
\section{Divergence and Curl}

\begin{enumerate}[1.]
\item {\it
 Let ${\bf F}=(y+z)^2\ii+y\jj+z\kk$. Find the divergence and curl of ${\bf F}$.}
 
 \medskip
$$
{\bf div\, F}=0+1+1=2,$$
$$
{\bf curl \, F} = 
  \left|\begin{matrix}
\ii & \jj & \kk\\
\partial_x& \partial_y& \partial_z\\
(y+z)^2& y& z\\      
            \end{matrix}\right|
=0\ii + 2(y+z)\jj -2(y+z)\kk
 $$
 
\bigskip


\item
{\it (a) Let ${\bf F}=x\cos y\ii + e^z \jj + \cos y \kk$. Find the curl and divergence of ${\bf F}$.}


$$
{\bf div\, F}= \cos y + 0 + 0 =\cos y$$
$$
{\bf curl \, F} = 
  \left|\begin{matrix}
\ii & \jj & \kk\\
\partial_x& \partial_y& \partial_z\\
x\cos y & e^z& \cos y\\      
            \end{matrix}\right|
=(-\sin y -e^z) \ii +  x\sin y \kk
 $$

\noindent
{\it (b) 
 Is it true that $\nabla\times  {\bf F}$ is always perpendicular to ${\bf F}$ for all $C^1$ vector fields ${\bf F}$
in $\mathbb{R}^3$? Explain why or why not. }


\medskip
For the vector field ${\bf F}$ from (a), we have
$${\bf F}\cdot (\nabla \times {\bf F})=x\cos y (-\sin y -e^z)  + x\sin y \cos y= -xe^z \cos y $$
which is clearly not always zero. Therefore the statement is false.


\bigskip


\item
{\it Prove that if $f$ and $g$ are two $C^2$ functions on an open set
$X\subset {\mathbb R}^3$, then $\nabla\cdot(\nabla f\times \nabla g)=0$.}

\medskip

We first compute
$$
\nabla f\times\nabla g=
\left|\begin{matrix}
\ii & \jj & \kk\\
\partial_xf& \partial_yf& \partial_zf\\
\partial_xg& \partial_yg& \partial_zg\\      
            \end{matrix}\right|
=(f_yg_z-f_zg_y)\ii-(f_xg_z-f_zg_x)\jj+(f_xg_y-f_yg_x)\kk,$$
so that
\begin{align*}
\nabla\cdot(\nabla f\times\nabla g)
&=(f_{xy}g_z+f_yg_{xz}-f_{xz}g_y-f_zg_{xy})
\\
&-(f_{xy}g_z+f_xg_{yz}-f_{yz}g_x-f_zg_{xy})
\\
&+(f_{xz}g_y+f_xg_{yz}-f_{yz}g_x-f_yg_{xz})\\
&=0.
\end{align*}

\bigskip




\item 
{\it 
Let ${\bf F}$ be a 2-dimensional $C^2$ vector field on an open set $U$ in the $xy$-plane. Let $P\in U$, and  let ${\bf r}(t)$ be the field line of ${\bf F}$ that passes through $P$.
Assume that:
\begin{itemize}
\item ${\bf F}$ has constant magnitude $1$ on a neighbourhood of $P$,
\item the field line ${\bf r}(t)$ has curvature $a$ at $P$.
\end{itemize}
 Find the curl of ${\bf F}$ at $P$. Prove your answer.}
 
 

 
 \bigskip

Let ${\bf F} = F_1\ii +F_2\jj$, and let ${\bf r}(t)$ be the field line as indicated.
We use the curvature formula
\begin{equation}\label{41hw2025e1}
a
=\frac{|\bfr'\times\bfr''|}{|\bfr'|^3}.
\end{equation}
For a field line, we have
\begin{align*}
\bfr'(t)&={\bf F}(\bfr)=\langle F_1(\bfr), F_2(\bfr)\rangle
\\
\bfr''(t)&= \langle \nabla F_1(\bfr)\cdot \bfr' , \nabla F_2(\bfr)\cdot \bfr'\rangle
\\
&= \langle \nabla F_1(\bfr)\cdot {\bf F}(\bfr) , \nabla F_2(\bfr)\cdot {\bf F}(\bfr)\rangle.
\end{align*}
We have $|\bfr'|=|{\bf F}|=1$ in a neighbourhood of $P$.
The cross product is
$$
{\bf r}'\times {\bf r}''=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
F_1(\bfr)& F_2(\bfr)&0\\
\nabla F_1(\bfr)\cdot {\bf F}(\bfr) & \nabla F_2(\bfr)\cdot {\bf F}(\bfr) &0\\
            \end{matrix}\right|
$$
$$= \Big( F_1(\nabla F_2(\bfr)\cdot {\bf F}(\bfr))-F_2( \nabla F_1(\bfr)\cdot {\bf F}(\bfr)\Big)\kk.$$
Therefore
\begin{equation}\label{41hw2025e2}
|{\bf r}'\times {\bf r}''|
= \left| F_1(F_{2,x}F_1+F_{2,y}F_2)-F_2(F_{1,x}F_1+F_{1,y}F_2) \right|,
\end{equation}
with everything evaluated at $\bfr$. Now we use that $|{\bf F}|=1$ is constant in the neighbourhood of $P$. Therefore in that neighbourhood,
\begin{align*}
0&=\frac{\partial}{\partial x}|{\bf F}|^2 = \frac{\partial}{\partial x}(F_1^2+F_2^2)
= F_1F_{1,x}+F_2 F_{2,x},
\\
0&=\frac{\partial}{\partial y}|{\bf F}|^2 = \frac{\partial}{\partial y}(F_1^2+F_2^2)
= F_1F_{1,y}+F_2 F_{2,y},
\end{align*}
so that $F_1F_{1,x}=-F_2F_{2,x}$ and  $F_2F_{2,y}=-F_1F_{1,y}$. Going back with this to (\ref{41hw2025e1}) and (\ref{41hw2025e2}), we get
\begin{align*}
a = |{\bf r}'\times {\bf r}''|
&= \left| F_1(F_{2,x}F_1-F_1F_{1,y})-F_2(-F_2F_{2,x}+F_{1,y}F_2) \right|
\\
&= \left| (F_1^2+F_2^2)F_{2,x}-(F_1^2+F_2^2)F_{1,y}\right|
\\
&= |{\bf F}|^2 |\hbox{\rm curl} {\bf F} |
\\
&= | \hbox{\rm curl} {\bf F} |
\end{align*}
Therefore
curl$\,{\bf F} (P)=\pm a\kk$, with the sign depending on which way the field line is turning.












\end{enumerate}


