\documentclass[a4paper,12pt]{article}
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\usepackage{amsmath}
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\usepackage{enumerate}
\usepackage{mathtools}
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\usepackage{siunitx}
\usetikzlibrary{shapes.geometric,angles,quotes}

\newcommand{\ud}{\,\mathrm{d}}
\newcommand{\unit}[1]{\hat{\textbf #1}}
\newcommand{\anonym}[2]{\left(#1 \mapsto #2\right)}
\newcommand{\tho}[3][]{\frac{\partial #1 #2}{\partial #3 #1}}
\newcommand{\leibniz}[3][]{\frac{\mathrm{d} #1 #2}{\mathrm{d} #3 #1}}
\newcommand{\chain}[3]{\tho{#1}{#2}\tho{#2}{#3}}
\newcommand{\exercise}[1]{\noindent\textbf{#1.}}

\title{Cuculutu Homework}
\author{Nguyễn Gia Phong}
\date{Summer 2019}

\begin{document}
\maketitle
\setcounter{section}{11}
\section{Vectors and the Geometry of Space}
\subsection{Three-Dimenstional Coordinate Systems}

\exercise{37} The region consisting of all points between the spheres of radius
$r$ and $R$ centered at origin:
\[r^2 < x^2 + y^2 + z^2 < R^2\qquad (r < R)\]

\subsection{Vectors}
\exercise{38} The gravitational force enacting the chain whose tension $T$ at
each end has magnitude 25 N and angle \ang{37} to the horizontal is
\[\mathbf{P} = 2\mathrm{proj}_{\unit P}\mathbf{T}
  = 2T\sin\ang{37}\unit{P} \approx 30\unit{P}\]

Therefore the weight of the chain is approximately 30 N.

\exercise{47} Given $\mathbf{r_0} = \langle x_0, y_0, z_0 \rangle$.

Let $\mathbf{r} = \langle x, y, z \rangle$,
\[\left|\mathbf{r} - \mathbf{r_0}\right| = 1
  \iff \left(x - x_0\right)^2 + \left(y - y_0\right)^2
     + \left(z - z_0\right)^2 = 1\]

Thus the set of all points $(x, y, z)$ is an unit sphere whose center is
$\left(x_0, y_0, z_0\right)$.

\subsection{The Dot Product}
\exercise{25} Given a triangle with vertices $P(1, -3, -2)$, $Q(2, 0, -4)$,
$R(6, -2, -5)$.

Since $\overrightarrow{PQ}\cdot\overrightarrow{QR}
= 1 \cdot 4 + 3(-2) + (-2)(-1) = 0$, $PQR$ is a right triangle.

\exercise{26} Given $\mathbf{u} = \langle 2, 1, -1 \rangle$ and
$\mathbf{v} = \langle 1, x, 0 \rangle$.
\begin{align*}
\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\cdot|\mathbf{v}|} = \cos\ang{45}
&\iff \frac{2 + x}{\sqrt{6\left(x^2 + 1\right)}} = \frac{1}{\sqrt 2}\\
&\iff 2x^2 + 8x + 8 = 6x^2 + 6\\
&\iff 4x^2 - 8x - 2 = 0\\
&\iff x = 1 \pm \sqrt\frac{3}{2}
\end{align*}

\exercise{27} Find a unit vector that is orthogonal to both
$\unit\i + \unit\j$ and $\unit\i + \unit k$.

A vector that is orthogonal to both of these vectors:
\[(\unit\i + \unit\j)\times(\unit\i + \unit k)
= \unit\i\times\unit\i + \unit\i\times\unit k
+ \unit\j\times\unit\i + \unit\j\times\unit k
= 0 - \unit\j - \unit k + \unit\i
= \unit\i - \unit\j - \unit k\]

Normalize the result we get the unit vector $\dfrac{1}{\sqrt 3}\left(\unit\i
- \unit\j - \unit k\right)$ which is orthogonal to both $\unit\i + \unit\j$
and $\unit\i + \unit k$.

\exercise{28} Find two unit vectors that make an angle of \ang{60}
with $\mathbf{v} = \langle 3, 4 \rangle$.

Let $\mathbf{u} = \langle x, y \rangle$ be an unit vector,
$|\mathbf{u}| = \sqrt{x^2 + y^2} = 1$.  \textbf{u} makes with
\textbf{v} an angle of \ang{60} if and only if
\[\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\cdot|\mathbf{v}|} = \cos\ang{60}
\iff \frac{3x + 4y}{\sqrt{3^2 + 4^2}} = \frac{1}{2}
\iff 6x + 8y = 5\]

Since $x^2 + y^2 = 1$, $\mathbf{u} = \Bigl<0.3 \pm 0.4\sqrt 3,
0.4 \mp 0.3\sqrt 3\Bigr>$.

\exercise{53} Given a point $P_1\left(x_1, y_1\right)$ and a line
$d: ax + by + c = 0$.

Let $P\left(x_0, y_0\right)$ be any point satisfying $ax_0 + by_0 + c = 0$,
$\mathrm{distance}\left(d, P_1\right)$ is component of $\mathbf{u} =
\overrightarrow{PP_1} = \langle x_1 - x_0, y_1 - y_0 \rangle$ along the normal
of the line $\mathbf{n} = \langle a, b \rangle$:
\begin{multline*}
  \mathrm{comp}_\mathbf{u}\mathbf{n}
= \frac{|\mathbf{n}\cdot\mathbf{u}|}{|\mathbf{n}|}
= \frac{\left|a\left(x_1 - x_0\right) + b\left(y_1 - y_0\right)\right|}
       {\sqrt{a^2 + b^2}}
= \frac{\left|ax_1 + by_1 + c\right|}{\sqrt{a^2 + b^2}}\\
\Longrightarrow \mathrm{distance}\left(3x - 4y + 5 = 0, (-2, 3)\right)
= \frac{\left|3(-2) + (-4)3 + 5\right|}{\sqrt{3^2 + (-4)^2}}
= \frac{13}{5}
\end{multline*}

\subsection{The Cross Product}
\exercise{18} Given $\mathbf{a} = \langle 1, 0, 1 \rangle$,
$\mathbf{b} = \langle 2, 1, -1 \rangle$ and
$\mathbf{c} = \langle 0, 1, 3 \rangle$.
\begin{multline*}
  \begin{cases}
    \mathbf{a}\times(\mathbf{b}\times\mathbf{c})
    = \langle 1, 0, 1 \rangle \times \langle 4, -6, 2 \rangle
    = \langle 6, 2, -6 \rangle\\
    (\mathbf{a}\times\mathbf{b})\times\mathbf{c}
    = \langle -1, 3, 1 \rangle \times \langle 0, 1, 3 \rangle
    = \langle 8, 3, -1 \rangle
  \end{cases}\\
  \Longrightarrow \mathbf{a}\times(\mathbf{b}\times\mathbf{c})
             \neq (\mathbf{a}\times\mathbf{b})\times\mathbf{c}
\end{multline*}

\exercise{38} Given $A(1, 3, 2)$, $B(3, -1, 6)$, $C(5, 2, 0)$ and $D(3, 6, -4)$.
\begin{align*}
   \overrightarrow{AB}\cdot
   \left(\overrightarrow{AC}\times\overrightarrow{AD}\right)
&= \langle 2, -4, 4 \rangle \cdot
   (\langle 4, -1, -2 \rangle \times \langle 2, 3, -6 \rangle)\\
&= \langle 2, -4, 4 \rangle \cdot \langle 12, 20, 14 \rangle\\
&= 24 - 80 + 56\\
&= 0
\end{align*}

Thus $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$
are coplanar, which means $A$, $B$, $C$ and $D$ are coplanar.

\exercise{39} The magnitude of the torque about $P$:
\begin{align*}
   |\boldsymbol\tau|
&= |\mathbf{r}\times\mathbf{F}|\\
&= |-\mathbf{r}\times-\mathbf{F}|\\
&= |\mathbf{r}|\cdot|\mathbf{F}|\cdot\sin\left(\ang{70}+\ang{10}\right)\\
&= 0.18 \cdot 60 \cdot \sin\ang{80}\\
&\approx 10.6\qquad(\mathrm{N}\cdot\mathrm{m})
\end{align*}

\setcounter{section}{13}
\section{Partial Derivatives}
\setcounter{subsection}{1}
\subsection{Limits et Continuity}
Determine the set of points at which the function is continuous.

\[F(x, y) = \frac{1 + x^2 + y^2}{1 - x^2 - y^2}\tag{31}\]

$F$ is a rational function, hence it is continuous on its domain
\[D_F = \left\{(x, y) \in \mathbb{R}^2 \,\middle|\, x^2 + y^2 \neq 1\right\}\]

\[H(x, y) = \frac{e^x + e^y}{e^{xy} - 1}\tag{32}\]

Since $H$ is a ratio of sums of exponential functions, it is continuous on its
domain \[D_H = \left\{(x, y) \in \mathbb{R}^2 \,\middle|\, xy \neq 0\right\}\]

\[f(x, y) = \begin{cases}
            \frac{x^2 y^3}{2x^2 + y^2}&\text{if }(x, y) \neq (0, 0)\\
            1&\text{if }(x, y) = (0, 0)
\end{cases}\tag{37}\]

On $\mathbb{R}^2 \backslash (0, 0)$, because $2x^2 + y^2 \geq 3x^2 |y|$
(AM-GM inequality)
\[0 \leq \left|\frac{x^2 y^3}{2x^2 + y^2}\right|
    \leq \left|\frac{x^2 y^3}{3x^2 |y|}\right| = \frac{y^2}{3}\]

Since $0 \to 0$ and $y^2 \to 0$ as $(x, y) \to (0, 0)$,
by applying the Squeeze Theorem, $|f(x, y)| \to 0$ as
$(x, y) \to (0, 0)$.

It is trivial on $\mathbb{R}^2 \backslash (0, 0)$ that
$-|f(x, y)| \leq f(x, y) \leq |f(x, y)|$. Thus by again applying the
Squeeze Theorem, we get
\[\lim_{x\to 0 \atop y\to 0}f(x, y) = 0 \neq 1 = f(0, 0)\]

Therefore, the rational function $f$ is only continuous on
$\mathbb{R}^2 \backslash (0, 0)$.

\subsection{Partial Derivatives}
\exercise{29} Find the first partial derivatives of the function
\begin{align*}
  F(x, y) &= \int_y^x\cos\left(e^t\right)\ud t\\
          &= \int_y^x\frac{1}{e^t}\ud\sin\left(e^t\right)\\
          &= \int_{e^y}^{e^x}\frac{1}{t}\ud\sin t\\
          &= \int_{e^y}^{e^x}\frac{\cos t}{t}\ud t\\
          &= \sum_{n=0}^\infty\int_{e^y}^{e^x}
             (-1)^n\frac{t^{2n-1}}{(2n)!}\ud t\\
          &= \left[\ln t + \sum_{n=1}^\infty
             \frac{(-t)^{2n}}{2n(2n)!}\right]_{e^y}^{e^x}\\
          &= x - y + \sum_{n=1}^\infty
             \frac{\left(-e^x\right)^{2n} - \left(-e^y\right)^{2n}}{2n(2n)!}
\end{align*}
\begin{align*}
  \tho{F}{x}&
= -1 + \sum_{n=1}^\infty\frac{2n\left(-e^x\right)^{2n}}{2n(2n)!}
= \sum_{n=0}^\infty\frac{\left(-e^x\right)^{2n}}{(2n)!}
= \cos\left(-e^x\right)
= \cos\left(e^x\right)\\
  \tho{F}{y}&
= 1 + \sum_{n=1}^\infty\frac{-2n\left(-e^y\right)^{2n}}{2n(2n)!}
= -\sum_{n=0}^\infty\frac{\left(-e^x\right)^{2n}}{(2n)!}
= -\cos\left(-e^x\right)
= -\cos\left(e^x\right)
\end{align*}

\exercise{48} Use implicit differentiation to find $\partial z/\partial x$
and $\partial z/\partial y$.
\[x^2 - y^2 + z^2 - 2z = 4\]
\begin{equation*}
  \begin{cases}
    \begin{aligned}
      2x + 2z\tho{z}{x} - 2\tho{z}{x} &= 0\\
      -2y + 2z\tho{z}{y} - 2\tho{z}{y} &= 0
    \end{aligned}
  \end{cases}
  \Longrightarrow
  \begin{cases}
    \begin{aligned}
      \tho{z}{x} &= \frac{x}{1 - z}\\
      \tho{z}{y} &= \frac{y}{z - 1}
    \end{aligned}
  \end{cases}
\end{equation*}

\exercise{65\&67} Find the indicated partial derivative.
\begin{align*}
   \frac{\partial^3}{\partial z\partial y\partial x}e^{xyz^2}
&= \frac{\partial^2}{\partial z\partial y}yz^2e^{xyz^2}\\
&= \frac{\partial}{\partial z}xyz^4e^{xyz^2}\\
&= 2x^2y^2z^5e^{xyz^2}\tag{65}
\end{align*}
\begin{align*}
   \frac{\partial^3}{\partial r^2\partial\theta}e^{r\theta}\sin\theta
&= \frac{\partial^2}{\partial r^2}
   \left(re^{r\theta}\sin\theta + e^{r\theta}\cos\theta\right)\\
&= \frac{\partial}{\partial r}
   \left(r\theta e^{r\theta}\sin\theta + \theta e^{r\theta}\cos\theta\right)\\
&= \theta^2e^{r\theta}(r\sin\theta + \cos\theta)\tag{67}
\end{align*}

\exercise{53} Find all the second partial derivatives of the function
$f(x, y) = x^3 y^5 + 2x^4 y$.

First partial derivatives of $f$:
\begin{align*}
  &f_x = 3x^2 y^5 + 8x^3 y\\
  &f_y = 5x^3 y^4 + 2x^4
\end{align*}

Second partial derivatives:
\begin{align*}
  &f_{xx} = 6xy^5 + 24x^2 y\\
  &f_{xy} = f_{yx} = 15x^2 y^4 + 8x^3\\
  &f_{yy} = 20x^3 y^3
\end{align*}

\exercise{80} Given $u = \exp\left(\sum_{i=1}^n a_i x_i\right)$,
where $\sum_{i=1}^n a_i^2 = 1$.
\[\sum_{i=1}^n\tho[^2]{u}{x_i}
= \sum_{i=1}^n\tho{a_i u}{x_i}
= \sum_{i=1}^n a_i^2 u
= u\]

\subsection{Tangent Planes}
Find an equation of the tangent plane to the given suface
at the specified point.

\[z = 3y^2 - 2x^2 + x,\qquad (2, -1, -3)\tag{1}\]
\begin{align*}
     &z + 3 = \tho{z}{x}(2,-1)(x-2) + \tho{z}{y}(2,-1)(y+1)\\
\iff &z + 3 = \anonym{(x,y)}{1-4x}(2,-1)(x-2) + \anonym{(x,y)}{6y}(2,-1)(y+1)\\
\iff &z + 3 = 17 - 8x - 6y - 6\\
\iff &8x + 6y + z = 8
\end{align*}

\[z = 3(x - 1)^2 + 2(y + 3)^2 + 7,\qquad (2, -2, 12)\tag{2}\]
\begin{align*}
     &z - 12 = \tho{z}{x}(2, -2)(x - 2) + \tho{z}{y}(2, -2)(y + 2)\\
\iff &z - 12 = \anonym{(x, y)}{6x - 6}(2, -2)(x - 2)
             + \anonym{(x, y)}{4y + 12}(2, -2)(y + 2)\\
\iff &z - 12 = 6x - 12 + 4y + 8\\
\iff &6x + 4y - z + 8 = 0
\end{align*}

\[z = \sqrt{xy},\qquad (1, 1, 1)\tag{3}\]
\begin{align*}
     &z - 1 = \tho{z}{x}(1, 1)(x - 1) + \tho{z}{y}(1, 1)(y - 1)\\
\iff &z - 1 = \anonym{(x, y)}{\sqrt\frac{y}{4x}}(1, 1)(x - 1)
            + \anonym{(x, y)}{\sqrt\frac{x}{4y}}(1, 1)(y - 1)\\
\iff &2z - 2 = x - 1 + y - 1\\
\iff &x + y - 2z = 0
\end{align*}

\subsection{The Chain Rule}
\exercise{4} Use the Chain Rule to find $\mathrm{d} z/\mathrm{d} t$.
\[z = \arctan\frac{y}{x},\qquad x = e^t,\qquad y = 1-e^{-t}\]
\begin{align*}
   \leibniz{z}{t}
&= \tho{z}{x}\cdot\leibniz{x}{t} + \tho{z}{y}\cdot\leibniz{y}{t}\\
&= \tho{\arctan(y/x)}{x}\cdot\leibniz{e^t}{t}
 + \tho{\arctan(y/x)}{y}\cdot\leibniz{\left(1 - e^{-t}\right)}{t}\\
&= \frac{x^2}{y^2 + x^2}\left(\tho{(y/x)}{x}e^t + \tho{(y/x)}{y}e^{-t}\right)\\
&= \frac{x^2}{y^2 + x^2}\left(\frac{-y}{x^2}e^t + \frac{1}{x}e^{-t}\right)\\
&= \frac{xe^{-t} - ye^t}{y^2 + x^2}\\
&= \frac{1 - e^t + 1}{e^{2t} + e^{-2t} - 2e^{-t} + 1}\\
&= \frac{e^{2t} - e^{3t}}{e^{4t} +e^{2t} - 2e^t + 1}
\end{align*}

\exercise{9\&11} Use the Chain Rule to find $\partial z/\partial s$ and
$\partial z/\partial t$.
\[z = \sin\theta\cos\phi,\qquad \theta = st^2,\qquad \phi = s^2t\tag{9}\]
\begin{align*}
& \tho{z}{s}
= \tho{z}{\theta}\tho{\theta}{s} + \tho{z}{\phi}\tho{\phi}{s}
= t^2\cos\theta\cos\phi - 2st\sin\theta\sin\phi\\
& \tho{z}{t}
= \tho{z}{\theta}\tho{\theta}{t} + \tho{z}{\phi}\tho{\phi}{t}
= 2st\cos\theta\cos\phi - t^2\sin\theta\sin\phi
\end{align*}

\[e^r\cos\theta,\qquad r = st,\qquad \theta = \sqrt{s^2 + t^2}\tag{11}\]
\begin{align*}
& \tho{z}{s}
= \tho{z}{r}\tho{r}{s} + \tho{z}{\theta}\tho{\theta}{s}
= e^rt\cos\theta - e^r\sin\theta\frac{s}{\sqrt{s^2 + t^2}}
= e^{st}\left(t\cos\theta - \frac{s\sin\theta}{\sqrt{s^2 + t^2}}\right)\\
& \tho{z}{t} = e^{st}\left(s\cos\theta - \frac{t\sin\theta}{\sqrt{s^2 + t^2}}\right)
\end{align*}

\exercise{13} Suppose $f$ is a differentiable function of $g(t)$ and $h(t)$,
satisfying
\begin{align*}
  g(3) &= 2\\
  \leibniz{g}{t}(3) &= 5\\
  \tho{f}{g}(2, 7) &= 6\\
  h(3) &= 7\\
  \leibniz{h}{t}(3) &= -4\\
  \tho{f}{h}(2, 7) &= -8
\end{align*}
\begin{align*}
   \leibniz{f}{t}(3)
&= \tho{f}{g}(g(3), h(3))\cdot\leibniz{g}{t}(3)
 + \tho{f}{h}(g(3), h(3))\cdot\leibniz{h}{t}(3)\\
&= \tho{f}{g}(2, 7) \cdot 5
 + \tho{f}{h}(2, 7) \cdot (-4)\\
&= 6 \cdot 5 + (-8)(-4)\\
&= 62
\end{align*}

\exercise{14} Let $W(s, t) = F(u(s, t), v(s, t))$, where $F$, $u$ and $v$ are
differentiable, and
\begin{align*}
  u(1, 0) &= 2\\
  u_s(1, 0) &= -2\\
  u_t(1, 0) &= 6\\
  F_u(2, 3) &= -1\\
  v(1, 0) &= 3\\
  v_s(1, 0) &= 5\\
  v_t(1, 0) &= 4\\
  F_v(2, 3) &= 10
\end{align*}
\begin{align*}
   W_s(1, 0)
&= F_u(u(1, 0), v(1, 0)) u_s(1, 0) + F_v(u(1, 0), v(1, 0)) v_s(1, 0)\\
&= F_u(2, 3) (-2) + F_v(2, 3) \cdot 5\\
&= (-1)(-2) + 10 \cdot 5\\
&= 22\\
   W_t(1, 0)
&= F_u(u(1, 0), v(1, 0)) u_t(1, 0) + F_v(u(1, 0), v(1, 0)) v_t(1, 0)\\
&= F_u(2, 3) \cdot 6 + F_v(2, 3) \cdot 4\\
&= -1 \cdot 6 + 10 \cdot 4\\
&= 34
\end{align*}

\exercise{17} Assume all functions are differentiable, write out the Chain Rule.
\[u = f(x(r, s, t), y(r, s, t))\]
\[\begin{dcases}
    \tho{u}{r} = \chain{u}{x}{r} + \chain{u}{y}{r}\\
    \tho{u}{r} = \chain{u}{x}{s} + \chain{u}{y}{s}\\
    \tho{u}{r} = \chain{u}{x}{t} + \chain{u}{y}{t}
  \end{dcases}\]

\exercise{23} Use the Chain Rule to find $\partial w/\partial r$ and
$\partial w/\partial\theta$ when $r = 2$ and $\theta = \pi/2$, given
\[w = xy + yz + zx,\qquad x = r\cos\theta,\qquad
  y = r\sin\theta,\qquad z = r\theta\]
\begin{align*}
& \begin{dcases}
    \tho{w}{r} = \chain{w}{x}{r} + \chain{w}{y}{r} + \chain{w}{z}{r}\\
    \tho{w}{\theta} = \chain{w}{x}{\theta} + \chain{w}{y}{\theta}
                    + \chain{w}{z}{\theta}
  \end{dcases}\\
\iff
& \begin{dcases}
    \tho{w}{r} = (y + z)\cos\theta + (x + z)\sin\theta + (y + x)\theta\\
    \tho{w}{\theta} = -(y + z)r\sin\theta + (x + z)r\cos\theta
                    + (y + x)r
  \end{dcases}
\end{align*}

For $(r, \theta) = (2, \pi/2)$
\begin{align*}
& \begin{dcases}
    \tho{w}{r} = x + z + (y + x)\frac{\pi}{2}\\
    \tho{w}{\theta} = 2x - 2z
  \end{dcases}\\
\iff
& \begin{dcases}
    \tho{w}{r} = 2\cos\frac{\pi}{2} + 2\frac{\pi}{2} + 2\left(\sin\frac{\pi}{2}
                 + \cos\frac{\pi}{2}\right)\frac{\pi}{2}\\
    \tho{w}{\theta} = 4\cos\frac{\pi}{2} - 4\frac{\pi}{2}
  \end{dcases}\\
\iff& \tho{w}{r} = -\tho{w}{\theta} = 2\pi
\end{align*}

\exercise{27} Find $\mathrm{d}y/\mathrm{d}x$.
\[y\cos x = x^2 + y^2
  \Longrightarrow
  \leibniz{y}{x}
= -\frac{\tho{}{x}\left(x^2 + y^2 - y\cos x\right)}
        {\tho{}{y}\left(x^2 + y^2 - y\cos x\right)}
= \frac{y\sin x + 2x}{\cos x - 2y}\]

\exercise{31} Find $\partial z/\partial x$ and $\partial z/\partial y$.
\[x^2 + 2y^2 + 3z^2 = 1
\Longrightarrow
\begin{dcases}
  \tho{z}{x} = -\frac{\tho{}{x}\left(x^2 + 2y^2 + 3z^2 - 1\right)}
                     {\tho{}{z}\left(x^2 + 2y^2 + 3z^2 - 1\right)}
             = -\frac{x}{3z}\\
  \tho{z}{x} = -\frac{\tho{}{y}\left(x^2 + 2y^2 + 3z^2 - 1\right)}
                     {\tho{}{z}\left(x^2 + 2y^2 + 3z^2 - 1\right)}
             = -\frac{2y}{3z}
\end{dcases}\]

\exercise{36} Wheat production $W$ in a given year depends on the average
temperature $T$ and the annual rainfall $R$. At current production levels,
$\partial W/\partial T = -2$ and $\partial W/\partial R = 8$. Estimate the
current rate of change of wheat production, given $\mathrm{d}T/\mathrm{d}t=0.15$
and $\mathrm{d}R/\mathrm{d}t=-0.1$.
\[\leibniz{W}{t}
= \tho{W}{T}\leibniz{T}{t} + \tho{W}{R}\leibniz{R}{t}
= (-1)0.15 + 8(-0.1)
= -0.95\]

\exercise{40} Use Ohm’s Law, $V = IR$, to find how the current $I$
is changing at the moment when $R = 400\,\mathrm\Omega$, $I = 0.08$ A,
$\mathrm{d}V/\mathrm{d}t = 0.01$ V/s,
and $\mathrm{d}R/\mathrm{d}t = 0.03\,\mathrm{\Omega/s}$.
\begin{align*}
   \leibniz{I}{t}
&= \tho{(V/R)}{V}\leibniz{V}{t} + \tho{(V/R)}{R}\leibniz{R}{t}\\
&= \frac{1}{R}(-0.01) - \frac{V}{R^2}0.03\\
&= \frac{-0.01}{400} - \frac{0.03I}{R}\\
&= \frac{-1}{40000} - \frac{0.03 \cdot 0.08}{400}\\
&= \frac{-31}{1000000}\,\mathrm{(A/t)}\\
&= -31\,\mathrm{(\mu A/t)}
\end{align*}

\exercise{42} The rate of change of production:
\begin{align*}
\leibniz{P}{t} &= \tho{\left(1.47L^{0.65}K^{0.35}\right)}{L}\leibniz{L}{t}
                + \tho{\left(1.47L^{0.65}K^{0.35}\right)}{K}\leibniz{K}{t}\\
               &= 0.9555\left(\frac{K}{L}\right)^{0.35} (-2)
                + 0.5145\left(\frac{L}{K}\right)^{0.65} \cdot 0.5\\
               &= -1.911\left(\frac{8}{30}\right)^{0.35}
                + 0.25725\left(\frac{30}{8}\right)^{0.65}\\
               &\approx -0.595832\text{ million dollars}\\
               &= -595832\text{ dollars}\\
\end{align*}

\exercise{47} Given $z = f(x - y)$.
\[\tho{z}{x} + \tho{z}{y}
= \leibniz{z}{(x - y)}\tho{(x - y)}{x} + \leibniz{z}{(x - y)}\tho{(x - y)}{y}
= \leibniz{z}{(x - y)}(1 - 1)
= 0\]

\subsection{Directional Derivatives and the Gradient Vector}
\exercise{5} Find the directional derivative of $f(x, y) = ye^{-x}$ at $(0, 4)$
in the direction indicated by the angle $\theta = 2\pi/3$.

Unit vector direction indicated by the angle $\theta = \frac{2\pi}{3}$
is $\mathbf{u} = \langle -1/2, \sqrt{3}/2 \rangle$.
\begin{align*}
  \mathrm{D}_\mathbf{u}f(0, 4)
&= \nabla f(0, 4)\cdot\mathbf{u}\\
&= \left<\tho{\left(ye^{-x}\right)}{x}(0, 4),
         \tho{\left(ye^{-x}\right)}{y}(0, 4)\right>
   \cdot \left<\frac{-1}{2}, \frac{\sqrt 3}{2}\right>\\
&= \left<\left((x, y) \mapsto -ye^{-x}\right)(0, 4),
         \left((x, y) \mapsto e^{-x}\right)(0, 4)\right>
   \cdot \left<\frac{-1}{2}, \frac{\sqrt 3}{2}\right>\\
&= \left<-4, 1\right> \cdot \left<\frac{-1}{2}, \frac{\sqrt 3}{2}\right>\\
&= 2 + \frac{\sqrt 3}{2}
\end{align*}

\exercise{7} Find the rate of change of $f(x, y) = \sin(2x + 3y)$ at $P(-6, 4)$
in the direction of the vector $\mathbf{u} = \frac{1}{2}(\sqrt{3}\unit\i - \unit\j)$.
\begin{align*}
  \mathrm{D}_\mathbf{u}f(-6, 4)
&= \nabla f(-6, 4)\cdot\mathbf{u}\\
&= \left<\tho{\sin(2x + 3y)}{x}(-6, 4),
         \tho{\sin(2x + 3y)}{y}(-6, 4)\right>
   \cdot \left<\frac{\sqrt 3}{2}, \frac{-1}{2}\right>\\
&= \left<2\cos(2(-6) + 3 \cdot 4),
         3\cos(2(-6) + 3 \cdot 4)\right>
   \cdot \left<\frac{\sqrt 3}{2}, \frac{-1}{2}\right>\\
&= \sqrt 3 - \frac{3}{2}
\end{align*}
\pagebreak

\exercise{11} Find the directional derivative of $f(x, y) = e^x\sin y$
at point $(0, \pi/3)$ in the direction of the vector
$\mathbf{v} = \langle -6, 8\rangle$
\begin{align*}
  \mathrm{comp}_\mathbf{v}\nabla f\left(0, \frac{\pi}{3}\right)
&= \frac{\nabla f\left(0, \frac{\pi}{3}\right)\cdot\mathbf{v}}{|\mathbf{v}|}\\
&= \left<\tho{(e^x\sin y)}{x}\left(0, \frac{\pi}{3}\right),
         \tho{(e^x\sin y)}{y}\left(0, \frac{\pi}{3}\right)\right>
   \cdot \frac{\langle -6, 8\rangle}{\sqrt{(-6)^2 + 8^2}}\\
&= \left<\frac{\sqrt 3}{2}, \frac{1}{2}\right>
   \cdot \left<\frac{-3}{5}, \frac{4}{5}\right>\\
&= \frac{2}{5} - \frac{3\sqrt 3}{10}
\end{align*}

\exercise{17} Find the directional derivative of
$h(r, s, t) = \ln(3r + 6s + 9t)$ at point $(1, 1, 1)$
in the direction of the vector $\mathbf{v} = \langle 4, 12, 6\rangle$.
\begin{align*}
  \mathrm{comp}_\mathbf{v}\nabla f(1, 1, 1)
&= \frac{\nabla f(1, 1, 1)\cdot\mathbf{v}}{|\mathbf{v}|}\\
&= \left<\frac{3}{3+6+9},\frac{6}{3+6+9},\frac{9}{3+6+9}\right>
   \cdot \left<\frac{2}{7},\frac{6}{7},\frac{3}{7}\right>\\
&= \left<\frac{1}{6},\frac{1}{3},\frac{1}{2}\right>
   \cdot \left<\frac{2}{7},\frac{6}{7},\frac{3}{7}\right>\\
&= \frac{23}{42}
\end{align*}

\exercise{21\&25} Find the maximum rate of change of $f$ at the given point and
the direction in which it occurs.
\[f(x, y) = 4y\sqrt{x},\qquad(4, 1)\tag{21}\]
\begin{align*}
   |\nabla f(4, 1)|
&= \left|\left<\tho{\left(4y\sqrt x\right)}{x}(4, 1),
               \tho{\left(4y\sqrt x\right)}{y}(4, 1)\right>\right|\\
&= \left|\left<1, 8\right>\right|\\
&= \sqrt{65}
\end{align*}

\[f(x, y, z) = \sqrt{x^2 + y^2 + z^2},\qquad(3, 6, -2)\tag{25}\]
\begin{align*}
   |\nabla f(3, 6, -2)|
&= \left|\left<\frac{3}{\sqrt{3^2 + 6^2 + (-2)^2}},
               \frac{6}{\sqrt{3^2 + 6^2 + (-2)^2}},
               \frac{-2}{\sqrt{3^2 + 6^2 + (-2)^2}}\right>\right|\\
&= 1
\end{align*}

\exercise{29} Find all points at which the direction of fastest change of the
function $f(x, y) = x^2 + y^2 - 2x - 4y$ is $\unit\i + \unit\j$.

The rate of change at point $(a, b)$ is maximum in direction $\unit\i + \unit\j$
if and only if $\nabla f(a, b)$ has the same direction:
\begin{align*}
  \nabla f(a, b) \times (\unit\i + \unit\j) = \mathbf{0}
&\iff ((2x-2)\unit\i + (2y-4)\unit\j) \times (\unit\i + \unit\j) = \mathrm{0}\\
&\iff 2(x - y + 1)\unit k = \mathrm{0}\\
&\iff x - y + 1 = 0
\end{align*}

Thus the points satisfying given the requirement is the line whose equation is
$x - y + 1 = 0$.

\exercise{32} The temperature at a point $(x, y, z)$ is given by
\[T(x, y, z) = 200e^{-x^2 - 3y^2 - 9z^2}\]

The rate of change of temperature at the point $P(2, -1, 2)$ in direction
$\mathbf{u}$ is
\begin{align*}
   \mathrm{D}_\mathbf{u}f(2, -1, 2)
&= \nabla f(2, -1, 2)\cdot\mathbf{u}\\
&= \left((x, y, z) \mapsto \frac{-400}{e^{x^2 + 3y^2 + 9z^2}}
         \langle x, 3y, 9z \rangle\right)(2, -1, 2)\cdot\mathbf{u}\\
&= \frac{-400}{e^{2^2 + 3(-1)^2 + 9 \cdot 2^2}}
   \langle 2, 3(-1), 9 \cdot 2 \rangle\cdot\mathbf{u}\\
&= \left<\frac{-800}{e^{43}}, \frac{1200}{e^{43}},
         \frac{-7200}{e^{43}}\right> \cdot \mathbf{u}
\end{align*}

For $\mathbf{u} = \left<1/\sqrt 6, -2/\sqrt 6, 1/\sqrt 6\right>$,
the rate of change is
\[\frac{-800}{e^{43}\sqrt 6} + \frac{400\sqrt 6}{e^{43}}
+ \frac{-1200\sqrt 6}{e^{43}} = \frac{-10400}{e^{43}\sqrt 6}\tag{a}\]

Temperature increases the fastest at the same direction as $\nabla f(2, -1, 2)$
\[\mathbf{u} = \left<\frac{-2}{\sqrt{337}}, \frac{3}{\sqrt{337}},
                     \frac{-18}{\sqrt{337}}\right>\tag{b}\]

In this direction, the rate of increase is
\[|\nabla f(2, -1, 2)| = \frac{400\sqrt{337}}{e^{43}}\tag{c}\]

\exercise{41} Find equations of the tangent plane and the normal line to the
surface $F(x, y, z) = 2(x - 2)^2 + (y - 1)^2 + (z - 3)^2 = 10$ at $(3, 3, 5)$.

Equation of the tangent plane:
\begin{align*}
     F_x(3, 3, 5)(x - 3) + F_y(3, 3, 5)(y - 3) + F_z(3, 3, 5)(z - 5) &= 0\\
\iff 4(3 - 2)(x - 3) + 2(3 - 1)(y - 3) + 2(5 - 3)(z - 5) &= 0\\
\iff x + y + z &= 11
\end{align*}

Equation of the normal line:
\[\frac{x - 3}{F_x(3, 3, 5)} = \frac{y - 3}{F_y(3, 3, 5)}
                             = \frac{z - 5}{F_z(3, 3, 5)}
\iff x - 3 = y - 3 = z - 5\]

\exercise{51} Given an ellipsoid
\[E(x, y, z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]

Its tangent plane at the point $(x_0, y_0, z_0)$ has the equation of
\begin{align*}
 &E_x(x_0, y_0, z_0)(x - x_0) + E_y(x_0, y_0, z_0)(y - y_0)
+ E_z(x_0, y_0, z_0)(z - z_0) = 0\\
\iff &\frac{2x_0}{a^2}(x - x_0) + \frac{2y_0}{b^2}(y - y_0)
    + \frac{2z_0}{c^2}(z - z_0) = 0\\
\iff &\frac{2xx_0}{a^2} + \frac{2yy_0}{b^2} + \frac{2zz_0}{c^2}
    = \frac{2x_0^2}{a^2} + \frac{2y_0^2}{b^2} + \frac{2z_0^2}{c^2}\\
\iff &\frac{2xx_0}{a^2} + \frac{2yy_0}{b^2} + \frac{2zz_0}{c^2} = 2\\
\iff &\frac{xx_0}{a^2} + \frac{yy_0}{b^2} + \frac{zz_0}{c^2} = 1
\end{align*}

\exercise{56} Consider an ellipsoid $3x^2 + 2y^2 + z^2 = 9$ and the sphere
$x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 0$. A point in their intersection must
satisfy the following equation
\begin{align*}
     &x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 9 - 3x^2 - 2y^2 - z^2\\
\iff &4x^2 - 8x + 4 + 3y^2 - 6y + 3 + 2z^2 - 8z + 8 = 0\\
\iff &4(x - 1)^2 + 3(y - 1)^2 + 2(z - 2)^2 = 0\\
\iff &\begin{cases}x = y = 1\\z = 2\end{cases}
\end{align*}

Thus the intersection is a subset of $\{(1, 1, 2)\}$. Since $P(1, 1, 2)$ lies
on both the ellipsoid and the sphere, it is the one and only intersection point
of the two. Therefore, they are tangent to each other at $P$.

\subsection{Minimum and Maximum Values}
\exercise{1} Suppose (1, 1) is a critical point of a function f with continuous
second derivatives.
\begin{multline}
  \begin{cases}
    \begin{vmatrix}
      f_{xx}(1, 1) & f_{xy}(1, 1)\\
      f_{yx}(1, 1) & f_{yy}(1, 1)
    \end{vmatrix} = 4 \cdot 2 - 1^2 = 7 > 0\\
    f_{xx}(1, 1) = 4 > 0
  \end{cases}\\
  \Longrightarrow f(1, 1)\text{ is a local minumum}\tag{a}
\end{multline}
\begin{multline}
  \begin{vmatrix}
    f_{xx}(1, 1) & f_{xy}(1, 1)\\
    f_{yx}(1, 1) & f_{yy}(1, 1)
  \end{vmatrix} = 4 \cdot 2 - 3^2 = -1 < 0\\
  \Longrightarrow (1, 1)\text{ is a saddle point of } f\tag{b}
\end{multline}

\exercise{7\&13\&15} Find the local maximum and minimum values and saddle points
of the function and graph the function.

For the next few exercises, $D$ is defined as
\[D(x, y) =
\begin{vmatrix}
  f_{xx}(x, y) & f_{xy}(x, y)\\
  f_{yx}(x, y) & f_{yy}(x, y)
\end{vmatrix}\]

\[f(x, y) = (x - y)(1 - xy) = xy^2 - x^2y + x - y\tag{7}\]
\begin{align*}
  f_x = f_y = 0
  &\iff y^2 - 2xy + 1 = 2xy - x^2 - 1 = 0\\
  &\iff x^2 = y^2 = 2xy - 1\\
  &\iff x = y = \pm 1
\end{align*}

As $f_{xx} = -2y$, $f_{yy} = 2x$ and $f_{xy} = f_{yx} = 2y - 2x$,
$D(x, y) = -4xy - (2y - 2x)^2$, thus $D(1, 1) = D(-1, -1) = -4 < 0$.
Therefore $(\pm 1, \pm 1)$ are saddle points of $f$.

\begin{tikzpicture}[domain=-2:2]
  \begin{axis}[xlabel={x}, ylabel={y}, zmin=-2, zmax=2]
    \addplot3[surf]{(x - y) * (1 - x*y)};
  \end{axis}
\end{tikzpicture}

\[f(x, y) = e^x\cos y\tag{13}\]

Since $f_x = f_y = 0 \iff e^x\cos y = -e^x\sin y = 0$ has no solution,
$f$ does not have any local minumum or maximum value.

\[f(x, y) = (x^2 + y^2)e^{y^2 - x^2}\tag{15}\]
\begin{align*}
  &f_x = f_y = 0\\
  \iff &e^{y^2 - x^2}(2x + (x^2 + y^2)(-2x))
      = e^{y^2 - x^2}(2y + (x^2 + y^2)2y) = 0\\
  \iff &x^3 + xy^2 - x = x^2y + y^3 + y = 0\\
  \iff &(x^2 + y^2 - 1)(x - y) = x^2y + y^3 + y = 0\\
  \iff &(x, y) \in \{(-1, 0), (0, 0), (1, 0)\}
\end{align*}

Second derivatives of $f$
\begin{align*}
  f_{xx} &= (4x^4 + 4x^2y^2 - 10x^2 - 2y^2 + 2)e^{y^2 - x^2}\\
  f_{xy} &= f_{yx} = -4xy(x^2 + y^2)e^{y^2 - x^2}\\
  f_{yy} &= (4x^2y^2 + 4y^4 + 2x^2 + 10y^2 + 2)e^{y^2 - x^2}
\end{align*}

From these we can calculate $D(0, 0) = 4 > 0$ and $D(\pm 1, 0) = -16/e^2 < 0$
and thus conclude that $f(0, 0) = 0$ is the only local minimum value of $f$.

\exercise{29\&34} Find the absolute maximum and minimum values of $f$
on the set $D$.
\[f = x^2 + y^2 - 2x,\qquad
D = \{(x, y) \,|\, x \geq 0, |x| + |y| \leq 2\}\tag{29}\]

The critical points of $f$ occur when
\[f_x = f_y = 0 \iff 2x - 2 = 2y = 0 \iff (x, y) = (1, 0)\]

The value of $f$ at the only critical point $(1, 0)$ is $f(1, 0) = 0$.

\begin{tikzpicture}
  \begin{axis}[
    axis x line=middle, axis y line=middle,
    xmin=-1.5, xmax=4.5, xlabel={x}, ymin=-3, ymax=3, ylabel={y},
    xlabel style={at=(current axis.right of origin), anchor=west},
    ylabel style={at=(current axis.above origin), anchor=south}]
    \addplot[red] plot coordinates {(0,-2) (0,2)};
    \addplot[green] plot coordinates {(0,-2) (2,0)};
    \addplot[blue] plot coordinates {(0,2) (2,0)};
    \legend{$L_0$, $L_1$, $L_2$}
  \end{axis}
\end{tikzpicture}

On $L_0$, we have $x = 0$ and
\[f(x, y) = f(0, y) = y^2, -2 \leq y \leq 2
\qquad\Longrightarrow 0 \leq f(x, y) \leq 4\]

On $L_1$, we have $0 \leq y = x - 2 \leq 2$ and thus
\[f(x, y) = f(x, x - 2) = 2x^2 - 6x + 4
\Longrightarrow 0 \leq f(x, y) \leq 24\]

On $L_2$, we have $0 \leq y = 2 - x \leq 2$ and thus
\[f(x, y) = f(x, 2 - x) = 2x^2 - 6x + 4
\Longrightarrow 0 \leq f(x, y) \leq 4\]

Therefore, on the boundary, the minimum value of $f$ is 0
and the maximum is 24.

\[f(x, y) = xy^2,\qquad
D = \{(x, y) \,|\, x \geq 0, y \geq 0, x^2 + y^2 \leq 3\}\tag{34}\]

The critical points of $f$ occur when
\[f_x = f_y = 0 \iff y^2 = 2xy = 0 \iff y = 0\]

\begin{tikzpicture}
  \begin{axis}[
    axis x line=middle, axis y line=middle,
    xmin=-1, xmax=3, xlabel={x}, ymin=-1, ymax=3, ylabel={y},
    xlabel style={at=(current axis.right of origin), anchor=west},
    ylabel style={at=(current axis.above origin), anchor=south}]
    \addplot[domain=0:1.732, red]{sqrt(3 - x^2)};
    \addplot[domain=0:1.732, red]{sin(x/pi*180)};
    \addplot[green] plot coordinates {(0,0) (1.732,0)};
    \addplot[blue] plot coordinates {(0,0) (0,1.732)};
    \legend{$C$, $L_0$, $L_1$}
  \end{axis}
\end{tikzpicture}

The critical points of $f$ are on $L_1$ and its values there are 0.
On $L_0$, the value of $f(x, y)$ is also always 0.

On $C$, $y^2 = 3 - x^2$ and $0 \leq x \leq \sqrt 3$, hence
$0 \leq f(x, y) = 3x - x^3 \leq 2$.

Thus, on the boundary, the minimum value of $f$ is 0
and the maximum is 2.\pagebreak

\exercise{41} Find all the points $P(a, b, c)$ on the cone $z^2 = x^2 + y^2$
that are closest to the point $Q(4, 2, 0)$.

Coordinates of $P$ satisfy $c = \sqrt{a^2 + b^2}$, thus
\begin{align*}
  PQ^2 &= (a - 4)^2 + (b - 2)^2 + a^2 + b^2\\
  &= 2a^2 - 8a + 2b^2 - 4b + 20\\
  &= 2(a - 2)^2 + 2(b - 1)^2 + 10 \leq 10
\end{align*}

Therefore the closest point to $Q$ on the cone is $\left(2, 1, \pm\sqrt 5\right)$.
The minumum distance is $\sqrt{10}$.

\exercise{49} Find the dimensions $(x, y, z)$ of a rectangular box of
maximum volume such that the sum of the lengths of its 12 edges is a constant
$c = 4(x + y + z)$.

By AM-GM inequality, the volume of the box is
\[V = xyz \leq \left(\frac{x + y + z}{3}\right)^2 = \frac{16c^2}{9}\]

Equality occurs when $x = y = z = c/12$.

\subsection{Lagrange Multipliers}
\exercise{1} It is estimated that the minumum of $f$ is 30
and the maximum value is 60.

\exercise{5\&8\&13}. Use Lagrange multipliers to find the maximum and minimum
values of the function subject to the given function.

\[f(x, y) = y^2 - x^2,\qquad \frac{x^2}{4} + y^2 = 1\tag{5}\]
\begin{align*}
  \begin{cases}
    \nabla f(x, y) = \lambda\nabla((x, y) \mapsto \frac{x^2}{4} + y^2)\\
    \frac{x^2}{4} + y^2 = 1
  \end{cases}
  &\iff
  \begin{cases}
    \left<-2x, 2y\right> = \lambda\left<\frac{x}{2}, 2y\right>\\
    \frac{x^2}{4} + y^2 = 1
  \end{cases}\\
  &\iff
  \begin{cases}
    -2x = \frac{\lambda x}{2}\\
    2y = 2\lambda y\\
    \frac{x^2}{4} + y^2 = 1
  \end{cases}\\
\end{align*}

For $x = 0$, $\lambda = 1$ and $y = \pm 1$; for $y = 0$, $\lambda = -4$
and $x = \pm 2$. Thus the minumum value of $f$ is $f(\pm 1, 0) = -1$
and the maximum value is $f(0, \pm 2) = 4$.

\[f(x, y, z) = x^2 + y^2 + z^2,\qquad x + y + z = 12\tag{8}\]
\begin{align*}
  \begin{cases}
    \nabla f(x, y, z) = \lambda\nabla((x, y, z) \mapsto x + y + z)\\
    x + y + z = 12
  \end{cases}
  &\iff
  \begin{cases}
    \left<2x, 2y, 2z\right> = \lambda\left<1, 1, 1\right>\\
    x + y + z = 12
  \end{cases}\\
  &\iff
  \begin{cases}
    x = y = z = \frac{\lambda}{2}\\
    x + y + z = 12
  \end{cases}\\
  &\iff
  \begin{cases}
    x = y = z = 4\\
    \lambda = 8
  \end{cases}
\end{align*}

Since $f(4, 4, 4) = 48 < f(12, 0, 0) = 144$, absolute minumum value of the
function subject to $x + y + z = 12$ is $f(4, 4, 4) = 48$.

\[f(x, y, z, t) = x + y + z + t,\qquad x^2 + y^2 + z^2 + t^2 = 1\tag{13}\]
\begin{align*}
  &\begin{cases}
    \nabla f(x, y, z, t) = \lambda\nabla((x, y, z, t) \mapsto x^2 + y^2 + z^2 + t^2)\\
    x^2 + y^2 + z^2 + t^2 = 1
  \end{cases}\\
  \iff
  &\begin{cases}
    \left<1, 1, 1, 1\right> = \lambda\left<2x, 2y, 2z, 2t\right>\\
    x^2 + y^2 + z^2 + t^2 = 1
  \end{cases}\\
  \iff
  &\begin{cases}
    x = y = z = t = \frac{1}{2\lambda}\\
    x^2 + y^2 + z^2 + t^2 = 1
  \end{cases}\\
  \iff
  &\begin{cases}
    x = y = z = t = \pm\frac{1}{2}\\
    \lambda = 1
  \end{cases}
\end{align*}

$f(-0.5, -0.5, -0.5, -0.5) = -2$ is the minumum value of $f$
and $f(0.5, 0.5, 0.5, 0.5) = 4$ is the maximum value.\pagebreak

\exercise{15} Find the extreme values of $f(x, y, z) = 2x + y$ subject to
$x + y + z = 1$ and $y^2 + z^2 = 4$.

Extreme values of $f$ occur when
\begin{align*}
  &\begin{cases}
    \nabla f(x, y, z) = \lambda\nabla((x, y, z) \mapsto x + y + z)
                      + \mu\nabla((x, y, z) \mapsto y^2 + z^2)\\
    x + y + z = 1\\
    y^2 + z^2 = 4
  \end{cases}\\
  \iff
  &\begin{cases}
    \left<2, 1, 0\right> = \lambda\left<1, 1, 1\right>
                         + \mu\left<0, 2y, 2z\right>\\
    x + y + z = 1\\
    y^2 + z^2 = 4
  \end{cases}\\
  \iff
  &\begin{cases}
    \lambda = 1\\
    \mu = \frac{1}{\sqrt 8}\\
    x = 1\\
    y = \pm \sqrt 2\\
    z = \mp \sqrt 2
  \end{cases}
\end{align*}

Thus the minumum value of $f$ on the given constraints is
$f(1, -\sqrt 2) = 2 - \sqrt 2$ and the maximum value is
$f(1, \sqrt 2) = 2 + \sqrt 2$.

\exercise{21} Find the extreme values of $f(x, y) = e^{-xy}$
on $x^2 + 4y^2 \leq 1$.

Critical points of $f$ occur when $f_x = f_y = 0 \iff x = y = 0$,
the value of $f$ there is $e^0 = 1$.

On the boundary $x^2 + 4y^2 = 1$ the minimum and maximum values can be
determined using the Lagrange Method:
\begin{align*}
  \begin{cases}
    \left<-ye^{-xy}, -xe^{-xy}\right> = \lambda\left<2x, 8y\right>\\
    x^2 + 4y^2 = 1
  \end{cases}
  &\Longrightarrow
  \begin{cases}
    x \in \left\{\frac{\pm 1}{\sqrt 2}\right\}\\
    y \in \left\{\frac{\pm 1}{\sqrt 8}\right\}
  \end{cases}
\end{align*}

Thus on the boundary the minumum value of $f$ is $e^{-1/4} = \sqrt[4]{1/e}$
and the maximum value is $\sqrt[4] e$. These are also the absolute extreme
values of $f$ in the ellipse.

\exercise{37} Given function $f$ on $\mathbb{R}_+^n$
\[f(x_1, x_2, \ldots, x_n) = \sqrt[n]{\prod_{i=1}^n x_i}\]

By Lagrange Method, its extreme values subject to $\sum_{i=1}^n x_i = c$ satisfy
\[\begin{cases}
  \nabla f = \lambda\nabla\sum_{i=1}^n x_i\\
  \sum_{i=1}^n x_i = c
\end{cases}
\iff
\begin{cases}
  \left<\frac{x_1^{1-2/n}}{n}, \ldots, \frac{x_i^{1-2/n}}{n}\right>f
  = \lambda\left<x_1, x_2, \ldots, x_n\right>\\
  \sum_{i=1}^n x_i = c
\end{cases}\]

\[\Longrightarrow
\begin{cases}
  x_1 = x_2 = \ldots = x_n\\
  \sum_{i=1}^n x_i = c
\end{cases}
\iff x_1 = x_2 = \ldots = x_n = \frac{c}{n}\]

At $x_1 = x_2 = \ldots = x_n = c/n$, $f(x_1, x_2, \ldots, x_n) = c/n$.
As $c/n > 0 = f(c, 0, \ldots, 0)$, $c/n$ is the maximum value of $f$
on the given constraint.

\exercise{48} By AM-GM inequality,
as $\sum_{i=1}^n x_i^2 = \sum_{i=1}^n y_i^2 = 1$,

\[\sum_{i=1}^n x_i y_i \leq \sum_{i=1}^n\frac{x_i^2 + y_i^2}{2} = 1\]
with equality when $\sum_{i=1}^n(x_i - y_i)^2 = 0$.

\subsection*{Problem Plus}
\exercise{1} A rectangle with length L and width W is cut into four smaller
rectangles by two lines parallel to the sides.

Let $x, y$ be two nonnegative numbers satisfying $x \leq L$ and $y \leq W$.
The sum of the squares of the areas of the smaller rectangles would then be
\begin{align*}
  f(x, y) &= x^2y^2 + x^2(W-y)^2 + (L-x)^2y^2 + (L-x)^2(W-y)^2\\
  &= (x^2 + (L-x)^2)(y^2 + (W-y)^2)\\
\end{align*}

By AM-GM inequality, $f(x, y) \geq 4x(L-x)y(W-y)$ with the equality
$f(x, y) = L^2W^2/4$ if and only if $x = L - x = L/2$ and $y = W - y = y/2$.

On the other hand,
\begin{align*}
  \begin{cases}
    0 \leq x \leq L\\
    0 \leq y \leq W
  \end{cases}
  &\Longrightarrow
  \begin{cases}
    2x(L - x) \geq 0\\
    2y(W - y) \geq 0
  \end{cases}
  \iff
  \begin{cases}
    L^2 \geq x^2 + (L-x)^2\\
    W^2 \geq y^2 + (W-y)^2
  \end{cases}\\
  &\Longrightarrow
  f(x, y) \leq L^2W^2
\end{align*}
with equality when $(x, y) \in \{(0, 0), (0, W), (L, W), (L, 0)\}$.

\exercise{3} A long piece of galvanized sheet metal with width $w$ is to be
bent into a symmetric form with three straight sides to make a rain gutter.

Cross-section area, with $0 \leq x \leq w/2$
and $0 \leq \theta \leq \max\left(\arccos\frac{2x-w}{2x}, \pi\right)$
\begin{align*}
  A(x, \theta) &= (w - 2x + x\cos\theta)x\sin\theta\\
  &= wx\sin\theta - x^2\left(2\sin\theta - \frac{\sin2\theta}{2}\right)
\end{align*}

First derivatives:
\begin{align*}
  A_x &= w\sin\theta - 2x\left(2\sin\theta - \frac{\sin2\theta}{2}\right)\\
  A_\theta &= wx\cos\theta - x^2(2\cos\theta - \cos2\theta)
\end{align*}

Critical points occur when
\[A_x = A_\theta = 0 \iff
\begin{cases}
  w\sin\theta = 2x\left(2\sin\theta - \dfrac{\sin2\theta}{2}\right)\\
  wx\cos\theta = x^2(2\cos\theta - \cos2\theta)
\end{cases}\tag{$*$}\]

\begin{tikzpicture}
  \begin{axis}[
    axis x line=middle, axis y line=middle,
    xmin=-0.15, xmax=0.75, xlabel={$\frac{x}{w}$},
    ymin=-0.7, ymax=3.8, ylabel={$\theta$},
    xlabel style={at=(current axis.right of origin), anchor=west},
    ylabel style={at=(current axis.above origin), anchor=south}]
    \addplot[domain=0.25:0.5, color=red]{acos(1 - 0.5/x)/57.3};
    \addplot[magenta] plot coordinates {(0.5,1.57) (0.5,0)};
    \addplot[blue] plot coordinates {(0,0) (0.5,0)};
    \addplot[cyan] plot coordinates {(0,0) (0,3.14)};
    \addplot[green] plot coordinates {(0,3.14) (0.25,3.14)};
    \legend{$C$, $L_0$, $L_1$, $L_2$, $L_3$}
  \end{axis}
\end{tikzpicture}

For $x = 0$ (along $L_2$), it is obvious that the area is 0. For $x \neq 0$,
\begin{align*}
  (*) &\iff
  \begin{cases}
    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
    w\sin\theta(2\cos\theta-\cos2\theta) = w\cos\theta(4\sin\theta-\sin2\theta)
  \end{cases}\\
  &\iff
  \begin{cases}
    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
    2\cos\theta - \cos2\theta = \cos\theta(4 - 2\cos\theta)
  \end{cases}\\
  &\iff
  \begin{cases}
    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
    -\cos2\theta = 2\cos\theta - 2\cos^2\theta
  \end{cases}\\
  &\iff
  \begin{cases}
    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
    1 = 2\cos\theta
  \end{cases}\\
  &\iff
  \begin{cases}
    x = \frac{w}{3}\\
    \theta = \frac{\pi}{3}
  \end{cases}
\end{align*}
At this point, $A(x, \theta) = w^2/4\sqrt3$.

Along $C$, $A\left(x, \arccos\frac{2x-w}{2x}\right)
= \frac{1}{4}\sqrt{w(4x-w)(w-2x)^2} \in \left[0, \frac{w^2}{12\sqrt3}\right]$.

Along $L_0$, $A(w/2, \theta)
= \frac{w^2}{8}\sin(\pi - 2\theta) \in [0, w^2/8]$.

Along $L_1$ and $L_3$, $A(x, \theta) = A(x, 0) = A(x, \pi) = 0$.

In conclusion, the maximum cross-section is $\frac{w^2}{4\sqrt3}$
at $(x, \theta) = (w/3, \pi/3)$.

\exercise{4} For what values of $r$ is the function
\[f(x, y, z) =
\begin{cases}
  \dfrac{(x + y + z)^r}{x^2 + y^2 + z^2}&\text{if }(x, y, z) \neq (0, 0, 0)\\
  0&\text{if }(x, y, z) = (0, 0, 0)\\
\end{cases}\]
continuous on $\mathbb{R}^3$?

Along $y = z = 0$, as $x \to 0$, $f(x, 0, 0) = x^{r-2} \to \infty$
(or the limit might not exist at all) for $r < 2$
and $f(x, 0, 0) = 1$ for $r = 2$.
Therefore for $r \leq 2$, $f$ is discontinuous at $(0, 0, 0)$.

It is not difficult to show that for $r > 2$, $f$ is continuous.
For every positive number $\varepsilon$,
let $\delta = (\varepsilon/3^r)^{1/(2r-2)}$, then from
\begin{align*}
  &0 < \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2} < \delta\\
  \iff &0 < \sqrt{x^2 + y^2 + z^2}
          < \left(\frac{\varepsilon}{3^r}\right)^\frac{1}{2r-2}\\
  \iff &0 < \frac{3^r(x^2 + y^2 + z^2)^r}{x^2 + y^2 + z^2} < \varepsilon
\end{align*}
and
\[(x + y + z)^2 \leq 3(x^2 + y^2 + z^2)
\iff |x + y + z|^r \leq 3^r(x^2 + y^2 + z^2)^r\]
we get
\[0 < \frac{|x + y + z|^r}{x^2 + y^2 + z^2} < \varepsilon
\iff |f(x, y, z) - 0| < \varepsilon\]

Thus by definition, for $r > 2$, $f(x, y, z) \to 0$ as $(x, y, z)\to(0, 0, 0)$,
hence $f$ is continuous on $\mathbb{R}^3$.

\exercise{5} Suppose $f$ is a differentiable function of one variable.
Show that all tangent planes to the surface $z = xf(y/x)$
intersect in a common point.

Let $t = y/x$,
\begin{align*}
\tho{z}{x} &= f(t) + x\tho{f(t)}{x}
            = f(t) + x\leibniz{f}{t}\tho{(y/x)}{x}
            = f(t) - t\leibniz{f}{t}\\
\tho{z}{y} &= x\tho{f(t)}{y}
            = x\leibniz{f}{t}\tho{(y/x)}{y}
            = \leibniz{f}{t}
\end{align*}

Equation of the tangent plane to the given surface at $P(a, b, af(b/a))$ is
\begin{align*}
     &z - af\left(\frac{b}{a}\right) = \left(f\left(\frac{b}{a}\right)
        - \frac{b}{a}\cdot\leibniz{f}{t}\left(\frac{b}{a}\right)\right)(x - a)
        + \leibniz{f}{t}\left(\frac{b}{a}\right)(y - b)\\
\iff &z = xf\left(\frac{b}{a}\right) + \leibniz{f}{t}\left(\frac{b}{a}\right)
          \left(y - \frac{bx}{a}\right)\\
\iff &\left(f\left(\frac{b}{a}\right)
            - \frac{b}{a}\cdot\leibniz{f}{t}\left(\frac{b}{a}\right)\right)x
      + \leibniz{f}{t}\left(\frac{b}{a}\right)y - z = 0
\end{align*}

Since the equation is homogenous, the tangent plane always goes through origin
$O(0, 0, 0)$.

\section{Multiple Integrals}
\subsection{Double Integrals over Rectangles}
\exercise{1} Use a Riemann sum with $m=3$ and $n=2$ to estimate the volume
of the solid that lies below the surface $z = xy$ and above the rectangle
$R = [0, 6] \times [0, 4]$.

Take the sample point to be the upper right corner of each square,
\[V \approx \sum_{i=1}^3\sum_{j=1}^2 ij \cdot 4 = 288\tag{a}\]

Take the sample point to be the center of each square,
\[V \approx \sum_{i=1}^3\sum_{j=1}^2 (2i-1)(2j-1)4 = 144\tag{b}\]

\exercise{13} Evaluate the double integral by first identifying it
as the volume of a solid.
\[\iint_{[-2,2]\times[1,6]}(4 - 2y)\ud A = 0\]

\subsection{Integrated Integrals}
Calculate the integrated integrals.
\[\int_1^4\int_0^2(6x^2 - 2x)\ud y\ud x
= \int_1^4(12x^2 - 4x)\ud x = 222\tag{3}\]
\[\int_{-3}^3\int_0^{\pi/2}(y + y^2\cos x)\ud x\ud y
= \int_{-3}^3 y^2\ud y = 0\tag{7}\]
\[\iint_{[0,\pi/2]^2}\sin(x - y)\ud A
= \int_0^{\pi/2}(\cos y - \sin y)\ud y = 0\tag{15}\]
\begin{align*}
  \iint_{[0,1]\times[-3,3]}\frac{xy^2}{x^2 + 1}\ud A
  &= \int_0^1\frac{x}{x^2 + 1}\ud x \cdot \int_{-3}^3 y^2\ud y\\
  &= \frac{1}{2}\int_0^1\frac{\ud x}{x+1}
     \cdot \left[\frac{y^3}{3}\right]_{-3}^3\\
  &= 9\ln(x + 1)\big]_0^1\\
  &= 9\ln 2\tag{17}
\end{align*}
\begin{align*}
  \iint_{[0,2]\times[0,3]}ye^{-xy}\ud A
  &= \int_0^3\int_0^2 ye^{-xy}\ud x\ud y\\
  &= \int_0^3(1 - e^{-2y})\ud y\\
  &= \left[y + \frac{e^{-2y}}{2}\right]_0^3\\
  &= \frac{1}{2e^6} + \frac{5}{2}\tag{21}
\end{align*}
\[\iint_{[-1,1]\times[-2,2]}\left(1-\frac{x^2}{4}-\frac{y^2}{9}\right)\ud A
= \int_{-1}^1\left(\frac{92}{27} - x^2\right)\ud x = \frac{166}{27}\tag{27}\]
\[\iint_{[0,4]\times[0,5]}(16 - x^2)\ud A
= \int_0^4(80 - 5x^2)\ud x = \frac{640}{3}\tag{30}\]

\exercise{40} Fubini's and Clairaut's theorems are similar in the way that
for continuous functions, order of variables are interchangeable in integration
and differentiation. By the Fundamental Theorem and these two theorems,
if $f(x, y)$ is continuous on $[a, b]\times[c, d]$ and
\[g(x, y) = \int_a^x\int_c^y g(s, t)\ud t\ud s\]
for $a < x < b$ and $c < y < d$, then $g_{xy} = g_{yx} = f(x, y)$.

\subsection{Double Integrals over General Regions}
Evaluate the iterated integral.
\[\int_0^1\int_0^{s^2}\cos s^3\ud t\ud s
= \int_0^1 s^2\cos s^3\ud s
= \left[\frac{\sin s^3}{3}\right]_0^1
= \frac{\sin 1}{3}\tag{5}\]
\[\int_0^\pi\int_0^{\sin x}x\ud y\ud x
= \int_0^\pi x\sin x\ud x
= [\sin x - x\cos x]_0^\pi
= \pi\tag{9}\]
\[\int_{-1}^2\int_{y^2}^{y+2}y\ud x\ud y
= \int_{-1}^2(2y + y^2 - y^3)\ud y
= \left[y^2 + \frac{y^3}{3} - \frac{y^4}{4}\right]_{-1}^2
= \frac{9}{4}\tag{15}\]
\[\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(2x - y)\ud y\ud x
= \int_{-2}^2 4x\sqrt{4 - x^2}\ud x
= 0\tag{21}\]
\begin{align*}
  \int_1^2\int_1^{7-3y}xy\ud x\ud y
  &= \int_1^2\left(\frac{9y^3}{2} - 21y^2 + 24y\right)\ud y\\
  &= \left[\frac{9y^4}{8} - 7y^3 + 12y^2\right]_1^2\\
  &= \frac{31}{8}\tag{25}
\end{align*}
\[\int_1^2\int_0^{\ln x} f(x, y)\ud y\ud x
= \int_0^{\ln 2}\int_{e^y}^2 f(x, y)\ud x\ud y\tag{47}\]
\[\int_0^1\int_{3y}^3 e^{x^2}\ud x\ud y
= \int_0^3\int_0^{x/3} e^{x^2}\ud y\ud x
= \int_0^3\frac{xe^{x^2}}{3}\ud x
= \left.\frac{e^{x^2}}{6}\right]_0^3
= \frac{e^9 - 1}{6}\tag{49}\]

\subsection{Double Integrals in Polar Coordinates}
Evaluate the given integral.
\[\int_0^{3\pi/2}\int_0^4 f(r\cos\theta, r\sin\theta)r\ud r\ud\theta\tag{1}\]
\begin{align*}
  \int_{\pi/4}^{\pi/2}\int_0^2(2\cos\theta - \sin\theta)r^2\ud r\ud\theta
  &= \int_{\pi/2}^{\pi/4}\frac{8}{3}(2\cos\theta - \sin\theta)\ud\theta\\
  &= \frac{8}{3}\left[2\sin\theta + \cos\theta\right]_{\pi/4}^{\pi/2}\\
  &= \frac{16}{3} - 4\sqrt 2\tag{8}
\end{align*}
\begin{align*}
  \int_{-\pi/2}^{\pi/2}\int_0^2 re^{-r^2}\ud r\ud\theta
  &= \int_{-\pi/2}^{\pi/2}\frac{1 - e^{-4}}{2}\ud\theta\\
  &= \pi\frac{1 - e^{-4}}{2}\tag{11}
\end{align*}
\begin{align*}
  \int_0^{2\pi}\int_0^{\sqrt{1/2}}\left(\sqrt{1 - r^2} - r\right)r\ud r\ud\theta
  &= \pi\int_0^{\sqrt{1/2}}\left(\sqrt{1 - r^2} - r\right)\ud r^2\\
  &= \pi\int_0^{1/2}(\sqrt{1 - x} - \sqrt x)\ud x\\
  &= \frac{\pi}{3}(2 - \sqrt 2)\tag{25}
\end{align*}
\begin{align*}
  \int_0^\pi\int_0^3 r\sin r^2\ud r\ud\theta
  &= \int_0^9\frac{\pi\sin x}{2}\ud x\\
  &= \left.\frac{\pi\cos x}{-2}\right]_0^9\\
  &= \frac{\pi}{2}(1 - \cos 9)\tag{29}
\end{align*}

\exercise{40} We define the improper integral
(over the entire plane $\mathbb{R}^2$)
\begin{align*}
  I &= \iint_{\mathbb{R}^2}\exp(-x^2-y^2)\ud A\\
  &= \int_{-\infty}^\infty\int_{-\infty}^\infty\exp(-x^2-y^2)\ud x\ud y\\
  &= \lim_{a\to\infty}\iint_{D_a}\exp(-x^2-y^2)\ud A
\end{align*}
where $D_a$ is the disk with radius $a$ and center the origin.

By changing to polar coordinates,
\begin{align*}
  I &= \lim_{a\to\infty}\int_0^{2\pi}\int_0^a\exp(-a^2)a\ud a\ud\theta\\
  &= \lim_{a\to\infty}\int_0^a-\pi\exp(-a^2)\ud-a^2\\
  &= -\pi\lim_{a\to\infty}\int_0^{-a^2}e^b\ud b\\
  &= -\pi\lim_{a\to\infty}\left.e^b\right]_0^{-a^2}\\
  &= \pi\lim_{a\to\infty}(1 - \exp(-a^2))\\
  &= \pi\tag{a}
\end{align*}

As $\exp(-x^2-y^2)$ is continuous on $\mathbb{R}^2$,
\[\int_{-\infty}^\infty\exp(-x^2)\ud x\int_{-\infty}^\infty\exp(-y^2)\ud y
= I = \pi\tag{b}\]

Thus $\int_{-\infty}^\infty\exp(-x^2)\ud x = \sqrt I = \sqrt\pi$ and
$\int_{-\infty}^\infty\exp(-x^2/2)\ud x = \sqrt{2\pi}$.

\subsection{Applications of Double Integrals}
\exercise{2} The total charge on the disk is
\[\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{x^2 + y^2}\ud y\ud x
= \int_0^{2\pi}\int_0^1 r^2\ud r\ud\theta
= \left.2\pi\frac{r^3}{3}\right]_0^1
= \frac{2\pi}{3}\]

\noindent Find the mass and center of mass of the lamina that occupies the
regions $D$ and has the given density function $\rho$.
\[D = [1, 3]\times[1, 4];\qquad\rho(x, y) = ky^2\tag{3}\]
\[m = \int_1^3\ud x \cdot \int_1^4 ky^2\ud y = 42k\]
\begin{align*}
  \bar x &= \frac{k}{m}\int_1^3\int_1^4 xy^2\ud y\ud x
= \frac{21k}{m}\int_1^3 x\ud x
= \frac{84k}{m}
= 2\\
  \bar y &= \frac{k}{m}\int_1^3\int_1^4 y^3\ud y\ud x
= \frac{2k}{m}\int_1^4 y^3\ud y
= \frac{255k}{m}
= \frac{85}{28}
\end{align*}

\[D = \{(x, y)\,|\,-1 \leq x \leq 1,\,0 \leq y \leq 1 - x^2\},\qquad
\rho(x, y) = ky\tag{7}\]
\[m = \int_{-1}^1\int_0^{1-x^2} ky\ud y\ud x
= \frac{k}{2}\int_{-1}^1 (x^4 - 2x^2 + 1)\ud x
= \frac{8k}{15}\]
\begin{align*}
  \bar x &= \frac{k}{m}\int_{-1}^1\int_0^{1-x^2} xy\ud y\ud x
  = \frac{15}{8}\int_{-1}^1 (x^5 - 2x^3 + x)\ud x
  = 0\\
  \bar y &= \frac{k}{m}\int_{-1}^1\int_0^{1-x^2} y^2\ud y\ud x
  = \frac{8}{45}\int_{-1}^1 (1 - x^2)^3\ud x
  = \frac{4}{7}
\end{align*}
\pagebreak

\[D = \left\{(x, y)\,\Big|\,0\leq y\leq\sin\frac{\pi x}{L},\,
0\leq x\leq L\right\},\qquad\rho(x, y) = y\tag{9}\]
\[m = \int_0^L\int_0^{\sin(\pi x/L)}y\ud y\ud x
= \int_0^L\frac{\sin^2(\pi x/L)}{2}\ud x
= \left[\frac{x}{4} - \frac{L}{8\pi}\sin\frac{2\pi x}{L}\right]_0^L
= \frac{L}{4}\]
\begin{align*}
  \bar x &= \int_0^L\int_0^{\sin(\pi x/L)}\frac{xy}{m}\ud y\ud x
= \int_0^L\frac{2x\sin^2(\pi x/L)}{L}\ud x
= \frac{L}{2}\\
  \bar y &= \int_0^L\int_0^{\sin(\pi x/L)}\frac{y^2}{m}\ud y\ud x
= \int_0^L\frac{4\sin^3(\pi x/L)}{3L}\ud x
= \frac{16}{9\pi}
\end{align*}

\[D = \{(x, y)\,|\,0\leq x\leq 1,\,0\leq y\leq\sqrt{1-x^2}\},\qquad
\rho(x, y) = ky\tag{11}\]
\[m = \int_0^1\int_0^{\sqrt{1-x^2}}ky\ud y\ud x
= \int_0^{\pi/2}\sin\theta\ud\theta\cdot\int_0^1 kr^2\ud r
= \frac{k}{3}\]
\begin{align*}
  \bar x &= \int_0^1\int_0^{\sqrt{1-x^2}}3xy\ud y\ud x
= \int_0^{\pi/2}\cos\theta\sin\theta\ud\theta\cdot\int_0^1 3r^3\ud r
  = \frac{3}{8}\\
  \bar y &= \int_0^1\int_0^{\sqrt{1-x^2}}3y^2\ud y\ud x
= \int_0^{\pi/2}\sin^2\theta\ud\theta\cdot\int_0^1 3r^3\ud r
= \frac{3\pi}{16}
\end{align*}

\subsection{Surface area}
Find the area of the surface.

\exercise{3} The part of the plane $3x + 2y + z = 6$
that lies in the first octant.
\begin{align*}
  A &= \int_0^2\int_0^{3-1.5x}\sqrt{1 + \left(\tho{z}{x}\right)^2
     + \left(\tho{z}{y}\right)^2}\ud y\ud x\\
    &= \int_0^2\int_0^{3-1.5x}\sqrt{14}\ud y\ud x\\
    &= \int_0^2\left(3 - \frac{3}{2}x\right)\sqrt{14}\ud x\\
    &= \left[3x\sqrt{14} - \frac{3x^2\sqrt{14}}{4}\right]_0^2\\
    &= 3\sqrt{14}
\end{align*}

\exercise{9} The part of the surface $z = xy$
that lies within the cylinder $x^2 + y^2 = 1$.
\begin{align*}
  A &= \iint_D\sqrt{1 + \left(\tho{xy}{x}\right)^2
     + \left(\tho{xy}{y}\right)^2}\ud A\\
    &= \int_0^{2\pi}\int_0^1 r\sqrt{1 + r^2}\ud r\ud\theta\\
    &= \pi\int_0^1\sqrt{1 + t}\ud t\\
    &= \left.\frac{2\pi\sqrt{(1 - t)^3}}{3}\right]_0^1\\
    &= \frac{2\pi}{3}\left(2\sqrt{2} - 1\right)
\end{align*}

\exercise{12} The part of the sphere $x^2 + y^2 + z^2 = 4z$
that lies inside the paraboloid $z = x^2 + y^2$,
in which it has the equation $z = 2 + \sqrt{4 - x^2 - y^2}$.
\begin{align*}
  A &= \iint_D\sqrt{1 + \left(\tho{}{x}\left(2 + \sqrt{4 - x^2 - y^2}\right)\right)^2
     + \left(\tho{}{y}\left(2 + \sqrt{4 - x^2 - y^2}\right)\right)^2}\ud A\\
    &= \iint_D\sqrt\frac{4}{4 - x^2 - y^2}\ud A\\
    &= \int_0^{2\pi}\int_0^{\sqrt 3}r\sqrt\frac{4}{4 - r^2}\ud r\ud\theta\\
    &= 2\pi\int_0^3\sqrt\frac{1}{4 - t}\ud t\\
    &= \left.-4\pi\sqrt{4 - t}\right]_0^3\\
    &= 4\pi
\end{align*}

\subsection{Triple Integrals}
Evaluate the integral.
\[\int_0^1\int_0^3\int_{-1}^2 xyz^2\ud y\ud z\ud x
= \int_0^1\int_0^3\frac{3xz^2}{2}\ud z\ud x
= \int_0^1\frac{27x}{2}\ud x
= \frac{27}{4}\tag{1}\]
\begin{align*}
  \int_0^2\int_0^{z^2}\int_0^{y-z}(2x - y)\ud x\ud y\ud z
  &= \int_0^2\int_0^{z^2}(z^2 - yz)\ud y\ud z\\
  &= \int_0^2\left(z^4 - \frac{z^5}{2}\right)\ud z\\
  &= \frac{16}{15}\tag{3}
\end{align*}
\[\int_0^3\int_0^x\int_{x-y}^{x+y}y\ud z\ud y\ud x
= \int_0^3\int_0^x 2y^2\ud y\ud x
= \int_0^3\frac{2x^3}{3}\ud x
= \frac{27}{2}\tag{9}\]
\begin{align*}
  \int_0^\pi\int_0^{\pi-x}\int_0^x\sin y\ud z\ud y\ud x
  &= \int_0^\pi\int_0^{\pi-x}x\sin y\ud y\ud x\\
  &= \int_0^\pi(x + x\cos y)\ud x\\
  &= \frac{\pi^2}{2} - 2\tag{12}
\end{align*}
\begin{align*}
  \int_0^1\int_0^{3x}\int_0^{\sqrt{9-y^2}}z\ud z\ud y\ud x
  &= \int_0^1\int_0^{3x}\frac{9 - y^2}{2}\ud y\ud x\\
  &= \int_0^1\frac{27x - 9x^3}{2}\ud x\\
  &= \frac{45}{8}\tag{18}
\end{align*}
\begin{align*}
  \int_0^2\int_0^{4-2x}\int_0^{4-2x-y}\ud z\ud y\ud x
  &= \int_0^2\int_0^{4-2x}(4 - 2x - y)\ud y\ud x\\
  &= \int_0^2\frac{(4 - 2x)^2}{2}\ud x\\
  &= \frac{16}{3}\tag{19}
\end{align*}
\begin{align*}
  \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{-1}^{4-z}\ud y\ud z\ud x
  &= \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(5 - z)\ud z\ud x\\
  &= \int_{-2}^2 10\sqrt{4 - x^2}\ud x\\
  &= 20\pi\tag{22}
\end{align*}

\subsection{Triple Integrals in Cylindrical Coordinates}
\exercise{1} Change from cylindrical coordinates to rectangular coordinates.
\begin{enumerate}[(a)]
  \item $\left(4, \frac{\pi}{3}, -2\right)
    \rightarrow \left(2, 2\sqrt 3, -2\right)$
  \item $\left(2, \frac{-\pi}{2}, 1\right) \rightarrow \left(0, -2, 1\right)$
\end{enumerate}

\exercise{3} Change from rectangular coordinates to cylindrical coordinates.
\begin{enumerate}[(a)]
  \item $\left(-1, 1, 1\right)
    \rightarrow \left(\sqrt 2, \frac{3\pi}{4}, 1\right)$
  \item $\left(-2, 2\sqrt 3, 3\right)
    \rightarrow \left(4, \frac{2\pi}{3}, 3\right)$
\end{enumerate}

\exercise{7} In cylindrical coordinates $(r, \theta, z)$, $z = 4 - r^2$
is the paraboloid $z = 4 - x^2 - y^2$ in Cartesian coordinates.

\exercise{15\&17\&21} Evaluate the integral.
\[\int_{-\pi/2}^{\pi/2}\int_0^2\int_0^{r^2}r\ud z\ud r\ud\theta
= \pi\int_0^2 r^3\ud r
= 4\pi\tag{15}\]
\[\iiint_E\sqrt{x^2 + y^2}\ud V
= \int_0^{2\pi}\int_0^4\int_{-5}^4 r^2\ud z\ud r\ud\theta
= 18\pi\left.\frac{r^3}{3}\right]_0^4
= 384\pi\tag{17}\]
\begin{align*}
  \iiint_E x^2\ud V
  &= \int_0^{2\pi}\int_0^2\int_{z/2}^1 r^3\cos^2\theta\ud r\ud z\ud\theta\\
  &= \int_0^{2\pi}\cos^2\theta\ud\theta\int_0^2\int_{z/2}^1 r^3\ud r\ud z\\
  &= \left[\frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2}\right]_0^{2\pi}
     \int_0^2\left(\frac{1}{4} - \frac{z^4}{64}\right)\ud z\\
  &= \frac{2\pi}{5}\tag{21}
\end{align*}

\section{Vector Calculus}
\setcounter{subsection}{1}
\subsection{Line Integrals}
Evaluate the integral.
\begin{align*}
&\int_{-\pi/2}^{\pi/2}4\cos t(4\sin t)^4
 \sqrt{\left(\leibniz{4\cos t}{t}\right)^2
     + \left(\leibniz{4\sin t}{t}\right)^2}\ud t\\
=\,&4096\int_{-\pi/2}^{\pi/2}\sin^4 t\ud\sin t
= 4096\int_{-1}^1 w^4\ud w
= \frac{8192}{5}\tag{3}
\end{align*}
\begin{align*}
  \int_{\left\{(x, y)\in[1,4]\times[1,2]\,|\,y=\sqrt x\right\}}
  \left(x^2 y^3 - \sqrt x\right)\ud y
  &= \int_1^2(t^7 - t)\leibniz{t}{t}\ud t\\
  &= \left[\frac{t^8}{8} - \frac{t^2}{2}\right]_1^2\\
  &= \frac{243}{8}\tag{5}
\end{align*}
\begin{align*}
& \int_0^2(x + x)\ud x + \int_2^3(x + 6 - 2x)\ud x
+ \int_0^1(2y)^2\ud y + \int_1^0(3-x)^2\ud y\\
=\,&4 + \frac72 + \frac43 - \frac{19}{3}
=\frac{5}{2}\tag{7}
\end{align*}
\begin{align*}
  &\int_2^0 x^2\ud x + \int_0^4 x^2\ud x + \int_0^2 y^2\ud y + \int_2^3\ud y\\
= &\int_2^4 x^2\ud x + \int_0^3 y^2\ud y
= \left.\frac{x^3}{3}\right]_2^4 + \left.\frac{y^3}{3}\right]_0^3
= 13\tag{8}
\end{align*}
\begin{align*}
   \int_0^1(11y^7\unit\i + 3t^6\unit\j)\ud(11t^4\unit\i + t^3\unit\j)
&= \int_0^1(11y^7\unit\i + 3t^6\unit\j)\cdot(44t^3\unit\i + 3t^2\unit\j)\ud t\\
  &= \int_0^1(484t^{10} + 9t^8)\ud t\\
&= \left[44t^11 + t^9\right]_0^1\\
&= 45\tag{19}
\end{align*}
\begin{align*}
 &\int_0^1(\sin t^3\unit\i + \cos t^2\unit\j + t^4\unit k)
  \ud(t^3\unit\i + t^2\unit\j + t\unit k)\\
=&\int_0^1\sin x\ud x + \int_0^1\cos y\ud y + \int_0^1 z^4\ud z\\
=&\,\frac{6}{5} - \cos 1 - \sin 1\tag{21}
\end{align*}
\begin{align*}
  &\int_0^{2\pi}(t - \sin t)\ud(t - \sin t) + (3 - \cos t)\ud(1 - \cos t)\\
= &\int_0^{2\pi}((t - \sin t)(1 - \cos t) + (3 - \cos t)\sin t)\ud t\\
= &\int_0^{2\pi}(t - t\cos t + 2\sin t)\ud t\\
=\,&\left[\frac{t^2}{2} - t\sin t - 3\cos t\right]_0^{2\pi}
= 2\pi^2\tag{39}
\end{align*}
\begin{align*}
  &\,2\int_0^{2\pi}\left(4 + \frac{x^2 - y^2}{100}\right)
  \sqrt{(-10\sin t)^2 + (10\cos t)^2}\ud t\\
= &\int_0^{2\pi}\left(800 + (10\cos t)^2 - (10\sin t)^2\right)\ud t\\
= &\,100\int_0^{2\pi}(8 + \cos 2t)\ud t\\
= &\,\left[8t - \frac{\sin 2t}{2}\right]_0^{2\pi}
= 16\pi\tag{48}
\end{align*}

\subsection{The Fundamental Theorem for Line Integrals}
Evaluate the integrals.
\begin{align*}
   &\int_C(x^2\unit\i + y^2\unit\j)\cdot\ud(x\unit\i + 2x^2\unit\j)\\
=\,&(f\mapsto f(2, 8) - f(-1, 2))\left((x, y)\mapsto\frac{x^3 + y^3}{3}\right)
= 513\tag{12}
\end{align*}
\begin{align*}
   &\int_C(xy^2\unit\i + x^2y\unit\j)\cdot\ud\mathbf{r}\\
=\,&(f\mapsto f(2, 1) - f(0, 1))\left((x, y)\mapsto\frac{x^2y^2}{2}\right)
= 2\tag{13}
\end{align*}

\subsection{Green's Theorem}
Evaluate the integrals.
\begin{align*}
  \int_C\left(y+e^{\sqrt x}\right)\ud x + (2x + \cos y^2)\ud y
  &= \int_0^1\int_{y^2}^{\sqrt y}\ud x\ud y\\
  &= \int_0^1(\sqrt y - y^2)\ud y\\
  &= \left[\frac{2\sqrt{y^3}}{3} - \frac{y^3}{3}\right]_0^1\\
  &= \frac{1}{3}\tag{7}
\end{align*}
\begin{align*}
  \int_{x^2+y^2=4}y^3\ud x - x^3\ud y
  &= \iint_{x^2+y^2=4}(-3x^2-3y^2)\ud A\\
  &= -3\int_0^{2\pi}\int_0^2 r^3\ud r\ud\theta\\
  &= -6\pi\left.\frac{r^4}{4}\right]_0^2\\
  &= -24\pi\tag{9}
\end{align*}
\begin{align*}
  \int_C(1-y^3)\ud x + (x^3+\exp y^2)\ud y
  &= \iint_D(3x^2 + 3y^2)\ud A\\
  &= 3\int_0^{2\pi}\int_2^3 r^3\ud r\ud\theta\\
  &= 6\pi\left.\frac{r^4}{4}\right]_2^3\\
  &= \frac{195}{8}\pi\tag{10}
\end{align*}
\begin{align*}
  &\int_C(y\cos x - xy\sin x)\ud x + (xy + x\cos x)\ud y\\
= &-\iint_D(y + \cos x - x\sin x - \cos x + x\sin x)\ud A\\
= &-\int_0^2\int_0^{4-2x}y\ud y\ud x = \frac{16}{-3}\tag{11}
\end{align*}
\begin{align*}
  &\int_C(\exp-x + y^2)\ud x + (\exp-y + x^2)\ud y\\
  =&-\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}(2x - 2y)\ud y\ud x\\
  =&-\int_{-\pi/2}^{\pi/2}(2x\cos x - \cos^2 x)\ud x
  =\frac\pi 2\tag{12}
\end{align*}
\[\int_0^1\int_0^{1-x}(y^2 - x)\ud y\ud x
= \int_0^1\left(\frac{(1-x)^3}{3} + x^2 - x\right)\ud x
= \frac{-1}{12}\tag{17}\]
\begin{align*}
  \int_\text{cycloid}y\ud x + \int_\text{segment}y\ud x
  &= \int_{2\pi}^0(1-\cos t)\ud(t-\sin t) + 0\\
  &= \int_{2\pi}^0\left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right)\ud t\\
  &= \left[\frac{3t}{2} - 2\sin t + \frac{\sin 2t}{4}\right]_{2\pi}^0
  = 3\pi\tag{19}
\end{align*}

\subsection{Curl and Divergence}
\exercise{19} Since the divergence of curl of $\mathbf G$ is $1 \neq 0$,
there does not exist a vector field $\mathbf G$ satisfying the given condition.

\subsection{Parametric Surfaces and Their Areas}
\exercise{19} One parametric representation for the surface $x + y + z = 0$ is
$\mathbf{r}(u, v) = \langle u, v, -u-v\rangle$.

\exercise{23} One parametric representation for the sphere $x^2 + y^2 + z^2 = 4$
above the cone $\sqrt{x^2 + y^2}$ is $\mathbf{r}(u, v) =
\langle 2\cos u\cos v, 2\cos u\sin v, 2\sin u\rangle$.

\exercise{39} The plane intersects with $Ox$ at $A(2, 0, 0)$, with $Oy$
at $B(0, 3, 0)$ and with $Oz$ at $C(0, 0, 6)$. The area of the triangle $ABC$
is $|\mathbf{AB}\times\mathbf{AC}|/2 = 3\sqrt{14}$.

\exercise{42} Surface of the cone $\sqrt{x^2 + y^2}$:
\[\iint_D\sqrt{1 + \left(\frac{x}{\sqrt{x^2+y^2}}\right)^2
+ \left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2}\ud A
= \iint_D\sqrt 2\ud A\]

For the part lying between $y = x$ and $y = x^2$, the area is
\[\int_0^1\int_{x^2}^x\sqrt{2}\ud y\ud x
= \sqrt 2\int_0^1(x - x^2)\ud x
= \frac{\sqrt 2}{6}\]

\exercise{43} Area of the surface:
\[\int_0^1\int_0^1\sqrt{1 + x + y}\ud y\ud x
= \frac{4 - 32\sqrt 2}{15} + \frac{12\sqrt 3}{5}\]

\exercise{45} Area of $z = xy$ within $x^2 + y^2 = 1$:
\[\iint_D\sqrt{1 + x^2 + y^2}\ud A
= \int_0^{2\pi}\int_0^1\sqrt{1 + r^2}r\ud r\ud\theta
= \pi\int_1^2\sqrt t\ud t
= \frac{2\pi}{3}\left(\sqrt 8 - 1\right)\]

\exercise{49} Area of the surface with given parametric equation
$\mathbf{r}(u, v) = \langle u^2, uv, v^2/2\rangle$ within $0 \leq u \leq 1$ and
$0 \leq v \leq 2$:
\[\iint_D|\mathbf{r}_u\times\mathbf{r}_v|\ud A
= \int_0^2\int_0^1(2u^2 + v^2)\ud u\ud v
= \int_0^2\left(\frac{2}{3} + v^2\right)\ud v = 4\]

\subsection{Surface Integrals}
Evaluate the surface integrals.
\begin{align*}
   \iint_S(x + y + z)\ud S
&= \int_0^2\int_0^1(4u + v + 1)\sqrt{14}\ud v\ud u\\
&= \int_0^2\left(4u + \frac{3}{2}\right)\sqrt{14}\ud u\\
&= 11\sqrt{14} \tag{5}
\end{align*}
\begin{align*}
   \int_0^2\int_0^3 x^2y(1+2x+3y)\sqrt{1 + 4 + 9}\ud x\ud y
&= \int_0^2\left(27y^2 + \frac{99}{2}y\right)\sqrt{14}\ud y\\
&= 171\sqrt{14}\tag{9}
\end{align*}
\begin{align*}
  &\int_0^1\int_0^1\left(xy\unit\i + yz\unit\j + zx\unit k\right)
  \cdot\left(\unit\i + 0\unit\j - 2x\unit k\right)
  \times\left(0\unit\i + \unit\j - 2y\unit k\right)\ud y\ud x\\
= &\int_0^1\int_0^1\left(xz + 2y^2z + 2x^2y\right)\ud y\ud x\\
= &\int_0^1\int_0^1((x + 2y^2)(4 - x^2 - y^2) + 2x^2y)\ud y\ud x\\
= &\int_0^1\int_0^1(4x - x^3 - xy^2 + 8y^2 - 2x^2y^2 - 2y^4 + 2x^2y)\ud y\ud x\\
= &\int_0^1\left(4x - x^3 - \frac{x}{3} + \frac{8}{3}
  - \frac{2x^2}{3} - \frac{2}{5} + x^2\right)\ud x\\
= &\,2 - \frac{1}{4} - \frac{1}{6} + \frac{8}{3}
  - \frac{2}{9} - \frac{2}{5} + \frac{1}{3}
= \frac{713}{180}\tag{23}
\end{align*}

\section{Second-Order Linear Equations}
\subsection{Homogeneous Linear Equations}
Solve the differential equation.

\[y'' - y' - 6y = 0\tag{1}\]

The auxiliary equation is $r^2 - r - 6 = 0$ whose roots are $r = -2, 3$.
Therefore, the general solution of the given differential equation is
\[y = \frac{c_1}{e^{2x}} + c_2 e^{3x}\]

\[y'' + 16y = 0\tag{3}\]

The auxiliary equation is $r^2 + 16 = 0$ whose roots are $r = \pm 4i$.
Therefore, the general solution of the given differential equation is
\[y = c_1\cos 4x + c_2\sin 4x\]

\[9y'' - 12y' + 4y = 0\tag{5}\]

The auxiliary equation is $9r^2 - 12r + 4 = 0$
whose roots are $r_1 = r_2 = 2/3$.
Therefore, the general solution of the given differential equation is
\[y = (c_1 + c_2 x)e^{2x/3}\]

\[2y'' = y'\tag{7}\]

The auxiliary equation is $2r^2 = r$ whose roots are $r = 0, 1/2$.
Therefore, the general solution of the given differential equation is
$y = c_1 + c_2\sqrt{e^x}$.

\[y'' - 6y' + 8y = 0,\qquad y(0) = 2,\qquad y'(0) = 2\tag{17}\]

The auxiliary equation is $r^2 - 6r + 8 = 0$ whose roots are $r = 2, 4$.
Therefore, the general solution of the given differential equation is
\[y = c_1 e^{2x} + c_2 e^{4x} \Longrightarrow y' = 2c_1 e^{2x} + 4c_2 e^{4x}\]

Since $y(0) = y'(0) = 2$,
\[c_1 + c_2 = 2c_1 + 4c_2 = 2 \iff (c_1, c_2) = (3, -1)
\iff y = 3e^{2x} - e^{4x}\]

\[9y'' + 12y' + 4y = 0,\qquad y(0) = 1,\qquad y'(0) = 0\tag{19}\]

The auxiliary equation is $9r^2 + 12r + 4 = 0$
whose roots are $r_1 = r_2 = -2/3$.
Therefore, the general solution of the given differential equation is
\[y = \frac{c_1 + c_2 x}{e^{2x/3}}
\Longrightarrow y' = \frac{c_2 - 2c_2 x/3 - 2c_1/3}{e^{2x/3}}\]

As $y(0) = 1$, $c_1 = 1$ and as $y'(0) = 0$, $c_2 = 2/3$, thus
\[y = \left(1 + \frac{2x}{3}\right)e^{-2x/3}\]

\subsection{Nonhomogeneous Linear Equations}
Solve the differential equation.

\[y'' - 2y' - 3y = \cos 2x\tag{1}\]

The auxiliary equation of $y'' - 2y' - 3y = 0$ is $r^2 - 2r - 3 = 0$
with roots $r = -1, 3$. So the solution of the complementary equation is
\[y_c = \frac{c_1}{e^x} + c_2 e^{3x}\]

Since $G(x) = \cos 2x$ is cosine function, we seek a particular solution
of the form $y_p = A\sin 2x + B\cos 2x$. Then $y_p' = 2A\cos 2x - 2B\sin 2x$
and $y_p'' = -4y$ so, substituting into the given differential equation,
we have
\begin{multline*}
  (4A - 7B)\cos 2x - (7A + 4B)\sin 2x = \cos 2x\\
\iff\begin{cases}
  4A - 7B = 1\\
  7A + 4B = 0
\end{cases}
\iff\begin{dcases}
  A = \frac{4}{65}\\
  B = \frac{-7}{65}
\end{dcases}
\end{multline*}

Thus the general solution of the given differential equation is
\[y = y_c + y_p
= \frac{c_1}{e^x} + c_2 e^{3x} + \frac{4\sin 2x}{65} - \frac{7\cos 2x}{65}\]

\[y'' + 9y = \frac{1}{e^{2x}}\tag{3}\]

The auxiliary equation of $y'' + 9y = 0$ is $r^2 + 9 = 0$
whose roots are $r = \pm 3i$.
Therefore, the general solution of the given differential equation is
\[y_c = c_1\cos 3x + c_2\sin 3x\]

Since $G(x) = e^{-2x}$ is an exponential function, we seek
a particular solution of an exponential function as well:
\[y_p = Ae^{-2x}
\Longrightarrow y_p' = -2Ae^{-2x}
\Longrightarrow y_p'' = 4Ae^{-2x}\]

Substituting these into the differential equation, we get
\[\frac{13A}{e^{2x}} = \frac{1}{e^{2x}}
\iff A = \frac{1}{13}
\iff y_p = \frac{1}{13e^{2x}}\]

Thus the general solution of the given differential equation is
\[y = y_c + y_p = c_1\cos 3x + c_2\sin 3x + \frac{1}{13e^{2x}}\]

\[y'' - 4y = e^x\cos x,\qquad y(0) = 1,\qquad y'(0) = 2\tag{8}\]

The auxiliary equation of $y'' + 4y = 0$ is $r^2 + 4 = 0$
whose roots are $r = \pm 2i$.
Therefore, the general solution of the given differential equation is
\[y_c = c_1\cos 2x + c_2\sin 2x\]

We seek a particular solution of the form $y_p = e^x(A\sin x + B\cos x)$.
Substituting this into the given differential equation we get
\begin{multline*}
  2e^x(A\cos x - B\sin x) + 4e^x(A\sin x + B\cos x) = e^x\cos x\\
  \iff\begin{cases}
    2A + 4B = 1\\
    4A - 2B = 0
  \end{cases}
  \iff\begin{cases}
    A = 0.1\\
    B = 0.2
  \end{cases}
\end{multline*}

Thus the general solution of the given differential equation is
\begin{align*}
  y &= y_c + y_p = c_1\cos 2x + c_2\sin 2x + e^x(0.1\sin x + 0.2\cos x)\\
  \Longrightarrow y' &= 2c_2\cos 2x - 2c_1\sin 2x + e^x(0.3\cos x - 0.1\sin x)
\end{align*}

From $y(0) = 1$ we obtain $c_1 = 0.8$ and from $y'(0) = 2$ we have $c_2 = 0.85$.
Thus the solution of the initial-value problem is
\[y = 0.8\cos 2x + 0.85\sin 2x + e^x(0.1\sin x + 0.2\cos x)\]

\[y'' - y' = xe^x,\qquad y(0) = 2,\qquad y'(0) = 1\tag{9}\]

The auxiliary equation of $y'' - y' = 0$ is $r^2 - r = 0$ with roots $r = 0, 1$.
So the solution of the complementary equation is
\[y_c = c_1 + c_2 e^x\]

Base on instinct, we seek a particular solution of the form $y_p = (A + x)e^x$.
Substituting this into the given differential equation we get
\[(2 + A + x)e^x + (1 + A + x)e^x = xe^x
\iff 3 + 2A = 0
\iff A = \frac{-3}{2}\]

Thus the general solution of the given differential equation is
\begin{align*}
y &= y_c + y_p
   = c_1 + c_2 e^x + \left(x - \frac{3}{2}\right)e^x
   = c_1 + (x + C)e^x\\
\Longrightarrow y' &= (x + C + 1)e^x
\end{align*}

From $y'(0) = 1$ we get $C = 0$ and from $y(0) = 2$ we get $c_1 = 2$.
Hence the solution of the initial-value problem is $y = c_1 + (x + C)e^x$.
\end{document}