Introduction to Vector Fields

Definition. A vector field in $\mathbb{R}^n$ is a function $\mathbf{V}$ that assigns a vector from each point in its domain. A vector field with domain $D$ in $\mathbb{R}^n$ has the form \begin{equation} {V}(x_1, \ldots, x_n) = \langle u_1 (x_1, \ldots,x_n), \ldots, u_n (x_1, \ldots, x_n) \rangle \end{equation} where the scalar functions $u_1,$,\ldots,$u_n$ are called the components of $\mathbf{V}.$

For example a vector field in $\mathbb{R}^2$ has the form \begin{equation} {V}(x,y)=u(x,y) {i}+v(x,y) {j}=\langle u,v \rangle \end{equation}
and in $\mathbb{R}^3$ has the form \begin{equation}{V}(x,y,z) = u(x,y,z) {i}+v(x,y,z) {j}+w(x,y,z) {k}=\langle u,v,w\rangle. \end{equation}

Common examples of vector fields include force fields, velocity fields, gravitational fields, magnetic fields, and electric fields. Vector fields can be used to quantify the amount of work done by a variable force acting on a moving body. Measuring the amount of force (fluid flow, electric charge, etc.) can sometimes be achieved by computing an integral of a vector field with respect to an orientable curve or surface.

Definition. Let $f$ be a differentiable function. The vector field obtained by applying the del operator to $f$ is called the gradient field of $f.$

Example. Find the gradient field of the function $f(x,y)=\sin x+e^{x y}.$

Solution. Since $$\nabla f(x,y)=\langle \cos x+y e^{x y},x e^{x y}\rangle$$ the gradient field of $f$ is \begin{equation} {V}(x,y)= \left(\cos x+y e^{x y} \right) {i}+ x e^{x y} {j}. \end{equation}

## Conservative Vector Fields

Definition. A vector field ${V}$ is said to be conservative in a region $D$ if ${V}=\nabla f$ for some scalar function $f$ in $D.$ The function $f$ is called a scalar potential of ${V}$ in $D.$

Example. Determine whether the vector field is conservative and if so, find a scalar potential function \begin{equation} {V}(x,y,z)=y^2{i}+\left(2x y+e^{3z}\right){j}+\left(3y e^{3z}\right){k} \end{equation}

Solution. If there is such a function $f$ then $f_x(x,y,z)=y^2$, $f_y(x,y,z)=2 x y+e^{3z}$, and $f_z(x,y,z)=3y e^{3z}.$ Integrating $f_x$ with respect to $x,$ $f(x,y,z)=x y^2+g(y,z).$ Then differentiating $f$ with respect to $y,$ we have $f_y(x,y,z)=2x y+g_y(y,z)$ and this yields $g_y(y,z)=e^{3z}.$ Thus $g(y,z)=y e^{3z}+h(z)$ and we have $$f(x,y,z)=x y^2+y e^{3z}+h(z).$$ Finally, differentiating $f$ with respect to $z$ and comparing, we obtain $h'(z)=0$ and therefore, $h(z)=K,$ a constant. The desired scalar function is $$f(x,y,z)=x y^2+y e^{3z}+K$$ with ${V}=\nabla f.$

Definition. A region $D$ in the plane is called connected (one piece) if it has the property: (i) any two points in the region can be connected by a piecewise smooth curve lying entirely within $D;$ and a simply connected region (no holes) is a connected region $D$ that has the property: (ii) every closed curve in $D$ encloses only points that are in $D.$

Theorem. (Conservative in Space) Suppose that the vector field ${V}$ and $\mathop{curl}{V}$ are both continuous in the simply connected region $D$ of $\mathbb{R}^3.$ Then ${V}$ is conservative in $D$ if and only if $\mathop{curl}{V}={0}.$

Theorem. (Conservative in the Plane) Consider the vector field $${V}(x,y)=u(x,y){i}+v(x,y){j}$$ where $u$ and $v$ have continuous first partials in the open simply connected region $D$ in the plane. Then ${V}(x,y)$ is conservative in $D$ if and only if \begin{equation} \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} \end{equation} on $D.$

Proof. Note that a vector field ${V}=u(x,y){i}+v(x,y){j}$ in $\mathbb{R}^2$ can be regarded as the vector field ${U} = u(x,y,0){i}+v(x,y,0){j}+0{k}$ in $\mathbb{R}^3.$ Since \begin{equation} \text{curl} \, {U}= \left| \begin{array}{ccc} {i} & {j} & {k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{ \partial }{\partial z} \\ u(x,y,0) & v(x,y,0) & 0 \end{array} \right| =0\, {i}+0\, {j}+\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right){k} \end{equation} we have $\text{curl} \, {U}=0$ if and only if $\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}.$

Example. Determine whether the vector field is conservative and if so, find a scalar potential function ${V}(x,y)=2 x y{i}+x y^3 {j}.$

Solution. Since $u(x,y)=2 x y$, $v(x,y)= x y ^3$, and $\frac{\partial u}{\partial x}=2x\neq y^3=\frac{\partial v}{\partial y},$ we see that ${V}$ is not a conservative vector field.

Example. Show that the vector field \begin{equation} {V}(x,y,z)=
\left(\frac{y}{1+x^2}+\tan ^{-1}z\right) {i} +\left(\tan ^{-1}x\right) {j}+\left(\frac{x}{1+z^2}\right) {k} \end{equation} is conservative and find a scalar potential function.

Solution. Since $\text{curl}{V}=0,$ it follows ${V}$ is conservative. Now we set out to find the scalar potential function $f.$ Since \begin{equation}\frac{\partial f}{\partial x}=\frac{y}{1+x^2}+\tan ^{-1}z\end{equation} we set \begin{equation}f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c(y,z).\end{equation} Since \begin{equation}\frac{\partial f}{\partial y}=\tan ^{-1}z=\frac{\partial }{\partial y}\left(y \tan ^{-1}x+x \tan ^{-1} z+c\right)=\tan ^{-1}x+\frac{\partial c}{\partial y},\end{equation} we find $\frac{\partial c}{\partial y}=0$ and $c=c_1(z)$ and so we set \begin{equation}f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c_1(z).\end{equation} Since \begin{equation}\frac{\partial f}{\partial z}=\frac{x}{1+z^2}=\frac{\partial }{\partial z}\left[y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\right] = \frac{x}{1+z^2} + c’_1(z),\end{equation} we then find $c_1 ‘(z)=0,$ $c_1=0$ and so we obtain \begin{equation} f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z \end{equation} as desired.

## Exercises on Vector Fields

Exercise. If ${F}(x,y)=u(x,y){i}+v(x,y){j}$ where $u$ and $v$ are differentiable functions, show that $\text{curl}{F}=0$ if and only if $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}.$

Exercise. Show that $\text{div}(f\nabla g)=f \text{div} \nabla g+\nabla f \cdot \nabla g.$

Exercise. Show that the curl of the gradient of a function is always $0.$

Exercise. Show that the divergence of the curl of a vector field is $0.$

Exercise. Let ${F}=.$ Either find a vector field ${G}$ such that ${F}=\text{curl}{G}$ or show that no such ${G}$ exists.

Exercise. Find the divergence and the curl for the following vector field.

$(1) \quad {F}(x,y)=x^2 {i}+ x y {j}+ z^3 {k}.$

$(2) \quad {F}(x,y,z)= z \, {i}-{j}+2y\, {k}.$

$(3) \quad {F}(x,y,z)= x y z\, {i}+y\, {j}+ x \, {k}.$

$(4) \quad {F}(x,y,z)= e^{-x}\sin y \, {i}+ e^{-x}\cos y\, {j}+{k}.$

$(5) \quad {F}(x,y)= x \, {i}+y \, {j}.$

$(6) \quad {F}(x,y)= x^2 \, {i}-y^2 \, {j}.$

$(7) \quad {F}(x,y,z)= x^2\, {i}+ y^2 \, {j}+ z^2 \, {k}.$

$(8) \quad {F}(x,y,z)= 2x z\, {i}+ y z^2\, {j}-\, {k}.$

$(9) \quad {F}(x,y,z)= z^2e^{-x}\, {i}+ y^3\ln z\, {j}+ x e^{-y} \, {k}.$

Exercise. Find the divergence of ${F}$ given that ${F}=\nabla f$ where $f(x,y,z)=x y^3z^2.$

Exercise. If ${F}(x,y,z)=x y \, {i}+y z \, {j}+z^2\, {k}$ and ${G}(x,y,z)=x \, {i}+y \, {j}-z \, {k}$ find $\text{curl}({F}\times {G}).$

Exercise. Determine whether or not the following vector fields are conservative.

$(1) \quad {F}(x,y)=y^2\, {i}+2 x y\, {j}$

$(2) \quad {F}(x,y)=2x y^3\, {i}+3y^2x^2\, {j}$

$(3) \quad {F}(x,y)=x e^{x y}\sin y \, {i}+ \left(e^{x y}\cos x y+y\right) {j}$

$(4) \quad {F}(x,y)=\left(-y+e^x\sin y\right){i}+\left((x+2)e^x\cos y\right){j}$

$(5) \quad {F}(x,y)=\left(y-x^2\right){i}+\left(2x+y^2\right){j}$

$(6) \quad {F}(x,y)= e^{2x}\sin y \, {i}+ e^{2x}\cos y {j}$