%% multivar.tex
%% Copyright 2021 Rebecca B. Turner.
%
% This work may be distributed and/or modified under the
% conditions of the LaTeX Project Public License, either version 1.3
% of this license or (at your option) any later version.
% The latest version of this license is in
%   http://www.latex-project.org/lppl.txt
% and version 1.3 or later is part of all distributions of LaTeX
% version 2005/12/01 or later.
%
% This work has the LPPL maintenance status `maintained'.
% 
% The Current Maintainer of this work is Rebecca B. Turner.
%
% This work consists of the files:
%     README.md
%     rbt-mathnotes.tex
%     rbt-mathnotes.sty
%     rbt-mathnotes.cls
%     rbt-mathnotes-util.sty
%     rbt-mathnotes-messages.sty
%     rbt-mathnotes-hw.cls
%     rbt-mathnotes-formula-sheet.cls
%     examples/cheat-sheet.tex
%     examples/multivar.tex
%     examples/topology-hw-1.tex
% and the derived files:
%     rbt-mathnotes.pdf
%     examples/cheat-sheet.pdf
%     examples/multivar.pdf
%     examples/topology-hw-1.pdf

\documentclass[knowledge]{rbt-mathnotes}
\title{Multivariable Calculus}
\mathnotes{
  instructor  = Prof.~Corey Bregman ,
  name        = Rebecca Turner ,
  email       = rebeccaturner@brandeis.edu ,
  course      = \textsc{math} 20a (multivariable calculus) ,
  institution = Brandeis University ,
  semester    = Fall 2019 ,
}

\ExplSyntaxOn
\NewDocumentCommand \normalized { m }
  { \frac { #1 } { \| #1 \| } }
\let \gr \grad
\def \ddx { \frac{d}{dx} }
% VL = vector literal
\NewDocumentCommand \vl { m } { \left\langle #1 \right\rangle }
\ExplSyntaxOff

% \makeatletter
% \@ifpackageloaded{knowledge}
%   {
\knowledge{notion, index={Derivative!Partial}}
  | partial derivative
\AtBeginDocument{\index{Partial derivative|see{Derivative, partial}}}

\knowledge{notion, index={Derivative!Directional}}
  | directional derivative
\AtBeginDocument{\index{Directional derivative|see{Directional, partial}}}

\knowledge{notion, index=Unit vector}
  | unit vector

\knowledge{notion, index=Limit}
  | limit

\knowledge{notion, index={Differentiable functions}}
  | differentiable

\knowledge{notion, index=Gradient}
  | gradient

\knowledge{notion, index={Neighborhood (topology)}}
  | neighborhood

\knowledge{notion, index=Local maximum}
  | local maximum
  | local maxima

\knowledge{notion, index=Local minimum}
  | local minimum
  | local minima

\knowledge{notion, index=Absolute maximum}
  | absolute maximum
  | absolute maxima

\knowledge{notion, index=Absolute minimum}
  | absolute minimum
  | absolute minima

\knowledge{notion, index=Local extremum}
  | local extremum
  | local extrema

\knowledge{notion, index=Critical point}
  | critical point

\knowledge{notion, index=Saddle point}
  | saddle point

\knowledge{notion, index={Lagrange multiplier}}
  | Lagrange multiplier
  | Lagrange multipliers
  | the method of Lagrange multipliers

\knowledge{url={https://en.wikipedia.org/wiki/Joseph-Louis_Lagrange}}
  | Joseph-Louis Lagrange
  | Lagrange

\knowledge{url={https://en.wikipedia.org/wiki/Marquis_de_Condorcet}}
    | Marquis de Condorcet

  \knowledge{url={https://en.wikipedia.org/wiki/Adrien-Marie_Legendre}}
    | Adrien-Marie Legendre
%   }
%   {}
% \makeatother
\date{2019-10-23}
\begin{document}
\maketitle
\tableofcontents

\chapter{Vectors}
I already know about vectors --- I've been taught them in about five
different courses so far. I'm skipping this.

\chapter{Partial derivatives}

If we have a function of multiple variables, say
\[f(a_1, a_2, a_3, \dots),\]
we might care about the change of $f$ with respect to only one variable. By
picking a fixed value for all but one of the variables, we can determine
this.

Say that we want to find the "partial derivative" of $f$ with respect to
$a_2$; then, by constructing $g(a_2) = f(c_1, a_2, c_3, \dots)$, we've
created a function of \emph{one} variable, which we can differentiate as
usual.
\begin{notation}
  We write the "partial derivative" of a function $f$ at a point $\vec p$ 
  with respect to a basis element $a$ of $\vec p$ as \fbox{$f_a(\vec p)$.}

  We may also use much more common notation
  \[\pd[f]{a},\]
  using the "partial derivative" symbol $\partial$, a stylized cursive
  ``d''.\footnote{Introduced by "Marquis de Condorcet" in 1770, who used it
  to represent a partial \emph{differential}, i.e.~the $dy$ or $dx$ in
  $dy/dx$, and then adapted in 1786 by "Adrien-Marie Legendre" for use as the
  partial derivative.}

  In the interest of completeness, I'll exhaustedly note that the book also
  uses, on occasion, the notation $D_a f$.
\end{notation}

We can also calculate ""higher partial derivatives"" --- similarly to the
higher ordinary derivatives. The notation is a fairly clear extension:
\[(f_x)_x = f_{xx} = \pd x \left( \pd[f]{x} \right) = \pd[^2 f]{x^2}.\]

\begin{thm}[Clairaut's Theorem]
  Suppose $f$ is defined on a neighborhood $N$ about a point $\vec p$.
  If $f_{xy}$ and $f_{yx}$ are continuous in $N$, then $f_{xy}(\vec p) =
  f_{yx}(\vec p)$.
\end{thm}

\section{Gradients}
\begin{notation}
  This ridiculous textbook denotes the "partial derivative" of a function
  $f(x, y) = z$ with respect to  $x$ as $f_x(x, y)$.
\end{notation}

\begin{defn}
  The ""directional derivative"" of a function $f$ at $(x_0, y_0)$ in the
  direction of a "unit vector" $\vec u = \langle a, b \rangle$ is
  \[D_{\vec u} f(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb) -
  f(x_0, y_0)}{h},\]
  if the "limit" exists.
\end{defn}

If $f : \Re^2 \mapsto \Re$ is a "differentiable" function, then $f$ has a
"directional derivative" in the direction of any "unit vector" $\vec u =
\langle a, b \rangle$ of
\begin{alignat*}{1}
  D_{\vec u} f(x, y) &= f_x(x, y) a + f_y(x, y) b.
\intertext{Or, if $\vec u = \langle \cos \theta, \sin \theta \rangle$, then}
  D_{\vec u} f(x, y) &= f_x(x, y) \cos \theta + f_y(x, y) \sin \theta.
\end{alignat*}

Noticing that the "directional derivative" of a function can be written as
the dot product of two vectors,
\begin{alignat*}{1}
  D_{\vec u} f(x, y) &= f_x(x, y) a + f_y(x, y) b \\
    &= \langle f_x(x, y), f_y(x, y) \rangle \cdot \langle a, b \rangle \\
    &= \langle f_x(x, y), f_y(x, y) \rangle \cdot \vec u,
\end{alignat*}
we call the first vector $\langle f_x(x, y), f_y(x, y) \rangle$ the
""gradient"" of $f$ and denote it as $\grad f$.

\begin{defn}
  The "gradient" of a function $f$ of two variables is defined as
  \[\grad f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle
    = \frac{\partial f}{\partial x} \vec i + \frac{\partial f}{\partial y}
    \vec j.\]
\end{defn}

Therefore, we can rewrite the "directional derivative" of a function $f$
as
\[D_{\vec u} f(x, y) = \grad f(x, y) \cdot \vec u.\]
It's intuitive, then, that the maximum value of the "directional derivative"
is $|\grad f(x, y)|$, when $\vec u$ is parallel to $\grad f(x, y)$.

\section{Maximum and minimum values}
\begin{defn}
  $f : A^k \mapsto B$ has a ""local maximum"" at $\vec a$ if for some
  "neighborhood" $N \subset A$ about $\vec a$, for all $\vec x \in N$,
  $f(\vec x) \le f(\vec a)$.

  Conversely, if $f(\vec x) \ge f(\vec a)$, then $f(\vec a)$ is a ""local
  \emph{minimum}@local minimum"".

  If the statement also holds true for $N = A$, then $\vec a$ is an
  ""absolute maximum"" (or ""absolute minimum"").
\end{defn}

If $f$ has a "local maximum" or "minimum@local minimum" at $\vec a$ and the
partials of $f$ exist at $\vec a$, then $\partial f/\partial x (\vec a) = 0$
and $f_y(a, b) = 0$; geometrically, the tangent plane to a maximum or
minimum must be horizontal.

\begin{defn}
  A point $\vec a$ is called a ""critical point"" of $f$ if $f_x(\vec a) =
  0$ or $f_x(\vec a)$ doesn't exist for all variables of $f$.
\end{defn}

\begin{defn}
  A ""saddle point"" of a function is a "critical point" which is not a
  "local extremum" of the function.
\end{defn}

If $(a, b)$ is a critical point of $f$, then let
\[D = D(a, b) = f_{xx} (a,b) \, f_{yy} (a,b) - (f_{xy}(a,b))^2.\]
If $D < 0$, then $(a, b)$ is a "saddle point" of $f$.

\section{Lagrange multipliers}
Often we want to find the "local extrema" of a function subject to
constraints, i.e.~maximizing the volume of an object while keeping its
surface area constant. The method of "Lagrange multipliers"\footnote{After
"Joseph-Louis Lagrange" (1736--1813), ``an Italian Enlightenment Era
mathematician and astronomer [who] made significant contributions to the
fields of analysis, number theory, and both classical and celestial
mechanics.''} is a strategy for doing this.

To find extrema of $f(\vec p)$ constrained with $g(\vec p) = k$,
we look for extrema of $f$ that are restricted to lie on the level curve
$g(\vec p) = k$; it happens that the largest $c$ such that $f(\vec p) = c$
intersects with  $g(\vec p) = k$ when the two level curves are tangent with
each other, i.e.~they have identical normals. In other words, for some
scalar $\lambda$, $\grad f(\vec p) = \lambda \grad g(\vec p)$.

More formally, suppose $f$ has an extrema at $\vec p_0$. Then, let the level
surface generated by the constraint $g(\vec p) = k$ be called $S$, where
$\vec p_0 \in S$. Then, let $C$ be the set of points given by $\vec r(t)$ such
that $C \subset S$ and $\vec p_0 \in C$. Further, let $t_0$ be a point such
that $\vec r(t_0) = \vec p_0$.

Then, $f \after \vec r$ gives the values of $f$ on the curve $C$. $f$ has an
extrema at $\vec p$, so $f \after \vec r$ must also, and $(f \after \vec
r)'(t_0) = 0$. If $f$ is "differentiable", we can use the chain rule to
write
\begin{alignat*}{1}
  0 &= (f \after \vec r)'(t_0) \\
  &= \grad f(\vec p_0) \cdot \vec r'(t_0).
\end{alignat*}
Therefore, the gradient of $f$ is orthogonal to the tangent of every such
curve $C$. We also know that $\grad g(\vec p_0)$ is orthogonal to $\vec
r'(t_0)$, so the gradients of $f$ and $g$ at $\vec p_0$ must be parallel.
Therefore, if $\grad g(\vec p_0) \ne 0$, there exists some $\lambda$ such
that
\begin{equation}
  \grad f(\vec p_0) = \lambda \grad g(\vec p_0),
\end{equation}
where the constant $\lambda$ is called a "Lagrange multiplier".

Then, the ""method of Lagrange multipliers"" gives us a process to
find the maximum and minimum values of a function $f(\vec p)$ subject to the
constraint  $g(\vec p) = k$, where $\vec p \in \Re^n$. To use the method of
Lagrange multipliers, we assume that the extreme values exist and that
$\grad g \ne 0$ on the level surface $g(\vec p) = k$.
\begin{enumerate}
  \item Find all values of $\vec p$ and $\lambda$ such that
  \begin{alignat*}{1}
    \grad f(\vec p) &=  \lambda \grad g(\vec p) \\
  \text{and}\qquad g(\vec p) &= k. \\
  \end{alignat*}

  \item Next, evaluate $f$ at all of the points found in the first step. The
  largest of these values is the maximum value of $f$, and the smallest of
  them is the minimum value.
\end{enumerate}

\chapter{Multiple integrals}
Single integrals are good for functions of one variable. To integrate
functions of multiple variables, we use multiple integrals. Straightforward
enough.

Multiple integrals allow us to calculate things like surface areas and
volumes of geometric objects.

In general, for some double integral
\[\underbrace{\int_a^b \overbrace{\int_c^d f(x,y)\,dx}^{\mathclap{\text{We
treat $y$ as constant while evaluating this.}}}\,dy,}_{\mathclap{\text{We've
eliminated $x$ from the equation before evaluating this.}}}\]
we do the opposite of partial differentiation and treat all variables other
than the one we're integrating for as constant, repeatedly, until we've
integrated with respect to all variables; each step in this process is
called, predictably, ""partial integration"".

\section{Double integrals}
For an axis-aligned rectangle $R$ on the $xy$-plane from $(x_0, y_0)$ to $(x_1,
y_1)$, the area of a function $f(x, y)$ under $R$ is given by the double
integral
\begin{alignat*}{1}
  \iint_{Y} f(x,y)\,dA &= \underbrace{\int_{x_0}^{x_1} \int_{y_0}^{y_1}
    f(x,y)\,dy\,dx}_{\mathclap{\text{This is the iterated form of the
    integral.}}} \\
    &= \int_{y_0}^{y_1} \int_{x_0}^{x_1} f(x,y)\,dx\,dy, \\
\end{alignat*}
where we use $\iint_R$ to mean ``integrating over the area of $R$'' and
``$dA$'' to mean ``with respect to area.''

The right-hand side of the equation above is called the ""iterated form"",
or an \reintro*"iterated integral".

We can also iterate over funkier regions if we're willing to play with the
limits of integration a bit. The easiest regions to integrate over are the
ones that are easily expressible as the region bounded above and below by
functions of one variable, e.g. ``the region under the line $y = 2x$ and
above the line $y = x^2$'' (note that this is bounded on the left at $x = 0$
and on the right at $x = 2$).

% Area[ImplicitRegion[x^2 < y \[And] y < 2 x, {x, y}
The area of that region is expressed by the integral
\begin{alignat*}{1}
  A &= \int_0^2 \int_{x^2}^{2x}\,dy\,dx \\
    &= \int_0^2 {\bigg[ x \bigg]}_{x^2}^{2x}\,dx \\
    &= \int_0^2 {\bigg( 2x-x^2 \bigg)}\,dx \\
    &= {\left[ x^2 - \frac{x^3}{3} \right]}_0^2 \\
    &= 4 - \frac{8}{3} = \frac{4}{3}.
\end{alignat*}

A more complicated region might be ``the region under the paraboloid $z =
x^2 + y^2$ and above the region in the $xy$-plane bounded by $y = \sqrt{x}$
and $y=1-\cos x$.''

We can build larger regions out of pieces, by summing smaller integrals.

\section{Polar coordinates}
Use the conversions
\begin{alignat*}{1}
  r &= \sqrt{x^2+y^2} \\
  x &= r\cos\theta \\
  y &= r\sin\theta \\
\intertext{for the coordinates and then we have that if $R$ is a ``polar
rectangle'' (arc-shaped region bounded by angles and radii) from $r=a$ to
$r=b$ and $\theta = \alpha$ to $\theta = \beta$, we have}
  \iint_R f(x,y)\,dA &= \int_\alpha^\beta \int_a^b f(r\cos\theta,
  r\sin\theta)r\,dr\,d\theta, \\
\intertext{which makes our lives easier for circly areas and volumes. Don't
forget to multiply by $r$.
\endgraf
For squiggly and varying radii, we can use functions $h_1(\theta)$ and
$h_2(\theta)$ instead of constants $a$ and $b$:}
  \iint_D f(x,y)\,dA &= \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,
  r\sin\theta)r\,dr\,d\theta. \\
\end{alignat*}

\section{Cylindrical coordinates}
Just add $z$.

\section{Spherical coordinates}
I can never remember how these work. If we have a point $P$, and we drop it
down to the $xy$-plane, the angle between the positive $x$-axis and the
segment from the origin to $P$ is $\theta$.

Next, the angle between the positive $z$-axis and the segment from the origin
to $P$ is $\phi$.

Finally, the length of the segment from the origin to $P$ is $\rho$.

The conversions
\begin{alignat*}{1}
  x &= \rho \sin \phi \cos \theta \\
  y &= \rho \sin \phi \sin \theta \\
  z &= \rho \cos \phi \\
\intertext{give us the integral-conversion for the spherical wedge bounded
by $a \le \rho \le b, \alpha \le \theta \le \beta, c \le \phi \le d$ as}
  \iiint_E f(x,y,z)\,dV &= \int_c^d \int_\alpha^\beta \int_a^b
    f(\rho \sin\phi \cos\theta,\,
      \rho \sin\phi \sin\theta,\,
      \rho \cos \theta)
    [\rho^2 \sin\phi]
    \,d\rho\,d\theta\,d\phi. \\
\end{alignat*}
Very gross!

\section{Surface area}

For $f(x,y)$ with $f_x$, $f_y$ continuous, the surface area of $f$ within
a region $D$ is
\[A = \iint_D \left(\sqrt{f_x(x,y)^2 + f_y(x,y)^2 + 1}\right) \,dA.\]

% Need notes on:
%   - Change of variables (in multiple integrals)

\chapter{Vector calculus}
A vector field is a mapping $\R^k \mapsto \R^n$; for each point in
$k$-dimensional Euclidean space, we associate an $n$-dimensional vector.
These vectors can represent velocity, distance, or anything else, and come
up in all sorts of applied fields.

We'll be mostly concerned with vector fields $\R^2\mapsto\R^2$ and
$\R^3\mapsto\R^3$.

If we have a plane curve given by the vector equation
\begin{alignat*}{1}
  \vec r(t) &= \left< x(t), y(t) \right> \qquad a \le t \le b, \\
\intertext{then the line integral of $f$ along $\vec r(t)$ from $a$ to $b$
is}
  & \int_a^b f(x(t), y(t)) \sqrt{{\left(\dd[x]{t}\right)}^2 +
  {\left(\dd[y]{t}\right)}^2}\,dt,
\end{alignat*}
i.e.\ the length of the curve multiplied, at each point, by the value of the
vector field $f$ at that point.

% Need notes on:
% 16.1, 16.2, 16.3, 16.4, 16.5,
% Need to update cheat sheet.

\appendix
\chapter{Common formulas for derivatives and integrals}
\backmatter
\section{Derivatives}

\begin{alignat*}{2}
  \ddx&\;& (f + g) &= f' + g' \\
  \ddx&& x^n &= nx^{n - 1} \\
  \ddx&& (fg) &= fg' + f'g \\
  \ddx&& \frac{h}{l} &= \frac{l h' - h l'}{l^2} \\
  \ddx&& f(g(x)) &= f'(g(x)) g'(x) \quad\text{(Chain rule.)} \\
  \ddx&& b^x &= b^x \ln b \\
  \ddx&& f^{-1}(x) &= \frac{1}{f'(f^{-1}(x)} \\
  \ddx&& c &= 0 \\
  \ddx&& c\,f &= c\,f' \\
  \ddx&& e^x &= e^x \\
  \ddx&& e^{f(x)} &= f'(x) e^{f(x)} \quad\text{(By the chain rule.)}  \\
  \ddx&& \ln x &= \frac{1}{x} \\
  \ddx&& \log_b x &= \frac{1}{x \ln b} \\
  \ddx&& [\vec u \cdot \vec v] &= \vec u' \cdot \vec v + \vec u \cdot \vec v' \\
  \ddx&& [\vec u \times \vec v] &= \vec u' \times \vec v + \vec u \times \vec v' \\
\end{alignat*}

\subsection{Trigenometric}
\begin{alignat*}{2}
  \ddx&\;& \sin x &= \cos x \\
  \ddx&& \cos x &= -\sin x \\
  \ddx&& \tan x &= \sec^2 x \\
  \ddx&& \cot x &= -\csc^2 x \\
  \ddx&& \sec x &=  \sec x \tan x \\
  \ddx&& \csc x &=  -\csc x \cot x \\
  % inverse
  \ddx&& \sin^{-1} x &= \frac{ 1}{\sqrt{1 - x^2}} \\
  \ddx&& \cos^{-1} x &= \frac{-1}{\sqrt{1 - x^2}} \\
  \ddx&& \tan^{-1} x &= \frac{ 1}{1 + x^2} \\
  \ddx&& \cot^{-1} x &= \frac{-1}{1 + x^2} \\
  \ddx&& \sec^{-1} x &= \frac{ 1}{|x| \sqrt{x^2 - 1}} \\
  \ddx&& \csc^{-1} x &= \frac{-1}{|x| \sqrt{x^2 - 1}} \\
\end{alignat*}

\section{Integrals}
See also:
\emph{\href{https://www.whitman.edu/mathematics/calculus/calculus_08_Techniques_of_Integration.pdf}{Techniques
of Integration}}.
\begin{alignat*}{1}
  \int x^n\,dx &= \frac{x^{n + 1}}{n + 1} + C \quad \text{when } n \ne -1 \\
  \int x^{-1}\,dx &= \ln|x| + C \\
  \int e^x\,dx &= e^x + C \\
  \dd{t} \int_{a(t)}^{b(t)} g(s)\,ds &= b'(t) g(b(t)) - a'(t) g(a(t))
  \quad\text{(Leibniz' rule.)} \\
  \int uv'\,dx &= uv - \int u'v\,dx \\
\end{alignat*}

\subsection{Trigenometric}
\begin{alignat*}{1}
  \int \sin x\,dx &= -\cos x + C \\
  \int \cos x\,dx &= \sin x + C \\
  \int \sec^2 x\,dx &= \tan x + C \\
  \int \sec x \tan x\,dx &= \sec x + C \\
  \int \frac{1}{1 + x^2}\,dx &= \tan^{-1} x + C \\
  \int \frac{1}{\sqrt{1 + x^2}}\,dx &= \sin^{-1} x + C \\
\end{alignat*}

\printindex
\end{document}