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% 28/01/2022 10AM

\begin{example*}
    Let $\bf{F}(\bf{x}) = (xe^y, z^2, xy)$. For
    $C_1$ let $\bf{x}(t) = (t, t, t)$, and for
    $C_2$ let $\bf{x}(t) = (t, t^2, t^3)$. We'll
    integrate from $(0,0,0)$ to $(1,1,1)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/5ab7844482c311ec.png}
    \end{center}
    \ul{For $C_1$}: $\bf{F}(t) = (te^{t^2}, t^6,
    t^3)$ and $\dot{\bf{x}}(t) = (1, 2t, 3t^2)$.
    \begin{align*}
        \implies \int_{C_1} \bf{F} \cdot \dd \bf{x}
        &= \int_0^1 \dd t (te^{t^2} + 2t^7 + 3t^5)
        \\
        &= \frac{1}{4} (1 + 2e)
    \end{align*}
    \ul{For $C_2$}: $\bf{F}(t) = (te^t, t^2, t^z)$
    and $\dot{\bf{x}}(t) = (1, 1, 1)$.
    \begin{align*}
        \implies \int_{C_2} \bf{F} \cdot \dd
        \bf{x}
        &= \int_0^1 \dd t(te^t + 2t^2) \\
        &= \frac{5}{3}
    \end{align*}
    \begin{note*}
        Answer depends on $C$.
    \end{note*}
\end{example*}

\noindent
Sometimes we will integrate along a closed path
$C$, with $\bf{a} = bf$. The line integral is the
\emph{circulation} of $\bf{F}$ around $C$, denoted
as
\[ \oint_C \bf{F} \cdot \dd \bf{x} \]
Sometimes we will have a \emph{piecewise smooth}
curve $C = C_1 + C_2$, and then we define
\[ \int_{C_1 + C_2} \bf{F} \cdot \dd \bf{x} =
\int_{C_1} \bf{F} \cdot \dd \bf{x} + \int_{C_2}
\bf{F} \cdot \dd \bf{x} .\]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/137c1c3882c411ec.png}
\end{center}
The curve $-C$ is the curve $C$ but with the
opposite orientation, so
\[ \int_{-C} \bf{F} \cdot \dd \bf{x} = -\int_C
\bf{F} \cdot \dd \bf{x} \]
for example let $C = C_1 - C_2$ in our example.
Then
\[ \oint_C \bf{F} \cdot \dd \bf{x} = \int_{C_1}
Fbf \cdot \dd \bf{x} - \int_{C_2} \bf{F} \cdot \dd
\bf{x} = \frac{1}{4}(1 + 2e) - \frac{5}{3} \]

\subsection{Conservative Fields}

\ul{Question}: Do there exist $\bf{F}$ such that
$\int_C \bf{F} \cdot \dd \bf{x}$ is independent of
the path chosen between two fixed end points
$\bf{a}$ and $\bf{b}$ i.e.
\[ \int_{C_1} \bf{F} \cdot \dd \bf{x} = \int_{C_2}
\bf{F} \cdot \dd \bf{x} ?\]
(for all $C_1$ and $C_2$ with the same end
points). \myskip
Equivalently, considering $C = C_1 - C_2$, this
would mean
\[ \oint_C \bf{F} \cdot \dd \bf{x} = 0 \]
for all closed paths $C$.

\subsubsection*{The gradient}

Consider a scalar field $\phi : \RR^n \to \RR$.
The \emph{partial derivatives} are defined to be
\[ \pfrac{\phi}{x^1} = \lim_{e \to 0} \frac{1}{e}
[\phi(x^1 + e, x^2, \dots, x^n) - \phi(x^1, x^2,
\dots, x^n)] \]
and similar for $\pfrac{\phi}{x^2}$ etc. The
function is differentiable if all $n$ partial
derivatives exist. We write:
\[ \partial_i \phi = \pfrac{\phi}{x^i} \qquad i =
1, \dots, n \]
Also, it's not uncommon to stress which variables
are held fixed by writing
\[ \left( \pfrac{\phi}{x^1} \right)_{x^2, \dots,
x^n} \]
Let $\{\bf{e}_i\}$ be orthonormal basis of
$\RR^n$. Then the \emph{gradient} of a scalar field
is vector field, defined as
\[ \nabla \phi = \pfrac{\phi}{x^i} \bf{e}_i \]

\begin{note*}
    Sometimes the $\nabla$ is written with bold or
    underline.
\end{note*}

If we want to compute how $\phi$ changes in some
direction $\hat{\bf{n}}$ with $|\hat{\bf{n}}| =
1$, then we compute the \emph{directional
derivative} $\hat{\bf{n}} \cdot \nabla \phi$.
\myskip
This is maximised at any point $\bf{x}$ by picking
$\hat{\bf{n}} \parallel \nabla \phi$. But this
means that $\nabla \phi(\bf{x})$ points in the
direction in which $\phi(\bf{x})$ increases most
quickly.

\subsubsection*{Back to Conservative Fields}

A vector field $\bf{F}$ is called
\emph{conservative} if it can be written as
\[ \bf{F} = \nabla \phi \]
for some $\phi$ called a \emph{potential}.

\begin{claim*}
    \[ \oint_C \bf{F} \cdot \dd \bf{x} = 0
    \,\,\forall\,\, C \]
    if and only if $\bf{F}$ is conservative.
\end{claim*}

\begin{proof}
    If $\bf{F} = \nabla \phi$ then along any
    \emph{open} curve $C$, parametrised by
    $\bf{x}(t)$, we have
    \begin{align*}
        \int_C \bf{F} \cdot \dd \bf{x}
        &= \int_C \nabla \phi \cdot \dd \bf{x} \\
        &= \int_{t_a}^{t_b} \pfrac{\phi}{x^i}
        \dfrac{x^i}{t} \dd t \\
        &= \int_{t_a}^{t_b} \dfrac{}{t}
        \phi(\bf{x}(t)) \dd t \\
        &= \phi(\bf{x}(t_b)) - \phi(\bf{x}(t_a))
    \end{align*}
    i.e. only depends on the end points.
    Conversely, suppose that
    \[ \oint_C \bf{F} \cdot \dd \bf{x} = 0 \]
    Let $\phi(\bf{0}) = 0$ and define
    \[ \phi(\bf{y}) = \int_{C(\bf{y})} \bf{F}
    \cdot \dd \bf{x} \]
    Then
    \begin{align*}
        \pfrac{\phi}{x^i} (\bf{y})
        &= \lim_{e \to 0} \frac{1}{e}
        \left[ \int_{C(\bf{y} +
        e \bf{e}_i} \bf{F} \cdot \dd \bf{x} -
        \int_{C(\bf{y})} \right] \\
        &= \lim_{e \to 0} \frac{1}{e}
        \int_{\bf{y}}^{\bf{y} + e \bf{e}_i} \bf{F}
        \cdot \dd \bf{x} \\
        &= \lim_{e \to 0} \frac{1}{e} (eF_i) \\
        &= F_i
    \end{align*}
\end{proof}