Electromagnetic Waves, Antennas, and RF: Gradient, Divergence, and Curl - Exercises and Complete Worked Solutions
AntennasElectromagneticWaves6
2026-09-16 01:20:03
2026-09-16 01:20:03
Example
Electromagnetic Waves, Antennas, and RF: Spatial Derivatives of Fields - Gradient, Divergence, and Curl
electromagnetic waves
vector calculus
gradient
divergence
curl
directional derivative
flux
circulation
nabla operator
scalar field
vector field
plane wave
wave vector
electric field
exercises
worked solutions
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\section*{Electromagnetic Waves, Antennas, and RF: Gradient, Divergence, and Curl - Exercises and Complete Worked Solutions}
This companion article provides self-study exercises for EM03, \emph{Spatial Derivatives of Fields: Gradient, Divergence, and Curl}. All exercises are stated first. Complete worked solutions follow in Part II.
The goal is not only to compute formulas mechanically, but also to connect each differential operator with its physical meaning:
\begin{itemize}
\item the gradient of a scalar field points in the direction of fastest increase,
\item the divergence of a vector field measures local source or sink strength, and
\item the curl of a vector field measures local oriented circulation.
\end{itemize}
The central definitions used throughout are
\begin{equation}
\boxed{
\nabla
=\hat{\mathbf x}\frac{\partial}{\partial x}
+\hat{\mathbf y}\frac{\partial}{\partial y}
+\hat{\mathbf z}\frac{\partial}{\partial z}.}
\end{equation}
For a scalar field $\phi(x,y,z)$,
\begin{equation}
\boxed{
\nabla \phi
=\hat{\mathbf x}\frac{\partial \phi}{\partial x}
+\hat{\mathbf y}\frac{\partial \phi}{\partial y}
+\hat{\mathbf z}\frac{\partial \phi}{\partial z}.}
\end{equation}
For a vector field $\mathbf A=A_x\hat{\mathbf x}+A_y\hat{\mathbf y}+A_z\hat{\mathbf z}$,
\begin{equation}
\boxed{
\nabla\cdot \mathbf A
=\frac{\partial A_x}{\partial x}
+\frac{\partial A_y}{\partial y}
+\frac{\partial A_z}{\partial z}}
\end{equation}
and
\begin{equation}
\boxed{
\nabla\times \mathbf A
=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)\hat{\mathbf x}
+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)\hat{\mathbf y}
+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\hat{\mathbf z}.}
\end{equation}
The directional derivative of a scalar field in the unit direction $\hat{\mathbf n}$ is
\begin{equation}
\boxed{
D_{\hat{\mathbf n}}\phi
=\nabla\phi\cdot\hat{\mathbf n}.}
\end{equation}
For plane-wave phase,
\begin{equation}
\boxed{
\theta(\mathbf r,t)=\mathbf k\cdot\mathbf r-\omega t,}
\end{equation}
so that
\begin{equation}
\boxed{\nabla\theta=\mathbf k.}
\end{equation}
These are the same conventions developed in EM03 and in standard references on vector calculus and electrodynamics \cite{Schey2005,Griffiths2017,OpenStaxCalc3,MIT1802}.
\section*{How to use this problem set}
Attempt all exercises in Part I before reading Part II. For each answer, keep three questions separate:
\begin{enumerate}
\item What is the operator acting on?
\item What kind of object should the result be: scalar or vector?
\item What local physical or geometric meaning does the result have?
\end{enumerate}
Many mistakes in vector calculus are not arithmetic errors but \emph{type errors}: for example, expecting a scalar from a gradient or expecting a vector from a divergence.
\section*{Part I: Exercises}
\section*{Exercise 1: partial derivatives of a scalar field}
Let
\begin{equation}
\phi(x,y,z)=x^2y+3yz-z^2.
\end{equation}
Compute
\begin{equation}
\frac{\partial \phi}{\partial x},
\qquad
\frac{\partial \phi}{\partial y},
\qquad
\frac{\partial \phi}{\partial z}.
\end{equation}
\section*{Exercise 2: gradient at a point}
Let
\begin{equation}
\phi(x,y,z)=x^2+2y^2+z.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] $\nabla\phi$;
\item[(b)] $\nabla\phi$ at the point $(1,-1,2)$;
\item[(c)] the magnitude $|\nabla\phi|$ at that point;
\item[(d)] the unit direction of fastest increase at that point.
\end{enumerate}
\begin{center}
\includegraphics{EM03E_fig01_gradient_normals.png}
\vspace{0.45em}
\textbf{Figure.}
The gradient is normal to level curves and points toward larger scalar-field values.
\end{center}
\section*{Exercise 3: directional derivative}
Using the scalar field from Exercise 2, find the directional derivative at $(1,-1,2)$ in the unit direction
\begin{equation}
\hat{\mathbf n}=\frac{1}{\sqrt{2}}\hat{\mathbf x}+\frac{1}{\sqrt{2}}\hat{\mathbf y}.
\end{equation}
Interpret the sign of the result.
\section*{Exercise 4: level surfaces and normals}
Consider the scalar field
\begin{equation}
\phi(x,y,z)=x+2y-z.
\end{equation}
Show that the surfaces $\phi=C$ are planes. Find a normal vector to these planes and explain its relation to the gradient.
\section*{Exercise 5: divergence of a linear field}
Let
\begin{equation}
\mathbf A(x,y,z)=x\hat{\mathbf x}-2y\hat{\mathbf y}+3z\hat{\mathbf z}.
\end{equation}
Compute $\nabla\cdot\mathbf A$ and state whether the field behaves locally like a net source, a net sink, or neither.
\section*{Exercise 6: zero-divergence field}
Let
\begin{equation}
\mathbf B(x,y,z)=-y\hat{\mathbf x}+x\hat{\mathbf y}.
\end{equation}
Compute $\nabla\cdot\mathbf B$ and interpret the result physically.
\begin{center}
\includegraphics{EM03E_fig02_divergence_box.png}
\vspace{0.45em}
\textbf{Figure.}
Divergence compares local outward and inward flux through a small closed box.
\end{center}
\section*{Exercise 7: divergence as flux density}
A vector field is
\begin{equation}
\mathbf F(x,y,z)=2x\hat{\mathbf x}+y\hat{\mathbf y}.
\end{equation}
At the point $(1,2,0)$:
\begin{enumerate}
\item[(a)] compute $\nabla\cdot\mathbf F$;
\item[(b)] estimate the net outward flux through a small rectangular box of volume
\begin{equation}
\Delta V=0.02\,\text{m}^3
\end{equation}
centered at that point using the local relation
\begin{equation}
\text{net outward flux}\approx (\nabla\cdot\mathbf F)\Delta V.
\end{equation}
\end{enumerate}
\section*{Exercise 8: curl of a rotational field}
Let
\begin{equation}
\mathbf C(x,y,z)=-y\hat{\mathbf x}+x\hat{\mathbf y}.
\end{equation}
Compute $\nabla\times\mathbf C$.
\section*{Exercise 9: curl of a gradient field}
Let
\begin{equation}
\psi(x,y,z)=x^2+y^2+z^2
\end{equation}
and let
\begin{equation}
\mathbf G=\nabla\psi.
\end{equation}
Find $\mathbf G$ and then compute $\nabla\times\mathbf G$.
\begin{center}
\includegraphics{EM03E_fig03_curl_loop.png}
\vspace{0.45em}
\textbf{Figure.}
Curl measures the oriented circulation tendency around a small loop.
\end{center}
\section*{Exercise 10: compare zero divergence and zero curl}
For the two vector fields
\begin{equation}
\mathbf U=x\hat{\mathbf x}+y\hat{\mathbf y}
\end{equation}
and
\begin{equation}
\mathbf V=-y\hat{\mathbf x}+x\hat{\mathbf y},
\end{equation}
compute both $\nabla\cdot$ and $\nabla\times$ for each field. Which field is source-like? Which field is rotational?
\section*{Exercise 11: gradient of plane-wave phase}
Let the phase be
\begin{equation}
\theta(x,y,z,t)=2x-3y+4z-\omega t.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] $\nabla\theta$;
\item[(b)] the wave vector $\mathbf k$;
\item[(c)] the magnitude $|\mathbf k|$.
\end{enumerate}
\begin{center}
\includegraphics{EM03E_fig04_wave_phase_gradient.png}
\vspace{0.45em}
\textbf{Figure.}
For a plane wave, the gradient of phase is the wave vector, normal to constant-phase planes.
\end{center}
\section*{Exercise 12: gradient of a scalar plane wave}
Let
\begin{equation}
\psi(x,t)=\psi_0\cos(kx-\omega t).
\end{equation}
Treat this as a three-dimensional scalar field that depends only on $x$.
Find:
\begin{enumerate}
\item[(a)] $\partial \psi/\partial x$;
\item[(b)] $\nabla\psi$;
\item[(c)] the direction of $\nabla\psi$.
\end{enumerate}
\section*{Exercise 13: divergence and curl of a transverse field}
Let
\begin{equation}
\mathbf E(z,t)=E_0\cos(kz-\omega t)\hat{\mathbf x}.
\end{equation}
Compute:
\begin{enumerate}
\item[(a)] $\nabla\cdot\mathbf E$;
\item[(b)] $\nabla\times\mathbf E$.
\end{enumerate}
State clearly whether each result is zero or nonzero.
\section*{Exercise 14: identify the operator type and meaning}
For each expression below, state whether the result is a scalar or a vector, and give its local physical meaning.
\begin{enumerate}
\item[(a)] $\nabla \phi$;
\item[(b)] $\nabla\cdot\mathbf A$;
\item[(c)] $\nabla\times\mathbf A$;
\item[(d)] $\nabla\theta$ for $\theta=\mathbf k\cdot\mathbf r-\omega t$.
\end{enumerate}
\section*{Part II: Complete worked solutions}
\section*{Solution 1: partial derivatives of a scalar field}
Given
\begin{equation}
\phi(x,y,z)=x^2y+3yz-z^2,
\end{equation}
we differentiate with respect to one variable at a time while holding the others fixed.
For $x$,
\begin{equation}
\frac{\partial \phi}{\partial x}=2xy.
\end{equation}
For $y$,
\begin{equation}
\frac{\partial \phi}{\partial y}=x^2+3z.
\end{equation}
For $z$,
\begin{equation}
\frac{\partial \phi}{\partial z}=3y-2z.
\end{equation}
Therefore,
\begin{equation}
\boxed{
\frac{\partial \phi}{\partial x}=2xy,
\qquad
\frac{\partial \phi}{\partial y}=x^2+3z,
\qquad
\frac{\partial \phi}{\partial z}=3y-2z.}
\end{equation}
\section*{Solution 2: gradient at a point}
The field is
\begin{equation}
\phi(x,y,z)=x^2+2y^2+z.
\end{equation}
Its partial derivatives are
\begin{equation}
\frac{\partial \phi}{\partial x}=2x,
\qquad
\frac{\partial \phi}{\partial y}=4y,
\qquad
\frac{\partial \phi}{\partial z}=1.
\end{equation}
Hence
\begin{equation}
\boxed{
\nabla\phi=2x\hat{\mathbf x}+4y\hat{\mathbf y}+\hat{\mathbf z}.}
\end{equation}
At $(1,-1,2)$,
\begin{equation}
\boxed{
\nabla\phi(1,-1,2)=2\hat{\mathbf x}-4\hat{\mathbf y}+\hat{\mathbf z}.}
\end{equation}
Its magnitude is
\begin{equation}
|\nabla\phi|=
\sqrt{2^2+(-4)^2+1^2}
=\sqrt{21}.
\end{equation}
Thus
\begin{equation}
\boxed{|\nabla\phi(1,-1,2)|=\sqrt{21}.}
\end{equation}
The unit direction of fastest increase is the normalized gradient:
\begin{equation}
\boxed{
\hat{\mathbf u}
=\frac{2\hat{\mathbf x}-4\hat{\mathbf y}+\hat{\mathbf z}}{\sqrt{21}}.}
\end{equation}
\section*{Solution 3: directional derivative}
From Solution 2,
\begin{equation}
\nabla\phi(1,-1,2)=2\hat{\mathbf x}-4\hat{\mathbf y}+\hat{\mathbf z}.
\end{equation}
The unit direction is
\begin{equation}
\hat{\mathbf n}=\frac{1}{\sqrt{2}}\hat{\mathbf x}+\frac{1}{\sqrt{2}}\hat{\mathbf y}.
\end{equation}
Therefore,
\begin{equation}
D_{\hat{\mathbf n}}\phi
=\nabla\phi\cdot\hat{\mathbf n}
=(2,-4,1)\cdot\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right).
\end{equation}
So
\begin{equation}
D_{\hat{\mathbf n}}\phi
=\frac{2}{\sqrt{2}}-\frac{4}{\sqrt{2}}
=-\frac{2}{\sqrt{2}}
=-\sqrt{2}.
\end{equation}
Thus
\begin{equation}
\boxed{D_{\hat{\mathbf n}}\phi=-\sqrt{2}.}
\end{equation}
The negative sign means the scalar field decreases as one moves locally in the direction $\hat{\mathbf n}$.
\section*{Solution 4: level surfaces and normals}
If
\begin{equation}
\phi(x,y,z)=x+2y-z,
\end{equation}
then the level surfaces $\phi=C$ satisfy
\begin{equation}
x+2y-z=C,
\end{equation}
which is the equation of a plane.
The gradient is
\begin{equation}
\nabla\phi
=\hat{\mathbf x}+2\hat{\mathbf y}-\hat{\mathbf z}.
\end{equation}
Therefore a normal vector to the level planes is
\begin{equation}
\boxed{\hat{\mathbf x}+2\hat{\mathbf y}-\hat{\mathbf z}.}
\end{equation}
This illustrates the general rule that the gradient is normal to a constant-value surface.
\section*{Solution 5: divergence of a linear field}
For
\begin{equation}
\mathbf A=x\hat{\mathbf x}-2y\hat{\mathbf y}+3z\hat{\mathbf z},
\end{equation}
we identify
\begin{equation}
A_x=x,
\qquad
A_y=-2y,
\qquad
A_z=3z.
\end{equation}
Then
\begin{equation}
\nabla\cdot\mathbf A
=\frac{\partial x}{\partial x}
+\frac{\partial (-2y)}{\partial y}
+\frac{\partial (3z)}{\partial z}
=1-2+3=2.
\end{equation}
Hence
\begin{equation}
\boxed{\nabla\cdot\mathbf A=2.}
\end{equation}
Because the divergence is positive, the field behaves locally like a net source.
\section*{Solution 6: zero-divergence field}
For
\begin{equation}
\mathbf B=-y\hat{\mathbf x}+x\hat{\mathbf y},
\end{equation}
the components are
\begin{equation}
B_x=-y,
\qquad
B_y=x,
\qquad
B_z=0.
\end{equation}
Therefore,
\begin{equation}
\nabla\cdot\mathbf B
=\frac{\partial (-y)}{\partial x}
+\frac{\partial x}{\partial y}
+\frac{\partial 0}{\partial z}
=0+0+0=0.
\end{equation}
So
\begin{equation}
\boxed{\nabla\cdot\mathbf B=0.}
\end{equation}
The field has no local net source or sink strength, even though it may still circulate.
\section*{Solution 7: divergence as flux density}
Given
\begin{equation}
\mathbf F=2x\hat{\mathbf x}+y\hat{\mathbf y},
\end{equation}
we compute
\begin{equation}
\nabla\cdot\mathbf F
=\frac{\partial (2x)}{\partial x}
+\frac{\partial y}{\partial y}
+\frac{\partial 0}{\partial z}
=2+1+0=3.
\end{equation}
Thus,
\begin{equation}
\boxed{\nabla\cdot\mathbf F=3.}
\end{equation}
Since the field is linear, this value is the same at $(1,2,0)$.
Using
\begin{equation}
\text{net outward flux}\approx (\nabla\cdot\mathbf F)\Delta V,
\end{equation}
with $\Delta V=0.02\,\text{m}^3$, we obtain
\begin{equation}
\text{net outward flux}\approx 3(0.02)=0.06.
\end{equation}
Therefore,
\begin{equation}
\boxed{\text{net outward flux}\approx 0.06}
\end{equation}
in the corresponding flux units of the field.
\section*{Solution 8: curl of a rotational field}
Let
\begin{equation}
\mathbf C=-y\hat{\mathbf x}+x\hat{\mathbf y}.
\end{equation}
Then
\begin{equation}
C_x=-y,
\qquad
C_y=x,
\qquad
C_z=0.
\end{equation}
Using the component formula for curl,
\begin{equation}
\nabla\times\mathbf C
=\left(\frac{\partial C_z}{\partial y}-\frac{\partial C_y}{\partial z}\right)\hat{\mathbf x}
+\left(\frac{\partial C_x}{\partial z}-\frac{\partial C_z}{\partial x}\right)\hat{\mathbf y}
+\left(\frac{\partial C_y}{\partial x}-\frac{\partial C_x}{\partial y}\right)\hat{\mathbf z}.
\end{equation}
The first two components are zero. The third is
\begin{equation}
\frac{\partial C_y}{\partial x}-\frac{\partial C_x}{\partial y}=1-(-1)=2.
\end{equation}
Hence
\begin{equation}
\boxed{\nabla\times\mathbf C=2\hat{\mathbf z}.}
\end{equation}
So the field has nonzero local circulation about the $z$ axis.
\section*{Solution 9: curl of a gradient field}
Given
\begin{equation}
\psi=x^2+y^2+z^2,
\end{equation}
its gradient is
\begin{equation}
\mathbf G=\nabla\psi=2x\hat{\mathbf x}+2y\hat{\mathbf y}+2z\hat{\mathbf z}.
\end{equation}
Now compute the curl:
\begin{equation}
\nabla\times\mathbf G
=\left(\frac{\partial (2z)}{\partial y}-\frac{\partial (2y)}{\partial z}\right)\hat{\mathbf x}
+\left(\frac{\partial (2x)}{\partial z}-\frac{\partial (2z)}{\partial x}\right)\hat{\mathbf y}
+\left(\frac{\partial (2y)}{\partial x}-\frac{\partial (2x)}{\partial y}\right)\hat{\mathbf z}.
\end{equation}
Every derivative above is zero, so
\begin{equation}
\boxed{\nabla\times\mathbf G=\mathbf 0.}
\end{equation}
This is a standard example that a gradient field is irrotational.
\section*{Solution 10: compare zero divergence and zero curl}
First consider
\begin{equation}
\mathbf U=x\hat{\mathbf x}+y\hat{\mathbf y}.
\end{equation}
Its divergence is
\begin{equation}
\nabla\cdot\mathbf U
=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}=1+1=2.
\end{equation}
Its curl is
\begin{equation}
\nabla\times\mathbf U
=\left(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y}\right)\hat{\mathbf z}
=0\hat{\mathbf z}.
\end{equation}
Thus
\begin{equation}
\boxed{\nabla\cdot\mathbf U=2,
\qquad
\nabla\times\mathbf U=\mathbf 0.}
\end{equation}
So $\mathbf U$ is source-like but not rotational.
Now consider
\begin{equation}
\mathbf V=-y\hat{\mathbf x}+x\hat{\mathbf y}.
\end{equation}
Its divergence is
\begin{equation}
\nabla\cdot\mathbf V
=\frac{\partial (-y)}{\partial x}+\frac{\partial x}{\partial y}=0.
\end{equation}
Its curl is
\begin{equation}
\nabla\times\mathbf V
=\left(\frac{\partial x}{\partial x}-\frac{\partial (-y)}{\partial y}\right)\hat{\mathbf z}
=(1-(-1))\hat{\mathbf z}=2\hat{\mathbf z}.
\end{equation}
Therefore,
\begin{equation}
\boxed{\nabla\cdot\mathbf V=0,
\qquad
\nabla\times\mathbf V=2\hat{\mathbf z}.}
\end{equation}
So $\mathbf V$ is rotational but not source-like.
\section*{Solution 11: gradient of plane-wave phase}
For
\begin{equation}
\theta(x,y,z,t)=2x-3y+4z-\omega t,
\end{equation}
we compute
\begin{equation}
\frac{\partial \theta}{\partial x}=2,
\qquad
\frac{\partial \theta}{\partial y}=-3,
\qquad
\frac{\partial \theta}{\partial z}=4.
\end{equation}
Hence
\begin{equation}
\boxed{\nabla\theta=2\hat{\mathbf x}-3\hat{\mathbf y}+4\hat{\mathbf z}.}
\end{equation}
By comparison with $\theta=\mathbf k\cdot\mathbf r-\omega t$, we identify
\begin{equation}
\boxed{\mathbf k=2\hat{\mathbf x}-3\hat{\mathbf y}+4\hat{\mathbf z}.}
\end{equation}
Its magnitude is
\begin{equation}
|\mathbf k|=\sqrt{2^2+(-3)^2+4^2}=\sqrt{29}.
\end{equation}
Thus
\begin{equation}
\boxed{|\mathbf k|=\sqrt{29}.}
\end{equation}
\section*{Solution 12: gradient of a scalar plane wave}
Let
\begin{equation}
\psi(x,t)=\psi_0\cos(kx-\omega t).
\end{equation}
Differentiate with respect to $x$:
\begin{equation}
\frac{\partial \psi}{\partial x}
=-k\psi_0\sin(kx-\omega t).
\end{equation}
Because the field depends only on $x$,
\begin{equation}
\frac{\partial \psi}{\partial y}=0,
\qquad
\frac{\partial \psi}{\partial z}=0.
\end{equation}
Therefore,
\begin{equation}
\boxed{
\nabla\psi
=-k\psi_0\sin(kx-\omega t)\hat{\mathbf x}.}
\end{equation}
So the gradient points purely in the $\pm x$ direction, depending on the sign of the sine factor.
\section*{Solution 13: divergence and curl of a transverse field}
The field is
\begin{equation}
\mathbf E(z,t)=E_0\cos(kz-\omega t)\hat{\mathbf x}.
\end{equation}
Its components are
\begin{equation}
E_x=E_0\cos(kz-\omega t),
\qquad
E_y=0,
\qquad
E_z=0.
\end{equation}
For the divergence,
\begin{equation}
\nabla\cdot\mathbf E
=\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}.
\end{equation}
Since $E_x$ depends on $z$ but not on $x$, and $E_y=E_z=0$,
\begin{equation}
\boxed{\nabla\cdot\mathbf E=0.}
\end{equation}
For the curl, only the $y$ component survives. More explicitly,
\begin{equation}
(\nabla\times\mathbf E)_y
=\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}
=\frac{\partial E_x}{\partial z}.
\end{equation}
However, from the standard component formula,
\begin{equation}
\nabla\times\mathbf E
=\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\hat{\mathbf x}
+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\hat{\mathbf y}
+\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\hat{\mathbf z}.
\end{equation}
Thus,
\begin{equation}
\frac{\partial E_x}{\partial z}
=-kE_0\sin(kz-\omega t),
\end{equation}
and therefore
\begin{equation}
\boxed{
\nabla\times\mathbf E
=-kE_0\sin(kz-\omega t)\hat{\mathbf y}.}
\end{equation}
So the divergence is zero while the curl is generally nonzero.
\section*{Solution 14: identify the operator type and meaning}
\begin{enumerate}
\item[(a)] $\nabla\phi$ is a \textbf{vector}. It points in the direction of fastest increase of the scalar field and has magnitude equal to the maximum local rate of increase.
\item[(b)] $\nabla\cdot\mathbf A$ is a \textbf{scalar}. It measures local source or sink strength, or equivalently the net outward flux per unit volume in the small-volume limit.
\item[(c)] $\nabla\times\mathbf A$ is a \textbf{vector}. It measures local oriented circulation, with the direction set by the right-hand rule.
\item[(d)] $\nabla\theta$ is a \textbf{vector}. For plane-wave phase $\theta=\mathbf k\cdot\mathbf r-\omega t$, it equals the wave vector $\mathbf k$, which is normal to constant-phase planes and points in the propagation direction.
\end{enumerate}
\section*{Concluding remarks}
The key lesson from EM03 and EM03E is that three different questions about spatial change lead to three different operators:
\begin{equation}
\boxed{\nabla\phi}
\end{equation}
asks how a scalar field rises,
\begin{equation}
\boxed{\nabla\cdot\mathbf A}
\end{equation}
asks whether a vector field spreads out or converges, and
\begin{equation}
\boxed{\nabla\times\mathbf A}
\end{equation}
asks whether a vector field circulates.
For plane-wave phase,
\begin{equation}
\boxed{\nabla(\mathbf k\cdot\mathbf r-\omega t)=\mathbf k}
\end{equation}
connects vector calculus directly to wave geometry.
The next lesson, EM04, combines these ideas further by introducing the Laplacian and the three-dimensional wave equation.
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\bibitem{FeynmanV2}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume II},
Addison-Wesley, 1964,
chapters on vector electromagnetic fields and Maxwell's equations.
\end{thebibliography}