From 70c413d6c86cb01df2a3a5dd8b2bc8a80c3d4317 Mon Sep 17 00:00:00 2001 From: Nguyễn Gia Phong Date: Tue, 18 Feb 2020 11:43:29 +0700 Subject: [usth] Organize math homework --- usth/MATH1.5/cursived.pdf | Bin 0 -> 304447 bytes usth/MATH1.5/cursived.tex | 1826 ++++++++++++++++++++++++++++++++++++ usth/MATH1.5/homework/cursived.pdf | Bin 304447 -> 0 bytes usth/MATH1.5/homework/cursived.tex | 1826 ------------------------------------ usth/MATH1.5/homework/review.pdf | Bin 237554 -> 0 bytes usth/MATH1.5/homework/review.tex | 818 ---------------- usth/MATH1.5/review.pdf | Bin 0 -> 237554 bytes usth/MATH1.5/review.tex | 818 ++++++++++++++++ usth/MATH2.1/problems.pdf | Bin 0 -> 651358 bytes usth/MATH2.1/solutions.pdf | Bin 0 -> 321054 bytes usth/MATH2.1/solutions.tex | 1633 ++++++++++++++++++++++++++++++++ 11 files changed, 4277 insertions(+), 2644 deletions(-) create mode 100644 usth/MATH1.5/cursived.pdf create mode 100644 usth/MATH1.5/cursived.tex delete mode 100644 usth/MATH1.5/homework/cursived.pdf delete mode 100644 usth/MATH1.5/homework/cursived.tex delete mode 100644 usth/MATH1.5/homework/review.pdf delete mode 100644 usth/MATH1.5/homework/review.tex create mode 100644 usth/MATH1.5/review.pdf create mode 100644 usth/MATH1.5/review.tex create mode 100644 usth/MATH2.1/problems.pdf create mode 100644 usth/MATH2.1/solutions.pdf create mode 100644 usth/MATH2.1/solutions.tex diff --git a/usth/MATH1.5/cursived.pdf b/usth/MATH1.5/cursived.pdf new file mode 100644 index 0000000..c7a2337 Binary files /dev/null and b/usth/MATH1.5/cursived.pdf differ diff --git a/usth/MATH1.5/cursived.tex b/usth/MATH1.5/cursived.tex new file mode 100644 index 0000000..59a3ac7 --- /dev/null +++ b/usth/MATH1.5/cursived.tex @@ -0,0 +1,1826 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{enumerate} +\usepackage{mathtools} +\usepackage{pgfplots} +\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 +\[\begin{cases} + \nabla f = \lambda\nabla\sum_{i=1}^n x_i\\ + \sum_{i=1}^n x_i = c +\end{cases} +\iff +\begin{cases} + \left<\frac{x_1^{1-2/n}}{n}, \ldots, \frac{x_i^{1-2/n}}{n}\right>f + = \lambda\left\\ + \sum_{i=1}^n x_i = c +\end{cases}\] + +\[\Longrightarrow +\begin{cases} + x_1 = x_2 = \ldots = x_n\\ + \sum_{i=1}^n x_i = c +\end{cases} +\iff x_1 = x_2 = \ldots = x_n = \frac{c}{n}\] + +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} diff --git a/usth/MATH1.5/homework/cursived.pdf b/usth/MATH1.5/homework/cursived.pdf deleted file mode 100644 index c7a2337..0000000 Binary files a/usth/MATH1.5/homework/cursived.pdf and /dev/null differ diff --git a/usth/MATH1.5/homework/cursived.tex b/usth/MATH1.5/homework/cursived.tex deleted file mode 100644 index 59a3ac7..0000000 --- a/usth/MATH1.5/homework/cursived.tex +++ /dev/null @@ -1,1826 +0,0 @@ -\documentclass[a4paper,12pt]{article} -\usepackage[utf8]{inputenc} -\usepackage[english,vietnamese]{babel} -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{enumerate} -\usepackage{mathtools} -\usepackage{pgfplots} -\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 -\[\begin{cases} - \nabla f = \lambda\nabla\sum_{i=1}^n x_i\\ - \sum_{i=1}^n x_i = c -\end{cases} -\iff -\begin{cases} - \left<\frac{x_1^{1-2/n}}{n}, \ldots, \frac{x_i^{1-2/n}}{n}\right>f - = \lambda\left\\ - \sum_{i=1}^n x_i = c -\end{cases}\] - -\[\Longrightarrow -\begin{cases} - x_1 = x_2 = \ldots = x_n\\ - \sum_{i=1}^n x_i = c -\end{cases} -\iff x_1 = x_2 = \ldots = x_n = \frac{c}{n}\] - -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} diff --git a/usth/MATH1.5/homework/review.pdf b/usth/MATH1.5/homework/review.pdf deleted file mode 100644 index 10f4b74..0000000 Binary files a/usth/MATH1.5/homework/review.pdf and /dev/null differ diff --git a/usth/MATH1.5/homework/review.tex b/usth/MATH1.5/homework/review.tex deleted file mode 100644 index 55ad0a0..0000000 --- a/usth/MATH1.5/homework/review.tex +++ /dev/null @@ -1,818 +0,0 @@ -\documentclass[a4paper,12pt]{article} -\usepackage[utf8]{inputenc} -\usepackage[english,vietnamese]{babel} -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{enumerate} -\usepackage{mathtools} -\usepackage{multicol} -\usepackage{pgfplots} -\usepackage{siunitx} -\usetikzlibrary{shapes.geometric,angles,quotes} - -\newcommand{\ud}{\,\mathrm{d}} -\newcommand{\curl}{\mathrm{curl}} -\newcommand{\del}{\mathrm{div}} -\newcommand{\unit}[1]{\hat{\textbf #1}} -\newcommand{\anonym}[2]{\left(#1 \mapsto #2\right)} -\newcommand{\tho}[3][]{\dfrac{\partial #1 #2}{\partial #3 #1}} -\newcommand{\leibniz}[3][]{\dfrac{\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 Review} -\author{Nguyễn Gia Phong} -\date{Summer 2019} - -\begin{document} -\maketitle -\setcounter{section}{13} -\section{Partial Derivatives} -\setcounter{subsection}{1} -\subsection{Limits et Continuity} -\exercise{37} Determine the set of points at which the function is continuous. -\[f(x, y) = \begin{dcases} - \frac{x^2 y^3}{2x^2 + y^2} &\text{if }(x, y) \neq (0, 0)\\ - 1 &\text{if }(x, y) = (0, 0) -\end{dcases}\] - -By AM-GM inequality, -\[x^2 + x^2 + y^2 \geq 3x^2|y| -\iff \frac{x^2|y^3|}{3x^2|y|} \geq \frac{x^2|y^3|}{2x^2 + y^2} \geq 0 -\iff \frac{-y^2}{3} \leq \frac{x^2 y^3}{2x^2 + y^2} \leq \frac{y^2}{3}\] - -Since $\pm y^2/3 \to 0$ as $y \to 0$, by the Squeeze Theorem, -\[\lim_{x\to 0\atop y\to 0}f(x, y) = 0 \neq f(0, 0)\] - -Therefore $f$ is discontiuous at (0, 0). On $\mathbb R^2\backslash\{0\}$, -$f$ is a rational function and thus is continuous on its domain. - -\exercise{44} Let -\[f(x, y) = \begin{dcases} - 0 &\text{if }y \leq 0\text{ or }y \geq x^4\\ - 1 &\text{if }0 < y < x^4 -\end{dcases}\] - -\begin{enumerate}[(a)] - \item For all paths of the form $y = mx^a$ with $a < 4 \iff 4 - a > 0$, - consider the function $g(x) = |y| - x^4 = |m|\cdot|x|^a - |x|^4$: - \[g(x) \geq 0 \iff |m|\cdot|x|^a \geq |x|^4 \iff |x| \leq \sqrt[4-a]{|m|}\] - When this condition is met, either $y \leq 0$ or $y = |y| \geq x^4$, - so $f(x, y) = 0$. Therefore $f(x, y) = 0 \to 0$ as $(x, y) \to (0, 0)$ on - \[\left\{(x, y)\,\Big|\, - x \in \left[-\sqrt[4-a]{|m|}, \sqrt[4-a]{|m|}\right] \cap D\right\}\] - which includes the point (0, 0) if the domain $D$ of $x \mapsto mx^a$ does. - \item It is trivial that $f(0, 0) = 0$. Along $y = x^4/2$, for $x \neq 0$, - \[x^4 - y = x^4 - \frac{x^4}{2} = \frac{x^4}{2} > 0 - \iff y < x^4 \Longrightarrow f(x, y) = 1\] - Hence \[\lim_{x\to 0\atop y\to 0} f\left(x, \frac{x^4}{2}\right) = 1 - \neq f(0, 0) = 0\] or $f$ is discontiuous on $y = x^4/2$ at (0, 0). - \item Using the same reasoning, one may also easily show that - $f$ is discontiuous on the entire curve $y = x^4/20$. -\end{enumerate} - -\subsection{Partial Derivatives} -\exercise{33} Find the first partial derivatives of the function. -\begin{align*} - w &= \ln(x + 2y + 3z)\\ - \tho{w}{x} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{x} - = \frac{1}{x + 2y + 3z}\\ - \tho{w}{y} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{y} - = \frac{2}{x + 2y + 3z}\\ - \tho{w}{z} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{z} - = \frac{3}{x + 2y + 3z} -\end{align*} - -\exercise{50} Use implicit differentiation to find -$\partial z/\partial x$ and $\partial z/\partial y$. -\[yz + x\ln y = z^2 -\Longrightarrow \begin{dcases} - y\tho{z}{x} + \ln y &= 2z\tho{z}{x}\\ - z + \frac{x}{y} &= 2z\tho{z}{y}\\ -\end{dcases} -\iff \begin{dcases} - \frac{\ln y}{2z - y} &= \tho{z}{x}\\ - 2 + \frac{x}{2yz} &= \tho{z}{y} -\end{dcases}\] - -\exercise{66} Find $g_{rst}$. -\begin{multline*} - g(r, s, t) = e^r\sin(st) \Longrightarrow g_r = e^r\sin(st)\\ - \Longrightarrow g_{rs} = se^r\cos(st) \Longrightarrow g_{rst} = -ste^r\sin(st) -\end{multline*} - -\exercise{101} Let -\[f(x, y) = \begin{dcases} - \frac{x^3y + xy^3}{x^2 + y^2} &\text{if } (x, y) \neq (0, 0)\\ - 0 &\text{if } (x, y) = (0, 0) -\end{dcases}\] -\begin{enumerate}[(a)] - \item Graph $f$. - - \begin{tikzpicture} - \begin{axis}[xlabel={x}, ylabel={y}] - \addplot3[surf]{(x^3 * y - x * y^3) / (x^2 + y^2)}; - \end{axis} - \end{tikzpicture} - \item Find the first partial derivatives of $f$ when $(x, y) \neq (0, 0)$. - \begin{align*} - \tho{f}{x} &= \frac{(x^2 + y^2)\tho{(x^3y - xy^3)}{x} - - (x^3y - xy^3)\tho{(x^2 + y^2)}{x}}{(x^2 + y^2)^2}\\ - &= \frac{(x^2 + y^2)(3x^2y - y^3) - 2x(x^3y - xy^3)}{(x^2 + y^2)^2}\\ - &= \frac{x^4y + 4x^2y^3 - y^5}{x^4 + 2x^2y^2 + y^4} - \end{align*} - \begin{align*} - \tho{f}{x} - &= \frac{(x^2 + y^2)(x^3 - 3xy^2) - 2y(x^3y - xy^3)}{(x^2 + y^2)^2}\\ - &= \frac{x^5 - 4x^3y^2 - xy^4}{x^4 + 2x^2y^2 + y^4} - \end{align*} - \item Find $f_x$, $f_y$ at (0, 0). - \begin{align*} - f_x(0, 0) &= \lim_{h\to 0}\frac{f(h, 0) - f(0, 0)}{h} - = \lim_{h\to 0}\frac{\frac{h^30 - h0^3}{h^2 + 0^2} - 0}{h} - = \lim_{h\to 0}0 = 0\\ - f_y(0, 0) &= \lim_{h\to 0}\frac{f(0, h) - f(0, 0)}{h} - = \lim_{h\to 0}\frac{0 - 0}{h} - = \lim_{h\to 0}0 = 0 - \end{align*} - \item Show that $f_{xy}(0, 0) = -1$ and $f_{yx}(0, 0) = 1$. - \begin{align*} - f_{xy}(0, 0) &= \lim_{h\to 0}\frac{f_x(0, h) - f_x(0, 0)}{h} - = \lim_{h\to 0}\frac{\frac{0 + 0 - h^5}{0 + 0 + h^4} - 0}{h} - = \lim_{h\to 0}-1 = -1\\ - f_{yx}(0, 0) &= \lim_{h\to 0}\frac{f_y(h, 0) - f_y(0, 0)}{h} - = \lim_{h\to 0}\frac{\frac{h^5 + 0 + 0}{h^4 + 0 + 0} - 0}{h} - = \lim_{h\to 0}1 = 1 - \end{align*} - \item The result of part (d) does not contradict Clairaut's Theorem, - which only covers the case $f_{xy}$ and $f_{yx}$ are continuous at (0, 0). - Using GeoGebra we get the second derivatives of $f$ on - $\mathbb R\backslash\{0\}$ as followed: - \[f_{xy} = f_{yx} = \frac{x^6 + 9x^4y^2 - 9x^2y^4 - y^6}{(x^2 + y^2)^3}\] - Since $f_{xy}(x, 0) = x^6/x^6 \to 1$ while $f_{xy} = -y^6/y^6\to -1$ - as $(x, y) \to (0, 0)$ the second derivative is discontinuous at origin. -\end{enumerate} - -\setcounter{subsection}{5} -\subsection{Directional Derivatives} -\exercise{17} Find the directional derivative of $h(r,s,t) = \ln(3r + 6s + 9t)$ -at (1, 1, 1) in the direction of $\mathbf v = 4\unit\i + 12\unit\j + 6\unit k$. - -From gradient of $h$ -\[\nabla h = \frac{3\unit\i + 6\unit\j + 9\unit k}{3r + 6s + 9t} -\Longrightarrow \nabla h(1, 1, 1) -= \frac{\unit\i}{6} + \frac{\unit\j}{3} + \frac{\unit k}{2}\] -and unit vector of $\mathbf v$ -\[\unit v = \frac{2\unit\i}{7} + \frac{6\unit\j}{7} + \frac{3\unit k}{7}\] -we can compute the direction derivative as -\[\mathrm D_{\unit v}(1, 1, 1) = \nabla h(1, 1, 1)\cdot\unit v -= \frac{1}{21} + \frac{4}{7} + \frac{3}{14} = \frac{23}{42}\] - -\subsection{Maximum and Minimum Values} -\exercise{18} Find the local maximum and minimum values and -saddle point(s) of the function. If you have three-dimensional -graphing software, graph the function with a domain and viewpoint -that reveal all the important aspects of the function. -\[f(x, y) = \sin x\sin y,\qquad -\pi < x < \pi,\qquad -\pi < y < \pi\] - -\begin{multicols}{2} - \begin{align*} - &\Longrightarrow\begin{cases} - f_x = \cos x\sin y\\ - f_y = \sin x\cos y - \end{cases}\\ - &\Longrightarrow\begin{cases} - f_{xx} = f_{yy} = -\sin x\sin y\\ - f_{xy} = f_{yx} = \sin x\sin y - \end{cases}\\ - &\Longrightarrow D = f_{xx}f_{yy} - f_{xy}^2 = 0 - \end{align*} - For $f_x = f_y = 0$, either $x = y = 0$ or $x, y \in \{\pm\pi/2\}$. - $D$ does not indicate if $f$ has local extreme values - at these critical points. - - \noindent\begin{tikzpicture}[domain=-pi:pi] - \begin{axis}[xlabel={x}, ylabel={y}] - \addplot3[surf]{sin(deg(x)) * sin(deg(y))}; - \end{axis} - \end{tikzpicture} -\end{multicols} - -It is clear that $f$ has 2 local maximums of 1 at $x = y = \pm\pi$ -and 2 local minimum of -1 at $x = -y = \pm\pi$, since these are -its absolute extreme values as well. - -Suppose $f(0, 0)$ is a local minimum. Then, by definition, -$f(a, b) \geq f(0, 0) = 0$ if $(a, b)$ is sufficiently close to origin -(say, at most within $[-\pi/2, \pi/2]^2$). However, for all $a$, $b$ -satisfying $ab < 0$, $f(a, b) = \sin a\sin b < 0$, thus our assumption -is incorrect. Similarly, $f$ does not has a local maximum at origin because -\[\forall a, b \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]: ab > 0, -\qquad f(a, b) = \sin a\sin b > 0 = f(0, 0)\] -Therefore (0, 0) is a saddle point. - -\exercise{35} Find the absolute extreme values of $f(x, y) = 2x^3 + y^4$ -on the unit disc. - -The critical points of $f$ occur when -\[f_x = f_y = 0 \iff 6x^2 = 4y^3 = 0 \iff x = y = 0\] -at which $f(x, y) = f(0, 0) = 0$. - -On the unit circle, as $y^2 = 1 - x^2$, let -\[g(x) = f(x, y) = 2x^3 + (1 - x^2)^2 = x^4 + 2x^3 - 2x^2 + 1\] -Within $[-1, 1]$, $g'(x) = 4x^3 + 6x^2 - 4x = 0$ if and only if -$x = 0$ or $x = 0.5$. Since $g(-1) = -2$, $g(0) = 1$, $g(0.5) = 0.8125$ -and $g(1) = 2$, the absolute minimum and maximum of $g$ on $[-1, 1]$ -are respectively $g(-1) = -2$ and $g(1) = 2$. - -Thus on the boundary, the minimum value of $f$ is -2 at $(-1, \pm 1)$ -and the maximum value is 2 at $(1, \pm 1)$. - -\exercise{46} Find the dimensions of the box with volume $1000\text{ cm}^3$ -that has minimal surface area. - -Let the dimensions of the box be $x, y, z$ in dm, $x, y, z$ are positive -and $xyz = 1$. Total surface area of the box would then be -\[S(x, y, z) = 2xy + 2yz + 2zx\] - -By AM-GM inequality, -\[S(x, y, z) \geq 2\cdot 3\sqrt{xy\cdot yz\cdot zx} = 6\] - -Thus $S$ has its absolute minumum of 6 at $x = y = z = 1$. - -\exercise{53} If the length of the diagonal of a rectangular box must be $L$, -what is the largest possible volume? - -Let the dimensions of the box be three positive numbers $x, y, z$, -$x^2 + y^2 + z^2 = L^2$. The volume of the box would then be -$V(x, y, z) = xyz$. By AM-GM inequality, -\[V(x, y, z) = \sqrt{x^2 y^2 z^2} \leq \frac{x^2 + y^2 + z^2}{3} -= \frac{L^2}{3}\] - -Thus $V$ has its absolute maximum of $L^2/3$ at $x = y = z = L/\sqrt 3$. - -\subsection{Lagrange Multipliers} -\exercise{12} Use Lagrange multipliers to find the maximum and minimum values -of $f(x, y, z) = x^4 + y^4 + z^4$ subject to $g(x, y, z) = x^2 + y^2 + z^2 = 1$. - -Extreme values of $f$ occur when -\[\begin{cases} - \nabla f = \lambda\nabla g\\ - g(x, y, z) = 1 -\end{cases} -\iff\begin{cases} - \langle 4x^3, 4y^3, 4z^3\rangle - = \lambda\langle 2x, 2y, 2z\rangle \neq \mathbf 0\\ - x^2 + y^2 + z^2 = 1 -\end{cases}\] - -\begin{enumerate} - \item For $\lambda = 2/3$, $x^2 = y^2 = z^2 = 1/3 = f(x, y, z)$. - \item For $\lambda = 1$ and $(x^2, y^2, z^2) \in \{(0, 1/2, 1/2), - (1/2, 0, 1/2), (1/2, 1/2, 0)\}$, $f(x, y, z) = 1/2$. - \item For $\lambda = 2$ and $(x^2, y^2, z^2) \in \{(1, 0, 0), - (0, 1, 0), (0, 0, 1)\}$, $f(x, y, z) = 1$. -\end{enumerate} - -Therefore, subject to the given constrain, $f$ has absolute maximum of 1 -and minimum of 1/3.\pagebreak - -\exercise{42} Find the maximum and minimum volumes of a rectangular box -whose surface area is $1500\text{ cm}^2$ and whose total edge length is 200 cm. - -Let the dimensions of the box be $x, y, z$ in dm, with $x, y, z$ are positive, -$2xy + 2yz + 2zx = 15$ and $4x + 4y + 4z = 20$. From these constrains, -we can easily obtain $x + y = 5 - z$ and -\[xy + (x + y)z = \frac{15}{2} \iff xy = \frac{15}{2} - 5z + z^2\] - -Thus with $0 < z < 5$ the volume of the box is -\[V = xyz = z^3 - 5z^2 + \frac{15z}{2}\] -whose critical points are -\[\leibniz{V}{z} = 3z^2 - 10z + \frac{15}{2} = 0 -\iff z = \frac{10 \pm \sqrt{10}}{6}\] -at which $V = \dfrac{175 \pm 5\sqrt{10}}{54}$. - -On the other hand, the constrains are equivalent to -\[\begin{cases} - x^2 + y^2 + z^2 = 10\\ - x + y + z = 5 -\end{cases}\] -or the intersection of a sphere and a plane, which result in a circle $C$. -Hence the range of $z$ would be between $a$ and $b$, whereas each of $z = a$ -and $z = b$ only has one point in common with $C$. Since all surfaces -$x^2 + y^2 + z^2 = 10$, $x + y + z = 5$, $z = a$ and $z = b$ has $x = y$ -as their plane of symmetry, these two points must be on $x = y$ as well: -\begin{align*} - \begin{cases} - 2x^2 + z^2 = 10\\ - 2x + z = 5 - \end{cases} - \iff&\begin{cases} - 2x^2 + (5 - 2x)^2 = 10\\ - z = 5 - 2x - \end{cases}\\ - \iff&\begin{cases} - 6x^2 - 20x + 15 = 0\\ - z = 5 - 2x - \end{cases}\\ - \iff&\begin{dcases} - x = \frac{10 \pm \sqrt{10}}{6}\\ - z = \frac{5 \pm \sqrt{10}}{3} - \end{dcases}\\ - \Longrightarrow &\,V = \dfrac{175 \pm 5\sqrt{10}}{54} -\end{align*} -These are the maximum and minimum volumes of the given box. - -\section{Multiple Integrals} -\setcounter{subsection}{1} -\subsection{Interated Integrals} -\exercise{19} Calculate the double integral. -\begin{align*} - \int_0^{\pi/6}\int_0^{\pi/3}x\sin(x + y)\ud y\ud x -&= \int_0^{\pi/6}\left[-x\cos(x + y)\right]_{y=0}^{y=\pi/3}\ud x\\ -&= \int_0^{\pi/6}x\left(\cos x - \cos\left(x + \frac\pi 3\right)\right)\ud x\\ -&= \int_0^{\pi/6}x\cos\left(x - \frac\pi 3\right)\ud x\\ -&= \int_0^{\pi/6}x\ud\cos\left(x - \frac\pi 3\right)\\ -&= \left[x\sin\left(x - \frac\pi 3\right)\right]_0^{\pi/6} - - \int_0^{\pi/6}\sin\left(x - \frac\pi 3\right)\ud x\\ -&= -\frac{\pi}{12} + \left[\cos\left(x - \frac\pi 3\right)\right]_0^{\pi/6}\\ -&= \frac{\sqrt 3}{2} - \frac{1}{2} - \frac{\pi}{12} -\end{align*} - -\exercise{28} Find the volume of the solid enclosed by the surface -$z = 1 + e^x\sin y$ and the planes $x = \pm 1$, $y = 0$, $y = \pi$ and $z = 0$. -\begin{align*} - \int_0^\pi\int_{-1}^1(1 + e^x\sin y)\ud x \ud y -&= \int_0^\pi\left[x + e^x\sin y\right]_{x=-1}^{x=1}\ud y\\ -&= \int_0^\pi\left(2 + \left(e - \frac{1}{e}\right)\sin y\right)\ud y\\ -&= \left[2x + \left(\frac{1}{e} - e\right)\cos y\right]_0^\pi\\ -&= 2\pi -\end{align*} - -\subsection{Double Integrals over General Regions} -\exercise{10} Evaluate the double integral. -\begin{align*} - \int_1^e\int_0^{\ln x}x^3\ud y\ud x -&= \int_1^e x^3\ln x\ud x\\ -&= \int_1^e\ln x\ud\frac{x^4}{4}\\ -&= \left.\frac{x^4\ln x}{4}\right]_1^e - \int_1^e\frac{x^4}{4}\ud\ln x\\ -&= e^4 - \int_1^e\frac{x^3}{4}\ud x\\ -&= e^4 - \left.\frac{x^4}{16}\right]_1^e\\ -&= \frac{15e^4 + 1}{16} -\end{align*} - -\exercise{16} Set up iterated integrals for both orders of integration. -Then evaluate the double integral using the easier order -and explain why it’s easier. -\begin{multline*} - I = \iint_D y^2 e^{xy}\ud A,\qquad D\text{ is bounded by }y = x, y = 4, x = 0\\ - \Longrightarrow I = \int_0^4\int_x^4 y^2 e^{xy}\ud y\ud x - = \int_0^4\int_0^y y^2 e^{xy}\ud x\ud y -\end{multline*} - -Since $y^2 e^{xy}$ is simply an exponential function of $x$, -it would be easier to evaluate -\begin{align*} -I &= \int_0^4\int_0^y y^2 e^{xy}\ud x\ud y\\ - &= \int_0^4\left[y^3 e^{xy}\right]_{x=0}^{x=y}\ud y\\ - &= \int_0^4 y^3 e^{y^2}\ud y - = \int_0^4 y^2\ud\frac{e^{y^2}}{2}\\ - &= \left.\frac{y^2 e^{y^2}}{2}\right]_0^4 - \int_0^4\frac{e^{y^2}}{2}\ud y^2\\ - &= 8e^{16} - \int_0^{16}\frac{e^z}{2}\ud z\\ - &= 8e^{16} - \left.\frac{e^z}{2}\right]_0^{16} - = \frac{15e^{16}}{2} -\end{align*} - -\exercise{31} Find the volume of the solid bounded by the cylinder -$x^2 + y^2 = 1$ and the plane $y = z$ in the first octant. -\[\int_0^1\int_0^{\sqrt{1-x^2}}y\ud y\ud x -= \int_0^1\frac{1 - x^2}{2}\ud x -= \frac{1}{3}\] - -\subsection{Double Integrals in Polar Coordinates} -\exercise{13} Evaluate the given integral by changing to polar coordinates. -\[I = \iint_R\arctan\frac{y}{x}\ud A,\qquad -\text{where }R = \{(x, y)\,|\,1 \leq x^2 + y^2 \leq 4, 0 \leq y \leq x\}\] - -In polar coordinates, -\[R = [1, 2]\times \left[0, \frac\pi 4\right]\] -thus -\begin{align*} -I &= \int_0^{\pi/4}\int_1^2\arctan\frac{r\sin\theta}{r\cos\theta} - r\ud r\ud\theta\\ - &= \int_0^{\pi/4}\int_1^2\arctan\tan\theta r\ud r\ud\theta\\ - &= \int_0^{\pi/4}\int_1^2\theta r\ud r\ud\theta\\ - &= \int_0^{\pi/4}\frac{3\theta}{2}\ud r\ud\theta\\ - &= \frac{3\pi^2}{64} -\end{align*} - - -\begin{multicols*}{2} - \noindent\begin{tikzpicture}[domain=-pi:pi] - \begin{axis}[legend pos=south east, xlabel={$\theta$}, ylabel={$r$}, - axis x line = middle, axis y line = middle, - enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] - \addplot[color=magenta]{1}; - \addplot[color=green]{2 * cos(deg(x))}; - \legend{$r = 1$, $r = 2\cos\theta$} - \end{axis} - \end{tikzpicture} - - \exercise{17} Use a double integral to find the area of the region - inside $C_1: (x - 1)^2 + y^2 = 1$ and outside $C_0: x^2 + y^2 = 1$.\\ - - In polar coordinates $C_1$ has the equation $r = 2\cos\theta$ and - the equation of $C_0$ is $r = 1$. Therefore the given region is within - $1 \leq r \leq 2\cos\theta$, whereas $\theta \in [-\pi, \pi]$. -\end{multicols*} - -Since on $[-\pi, \pi]$, $2\cos\theta \geq 1 \iff -\pi/3 \leq \theta \leq \pi/3$, -the area of the given region is -\begin{align*} - \int_{-\pi/3}^{\pi/3}\int_1^{2\cos\theta}r\ud r\ud\theta - &= \int_{-\pi/3}^{\pi/3}\frac{4\cos^2\theta - 1}{2}\ud\theta\\ - &= \int_{-\pi/3}^{\pi/3}\left(2\cos^2\theta - 1 + \frac{1}{2}\right)\ud\theta\\ - &= \int_{-\pi/3}^{\pi/3}\left(\cos 2\theta + \frac{1}{2}\right)\ud\theta\\ - &= \left[\frac{\sin 2\theta + \theta}{2}\right]_{-\pi/3}^{\pi/3}\\ - &= \frac{\sqrt 3}{2} + \frac\pi 3 -\end{align*} - -\subsection{Applications of Double Integrals} -\exercise{5} Find the mass and center of mass of the lamina that occupies -the region triangular $D$ with vertices (0, 0), (2, 1), (0, 3) -and has the given density function $\rho(x, y) = x + y$. -\begin{align*} -m &= \iint_D(x + y)\ud A\\ - &= \int_0^2\int_{x/2}^{3-x}(x+y)\ud y\ud x\\ - &= \int_0^2\frac{36 - 9x^2}{8}\ud x\\ - &= 9 - 3 = 6 -\end{align*} - -\begin{align*} - \bar x &= \iint_D\frac{x(x + y)}{m}\ud A& - \bar y &= \iint_D\frac{y(x + y)}{m}\ud A\\ - &= \int_0^2\int_{x/2}^{3-x}\frac{x^2 + xy}{6}\ud y\ud x& - &= \int_0^2\int_{x/2}^{3-x}\frac{xy + y^2}{6}\ud y\ud x\\ - &= \int_0^2\frac{12x - 3x^3}{16}\ud x& - &= \int_0^2\frac{6 - 3x}{4}\ud x\\ - &= \frac{3}{4}& - &= \frac{3}{2} -\end{align*} - -\subsection{Surface Area} -\exercise{7} Find the area of the part of -the hyperbolic paraboloid $z = y^2 - x^2$ -that lies between the cylinders $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$. -\begin{align*} - &\iint_D\sqrt{1 + \left(\tho{(y^2 - x^2)}{x}\right)^2 - + \left(\tho{(y^2 - x^2)}{y}\right)^2}\ud A\\ -= &\iint_D\sqrt{1 + 4x^2 + 4y^2}\ud A\\ -= &\int_0^{2\pi}\int_1^2 r\sqrt{1 + 4r^2\cos^2\theta + 4r^2\sin^2\theta} - \ud r\ud\theta\\ -= &\int_1^2\pi\sqrt{1 + 4r^2}\ud r^2\\ -= &\int_1^4\pi\sqrt{1 + 4t}\ud t\\ -= &\,\pi\left[\frac{(1 + 4t)^{1.5}}{6}\right]_1^4\\ -= &\,\frac{17^{1.5} - 5^{1.5}}{6}\pi -\end{align*} - -\section{Vector Calculus} -\setcounter{subsection}{1} -\subsection{Line Integrals} -\exercise{12} Evaluate the integral, where $C$ is the given curve. - -\[I = \int_C(x^2 + y^2 + z^2)\ud s,\qquad -C: x = t, y = \cos 2t, z = \sin 2t, 0 \leq t \leq 2\pi\] -\begin{align*} - I &= \int_0^{2\pi}(x^2 + y^2 + z^2)\sqrt{\left(\leibniz{x}{t}\right)^2 - + \left(\leibniz{z}{t}\right)^2 + \left(\leibniz{z}{t}\right)^2}\ud t\\ - &= \int_0^{2\pi}(t^2 + \cos^2 2t + \sin^2 2t) - \sqrt{\left(\leibniz{t}{t}\right)^2 + \left(\leibniz{\cos 2t}{t}\right)^2 - + \left(\leibniz{\sin 2t}{t}\right)^2}\ud t\\ - &= \int_0^{2\pi}(t^2 + 1)\sqrt 2\ud t = \frac{8\pi\sqrt 2}{3} + 2\pi\sqrt 2 -\end{align*} - -\exercise{15} With $C$ is the line segment from (1, 0, 0) to (4, 1, 2), -$x = 3t + 1$, $y = t$, $z = 2t$, whereas $0 \leq t \leq 1$ and -\begin{align*} - J &= \int_C z^2\ud x + x^2\ud y + y^2\ud z\\ - &= \int_0^1 z^2\leibniz{x}{t}\ud t + x^2\leibniz{y}{t}\ud t - + y^2\leibniz{z}{t}\ud t\\ - &= \int_0^1(x^2 + 2y^2 + 3z^2)\ud t\\ - &= \int_0^1(9t^2 + 6t + 1 + 2t^2 + 12t^2)\ud t\\ - &= \int_0^1(23t^2 + 6t + 1)\ud t\\ - &= \left[\frac{23t^3}{3} + 3t^2 + t\right]_0^1 = \frac{35}{3} -\end{align*} - -\exercise{39} Find the work done by the force field -$\mathbf F(x, y) = \langle x, y + 2\rangle$ is moving an object along an arch -of the cycloid $\mathbf r(t) = \langle t-\sin t, 1-\cos t\rangle$, -$0 \leq t \leq 2\pi$. -\begin{align*} - W &= \int_C\mathbf F\cdot\ud\mathbf r\\ - &= \int_0^{2\pi}\mathbf F\cdot\leibniz{\mathbf r}{t}\ud t\\ - &= \int_0^{2\pi}\langle x, y+2\rangle\cdot - \left<\leibniz{x}{t}, \leibniz{y}{t}\right>\ud t\\ - &= \int_0^{2\pi}\langle t-\sin t, 3-\cos t\rangle\cdot - \left<1-\cos t, \sin t\right>\ud t\\ - &= \int_0^{2\pi}\left(t - t\cos t + 2\sin t\right)\ud t\\ - &= \left[\frac{t^2}{2} - t\sin t - 3\cos t\right]_0^{2\pi} = 2\pi^2 -\end{align*} - -\subsection{The Fundamental Theorem for Line Integral} -\exercise{19} Show that the line integral is independent -from any path $C$ from (1, 0) to (2, 1) and evaluate the integral. -\[\int_C\frac{2x}{e^y}\ud x + \left(2y - \frac{x^2}{e^y}\right)\ud y -= \int_C\left(\frac{2x}{e^y}\unit\i + 2y\unit\j - \frac{x^2}{e^y}\unit\j\right) -\cdot\ud(x\unit\i + y\unit\j)\] - -Since on $\mathbb R^2$ -\[\tho{}{y}\frac{2x}{e^y} = \frac{-2x}{e^y} -= \tho{}{x}\left(2y - \frac{x^2}{e^y}\right)\] -the function -\[\mathbf F(x, y) = \frac{2x}{e^y}\unit\i + 2y\unit\j - \frac{x^2}{e^y}\unit\j\] -is conservative and thus the given line integral is independent from path. - -Let $f$ be a differentiable of $(x, y)$ that $\nabla f = \mathbf F$. -One function satisfying this is \[f(x, y) = y^2 + \frac{x^2}{e^y}\] -By the fundamental theorem for line integrals, -\[\int_C\mathbf F\cdot\ud\mathbf r = f(2, 1) - f(1, 0) = \frac{4}{e}\] - -\subsection{Green's Theorem} -\exercise{6} Use Green’s Theorem to evaluate the line integral along the given -positively oriented rectangle with vertices (0, 0), (5, 0), (5, 2) and (0, 2). -\begin{align*} - \int_C\cos y\ud x + x^2\sin y\ud y -&= \int_0^5\int_0^2\left(\tho{x^2\sin y}{x} - \tho{\cos y}{y}\right)\ud y\ud x\\ -&= \int_0^5\int_0^2(2x\sin y + \sin y)\ud y\ud x\\ -&= \int_0^5(2x + 1)(1 - \cos 2)\ud x\\ -&= 30 - 30\cos 2 -\end{align*} - -\begin{multicols}{2} - \noindent\begin{tikzpicture}[domain=-pi/2:pi/2] - \begin{axis}[legend pos=north east, xlabel={$x$}, ylabel={$y$}, - axis x line = middle, axis y line = middle, - enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] - \addplot[->,>=stealth,color=blue]{cos(deg(x))}; - \addplot[<-,>=stealth,color=orange]{0}; - \legend{$r = \cos x$, $r = 0$} - \end{axis} - \end{tikzpicture} - - \exercise{12} Use Green's Theorem to evaluate the line integral along the - path $C$ including the curve $y = \cos x$ from $(-\pi/2, 0)$ to $(\pi/2, 0)$ - and the line segment connecting these two points. - - Since the curve is negatively oriented, by Green's Theorem, -\end{multicols} -\begin{align*} - &\int_C(e^{-x} + y^2)\ud x + (e^{-y} + x^2)\ud y\\ -= &\,-\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}\left( - \tho{}{x}(e^{-y} + x^2) - \tho{}{y}(e^{-x} + y^2)\right)\ud y\ud x\\ -= &\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 1 2\int_{-\pi/2}^{\pi/2}(\cos 2x + 1)\ud x - - \int_{-\pi/2}^{\pi/2}2x\ud\sin x\\ -= &\left[\frac{\sin 2x}{4} + \frac{x}{2} - - 2x\sin x - 2\cos x\right]_{-\pi/2}^{\pi/2} = \frac\pi 2 -\end{align*} - -\subsection{Curl and Divergence} - -This section is to aid my revision of Electromagnetism. First, on $\mathbb R^3$, -we define -\[\nabla = \unit\i\tho{}{x} + \unit\j\tho{}{y} + \unit k\tho{}{z}\] -then the curl of vector field $\mathbf F = P\unit\i + Q\unit\j + R\unit k$ is -\begin{multline*} - \curl\mathbf F -= \nabla\times\mathbf F -= \begin{vmatrix} - \unit\i & \unit\j & \unit k\\ - \tho{}{x} & \tho{}{y} & \tho{}{z}\\ - P & Q & R -\end{vmatrix}\\ -= \unit\i\left(\tho{R}{y} - \tho{Q}{z}\right) -+ \unit\j\left(\tho{P}{z} - \tho{R}{x}\right) -+ \unit k\left(\tho{Q}{x} - \tho{P}{y}\right) -\end{multline*} - -If $f$ is a function of three variables that has continuous second-order -partial derivatives, then $\curl(\nabla f) = \mathbf 0$. - -On the other hand, if $\curl\mathbf F = \mathbf 0$ -then $\mathbf F$ is a conservative vector field (preconditions: -$P$, $Q$ and $R$ must be partially differentiable). - -Similarly, the divergence of vector field $\mathbf F$ is defined as -\[\del\mathbf{F} = \nabla\cdot\mathbf F -= \unit\i\tho{P}{x} + \unit\j\tho{Q}{y} + \unit k\tho{R}{z}\] - -Trivially, $\nabla\cdot(\nabla\times\mathbf F) = 0$ -because the terms cancel in pairs by Clairaut's Theorem. - -The cool thing about operators is that they can be weirdly combined, -e.g. $\del(\nabla f) = \nabla\cdot\nabla f = \nabla^2 f$ -and $\nabla^2 F = \nabla\cdot\nabla\cdot\mathbf F$. - -Now we are able to write Green's Theorem in the vector form -\[\oint_{\partial S}\mathbf F\cdot\ud\mathbf r -= \iint_S(\curl\mathbf F)\cdot\unit k\ud A\] -whereas $\mathbf r(t) = x(t)\unit\i + y(t)\unit\j$. -The outward normal vector to the contour is given by -$\mathbf n(t) = \leibniz{y}{t}\unit\i - \leibniz{x}{t}\unit\j$. -So we have the second vector form of Green's Theorem. -\[\oint_{\partial S}\mathbf F\cdot\unit n\ud s = \iint_S\del\mathbf F\ud A\] - -\subsection{Parametric Surfaces and Their Areas} -\exercise{42} Find the area of the part of the cone $z = \sqrt{x^2 + y^2}$ -that lies between the plane $y = x$ and the cylinder $y = x^2$. -\begin{align*} - &\int_0^1\int_{x^2}^x\sqrt{ - 1 + \left(\tho{z}{x}\right)^2 + \left(\tho{z}{y}\right)_2}\ud y\ud x\\ -=&\int_0^1\int_{x^2}^x\sqrt 2\ud y\ud x -= \int_0^1(x - x^2)\sqrt 2\ud y\ud x\\ -=&\,\frac{1}{2} - \frac{1}{3} -= \frac{1}{6} -\end{align*} - -\section{Second-Order Differential Equations} -\subsection{Homogeneous Linear Equations} -\exercise{11} Solve the differential equation. -\[2\leibniz{^2 y}{t^2} + 2\leibniz{y}{t} - y = 0\] - -Since the auxiliary equation $2r^2 + 2x - 1 = 0$ has two real and distinct -roots $\dfrac{\pm\sqrt 3 - 1}{2}$, the general solution is -\[y = c_1\exp\frac{\sqrt 3 - 1}{2}t + c_2\exp\frac{-\sqrt 3 - 1}{2}t\] - -\exercise{21} Solve the initial value problem. -\[y'' - 6y' + 10y = 0,\qquad y(0) = 2,\qquad y''(0) = 3\] - -Since the auxiliary equation $r^2 - 6x + 10 = 0$ has two complex roots -$3\pm i$, the general solution is -\[y = e^{3x}(c_1\cos x + c_2\sin x) -\Longrightarrow y' = e^{3x}((3c_1 + c_2)\cos x + (3c_2 - c_1)\sin x)\] - -As $y(0) = 2$, $c_1 = 2$. Similarly, from $y'(0) = 3$, -we can obtain $3c_1 - c_2 = 3 \Longrightarrow c_2 = 3$. Thus the solution -of the initial value problem is $y = e^{3x}(3\cos x + 2\sin x)$. - -\subsection{Nonhomogeneous Linear Equations} -Solve the differential equation or initial-value problem using the method of -undetermined coefficients. - -\[y'' - 4y' + 5y = e^{-x}\tag{5}\] - -The auxiliary equation of $y'' - 4y' + 5y = 0$ is $r^2 - 4r + 5 = 0$ with roots -$r = 2\pm i$. Hence the solution to the complementary equation is -\[y_c = e^{2x}(c_1\cos x + c_2\sin x)\] - -Since $G(x) = e^{-x}$ is an exponential function, we seek a particular solution -of an exponential function as well: -\[y_p = Ae^{-x} -\Longrightarrow y_p' = -Ae^{-x} -\Longrightarrow y_p'' = Ae^{-x}\] - -Substituting these into the differential equation, we get -\[Ae^{-x} - 4Ae^{-x} + 5Ae^{-x} = e^{-x} \iff A = \frac{1}{10}\] - -Thus the general solution of the exponential equation is -\[y = y_c + y_p = e^{2x}(c_1\cos x + c_2\sin x) + \frac{1}{10e^x}\] - -\[y'' + y' - 2y = x + \sin 2x,\qquad y(0) = 1,\qquad y'(0) = 0\tag{10}\] - -The auxiliary equation of $y'' + y' - 2y = 0$ is $r^2 + r - 2 = 0$ with roots -$r = -2, 1$. Thus the solution to the complementary equation is -\[y_c = c_1 e^x + \frac{c_2}{e^{2x}}\] - -We seek a particular solution of the form -\begin{multline*} - y_p = Ax + B + C\cos 2x + D\sin 2x\\ - \Longrightarrow y_p' = A - 2C\sin 2x + 2D\cos 2x\\ - \Longrightarrow y_p'' = -4C\cos 2x - 4D\sin 2x -\end{multline*} - -Substituting these into the differential equation, we get -\begin{multline*} - (-4C + 2D - 2C)\cos 2x + (-4D - 2C - 2D)\sin 2x + A - 2B - 2Ax = x + \sin 2x\\ - \iff\begin{cases} - -4C + 2D - 2C = 0\\ - -4D - 2C - 2D = 1\\ - A - 2B = 0\\ - -2A = 1 - \end{cases} - \iff\begin{cases} - A = -1/2\\ - B = -1/4\\ - C = -1/20\\ - D = -3/20 - \end{cases} -\end{multline*} - -Thus the general solution of the exponential equation is -\begin{multline*} - y = y_c + y_p = c_1 e^x + \frac{c_2}{e^{2x}} - \frac{x}{2} - \frac{1}{4} - - \frac{\cos 2x}{20} - \frac{3\sin 2x}{20}\\ - \Longrightarrow y' = c_1 e^x - \frac{2c_2}{e^{2x}} - \frac{1}{2} - + \frac{\sin 2x}{10} - \frac{3\cos 2x}{10} -\end{multline*} - -Since $y(0) = 1$ and $y'(0) = 0$, -\[\begin{dcases} - c_1 + c_2 - \frac{1}{4} - \frac{1}{20} = 1\\ - c_1 - 2c_2 - \frac{3}{10} = 0 -\end{dcases} -\iff \begin{dcases} - c_1 + c_2 = \frac{13}{10}\\ - c_1 - 2c_2 = \frac{3}{10} -\end{dcases} -\iff \begin{dcases} - c_1 = \frac{29}{30}\\ - c_2 = \frac{1}{3} -\end{dcases}\] - -Therefore the solution to the initial value problem is -\[y = \frac{29e^x}{30} + \frac{1}{3e^{2x}} - \frac{x}{2} - \frac{1}{4} - - \frac{\cos 2x}{20} - \frac{3\sin 2x}{20}\] - -\subsection{Applications} -\exercise{3} A spring with a mass of 2 kg has damping constant 14, and a force -of 6 N is required to keep the spring stretched 0.5 m beyond its natural -length. The spring is stretched 1 m beyond its natural length and then -released with zero velocity. Find the position of the mass at any time $t$. - -By Hooke's law, -\[k(0.5) = 6 \iff k = 12\] - -By Newton's second law of motion, -\[2\leibniz{^2 x}{t^2} + 14\leibniz{x}{t} + 12x = 0\] - -Since the auxiliary equation $2r^2 + 14r + 12 = 0$ has two real -and distinct roots $r = -6, -1$, the general solution is -\[x = \frac{c_1}{e^t} + \frac{c_2}{e^{6t}} -\Longrightarrow \leibniz{x}{t} = \frac{-c_1}{e^t} - \frac{6c_2}{e^{6t}}\] - -From $x(0) = 1$ and $x'(0) = 0$ we get -\[\begin{cases} - c_1 + c_2 = 1\\ - -c_1 - 6c_2 = 0 -\end{cases} -\iff\begin{cases} - c_1 = 6/5\\ - c_2 = -1/5 -\end{cases}\] - -Therefore the position at any time $t$ is -\[x = \frac{6}{5e^t} - \frac{c_2}{5e^{6t}}\] - -\setcounter{section}{8} -\section{First-Order Differential Equations} -\setcounter{subsection}{2} -\subsection{Separable Equations} -\exercise{8} Solve the differential equation. -\begin{align*} - \leibniz{y}{\theta} &= \frac{e^y\sin^2\theta}{y\sec\theta}\\ - \iff \int\frac{y}{e^y}\ud y &= \int\sin\theta\cos\theta\ud\theta\\ - \iff \int-y\ud e^{-y} &= \int\sin^2\theta\ud\sin\theta\\ - \iff \int e^{-y}\ud y - \frac{y}{e^y} &= \frac{\sin^3\theta}{3}\\ - \iff \frac{1 + y}{e^y} &= C - \frac{\sin^3\theta}{3} -\end{align*} -\setcounter{subsection}{4} -\subsection{Linear Equations} -\exercise{28} In a damped RL circuit, the generator supplies a voltage of -$E(t) = 40\sin 60t$ volts, the inductance is 1 H, the resistance is 10 $\Omega$ -and $I(0) = 1$ A. -\begin{align*} - &E - L\leibniz{I}{t} - RI = 0\\ - \iff &\frac{40}{L}\sin 60t = \leibniz{I}{t} + \frac{RI}{L}\\ - \iff &\frac{40e^{tR/L}}{L}\sin 60t - = \frac{RI}{L}e^{tR/L} + \leibniz{I}{t}e^{tR/L}\\ - \iff &\frac{40}{L}\int e^{tR/L}\sin 60t\ud t = \int\ud Ie^{tR/L}\tag{$*$} -\end{align*} - -Let $J = \int e^{tR/L}\sin 60t\ud t$, -\begin{align*} -J &= \frac{-1}{60}\int e^{tR/L}\ud\cos 60t\\ - &= \frac{1}{60}\int\cos 60t\ud e^{tR/L} - \frac{e^{tR/L}\cos 60t}{60}\\ - &= \frac{R}{3600L}\int e^{tR/L}\ud\sin 60t - \frac{e^{tR/L}\cos 60t}{60}\\ - &= \frac{R}{3600L}e^{tR/L}\sin 60t - \frac{R}{3600L}\int\sin 60t\ud e^{tR/L} - - \frac{e^{tR/L}\cos 60t}{60}\\ - &= \frac{R}{3600L}e^{tR/L}\sin 60t - \frac{R^2}{3600L^2}J - - \frac{e^{tR/L}\cos 60t}{60} -\end{align*} - -Hence $J = \dfrac{e^{tR/L}(RL\sin 60t - 60L^2\cos 60t)}{R^2+3600L^2}$ -and $(*)$ is equivalent to -\begin{multline*} - \frac{40e^{tR/L}(R\sin 60t - 60L\cos 60t)}{R^2+3600L^2} = Ie^{tR/L} - C\\ - \iff I = \frac{40R\sin 60t - 2400L\cos 60t}{R^2+3600L^2} - + \frac{C}{e^{tR/L}}\\ - \iff I = \frac{\sin 60t - 3\cos 60t}{5} - + \frac{C}{e^{t/20}} -\end{multline*} - -Since $I = 1$ at $t = 0$, -\[1 = \frac{\sin 0 - 3\cos 0}{5} + \frac{C}{e^0} \iff C = \frac 8 5\] -and thus $I = \dfrac{\sin 60t - 3\cos 60t}{5} + \dfrac{8}{5}\exp\dfrac{-t}{20}$. - -At $t = 0.1$, $I = (\sin 6 - 3\cos 6)/5 + 1.6e^{-1/200} \approx 2.11$ A. -\end{document} diff --git a/usth/MATH1.5/review.pdf b/usth/MATH1.5/review.pdf new file mode 100644 index 0000000..10f4b74 Binary files /dev/null and b/usth/MATH1.5/review.pdf differ diff --git a/usth/MATH1.5/review.tex b/usth/MATH1.5/review.tex new file mode 100644 index 0000000..55ad0a0 --- /dev/null +++ b/usth/MATH1.5/review.tex @@ -0,0 +1,818 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{enumerate} +\usepackage{mathtools} +\usepackage{multicol} +\usepackage{pgfplots} +\usepackage{siunitx} +\usetikzlibrary{shapes.geometric,angles,quotes} + +\newcommand{\ud}{\,\mathrm{d}} +\newcommand{\curl}{\mathrm{curl}} +\newcommand{\del}{\mathrm{div}} +\newcommand{\unit}[1]{\hat{\textbf #1}} +\newcommand{\anonym}[2]{\left(#1 \mapsto #2\right)} +\newcommand{\tho}[3][]{\dfrac{\partial #1 #2}{\partial #3 #1}} +\newcommand{\leibniz}[3][]{\dfrac{\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 Review} +\author{Nguyễn Gia Phong} +\date{Summer 2019} + +\begin{document} +\maketitle +\setcounter{section}{13} +\section{Partial Derivatives} +\setcounter{subsection}{1} +\subsection{Limits et Continuity} +\exercise{37} Determine the set of points at which the function is continuous. +\[f(x, y) = \begin{dcases} + \frac{x^2 y^3}{2x^2 + y^2} &\text{if }(x, y) \neq (0, 0)\\ + 1 &\text{if }(x, y) = (0, 0) +\end{dcases}\] + +By AM-GM inequality, +\[x^2 + x^2 + y^2 \geq 3x^2|y| +\iff \frac{x^2|y^3|}{3x^2|y|} \geq \frac{x^2|y^3|}{2x^2 + y^2} \geq 0 +\iff \frac{-y^2}{3} \leq \frac{x^2 y^3}{2x^2 + y^2} \leq \frac{y^2}{3}\] + +Since $\pm y^2/3 \to 0$ as $y \to 0$, by the Squeeze Theorem, +\[\lim_{x\to 0\atop y\to 0}f(x, y) = 0 \neq f(0, 0)\] + +Therefore $f$ is discontiuous at (0, 0). On $\mathbb R^2\backslash\{0\}$, +$f$ is a rational function and thus is continuous on its domain. + +\exercise{44} Let +\[f(x, y) = \begin{dcases} + 0 &\text{if }y \leq 0\text{ or }y \geq x^4\\ + 1 &\text{if }0 < y < x^4 +\end{dcases}\] + +\begin{enumerate}[(a)] + \item For all paths of the form $y = mx^a$ with $a < 4 \iff 4 - a > 0$, + consider the function $g(x) = |y| - x^4 = |m|\cdot|x|^a - |x|^4$: + \[g(x) \geq 0 \iff |m|\cdot|x|^a \geq |x|^4 \iff |x| \leq \sqrt[4-a]{|m|}\] + When this condition is met, either $y \leq 0$ or $y = |y| \geq x^4$, + so $f(x, y) = 0$. Therefore $f(x, y) = 0 \to 0$ as $(x, y) \to (0, 0)$ on + \[\left\{(x, y)\,\Big|\, + x \in \left[-\sqrt[4-a]{|m|}, \sqrt[4-a]{|m|}\right] \cap D\right\}\] + which includes the point (0, 0) if the domain $D$ of $x \mapsto mx^a$ does. + \item It is trivial that $f(0, 0) = 0$. Along $y = x^4/2$, for $x \neq 0$, + \[x^4 - y = x^4 - \frac{x^4}{2} = \frac{x^4}{2} > 0 + \iff y < x^4 \Longrightarrow f(x, y) = 1\] + Hence \[\lim_{x\to 0\atop y\to 0} f\left(x, \frac{x^4}{2}\right) = 1 + \neq f(0, 0) = 0\] or $f$ is discontiuous on $y = x^4/2$ at (0, 0). + \item Using the same reasoning, one may also easily show that + $f$ is discontiuous on the entire curve $y = x^4/20$. +\end{enumerate} + +\subsection{Partial Derivatives} +\exercise{33} Find the first partial derivatives of the function. +\begin{align*} + w &= \ln(x + 2y + 3z)\\ + \tho{w}{x} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{x} + = \frac{1}{x + 2y + 3z}\\ + \tho{w}{y} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{y} + = \frac{2}{x + 2y + 3z}\\ + \tho{w}{z} &= \frac{1}{x + 2y + 3z}\cdot\tho{(x + 2y + 3z)}{z} + = \frac{3}{x + 2y + 3z} +\end{align*} + +\exercise{50} Use implicit differentiation to find +$\partial z/\partial x$ and $\partial z/\partial y$. +\[yz + x\ln y = z^2 +\Longrightarrow \begin{dcases} + y\tho{z}{x} + \ln y &= 2z\tho{z}{x}\\ + z + \frac{x}{y} &= 2z\tho{z}{y}\\ +\end{dcases} +\iff \begin{dcases} + \frac{\ln y}{2z - y} &= \tho{z}{x}\\ + 2 + \frac{x}{2yz} &= \tho{z}{y} +\end{dcases}\] + +\exercise{66} Find $g_{rst}$. +\begin{multline*} + g(r, s, t) = e^r\sin(st) \Longrightarrow g_r = e^r\sin(st)\\ + \Longrightarrow g_{rs} = se^r\cos(st) \Longrightarrow g_{rst} = -ste^r\sin(st) +\end{multline*} + +\exercise{101} Let +\[f(x, y) = \begin{dcases} + \frac{x^3y + xy^3}{x^2 + y^2} &\text{if } (x, y) \neq (0, 0)\\ + 0 &\text{if } (x, y) = (0, 0) +\end{dcases}\] +\begin{enumerate}[(a)] + \item Graph $f$. + + \begin{tikzpicture} + \begin{axis}[xlabel={x}, ylabel={y}] + \addplot3[surf]{(x^3 * y - x * y^3) / (x^2 + y^2)}; + \end{axis} + \end{tikzpicture} + \item Find the first partial derivatives of $f$ when $(x, y) \neq (0, 0)$. + \begin{align*} + \tho{f}{x} &= \frac{(x^2 + y^2)\tho{(x^3y - xy^3)}{x} + - (x^3y - xy^3)\tho{(x^2 + y^2)}{x}}{(x^2 + y^2)^2}\\ + &= \frac{(x^2 + y^2)(3x^2y - y^3) - 2x(x^3y - xy^3)}{(x^2 + y^2)^2}\\ + &= \frac{x^4y + 4x^2y^3 - y^5}{x^4 + 2x^2y^2 + y^4} + \end{align*} + \begin{align*} + \tho{f}{x} + &= \frac{(x^2 + y^2)(x^3 - 3xy^2) - 2y(x^3y - xy^3)}{(x^2 + y^2)^2}\\ + &= \frac{x^5 - 4x^3y^2 - xy^4}{x^4 + 2x^2y^2 + y^4} + \end{align*} + \item Find $f_x$, $f_y$ at (0, 0). + \begin{align*} + f_x(0, 0) &= \lim_{h\to 0}\frac{f(h, 0) - f(0, 0)}{h} + = \lim_{h\to 0}\frac{\frac{h^30 - h0^3}{h^2 + 0^2} - 0}{h} + = \lim_{h\to 0}0 = 0\\ + f_y(0, 0) &= \lim_{h\to 0}\frac{f(0, h) - f(0, 0)}{h} + = \lim_{h\to 0}\frac{0 - 0}{h} + = \lim_{h\to 0}0 = 0 + \end{align*} + \item Show that $f_{xy}(0, 0) = -1$ and $f_{yx}(0, 0) = 1$. + \begin{align*} + f_{xy}(0, 0) &= \lim_{h\to 0}\frac{f_x(0, h) - f_x(0, 0)}{h} + = \lim_{h\to 0}\frac{\frac{0 + 0 - h^5}{0 + 0 + h^4} - 0}{h} + = \lim_{h\to 0}-1 = -1\\ + f_{yx}(0, 0) &= \lim_{h\to 0}\frac{f_y(h, 0) - f_y(0, 0)}{h} + = \lim_{h\to 0}\frac{\frac{h^5 + 0 + 0}{h^4 + 0 + 0} - 0}{h} + = \lim_{h\to 0}1 = 1 + \end{align*} + \item The result of part (d) does not contradict Clairaut's Theorem, + which only covers the case $f_{xy}$ and $f_{yx}$ are continuous at (0, 0). + Using GeoGebra we get the second derivatives of $f$ on + $\mathbb R\backslash\{0\}$ as followed: + \[f_{xy} = f_{yx} = \frac{x^6 + 9x^4y^2 - 9x^2y^4 - y^6}{(x^2 + y^2)^3}\] + Since $f_{xy}(x, 0) = x^6/x^6 \to 1$ while $f_{xy} = -y^6/y^6\to -1$ + as $(x, y) \to (0, 0)$ the second derivative is discontinuous at origin. +\end{enumerate} + +\setcounter{subsection}{5} +\subsection{Directional Derivatives} +\exercise{17} Find the directional derivative of $h(r,s,t) = \ln(3r + 6s + 9t)$ +at (1, 1, 1) in the direction of $\mathbf v = 4\unit\i + 12\unit\j + 6\unit k$. + +From gradient of $h$ +\[\nabla h = \frac{3\unit\i + 6\unit\j + 9\unit k}{3r + 6s + 9t} +\Longrightarrow \nabla h(1, 1, 1) += \frac{\unit\i}{6} + \frac{\unit\j}{3} + \frac{\unit k}{2}\] +and unit vector of $\mathbf v$ +\[\unit v = \frac{2\unit\i}{7} + \frac{6\unit\j}{7} + \frac{3\unit k}{7}\] +we can compute the direction derivative as +\[\mathrm D_{\unit v}(1, 1, 1) = \nabla h(1, 1, 1)\cdot\unit v += \frac{1}{21} + \frac{4}{7} + \frac{3}{14} = \frac{23}{42}\] + +\subsection{Maximum and Minimum Values} +\exercise{18} Find the local maximum and minimum values and +saddle point(s) of the function. If you have three-dimensional +graphing software, graph the function with a domain and viewpoint +that reveal all the important aspects of the function. +\[f(x, y) = \sin x\sin y,\qquad -\pi < x < \pi,\qquad -\pi < y < \pi\] + +\begin{multicols}{2} + \begin{align*} + &\Longrightarrow\begin{cases} + f_x = \cos x\sin y\\ + f_y = \sin x\cos y + \end{cases}\\ + &\Longrightarrow\begin{cases} + f_{xx} = f_{yy} = -\sin x\sin y\\ + f_{xy} = f_{yx} = \sin x\sin y + \end{cases}\\ + &\Longrightarrow D = f_{xx}f_{yy} - f_{xy}^2 = 0 + \end{align*} + For $f_x = f_y = 0$, either $x = y = 0$ or $x, y \in \{\pm\pi/2\}$. + $D$ does not indicate if $f$ has local extreme values + at these critical points. + + \noindent\begin{tikzpicture}[domain=-pi:pi] + \begin{axis}[xlabel={x}, ylabel={y}] + \addplot3[surf]{sin(deg(x)) * sin(deg(y))}; + \end{axis} + \end{tikzpicture} +\end{multicols} + +It is clear that $f$ has 2 local maximums of 1 at $x = y = \pm\pi$ +and 2 local minimum of -1 at $x = -y = \pm\pi$, since these are +its absolute extreme values as well. + +Suppose $f(0, 0)$ is a local minimum. Then, by definition, +$f(a, b) \geq f(0, 0) = 0$ if $(a, b)$ is sufficiently close to origin +(say, at most within $[-\pi/2, \pi/2]^2$). However, for all $a$, $b$ +satisfying $ab < 0$, $f(a, b) = \sin a\sin b < 0$, thus our assumption +is incorrect. Similarly, $f$ does not has a local maximum at origin because +\[\forall a, b \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]: ab > 0, +\qquad f(a, b) = \sin a\sin b > 0 = f(0, 0)\] +Therefore (0, 0) is a saddle point. + +\exercise{35} Find the absolute extreme values of $f(x, y) = 2x^3 + y^4$ +on the unit disc. + +The critical points of $f$ occur when +\[f_x = f_y = 0 \iff 6x^2 = 4y^3 = 0 \iff x = y = 0\] +at which $f(x, y) = f(0, 0) = 0$. + +On the unit circle, as $y^2 = 1 - x^2$, let +\[g(x) = f(x, y) = 2x^3 + (1 - x^2)^2 = x^4 + 2x^3 - 2x^2 + 1\] +Within $[-1, 1]$, $g'(x) = 4x^3 + 6x^2 - 4x = 0$ if and only if +$x = 0$ or $x = 0.5$. Since $g(-1) = -2$, $g(0) = 1$, $g(0.5) = 0.8125$ +and $g(1) = 2$, the absolute minimum and maximum of $g$ on $[-1, 1]$ +are respectively $g(-1) = -2$ and $g(1) = 2$. + +Thus on the boundary, the minimum value of $f$ is -2 at $(-1, \pm 1)$ +and the maximum value is 2 at $(1, \pm 1)$. + +\exercise{46} Find the dimensions of the box with volume $1000\text{ cm}^3$ +that has minimal surface area. + +Let the dimensions of the box be $x, y, z$ in dm, $x, y, z$ are positive +and $xyz = 1$. Total surface area of the box would then be +\[S(x, y, z) = 2xy + 2yz + 2zx\] + +By AM-GM inequality, +\[S(x, y, z) \geq 2\cdot 3\sqrt{xy\cdot yz\cdot zx} = 6\] + +Thus $S$ has its absolute minumum of 6 at $x = y = z = 1$. + +\exercise{53} If the length of the diagonal of a rectangular box must be $L$, +what is the largest possible volume? + +Let the dimensions of the box be three positive numbers $x, y, z$, +$x^2 + y^2 + z^2 = L^2$. The volume of the box would then be +$V(x, y, z) = xyz$. By AM-GM inequality, +\[V(x, y, z) = \sqrt{x^2 y^2 z^2} \leq \frac{x^2 + y^2 + z^2}{3} += \frac{L^2}{3}\] + +Thus $V$ has its absolute maximum of $L^2/3$ at $x = y = z = L/\sqrt 3$. + +\subsection{Lagrange Multipliers} +\exercise{12} Use Lagrange multipliers to find the maximum and minimum values +of $f(x, y, z) = x^4 + y^4 + z^4$ subject to $g(x, y, z) = x^2 + y^2 + z^2 = 1$. + +Extreme values of $f$ occur when +\[\begin{cases} + \nabla f = \lambda\nabla g\\ + g(x, y, z) = 1 +\end{cases} +\iff\begin{cases} + \langle 4x^3, 4y^3, 4z^3\rangle + = \lambda\langle 2x, 2y, 2z\rangle \neq \mathbf 0\\ + x^2 + y^2 + z^2 = 1 +\end{cases}\] + +\begin{enumerate} + \item For $\lambda = 2/3$, $x^2 = y^2 = z^2 = 1/3 = f(x, y, z)$. + \item For $\lambda = 1$ and $(x^2, y^2, z^2) \in \{(0, 1/2, 1/2), + (1/2, 0, 1/2), (1/2, 1/2, 0)\}$, $f(x, y, z) = 1/2$. + \item For $\lambda = 2$ and $(x^2, y^2, z^2) \in \{(1, 0, 0), + (0, 1, 0), (0, 0, 1)\}$, $f(x, y, z) = 1$. +\end{enumerate} + +Therefore, subject to the given constrain, $f$ has absolute maximum of 1 +and minimum of 1/3.\pagebreak + +\exercise{42} Find the maximum and minimum volumes of a rectangular box +whose surface area is $1500\text{ cm}^2$ and whose total edge length is 200 cm. + +Let the dimensions of the box be $x, y, z$ in dm, with $x, y, z$ are positive, +$2xy + 2yz + 2zx = 15$ and $4x + 4y + 4z = 20$. From these constrains, +we can easily obtain $x + y = 5 - z$ and +\[xy + (x + y)z = \frac{15}{2} \iff xy = \frac{15}{2} - 5z + z^2\] + +Thus with $0 < z < 5$ the volume of the box is +\[V = xyz = z^3 - 5z^2 + \frac{15z}{2}\] +whose critical points are +\[\leibniz{V}{z} = 3z^2 - 10z + \frac{15}{2} = 0 +\iff z = \frac{10 \pm \sqrt{10}}{6}\] +at which $V = \dfrac{175 \pm 5\sqrt{10}}{54}$. + +On the other hand, the constrains are equivalent to +\[\begin{cases} + x^2 + y^2 + z^2 = 10\\ + x + y + z = 5 +\end{cases}\] +or the intersection of a sphere and a plane, which result in a circle $C$. +Hence the range of $z$ would be between $a$ and $b$, whereas each of $z = a$ +and $z = b$ only has one point in common with $C$. Since all surfaces +$x^2 + y^2 + z^2 = 10$, $x + y + z = 5$, $z = a$ and $z = b$ has $x = y$ +as their plane of symmetry, these two points must be on $x = y$ as well: +\begin{align*} + \begin{cases} + 2x^2 + z^2 = 10\\ + 2x + z = 5 + \end{cases} + \iff&\begin{cases} + 2x^2 + (5 - 2x)^2 = 10\\ + z = 5 - 2x + \end{cases}\\ + \iff&\begin{cases} + 6x^2 - 20x + 15 = 0\\ + z = 5 - 2x + \end{cases}\\ + \iff&\begin{dcases} + x = \frac{10 \pm \sqrt{10}}{6}\\ + z = \frac{5 \pm \sqrt{10}}{3} + \end{dcases}\\ + \Longrightarrow &\,V = \dfrac{175 \pm 5\sqrt{10}}{54} +\end{align*} +These are the maximum and minimum volumes of the given box. + +\section{Multiple Integrals} +\setcounter{subsection}{1} +\subsection{Interated Integrals} +\exercise{19} Calculate the double integral. +\begin{align*} + \int_0^{\pi/6}\int_0^{\pi/3}x\sin(x + y)\ud y\ud x +&= \int_0^{\pi/6}\left[-x\cos(x + y)\right]_{y=0}^{y=\pi/3}\ud x\\ +&= \int_0^{\pi/6}x\left(\cos x - \cos\left(x + \frac\pi 3\right)\right)\ud x\\ +&= \int_0^{\pi/6}x\cos\left(x - \frac\pi 3\right)\ud x\\ +&= \int_0^{\pi/6}x\ud\cos\left(x - \frac\pi 3\right)\\ +&= \left[x\sin\left(x - \frac\pi 3\right)\right]_0^{\pi/6} + - \int_0^{\pi/6}\sin\left(x - \frac\pi 3\right)\ud x\\ +&= -\frac{\pi}{12} + \left[\cos\left(x - \frac\pi 3\right)\right]_0^{\pi/6}\\ +&= \frac{\sqrt 3}{2} - \frac{1}{2} - \frac{\pi}{12} +\end{align*} + +\exercise{28} Find the volume of the solid enclosed by the surface +$z = 1 + e^x\sin y$ and the planes $x = \pm 1$, $y = 0$, $y = \pi$ and $z = 0$. +\begin{align*} + \int_0^\pi\int_{-1}^1(1 + e^x\sin y)\ud x \ud y +&= \int_0^\pi\left[x + e^x\sin y\right]_{x=-1}^{x=1}\ud y\\ +&= \int_0^\pi\left(2 + \left(e - \frac{1}{e}\right)\sin y\right)\ud y\\ +&= \left[2x + \left(\frac{1}{e} - e\right)\cos y\right]_0^\pi\\ +&= 2\pi +\end{align*} + +\subsection{Double Integrals over General Regions} +\exercise{10} Evaluate the double integral. +\begin{align*} + \int_1^e\int_0^{\ln x}x^3\ud y\ud x +&= \int_1^e x^3\ln x\ud x\\ +&= \int_1^e\ln x\ud\frac{x^4}{4}\\ +&= \left.\frac{x^4\ln x}{4}\right]_1^e - \int_1^e\frac{x^4}{4}\ud\ln x\\ +&= e^4 - \int_1^e\frac{x^3}{4}\ud x\\ +&= e^4 - \left.\frac{x^4}{16}\right]_1^e\\ +&= \frac{15e^4 + 1}{16} +\end{align*} + +\exercise{16} Set up iterated integrals for both orders of integration. +Then evaluate the double integral using the easier order +and explain why it’s easier. +\begin{multline*} + I = \iint_D y^2 e^{xy}\ud A,\qquad D\text{ is bounded by }y = x, y = 4, x = 0\\ + \Longrightarrow I = \int_0^4\int_x^4 y^2 e^{xy}\ud y\ud x + = \int_0^4\int_0^y y^2 e^{xy}\ud x\ud y +\end{multline*} + +Since $y^2 e^{xy}$ is simply an exponential function of $x$, +it would be easier to evaluate +\begin{align*} +I &= \int_0^4\int_0^y y^2 e^{xy}\ud x\ud y\\ + &= \int_0^4\left[y^3 e^{xy}\right]_{x=0}^{x=y}\ud y\\ + &= \int_0^4 y^3 e^{y^2}\ud y + = \int_0^4 y^2\ud\frac{e^{y^2}}{2}\\ + &= \left.\frac{y^2 e^{y^2}}{2}\right]_0^4 - \int_0^4\frac{e^{y^2}}{2}\ud y^2\\ + &= 8e^{16} - \int_0^{16}\frac{e^z}{2}\ud z\\ + &= 8e^{16} - \left.\frac{e^z}{2}\right]_0^{16} + = \frac{15e^{16}}{2} +\end{align*} + +\exercise{31} Find the volume of the solid bounded by the cylinder +$x^2 + y^2 = 1$ and the plane $y = z$ in the first octant. +\[\int_0^1\int_0^{\sqrt{1-x^2}}y\ud y\ud x += \int_0^1\frac{1 - x^2}{2}\ud x += \frac{1}{3}\] + +\subsection{Double Integrals in Polar Coordinates} +\exercise{13} Evaluate the given integral by changing to polar coordinates. +\[I = \iint_R\arctan\frac{y}{x}\ud A,\qquad +\text{where }R = \{(x, y)\,|\,1 \leq x^2 + y^2 \leq 4, 0 \leq y \leq x\}\] + +In polar coordinates, +\[R = [1, 2]\times \left[0, \frac\pi 4\right]\] +thus +\begin{align*} +I &= \int_0^{\pi/4}\int_1^2\arctan\frac{r\sin\theta}{r\cos\theta} + r\ud r\ud\theta\\ + &= \int_0^{\pi/4}\int_1^2\arctan\tan\theta r\ud r\ud\theta\\ + &= \int_0^{\pi/4}\int_1^2\theta r\ud r\ud\theta\\ + &= \int_0^{\pi/4}\frac{3\theta}{2}\ud r\ud\theta\\ + &= \frac{3\pi^2}{64} +\end{align*} + + +\begin{multicols*}{2} + \noindent\begin{tikzpicture}[domain=-pi:pi] + \begin{axis}[legend pos=south east, xlabel={$\theta$}, ylabel={$r$}, + axis x line = middle, axis y line = middle, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] + \addplot[color=magenta]{1}; + \addplot[color=green]{2 * cos(deg(x))}; + \legend{$r = 1$, $r = 2\cos\theta$} + \end{axis} + \end{tikzpicture} + + \exercise{17} Use a double integral to find the area of the region + inside $C_1: (x - 1)^2 + y^2 = 1$ and outside $C_0: x^2 + y^2 = 1$.\\ + + In polar coordinates $C_1$ has the equation $r = 2\cos\theta$ and + the equation of $C_0$ is $r = 1$. Therefore the given region is within + $1 \leq r \leq 2\cos\theta$, whereas $\theta \in [-\pi, \pi]$. +\end{multicols*} + +Since on $[-\pi, \pi]$, $2\cos\theta \geq 1 \iff -\pi/3 \leq \theta \leq \pi/3$, +the area of the given region is +\begin{align*} + \int_{-\pi/3}^{\pi/3}\int_1^{2\cos\theta}r\ud r\ud\theta + &= \int_{-\pi/3}^{\pi/3}\frac{4\cos^2\theta - 1}{2}\ud\theta\\ + &= \int_{-\pi/3}^{\pi/3}\left(2\cos^2\theta - 1 + \frac{1}{2}\right)\ud\theta\\ + &= \int_{-\pi/3}^{\pi/3}\left(\cos 2\theta + \frac{1}{2}\right)\ud\theta\\ + &= \left[\frac{\sin 2\theta + \theta}{2}\right]_{-\pi/3}^{\pi/3}\\ + &= \frac{\sqrt 3}{2} + \frac\pi 3 +\end{align*} + +\subsection{Applications of Double Integrals} +\exercise{5} Find the mass and center of mass of the lamina that occupies +the region triangular $D$ with vertices (0, 0), (2, 1), (0, 3) +and has the given density function $\rho(x, y) = x + y$. +\begin{align*} +m &= \iint_D(x + y)\ud A\\ + &= \int_0^2\int_{x/2}^{3-x}(x+y)\ud y\ud x\\ + &= \int_0^2\frac{36 - 9x^2}{8}\ud x\\ + &= 9 - 3 = 6 +\end{align*} + +\begin{align*} + \bar x &= \iint_D\frac{x(x + y)}{m}\ud A& + \bar y &= \iint_D\frac{y(x + y)}{m}\ud A\\ + &= \int_0^2\int_{x/2}^{3-x}\frac{x^2 + xy}{6}\ud y\ud x& + &= \int_0^2\int_{x/2}^{3-x}\frac{xy + y^2}{6}\ud y\ud x\\ + &= \int_0^2\frac{12x - 3x^3}{16}\ud x& + &= \int_0^2\frac{6 - 3x}{4}\ud x\\ + &= \frac{3}{4}& + &= \frac{3}{2} +\end{align*} + +\subsection{Surface Area} +\exercise{7} Find the area of the part of +the hyperbolic paraboloid $z = y^2 - x^2$ +that lies between the cylinders $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$. +\begin{align*} + &\iint_D\sqrt{1 + \left(\tho{(y^2 - x^2)}{x}\right)^2 + + \left(\tho{(y^2 - x^2)}{y}\right)^2}\ud A\\ += &\iint_D\sqrt{1 + 4x^2 + 4y^2}\ud A\\ += &\int_0^{2\pi}\int_1^2 r\sqrt{1 + 4r^2\cos^2\theta + 4r^2\sin^2\theta} + \ud r\ud\theta\\ += &\int_1^2\pi\sqrt{1 + 4r^2}\ud r^2\\ += &\int_1^4\pi\sqrt{1 + 4t}\ud t\\ += &\,\pi\left[\frac{(1 + 4t)^{1.5}}{6}\right]_1^4\\ += &\,\frac{17^{1.5} - 5^{1.5}}{6}\pi +\end{align*} + +\section{Vector Calculus} +\setcounter{subsection}{1} +\subsection{Line Integrals} +\exercise{12} Evaluate the integral, where $C$ is the given curve. + +\[I = \int_C(x^2 + y^2 + z^2)\ud s,\qquad +C: x = t, y = \cos 2t, z = \sin 2t, 0 \leq t \leq 2\pi\] +\begin{align*} + I &= \int_0^{2\pi}(x^2 + y^2 + z^2)\sqrt{\left(\leibniz{x}{t}\right)^2 + + \left(\leibniz{z}{t}\right)^2 + \left(\leibniz{z}{t}\right)^2}\ud t\\ + &= \int_0^{2\pi}(t^2 + \cos^2 2t + \sin^2 2t) + \sqrt{\left(\leibniz{t}{t}\right)^2 + \left(\leibniz{\cos 2t}{t}\right)^2 + + \left(\leibniz{\sin 2t}{t}\right)^2}\ud t\\ + &= \int_0^{2\pi}(t^2 + 1)\sqrt 2\ud t = \frac{8\pi\sqrt 2}{3} + 2\pi\sqrt 2 +\end{align*} + +\exercise{15} With $C$ is the line segment from (1, 0, 0) to (4, 1, 2), +$x = 3t + 1$, $y = t$, $z = 2t$, whereas $0 \leq t \leq 1$ and +\begin{align*} + J &= \int_C z^2\ud x + x^2\ud y + y^2\ud z\\ + &= \int_0^1 z^2\leibniz{x}{t}\ud t + x^2\leibniz{y}{t}\ud t + + y^2\leibniz{z}{t}\ud t\\ + &= \int_0^1(x^2 + 2y^2 + 3z^2)\ud t\\ + &= \int_0^1(9t^2 + 6t + 1 + 2t^2 + 12t^2)\ud t\\ + &= \int_0^1(23t^2 + 6t + 1)\ud t\\ + &= \left[\frac{23t^3}{3} + 3t^2 + t\right]_0^1 = \frac{35}{3} +\end{align*} + +\exercise{39} Find the work done by the force field +$\mathbf F(x, y) = \langle x, y + 2\rangle$ is moving an object along an arch +of the cycloid $\mathbf r(t) = \langle t-\sin t, 1-\cos t\rangle$, +$0 \leq t \leq 2\pi$. +\begin{align*} + W &= \int_C\mathbf F\cdot\ud\mathbf r\\ + &= \int_0^{2\pi}\mathbf F\cdot\leibniz{\mathbf r}{t}\ud t\\ + &= \int_0^{2\pi}\langle x, y+2\rangle\cdot + \left<\leibniz{x}{t}, \leibniz{y}{t}\right>\ud t\\ + &= \int_0^{2\pi}\langle t-\sin t, 3-\cos t\rangle\cdot + \left<1-\cos t, \sin t\right>\ud t\\ + &= \int_0^{2\pi}\left(t - t\cos t + 2\sin t\right)\ud t\\ + &= \left[\frac{t^2}{2} - t\sin t - 3\cos t\right]_0^{2\pi} = 2\pi^2 +\end{align*} + +\subsection{The Fundamental Theorem for Line Integral} +\exercise{19} Show that the line integral is independent +from any path $C$ from (1, 0) to (2, 1) and evaluate the integral. +\[\int_C\frac{2x}{e^y}\ud x + \left(2y - \frac{x^2}{e^y}\right)\ud y += \int_C\left(\frac{2x}{e^y}\unit\i + 2y\unit\j - \frac{x^2}{e^y}\unit\j\right) +\cdot\ud(x\unit\i + y\unit\j)\] + +Since on $\mathbb R^2$ +\[\tho{}{y}\frac{2x}{e^y} = \frac{-2x}{e^y} += \tho{}{x}\left(2y - \frac{x^2}{e^y}\right)\] +the function +\[\mathbf F(x, y) = \frac{2x}{e^y}\unit\i + 2y\unit\j - \frac{x^2}{e^y}\unit\j\] +is conservative and thus the given line integral is independent from path. + +Let $f$ be a differentiable of $(x, y)$ that $\nabla f = \mathbf F$. +One function satisfying this is \[f(x, y) = y^2 + \frac{x^2}{e^y}\] +By the fundamental theorem for line integrals, +\[\int_C\mathbf F\cdot\ud\mathbf r = f(2, 1) - f(1, 0) = \frac{4}{e}\] + +\subsection{Green's Theorem} +\exercise{6} Use Green’s Theorem to evaluate the line integral along the given +positively oriented rectangle with vertices (0, 0), (5, 0), (5, 2) and (0, 2). +\begin{align*} + \int_C\cos y\ud x + x^2\sin y\ud y +&= \int_0^5\int_0^2\left(\tho{x^2\sin y}{x} - \tho{\cos y}{y}\right)\ud y\ud x\\ +&= \int_0^5\int_0^2(2x\sin y + \sin y)\ud y\ud x\\ +&= \int_0^5(2x + 1)(1 - \cos 2)\ud x\\ +&= 30 - 30\cos 2 +\end{align*} + +\begin{multicols}{2} + \noindent\begin{tikzpicture}[domain=-pi/2:pi/2] + \begin{axis}[legend pos=north east, xlabel={$x$}, ylabel={$y$}, + axis x line = middle, axis y line = middle, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] + \addplot[->,>=stealth,color=blue]{cos(deg(x))}; + \addplot[<-,>=stealth,color=orange]{0}; + \legend{$r = \cos x$, $r = 0$} + \end{axis} + \end{tikzpicture} + + \exercise{12} Use Green's Theorem to evaluate the line integral along the + path $C$ including the curve $y = \cos x$ from $(-\pi/2, 0)$ to $(\pi/2, 0)$ + and the line segment connecting these two points. + + Since the curve is negatively oriented, by Green's Theorem, +\end{multicols} +\begin{align*} + &\int_C(e^{-x} + y^2)\ud x + (e^{-y} + x^2)\ud y\\ += &\,-\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}\left( + \tho{}{x}(e^{-y} + x^2) - \tho{}{y}(e^{-x} + y^2)\right)\ud y\ud x\\ += &\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 1 2\int_{-\pi/2}^{\pi/2}(\cos 2x + 1)\ud x + - \int_{-\pi/2}^{\pi/2}2x\ud\sin x\\ += &\left[\frac{\sin 2x}{4} + \frac{x}{2} + - 2x\sin x - 2\cos x\right]_{-\pi/2}^{\pi/2} = \frac\pi 2 +\end{align*} + +\subsection{Curl and Divergence} + +This section is to aid my revision of Electromagnetism. First, on $\mathbb R^3$, +we define +\[\nabla = \unit\i\tho{}{x} + \unit\j\tho{}{y} + \unit k\tho{}{z}\] +then the curl of vector field $\mathbf F = P\unit\i + Q\unit\j + R\unit k$ is +\begin{multline*} + \curl\mathbf F += \nabla\times\mathbf F += \begin{vmatrix} + \unit\i & \unit\j & \unit k\\ + \tho{}{x} & \tho{}{y} & \tho{}{z}\\ + P & Q & R +\end{vmatrix}\\ += \unit\i\left(\tho{R}{y} - \tho{Q}{z}\right) ++ \unit\j\left(\tho{P}{z} - \tho{R}{x}\right) ++ \unit k\left(\tho{Q}{x} - \tho{P}{y}\right) +\end{multline*} + +If $f$ is a function of three variables that has continuous second-order +partial derivatives, then $\curl(\nabla f) = \mathbf 0$. + +On the other hand, if $\curl\mathbf F = \mathbf 0$ +then $\mathbf F$ is a conservative vector field (preconditions: +$P$, $Q$ and $R$ must be partially differentiable). + +Similarly, the divergence of vector field $\mathbf F$ is defined as +\[\del\mathbf{F} = \nabla\cdot\mathbf F += \unit\i\tho{P}{x} + \unit\j\tho{Q}{y} + \unit k\tho{R}{z}\] + +Trivially, $\nabla\cdot(\nabla\times\mathbf F) = 0$ +because the terms cancel in pairs by Clairaut's Theorem. + +The cool thing about operators is that they can be weirdly combined, +e.g. $\del(\nabla f) = \nabla\cdot\nabla f = \nabla^2 f$ +and $\nabla^2 F = \nabla\cdot\nabla\cdot\mathbf F$. + +Now we are able to write Green's Theorem in the vector form +\[\oint_{\partial S}\mathbf F\cdot\ud\mathbf r += \iint_S(\curl\mathbf F)\cdot\unit k\ud A\] +whereas $\mathbf r(t) = x(t)\unit\i + y(t)\unit\j$. +The outward normal vector to the contour is given by +$\mathbf n(t) = \leibniz{y}{t}\unit\i - \leibniz{x}{t}\unit\j$. +So we have the second vector form of Green's Theorem. +\[\oint_{\partial S}\mathbf F\cdot\unit n\ud s = \iint_S\del\mathbf F\ud A\] + +\subsection{Parametric Surfaces and Their Areas} +\exercise{42} Find the area of the part of the cone $z = \sqrt{x^2 + y^2}$ +that lies between the plane $y = x$ and the cylinder $y = x^2$. +\begin{align*} + &\int_0^1\int_{x^2}^x\sqrt{ + 1 + \left(\tho{z}{x}\right)^2 + \left(\tho{z}{y}\right)_2}\ud y\ud x\\ +=&\int_0^1\int_{x^2}^x\sqrt 2\ud y\ud x += \int_0^1(x - x^2)\sqrt 2\ud y\ud x\\ +=&\,\frac{1}{2} - \frac{1}{3} += \frac{1}{6} +\end{align*} + +\section{Second-Order Differential Equations} +\subsection{Homogeneous Linear Equations} +\exercise{11} Solve the differential equation. +\[2\leibniz{^2 y}{t^2} + 2\leibniz{y}{t} - y = 0\] + +Since the auxiliary equation $2r^2 + 2x - 1 = 0$ has two real and distinct +roots $\dfrac{\pm\sqrt 3 - 1}{2}$, the general solution is +\[y = c_1\exp\frac{\sqrt 3 - 1}{2}t + c_2\exp\frac{-\sqrt 3 - 1}{2}t\] + +\exercise{21} Solve the initial value problem. +\[y'' - 6y' + 10y = 0,\qquad y(0) = 2,\qquad y''(0) = 3\] + +Since the auxiliary equation $r^2 - 6x + 10 = 0$ has two complex roots +$3\pm i$, the general solution is +\[y = e^{3x}(c_1\cos x + c_2\sin x) +\Longrightarrow y' = e^{3x}((3c_1 + c_2)\cos x + (3c_2 - c_1)\sin x)\] + +As $y(0) = 2$, $c_1 = 2$. Similarly, from $y'(0) = 3$, +we can obtain $3c_1 - c_2 = 3 \Longrightarrow c_2 = 3$. Thus the solution +of the initial value problem is $y = e^{3x}(3\cos x + 2\sin x)$. + +\subsection{Nonhomogeneous Linear Equations} +Solve the differential equation or initial-value problem using the method of +undetermined coefficients. + +\[y'' - 4y' + 5y = e^{-x}\tag{5}\] + +The auxiliary equation of $y'' - 4y' + 5y = 0$ is $r^2 - 4r + 5 = 0$ with roots +$r = 2\pm i$. Hence the solution to the complementary equation is +\[y_c = e^{2x}(c_1\cos x + c_2\sin x)\] + +Since $G(x) = e^{-x}$ is an exponential function, we seek a particular solution +of an exponential function as well: +\[y_p = Ae^{-x} +\Longrightarrow y_p' = -Ae^{-x} +\Longrightarrow y_p'' = Ae^{-x}\] + +Substituting these into the differential equation, we get +\[Ae^{-x} - 4Ae^{-x} + 5Ae^{-x} = e^{-x} \iff A = \frac{1}{10}\] + +Thus the general solution of the exponential equation is +\[y = y_c + y_p = e^{2x}(c_1\cos x + c_2\sin x) + \frac{1}{10e^x}\] + +\[y'' + y' - 2y = x + \sin 2x,\qquad y(0) = 1,\qquad y'(0) = 0\tag{10}\] + +The auxiliary equation of $y'' + y' - 2y = 0$ is $r^2 + r - 2 = 0$ with roots +$r = -2, 1$. Thus the solution to the complementary equation is +\[y_c = c_1 e^x + \frac{c_2}{e^{2x}}\] + +We seek a particular solution of the form +\begin{multline*} + y_p = Ax + B + C\cos 2x + D\sin 2x\\ + \Longrightarrow y_p' = A - 2C\sin 2x + 2D\cos 2x\\ + \Longrightarrow y_p'' = -4C\cos 2x - 4D\sin 2x +\end{multline*} + +Substituting these into the differential equation, we get +\begin{multline*} + (-4C + 2D - 2C)\cos 2x + (-4D - 2C - 2D)\sin 2x + A - 2B - 2Ax = x + \sin 2x\\ + \iff\begin{cases} + -4C + 2D - 2C = 0\\ + -4D - 2C - 2D = 1\\ + A - 2B = 0\\ + -2A = 1 + \end{cases} + \iff\begin{cases} + A = -1/2\\ + B = -1/4\\ + C = -1/20\\ + D = -3/20 + \end{cases} +\end{multline*} + +Thus the general solution of the exponential equation is +\begin{multline*} + y = y_c + y_p = c_1 e^x + \frac{c_2}{e^{2x}} - \frac{x}{2} - \frac{1}{4} + - \frac{\cos 2x}{20} - \frac{3\sin 2x}{20}\\ + \Longrightarrow y' = c_1 e^x - \frac{2c_2}{e^{2x}} - \frac{1}{2} + + \frac{\sin 2x}{10} - \frac{3\cos 2x}{10} +\end{multline*} + +Since $y(0) = 1$ and $y'(0) = 0$, +\[\begin{dcases} + c_1 + c_2 - \frac{1}{4} - \frac{1}{20} = 1\\ + c_1 - 2c_2 - \frac{3}{10} = 0 +\end{dcases} +\iff \begin{dcases} + c_1 + c_2 = \frac{13}{10}\\ + c_1 - 2c_2 = \frac{3}{10} +\end{dcases} +\iff \begin{dcases} + c_1 = \frac{29}{30}\\ + c_2 = \frac{1}{3} +\end{dcases}\] + +Therefore the solution to the initial value problem is +\[y = \frac{29e^x}{30} + \frac{1}{3e^{2x}} - \frac{x}{2} - \frac{1}{4} + - \frac{\cos 2x}{20} - \frac{3\sin 2x}{20}\] + +\subsection{Applications} +\exercise{3} A spring with a mass of 2 kg has damping constant 14, and a force +of 6 N is required to keep the spring stretched 0.5 m beyond its natural +length. The spring is stretched 1 m beyond its natural length and then +released with zero velocity. Find the position of the mass at any time $t$. + +By Hooke's law, +\[k(0.5) = 6 \iff k = 12\] + +By Newton's second law of motion, +\[2\leibniz{^2 x}{t^2} + 14\leibniz{x}{t} + 12x = 0\] + +Since the auxiliary equation $2r^2 + 14r + 12 = 0$ has two real +and distinct roots $r = -6, -1$, the general solution is +\[x = \frac{c_1}{e^t} + \frac{c_2}{e^{6t}} +\Longrightarrow \leibniz{x}{t} = \frac{-c_1}{e^t} - \frac{6c_2}{e^{6t}}\] + +From $x(0) = 1$ and $x'(0) = 0$ we get +\[\begin{cases} + c_1 + c_2 = 1\\ + -c_1 - 6c_2 = 0 +\end{cases} +\iff\begin{cases} + c_1 = 6/5\\ + c_2 = -1/5 +\end{cases}\] + +Therefore the position at any time $t$ is +\[x = \frac{6}{5e^t} - \frac{c_2}{5e^{6t}}\] + +\setcounter{section}{8} +\section{First-Order Differential Equations} +\setcounter{subsection}{2} +\subsection{Separable Equations} +\exercise{8} Solve the differential equation. +\begin{align*} + \leibniz{y}{\theta} &= \frac{e^y\sin^2\theta}{y\sec\theta}\\ + \iff \int\frac{y}{e^y}\ud y &= \int\sin\theta\cos\theta\ud\theta\\ + \iff \int-y\ud e^{-y} &= \int\sin^2\theta\ud\sin\theta\\ + \iff \int e^{-y}\ud y - \frac{y}{e^y} &= \frac{\sin^3\theta}{3}\\ + \iff \frac{1 + y}{e^y} &= C - \frac{\sin^3\theta}{3} +\end{align*} +\setcounter{subsection}{4} +\subsection{Linear Equations} +\exercise{28} In a damped RL circuit, the generator supplies a voltage of +$E(t) = 40\sin 60t$ volts, the inductance is 1 H, the resistance is 10 $\Omega$ +and $I(0) = 1$ A. +\begin{align*} + &E - L\leibniz{I}{t} - RI = 0\\ + \iff &\frac{40}{L}\sin 60t = \leibniz{I}{t} + \frac{RI}{L}\\ + \iff &\frac{40e^{tR/L}}{L}\sin 60t + = \frac{RI}{L}e^{tR/L} + \leibniz{I}{t}e^{tR/L}\\ + \iff &\frac{40}{L}\int e^{tR/L}\sin 60t\ud t = \int\ud Ie^{tR/L}\tag{$*$} +\end{align*} + +Let $J = \int e^{tR/L}\sin 60t\ud t$, +\begin{align*} +J &= \frac{-1}{60}\int e^{tR/L}\ud\cos 60t\\ + &= \frac{1}{60}\int\cos 60t\ud e^{tR/L} - \frac{e^{tR/L}\cos 60t}{60}\\ + &= \frac{R}{3600L}\int e^{tR/L}\ud\sin 60t - \frac{e^{tR/L}\cos 60t}{60}\\ + &= \frac{R}{3600L}e^{tR/L}\sin 60t - \frac{R}{3600L}\int\sin 60t\ud e^{tR/L} + - \frac{e^{tR/L}\cos 60t}{60}\\ + &= \frac{R}{3600L}e^{tR/L}\sin 60t - \frac{R^2}{3600L^2}J + - \frac{e^{tR/L}\cos 60t}{60} +\end{align*} + +Hence $J = \dfrac{e^{tR/L}(RL\sin 60t - 60L^2\cos 60t)}{R^2+3600L^2}$ +and $(*)$ is equivalent to +\begin{multline*} + \frac{40e^{tR/L}(R\sin 60t - 60L\cos 60t)}{R^2+3600L^2} = Ie^{tR/L} - C\\ + \iff I = \frac{40R\sin 60t - 2400L\cos 60t}{R^2+3600L^2} + + \frac{C}{e^{tR/L}}\\ + \iff I = \frac{\sin 60t - 3\cos 60t}{5} + + \frac{C}{e^{t/20}} +\end{multline*} + +Since $I = 1$ at $t = 0$, +\[1 = \frac{\sin 0 - 3\cos 0}{5} + \frac{C}{e^0} \iff C = \frac 8 5\] +and thus $I = \dfrac{\sin 60t - 3\cos 60t}{5} + \dfrac{8}{5}\exp\dfrac{-t}{20}$. + +At $t = 0.1$, $I = (\sin 6 - 3\cos 6)/5 + 1.6e^{-1/200} \approx 2.11$ A. +\end{document} diff --git a/usth/MATH2.1/problems.pdf b/usth/MATH2.1/problems.pdf new file mode 100644 index 0000000..4889f5c Binary files /dev/null and b/usth/MATH2.1/problems.pdf differ diff --git a/usth/MATH2.1/solutions.pdf b/usth/MATH2.1/solutions.pdf new file mode 100644 index 0000000..12d4b19 Binary files /dev/null and b/usth/MATH2.1/solutions.pdf differ diff --git a/usth/MATH2.1/solutions.tex b/usth/MATH2.1/solutions.tex new file mode 100644 index 0000000..46c6915 --- /dev/null +++ b/usth/MATH2.1/solutions.tex @@ -0,0 +1,1633 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{enumerate} +\usepackage{lmodern} +\usepackage{mathtools} +\usepackage{pgfplots} +\usepackage{venndiagram} + +\newcommand{\E}{\mathbf E} +\newcommand{\C}{\mathrm C} +\newcommand{\N}{\mathcal N} +\newcommand{\ud}{\,\mathrm{d}} +\newcommand{\var}{\mathrm{var}} +\newcommand{\cov}{\mathrm{cov}} +\newcommand{\exercise}[1]{\noindent\textbf{#1.}} +\renewcommand{\thesubsection}{\Alph{subsection}} +\renewcommand{\thefootnote}{\fnsymbol{footnote}} + +\title{Probability Homework} +\author{Nguyễn Gia Phong} +\date{Fall 2019} + +\begin{document} +\maketitle +\section{Basic Probability 1} + +\exercise 1 Problems regarding de Morgan's law +\begin{enumerate}[(a)] + \item Consider rolling a six-sided die, where + \begin{align*} + &\begin{cases} + A = \{2, 4, 6\}\iff A^\C = \{1, 3, 5\}\\ + B = \{4, 5, 6\}\iff B^\C = \{1, 2, 3\} + \end{cases}\\ + \Longrightarrow &\begin{cases} + (A \cup B)^\C = \{2, 4, 5, 6\}^\C = \{1, 3\} = A^\C \cap B^\C\\ + (A \cap B)^\C = \{4, 6\}^\C = \{1, 2, 3, 5\} = A^\C \cup B^\C\\ + \end{cases} + \end{align*} + \item By de Morgan's law, + \begin{align*} + P\left(A^\C \cap B^\C\right) &= P\left((A \cup B)^\C\right)\\ + &= 1 - P\left(A \cup \left(A^\C \cap B\right)\right)\\ + &= 1 - P(A) - P\left(A^\C \cap B\right) + \tag{since $A \cap \left(A^\C \cap B\right) = \varnothing$}\\ + &= 1 -P(A) -P\left((A\cap B)\cup\left(A^\C\cap B\right)\right) +P(A\cap B) + \tag{since $(A\cap B)\cap\left(A^\C\cap B\right) = \varnothing$}\\ + &= 1 - P(A) - P(B) + P(A\cap B) + \end{align*}\label{1.b} + \item Consider events A and B such that $P(A) = 1/2$, $P(A\cup B) = 3/4$, + $P\left(B^\C\right) = 5/8$. + + \begin{align*} + & P\left(A^\C\cap B\right) = P(A\cup B) - P(A) + = \frac 3 4 - \frac 1 2 = \frac 1 4\\ + & P\left(A^\C\cap B^\C\right) = P\left((A\cup B)^\C\right) + = P(\Omega) - P(A\cup B) = 1 - \frac 3 4 = \frac 1 4\\ + & P\left(A\cap B^\C\right) = P\left(B^\C\right)-P\left((A\cup B)^\C\right) + = \frac 5 8 - \frac 1 4 = \frac 3 8\\ + & P(A\cap B) = P(A) - P\left(A\cap B^\C\right) + = \frac 1 2 - \frac 3 8 = \frac 1 8\\ + & P\left(A^\C\cup B^\C\right) = P\left((A\cap B)^\C\right) + = P(\Omega) - P(A\cap B) = 1 - \frac 1 8 = \frac 7 8 + \end{align*} +\end{enumerate} + +\exercise 2 A four-sided die is rolled repeatedly, +until the first time (if ever) that an even number is obtained. +What is the sample space for this experiment? + +Let the outcome be a n-dimensional vector, whose elements are values +of each roll in chronological order. The sample space would then be +\[\Omega = \{v \in \{1, 3\}^m\times\{2, 4\} \mid m \in \mathbb N\}\] + +\exercise 3 A ball is drawn at random from a box containing 6 red balls, +4 white balls, and 5 blue balls. + +Let $\Omega$ be the sample space then $n(\Omega) = 6 + 4 + 5 = 15$. +Let the R, W and B be the event where a red, white and blue ball is drawn +respectively, each of these events are mutually exclusive. Suppose each ball +is equally likely to be drawn, we get +\[\begin{cases} + n(R) = 6\\ + n(W) = 4\\ + n(B) = 5 +\end{cases} +\Longrightarrow\begin{dcases} + P(R) = \frac{n(R)}{n(\Omega)} = \frac{2}{5}\\ + P(W) = \frac{n(W)}{n(\Omega)} = \frac{4}{15}\\ + P(B) = \frac{n(B)}{n(\Omega)} = \frac{1}{3} +\end{dcases}\] + +\begin{enumerate}[(a)] + \item For a ball that is not red to be drawn, the probability is + \[P\left(R^\C\right) = P(\Omega) - P(R) = 1 - \frac 2 5 = \frac 3 5\] + \item For a ball that is either red or white to be drawn, the probability is + \[P(R\cup W) = P(R) + P(W) = \frac{2}{5} + \frac{4}{15} = \frac 2 3\] +\end{enumerate} + +\exercise 4 Given $P(C_a) = 0.8$, $P(C_b) = 0.6$ and $P(C_a\cap C_b) = 0.5$. + +We can easily prove that $P(C_a\cup C_b) = P(C_a) + P(C_b) - P(C_a\cap C_b)$ +(similar to what we did in exercise 1.b). Thus the probability that the student +will get at least one offer from these two companies is $0.8 + 0.6 - 0.5 = 0.9$. + +\exercise 5 Let G and C be the events that the selected student is a genius and +is a chocolate lover, respectively, then $P(G) = 0.6$, $P(C) = 0.7$ and +$P(G\cap C) = 0.4$. The probability that a randomly selected student is +neither a genius nor a chocolate lover is +\[P\left((G\cup C)^\C\right) = 1 - P(G) - P(C) + P(G\cap C) += 1 - 0.6 - 0.7 + 0.4 = 0.1\] + +\exercise 6 First, consider Rick's choice of entrance. We denote +the outcome that he chooses each gate as $R_A$, $R_B$, $R_C$ and $R_D$, +then $P\{R_A\} = 1/3$ and $P\{R_B\} = P\{R_C\} = P\{R_D\} = 2/9$. +The sample space is $\Omega_R = \{R_A, R_B, R_C, R_D\}$. + +Similarly, denote Brenda's and Ali's choices as $B_Y$ and $A_X$ respectively, +where $X$, $Y$ (and later $Z$) are one of the four entrances +$\omega = \{A, B, C, D\}$, we get +\[\begin{dcases} + P\{B_A\} = P\{B_B\} = P\{B_C\} = P\{B_D\} = \frac 1 4\\ + P\{A_A\} = P\{A_B\} = \frac{2}{35}\\ + P\{A_C\} = \frac 2 7\\ + P\{A_D\} = \frac 3 5 +\end{dcases}\] +The sample spaces of these two models are $\Omega_B = \{B_A, B_B, B_C, B_D\}$ +and $\Omega_A = \{A_A, A_B, A_C, A_D\}$. + +Now consider the probability model of the choices of the three friends. +The sample space is $\Omega = \Omega_R\times\Omega_B\times\Omega_A$. +Since the three friends chooses their entrance independently, +for all $\mathbf v = \langle R_Z, B_Y, A_X\rangle$ in $\Omega$, +\[P\{\mathbf v\} = P\{R_Z\} \cdot P\{B_Y\} \cdot P\{A_X\}\] + +\begin{enumerate}[(a)] + \item The event that at least two friends choose entrance B is + \[a = (\Omega_R \times \{B_B\} \times \{A_B\}) + \cup (\{R_B\} \times \Omega_B \times \{A_B\}) + \cup (\{R_B\} \times \{B_B\} \times \Omega_A)\] + + Notice that + \begin{align*} + &(\Omega_R \times \{B_B\} \times \{A_B\}) + \cap (\{R_B\} \times \Omega_B \times \{A_B\})\\ + =\,&(\Omega_R \times \{B_B\} \times \{A_B\}) + \cap (\{R_B\} \times \Omega_B \times \{A_B\}) + \cap (\{R_B\} \times \{B_B\} \times \Omega_A)\\ + =\,&\{R_B, B_B, A_B\} + \end{align*} + + Therefore the probability of this event is + \begin{align*} + P(a) &= P(\Omega_A \times \{B_B\} \times \{A_B\})\\ + &+ P(\{R_B\} \times \Omega_B \times \{A_B\})\\ + &- P\{R_B, B_B, A_B\}\\ + &+ P(\{R_B\} \times \{B_B\} \times \Omega_A)\\ + &- P\{R_B, B_B, A_B\}\\ + &= P\{B_B\} \cdot P\{A_B\}\\ + &+ P\{R_B\} \cdot P\{A_B\}\\ + &+ P\{R_B\} \cdot P\{B_B\}\\ + &- 2 \cdot P\{R_B\} \cdot P\{B_B\} \cdot P\{A_B\}\\ + &= \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} + - \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35} + = \frac{8}{105} + \end{align*} + \item The only four cases where all friends choose the same entrance are + $\{\langle R_X, B_X, A_X\rangle \mid X = \omega\}$. + Hence the probability of this event is + \begin{align*} + P(b) &= \sum_{X \in \omega} P\{R_X\}\cdot P\{B_X\}\cdot P\{A_X\}\\ + &= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{7} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{3}{5} + = \frac{2}{35} + \end{align*} +\end{enumerate} + +\exercise 7 We roll two fair six-sided dice. +\begin{enumerate}[(a)] + \item The event that doubles are rolled has six outcomes, thus its probability + is 6/36 = 1/6. + \item Among the six outcomes where the result is four or less + (\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}), + there are two that are doubles, hence the probability would then be 1/3. + \item Let $\omega = \{1, 2, 3, 4, 5, 6\}$, + the sample space is $\Omega = \omega^2$. For one die roll is a six, + the event is $C = (\{6\}\times\omega) \cup (\omega\times\{6\})$. + Since $(\{6\}\times\omega) \cap (\omega\times\{6\}) = \{(6, 6)\}$, + \begin{align*} + P(C) &= P(\{6\}\times\omega) + P(\omega\times\{6\}) - P\{(6, 6)\}\\ + &= \frac{n(\{6\}\times\omega) + n(\omega\times\{6\}) - n\{(6, 6)\}} + {n(\Omega)}\\ + &= \frac{6 + 6 - 1}{36} = \frac{11}{36} + \end{align*} +\end{enumerate} + +\exercise 8 A baby rolls two six-sided dice. Assumed that the dice are fair. +Let $\omega = \{1, 2, 3, 4, 5, 6\}$, the sample space is $\Omega = \omega^2$. +\begin{enumerate}[(a)] + \item There are six outcomes where the result of seven: + \[A = \{(m, 7 - m) \mid m \in \omega\}\] + hence this event's probability is + \[P(A) = \frac{n(A)}{n(\Omega)} = \frac{6}{36} = \frac{1}{6}\] + \item There are two outcomes where the result of eleven: + $B = \{(5, 6), (6, 5)\}$, thus $P(B) = 1/18$. + As $A$ and $B$ are disjoint, the probability of not getting + a sum of seven or eleven is + \begin{align*} + P\left((A\cup B)^\C\right) &= P(\Omega) - P(A\cup B)\\ + &= 1 - (P(A) + P(B))\\ + &= 1 - \frac{1}{6} - \frac{1}{18} = \frac{7}{9} + \end{align*} +\end{enumerate} + +\exercise 9 Given $n(\Omega) = 25$, $n(C) = 9$, $n(D) = 8$ +and $n\left((C\cup D)^\C\right) = 10$. + +By the Venn diagram, $a = C\cap D^\C$, $b = C\cap D$ and $c = C^\C\cap D$. +Let $E = C\cup D$ and $d = E^\C$, $n(d) = 10$ and $n(E) = n(\Omega)-n(d) = 15$. +Assume that the boy fairly randomly selected, +\begin{align*} + P(E) &= \frac{n(E)}{n(\Omega)} + = \frac{15}{25} + = \frac{3}{5}\\ + P(a) &= P\left((D\cap d)^\C\right) + = 1 - P(D\cap d) + = 1 - P(D) - P(d)\\ + &= 1 - \frac{n(D) + n(d)}{n(\Omega)} + = 1 - \frac{8 + 10}{25} + = \frac{7}{25}\\ + P(c) &= P\left((C\cap d)^\C\right) + = 1 - P(C\cap d) + = 1 - P(C) - P(d)\\ + &= 1 - \frac{n(C) + n(d)}{n(\Omega)} + = 1 - \frac{9 + 10}{25} + = \frac{6}{25}\\ + P(b) &= \frac{n(b)}{n(\Omega)} + = \frac{n(E) - n(a) - n(c)}{n(\Omega)} + = P(E) - P(a) - P(c)\\ + &= \frac{3}{5} - \frac{7}{25} - \frac{6}{25} + = \frac{2}{25} +\end{align*} +\pagebreak + +\exercise{10} +\begin{enumerate}[(a)] + \item Venn diagram: + \begin{venndiagram3sets}[ + labelOnlyB={5}, + labelABC={2}, + labelNotABC={0}, + overlap=1cm] + \setpostvennhook{ + \draw[-stealth] (labelA) -- ++(135:2.5cm) node[left]{15}; + \draw[-stealth] (labelB) -- ++(45:2.5cm) node[right]{8}; + \draw[-stealth] (labelC) -- ++(-90:2cm) node[below]{12}; + \draw[-stealth] (0,0) -- ++(-135:1.5cm) node[below]{27}; + \draw[-stealth] (1.4,1.9) -- ++(180:2.5cm) node[left]{4}; + \draw[-stealth] (2.4,3.6) -- ++(-157:3.8cm); + \draw[-stealth] (4,3) -- ++(-70:4cm) node[below]{21}; + \draw[-stealth] (1,2.5) -- ++(-40:5.3cm); + \draw[-stealth] (3.3,1.5) -- ++(-50:3cm); + } + \end{venndiagram3sets} + \item The number of tourists who had not visited Burundi: + \[n\left(B^\C\right) = n(\Omega) - n(B) = 27 - 8 = 19\] + \item The number of tourists who had not visited Cameroon + unless they had visited all three countries: + \[n\left((A\cap B)\cup C^\C\right) + = n(\Omega) - n(C) + n(A\cap B\cap C) + = 27 - 12 + 2 = 17\] + \item For the randomly selected tourist to have visited at least + two countries, that person must not visited only one country. + Thus the event can be denoted as + \[d = \Omega\setminus((A\setminus B\setminus C)\cup + (B\setminus C\setminus A)\cup(C\setminus A\setminus B))\] + Since the selection is random, the event's probability can be calculated as + \[P(d) = 1 - \frac{n((A\setminus B\setminus C)\cup + (B\setminus C\setminus A)\cup(C\setminus A\setminus B))}{n(\Omega)} + = 1 - \frac{21}{27} = \frac{2}{9}\] +\end{enumerate} +\pagebreak + +\section{Basic Probability 2} +\exercise 1 Let $A$ be the event that the chosen transistor is defective, +$B$ be the event that the chosen one is partially defective +and $C$ be the event that the chosen one is acceptable. +$A$, $B$ and $C$ are disjoint and $A\cup B\cup C = \Omega$, thus +\[n(\Omega) = n(A) + n(B) + n(C) = 5 + 10 + 25 = 40\] + +The probability that the chosen transistor does not immediately fail is +$P\left(A^\C\right) = 1 - P(A) = 1 - n(A)/n(\Omega) = 1 - 5/40 = 7/8$. + +Given this condition, the probability the chosen transistor is acceptable is +\[P\left(C|A^\C\right) = \frac{P\left(C\cap A^\C\right)}{P\left(A^\C\right)} += \frac{P(C)}{7/8} += \frac{8n(C)}{7n(\Omega)} = \frac{8\cdot 25}{7\cdot 40} = \frac 5 7\] + +\exercise 2 Denote the outcomes of tossing a coin as $H$ (head) and $T$ (tail). +\begin{enumerate}[(a)] + \item Consider tossing a coin $n$ times, the sample space is + $\Omega = \{H, T\}^n$. Let $A$ be the event of getting at least a head, + $A^\C$ would then be getting all tails ($\{T\}^n$). Suppose the chance of getting + head and tail are equal, + \[P\left(A^\C\right) = \frac{n\left(A^\C\right)}{n(\Omega)} = \frac{1}{2^n} + \Longrightarrow P(A) = 1 - P\left(A^\C\right) = \frac{2^n - 1}{2^n}\] + \item For $n = 4$, $P(A) = \dfrac{2^4 - 1}{2^4} = \dfrac{15}{16}$. + \item Consider rolling a six-sided die $n$ times, the sample space is + \[\Omega = \{1, 2, 3, 4, 5, 6\}^n\] + Let $B$ be the event of getting a six, $B^\C = \{1, 2, 3, 4, 5\}^n$. + Therefore the probability of $B$ is + \[P(B) = 1 - P\left(B^\C\right) + = 1 - \frac{n\left(B^\C\right)}{n(\Omega)} = 1 - \frac{5^n}{6^n}\] + For $n = 4$, $P(B) = 671/1296$. + \item For $P(B) = 0.99$, + \[1 - \left(\frac{5}{6}\right)^n = 0.99 + \iff \left(\frac{5}{6}\right)^n = 0.01 + \iff n = \log_{5/6}0.01 \approx 25\] +\end{enumerate} + +\exercise 3 Let $B$ be the event that the woman rides the bicycle to work, +$B^\C$ would be that she ride the scooter. Let $L$ be that she is late, +\begin{align*} + P(B) &= 0.7\\ + P\left(B^\C\right) &= 0.3\\ + P(L|B) &= 0.03\\ + P\left(L|B^\C\right) &= 0.02 +\end{align*} +\begin{enumerate}[(a)] + \item By Total Probability Theorem, the probability the woman + is late for work is + \[P(L) = P(B)\cdot P(L|B) + P\left(B^\C\right)\cdot P\left(L|B^\C\right) + = 0.7\cdot 0.03 + 0.3\cdot 0.02 = 0.027\] + \item The probability she is not late for work is + \[P\left(L^\C\right) = 1 - P(L) = 1 - 0.027 = 0.973\] + Since the woman is expected to be on time roughly 223 days a year, + she goes to work $223 / P\left(L^\C\right) \approx 229$ days a year. +\end{enumerate} + +\exercise 4 Consider flipping the coin twice, the sample space is +\[\Omega = \{(H, H), (H, T), (T, H), (T, T)\}\] +where $H$ stands for head and $T$ stands for tail. + +Denote getting a head from the first flip as $H_1$ and getting a head +from the second one as $H_2$. Assume that $P(H_1) = P(H_2) = 0.6$. +It is obvious that these two events are independent, or in other words +\[P(H_1\cap H_2) = P(H_1)\cdot P(H_2)\] + +Similarly, +\begin{align*} + P\{(H, T)\} &= P\left(H_1\cap H_2^\C\right) + = P\left(H_1\right)\cdot\left(1 - P\left(H_2\right)\right) = 0.24\\ + P\{(T, H)\} &= P\left(H_1^\C\cap H_2\right) + = \left(1 - P\left(H_1\right)\right)\cdot P\left(H_2\right) = 0.24 +\end{align*} + +Therefore if Minh and Nam flip the coin twice for both head and tail +and choose K-pop when they get a head first and US music otherwise, +the genre would be chosen equally even. + +\exercise 5 Place three maths, two history and four biology book on a shelf. +\begin{enumerate}[(a)] + \item There would be $(3 + 2 + 4)! = 362880$ ways to do it + without any further restriction. + \item If each subject needs to stay together, + there are $3! 2! 4! 3! = 1728$ ways. + \item If only biology books must stay together, + we can do it in $4!(3 + 2 + 1)!$ or 17280 ways. +\end{enumerate} + +\exercise 6 Seat six people around a table. +\begin{enumerate}[(a)] + \item If they can sit anywhere, there are $6!/6 = 120$ arrangements. + \item If two particular people must sit next to each other, + there are $2\cdot 5!/5$ or 48 arrangements; thus if those two cannot + sit side-by-side, the figure is $120 - 48 = 72$. +\end{enumerate} + +\exercise 7 Let $\omega$ be the set of cards in a standard 52-card deck. +Shuffle the deck an draw seven cards, +the sample space of this probability model is $\Omega = \binom{\omega}{7}$, +$n(\Omega) = \binom{52}{7}$. +\begin{enumerate}[(a)] + \item Let $A$ be the event that exactly three of the drawn ones are aces, + \[n(A) = \binom{4}{3}\binom{48}{4} + \Longrightarrow P(A) = \frac{n(A)}{n(\Omega)} = \frac{9}{1547}\] + \item Let $K$ be the event that exactly two of the drawn ones are kings, + \[n(K) = \binom{4}{2}\binom{48}{5} + \Longrightarrow P(K) = \frac{n(K)}{n(\Omega)} = \frac{594}{7735}\] + \item The probability that exactly three aces and two kings are drawn is + \[n(A\cap K) = \binom{4}{3}\binom{4}{2}\binom{44}{2} + \Longrightarrow P(A\cap K) = \frac{n(A\cap K)}{n(\Omega)} + = \frac{1419}{8361535}\] + + Thus probability that either exactly three aces or two kings are drawn is + \[P(A\cup K) = P(A) + P(K) - P(A\cap K) = \frac{137868}{1672307}\] +\end{enumerate} + +\exercise 8 Let $M$ be the event that a red marble is picked +and $C$ be the event of getting head from tossing the coin, we have +\begin{align*} + P(M|C) &= 0.6\\ + P\left(M|C^\C\right) &= 0.2\\ + P(C) = P\left(C^\C\right) &= 0.5 +\end{align*} + +\begin{enumerate}[(a)] + \item By Total Probability Theorem, the probability a red marble is picked is + \[P(M) = P(C)\cdot P(M|C) + P\left(C^\C\right)\cdot P\left(M|C^\C\right) + = 0.4\] + \item The probability that a blue marble is picked is + \[P\left(M^\C\right) = 1 - P(M) = 0.6\] + \item The probability of getting a head if the red marble is picked is + \[P(C|M) = \frac{P(C\cap M)}{P(M)} = \frac{P(C)\cdot P(M|C)}{P(M)} + = \frac{0.5\cdot 0.6}{0.4} = 0.75\] +\end{enumerate} + +\exercise 9 Consider $n$ random people and their birthdays, assuming +that all 366 birthdays are equally likely\footnote{If you are wondering +how one could be equally likely to be born on the leap day, then well, +the distribution of birthdays on other days is in fact not uniform either. +\textit{Don't complicate it, don't drive yourself insane!}}. +The size of the sample space is $n(\Omega_n) = 366^n$. + +Let $A_n$ be the event that no two of these $n$ people to celebrate +their birthday on the same day, $n(A_n) = \prod_{i=0}^{n-1}(366-i)$. +Thus the probability of this is +\[P(A_n) = \frac{n(A_n)}{n(\Omega_n)} = \prod_{i=1}^{n-1}\frac{366 - i}{366}\] + +Since $P(A_{23}) < 0.5 < P(A_{22})$, $n$ needs to be at least 23 +for the probability to be less than 0.5. + +\exercise{10} The reasoning is not correct because: +\begin{itemize} + \item If he is not to be released, the answer from the guard will be + both of other prisoners, and everyones' fate will be known. + \item Otherwise, in case the guard only gives one name, + our protagonist will sure be released. +\end{itemize} +\pagebreak + +\section{Discrete Random Variable 1} +\subsection{Discrete Random Variable and PMF} +\exercise 1 Consider a fair coin. +\begin{enumerate}[(a)] + \item Toss it twice and let $X$ be the number of heads, + $X$ would be a binomial random variable + \begin{align*} + X\colon \Omega &\to \{0, 1, 2\}\\ + \omega &\mapsto x + \end{align*} + whose probability mass function is + \[p_X(x) = \binom{2}{x}\cdot 0.5^x\cdot 0.5^{2-x} + = \frac{1}{2x!(2-x)!}\] + + Therefore PMF of $X$ for each case is + \begin{align*} + p_X(0) = p_X(2) &= \frac{1}{2\cdot 0!2!} = \frac{1}{4}\\ + p_X(1) &= \frac{1}{2\cdot 1!1!} = \frac{1}{2} + \end{align*} + \item Toss it thrice and let $Y$ be the number of heads, + $Y$ would be a binomial random variable + \begin{align*} + Y\colon \Omega &\to \{0, 1, 2, 3\}\\ + \omega &\mapsto y + \end{align*} + whose probability mass function is + \[p_Y(y) = \binom{3}{y}\cdot 0.5^y\cdot 0.5^{3-y} + = \frac{3}{4y!(3-y)!}\] + + Therefore PMF of $Y$ for each case is + \begin{align*} + p_Y(0) = p_Y(3) &= \frac{3}{4\cdot 0!3!} = \frac{1}{8}\\ + p_Y(1) = p_Y(2) &= \frac{3}{4\cdot 1!2!} = \frac{3}{8} + \end{align*} +\end{enumerate} + +\exercise 2 Toss a pair of fair siz-sided dice +and let $X$ be the sum of the points +\begin{align*} + X\colon \Omega &\to [2, 12]\cap\mathbb Z\\ + \omega &\mapsto x +\end{align*} +with $\Omega = S^2 = \{1, 2, 3, 4, 5, 6\}^2 \Longrightarrow n(\Omega) = 36$. + +\begin{enumerate}[(a)] + \item $X$ is a random variable whose PMF is + \begin{align*} + p_X(2) &= P\{(1,1)\} = \frac{1}{36}\\ + p_X(3) &= P\{(1,2), (2,1)\} = \frac{1}{18}\\ + p_X(4) &= P\{(1,3), (2,2), (3,1)\} = \frac{1}{12}\\ + p_X(5) &= P\{(1,4), (2,3), (3,2), (4,1)\} = \frac{1}{9}\\ + p_X(6) &= P\{(1,5), (2,4), (3,3), (4,2), (5,1)\} = \frac{5}{36}\\ + p_X(7) &= P\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} = \frac{1}{6}\\ + p_X(8) &= P\{(2,6), (3,5), (4,4), (5,3), (6,2)\} = \frac{5}{36}\\ + p_X(9) &= P\{(3,6), (4,5), (5,4), (6,3)\} = \frac{1}{9}\\ + p_X(10) &= P\{(4,6), (5,5), (6,4)\} = \frac{1}{12}\\ + p_X(11) &= P\{(5,6), (6,5)\} = \frac{1}{18}\\ + p_X(12) &= P\{(6,6)\} = \frac{1}{36} + \end{align*} + \item The graph of $p_X(x)$: + + \begin{tikzpicture} + \begin{axis}[xlabel={$x$}, ylabel={$p_X(x)$}] + \addplot[ycomb, samples at={2,3,...,12}]{1/6-abs(x-7)/36}; + \end{axis} + \end{tikzpicture} +\end{enumerate} +\pagebreak + +\exercise 3 Denote the event of winning, tying and losing the first game as +$A_2$, $A_1$ and $A_0$ respectively, we get $P(A_2) = P(A_1) = 0.2$ +and $P(A_0) = 0.6$. Similarly, let $B_2$, $B_1$ and $B_0$ in that order +be the event MIT soccer team winning, tying and losing the second game, +we get $P(B_2) = P(B_1) = 0.35$ and $P(B_0) = 0.3$. + +Let $A$ and $B$ be the random variable satisfying +\begin{align*} + A &= \begin{cases} + 2\text{ if }A_2\\ + 1\text{ if }A_1\\ + 0\text{ if }A_0 + \end{cases} + \Longrightarrow\begin{cases} + p_A(2) = p_A(1) = 0.2\\ + p_A(0) = 0.6 + \end{cases}\\ + B &= \begin{cases} + 2\text{ if }B_2\\ + 1\text{ if }B_1\\ + 0\text{ if }B_0 + \end{cases} + \Longrightarrow\begin{cases} + p_B(2) = p_B(1) = 0.35\\ + p_B(0) = 0.3 + \end{cases} +\end{align*} +then the number of points the team earns over the weekend is $X = A + B$. + +Since the outcome of the two games are independent, +\begin{align*} + p_X(0) &= p_A(0)\cdot p_B(0) = 0.18\\ + p_X(1) &= p_A(0)\cdot p_B(1) + p_A(1)\cdot p_B(0) = 0.27\\ + p_X(2) &= p_A(0)\cdot p_B(2) + p_A(1)\cdot p_B(1) + p_A(2)\cdot p_B(0) + = 0.34\\ + p_X(3) &= p_A(1)\cdot p_B(2) + p_A(2)\cdot p_B(1) = 0.14\\ + p_X(4) &= p_A(2)\cdot p_B(2) = 0.07 +\end{align*} + +\subsection{Expectation of Random Variables} +\exercise 4 Given a random variable +\[X = \begin{cases} + -2&\text{ with probability of 1/3}\\ + 3&\text{ with probability of 1/2}\\ + 1&\text{ with probability of 1/6} +\end{cases}\] +\begin{align*} + \E[X] &= \sum_{x\in\{-2, 1, 3\}}xp_X(x) = 1\\ + \E[2X+5] &= \sum_{x\in\{-2, 1, 3\}}(2x + 5)p_X(x) = 7\\ + \E\left[X^2\right] &= \sum_{x\in\{-2, 1, 3\}}x^2 p_X(x) = 6 +\end{align*} + +\exercise 5 Consider the genders of the three children, and assume +that both genders\footnote{You SJWs really need to calm down. +This is just a mathematical problem.} are equally likely. + +Let $X$ be the number of girls, $X$ is a binomial random variable +whose probability mass function is +\begin{multline*} + p_X(x) = \binom{3}{x}\cdot 0.5^x\cdot 0.5^{3-x} = \frac{3}{4x!(3-x)!}\\ + \Longrightarrow \E[X] = \sum_{x=0}^3\frac{3x}{4x!(3-x)!} = \frac 3 2 +\end{multline*} + +\exercise 6 Consider rolling a fair six-sided die, the sample space is +$\Omega = \{1, 2, 3, 4, 5, 6\}$. Let $X$ be a random variable given by +\begin{multline*} + X(\omega) = \begin{cases} + -1&\text{ if }\omega \in \{1, 2, 3\}\\ + 2&\text{ if }\omega \in \{4, 5\}\\ + 8&\text{ if }\omega = 6 + \end{cases} + \Longrightarrow\begin{dcases} + p_X(-1) = \frac{3}{6} = \frac{1}{2}\\ + p_X(2) = \frac{2}{6} = \frac{1}{3}\\ + p_X(8) = \frac{1}{6} + \end{dcases}\\ + \Longrightarrow \E[X] = \frac{-1}{2} + \frac{2}{3} + \frac{4}{3} = \frac{3}{2} +\end{multline*} + +Practically, this means that at the end of the day, +it is very unlikely that the house will win. + +\exercise 7 Let $X$ be the prize in dollars on a randomly chosen +lottery ticket, its PMF is +\begin{align*} + p_X(100) &= \frac{5}{10\,000} = \frac{1}{2\,000}\\ + p_X(25) &= \frac{20}{10\,000} = \frac{1}{5\,000}\\ + p_X(5) &= \frac{200}{10\,000} = \frac{1}{500}\\ + p_X(0) &= \frac{10\,000-200-20-5}{10\,000} = \frac{391}{400} +\end{align*} + +Thus the expected value for a ticket's value in dollars is +\[\E[X] = \sum_x x\cdot p_X(x) = \frac{13}{200}\] +or 6.5 cents. + +\exercise 8 Let $X$ be the prize in dollars on a randomly chosen +raffle ticket, its PMF is +\begin{align*} + & p_X(1998) = p_X(999) = \frac{1}{5000}\\ + & p_X(498) = \frac{2}{5000} = \frac{1}{2500}\\ + & p_X(98) = \frac{5}{5000} = \frac{1}{1000}\\ + & p_X(-2) = \frac{5000-5-2-1-1}{5000} = \frac{4991}{5000} +\end{align*} + +Thus the expected value in dollars to get when buying a ticket is +\[\E[X] = \sum_x x\cdot p_X(x) = \frac{-11}{10}\] +or to lose \$1.1. + +\subsection{Variance and Standard Deviation} +\exercise{9} Given the outcome $X$ from rolling a fair six-sided die. +\begin{align*} + \E[X] &= \frac{1+2+3+4+5+6}{6} = \frac{7}{2}\\ + \Longrightarrow \var(X) &= \E\left[(X - \E[X])^2\right] + = \sum_x \frac{(x - \frac{7}{2})^2}{6} = \frac{35}{24}\\ + \Longrightarrow \sigma_X &= \sqrt{\var(X)} = \sqrt\frac{35}{24} +\end{align*} + +\exercise{10} Based on the result of exercise 2, +\[E[X] = 7,\qquad\var(X) = \frac{35}{6},\qquad\sigma_X = \sqrt\frac{35}{6}\] + +\exercise{11} Given the integral random variable $X$ with PMF +\[p_X(x) = \begin{dcases} + \frac{1}{9} &\text{ if } x \in [-4, 4]\\ + 0 &\text{ otherwise} +\end{dcases}\] + +Let $S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$, +\begin{multline*} + \E[X] = \sum_{x\in\mathbb Z}x\cdot p_X(x) + = \sum_{x\in S}\frac{x}{9} + \sum_{x\in\mathbb Z\setminus S}x\cdot 0 = 0\\ + \Longrightarrow \var(X) = \E\left[X^2\right] + = \sum_{x\in S}\frac{x^2}{9} = \frac{20}{3} +\end{multline*} + +\exercise{12} Given the integral random variable $X$ with PMF +\[p_X(x) = \begin{dcases} + \frac{x^2}{a} &\text{ if } x \in [-3, 3]\\ + 0 &\text{ otherwise} +\end{dcases}\] +\begin{enumerate}[(a)] + \item Let $S = \{-3, -2, -1, 0, 1, 2, 3\}$. Since + \begin{multline*} + \sum_{x\in\mathbb Z}p_X(x) = 1 + \iff \sum_{x\in S}\frac{x^2}{a} + \sum_{x\in\mathbb Z\setminus S}0 = 1 + \iff a = \sum_{x\in S}x^2 = 28\\ + \Longrightarrow \E[X] = \sum_{x\in S}\frac{x^3}{28} = 0 + \end{multline*} + \item Let $Z = (X - \E[X])^2 = X^2$, + the range of $Z$ is $\{z^2\mid z\in\mathbb Z\}$. + For all $z > 9$, it is trivial that $p_Z(z) = 0$. Otherwise, + \begin{align*} + p_Z(0) &= P(Z = 0) = P(X=0) = p_X(0) = \frac{0^2}{28} = 0\\ + p_Z(1) &= P(X = \pm 1) = p_X(-1) + p_X(1) + = \frac{(-1)^2}{28} + \frac{1^2}{28} = \frac{1}{14}\\ + p_Z(4) &= P(X = \pm 2) = p_X(-2) + p_X(2) + = \frac{(-2)^2}{28} + \frac{2^2}{28} = \frac{2}{7}\\ + p_Z(9) &= P(X = \pm 3) = p_X(-3) + p_X(3) + = \frac{(-3)^2}{28} + \frac{3^2}{28} = \frac{9}{14} + \end{align*} + \item The variance of $X$ is + \[\var(X) = \E\left[(X - \E[X])^2\right] + = \E[Z] = 1\cdot\frac{1}{14} + 4\cdot\frac{2}{7} + 9\cdot\frac{9}{14} = 7\] +\end{enumerate} +\pagebreak + +\section{Discrete Random Variable 2} +\subsection{Conditional PMF and Expectation} +\exercise 1 Compute conditional PMF: +\begin{enumerate}[(a)] + \item Let $X$ be the roll if a fair six-sided die and $A$ be the event that + the roll is an number greater or equal to 4, we have $A = \{X \ge 4\}$ + and $P(A) = 0.5$, thus + \[p_{X|A}(x) = \frac{P(\{X = x\}\cap\{X \ge 4\})}{P(A)}\] + + For $x \in \{1, 2, 3\}$, $\{X = x\}\cap\{X \ge 4\} = \varnothing$ + so $p_{X|A}(x) = 0/0.5 = 0$. + + For $x \in \{4, 5, 6\}$, $\{X = x\}\cap\{X \ge 4\} = \{x\}$, + \[p_{X|A}(x) = \frac{1/6}{0.5} = \frac{1}{3}\] + \item Let $X$ represent number of heads from the three-time toss + of a fair coin and $B = \{X \ge 2\}$, + $P(B) = \binom{3}{2}0.5^3 + \binom{3}{3}0.5^3 = 0.5$. + \begin{multline*} + p_{X|B}(x) = \frac{P(\{X = x\}\cap B)}{P(B)} + = \frac{P(\{X = x\}\cap\{X \ge 2\})}{0.5}\\ + \Longrightarrow\begin{dcases} + p_{X|B}(0) = p_{X|B}(1) = 0\\ + p_{X|B}(2) = \frac{\binom{3}{2}0.5^3}{0.5} = \frac{3}{4}\\ + p_{X|B}(3) = \frac{\binom{3}{3}0.5^3}{0.5} = \frac{1}{4} + \end{dcases} + \end{multline*} + \item Let $X$ be the roll of a pair of fair dice and $C = \{X = 7\}$. + As shown in the previous section, $P(C) = 1/6$ and thus + \[p_{X|C}(x) = \begin{cases} + 1\text{ if }x = 7\\ + 0\text{ if }x \neq 7 + \end{cases}\] +\end{enumerate} + +\exercise 2 Consider the destination of the message and denote the event +it arrives at Liberty City, Chicago and San Fierro as $B$, $C$ and $F$ +respectively, we have $B\cup C\cup F = \Omega$. The expected transit time is +\begin{align*} + \E[X] &= P(B)\E[X|B] + P(C)\E[X|C] + P(F)\E[X|F]\\ + &= 0.5\cdot 0.05 + 0.3\cdot 0.1 + 0.2\cdot 0.3 = 0.115 +\end{align*} + +\exercise 3 Let $V$ and $T$ be respectively the speed (in mph) and time +(in hours) Alyssa get to class. Denote the event she walk to class as $W$, +$P(W) = 0.6$, +\[\begin{cases} + \E[V|W] = 5\\ + \E\left[V|W^\C\right] = 30 +\end{cases} +\Longrightarrow\begin{dcases} + \E[T|W] = \frac{2}{5}\\ +\E\left[T|W^\C\right] = \frac{1}{15} +\end{dcases}\] +\begin{enumerate}[(a)] + \item The expected value of Alyssa's speed is + \[\E[V] = P(W)\E[V|W] + P\left(W^\C\right)\E\left[V|W^\C\right] + = 0.6\cdot 5 + (1 - 0.6)30 = 15\] + \item The expected value of the time Alyssa she takes to get to class is + \[\E[T] = P(W)\E[T|W] + P\left(W^\C\right)\E\left[T|W^\C\right] + = 0.6\cdot\frac{2}{5} + (1 - 0.6)\frac{1}{15} = \frac{4}{15}\] +\end{enumerate} + +\exercise 4 With $X$ being the number of tries until the program works +correctly and $p$ being the probability each try succeed, +we have $\mathrm{range}(X) = \mathbb N^*$ and $p_X(x) = (1 - p)^{x - 1}p$. +The mean of $X$ is +\begin{align*} + \E[X] &= \sum_{x\in\mathbb N^*}x\cdot p_X(x)\\ + &= \sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\\ + &= -p\sum_{x\in\mathbb N^*}\frac{\ud (1 - p)^x}{\ud p}\\ + &= -p\frac{\ud}{\ud p}\left(\sum_{x\in\mathbb N}(1 - p)^{x} - 1\right)\\ + &= -p\frac{\ud}{\ud p}\left(\frac{1}{p} - 1\right)\\ + &= \frac{1}{p} +\end{align*} + +Similarly, +\begin{align*} + \E\left[X^2\right] &= p\sum_{x\in\mathbb N^*}x^2(1 - p)^{x - 1}\\ + &= -p\frac{\ud}{\ud p} + \left(\frac{1 - p}{p}\sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\right)\\ + &= -p\frac{\ud}{\ud p}\left(\frac{1}{p^2} - \frac{1}{p}\right)\\ + &= \frac{2}{p^2} - \frac{1}{p} +\end{align*} + +Therefore the variance of $X$ is +\[\var(X) = \E\left[X^2\right] - (\E[X])^2 = \frac{1}{p^2} - \frac{1}{p}\] + +\subsection{Joint PMF and independent variables} +\exercise 5 Consider two independent coin tosses, +each with a 3/4 probability of a head, +and let $X$ be the number of heads obtained, +$X$ is a binomial random variable. +\[\E[X] = 0p_X(0) + 1p_X(1) + 2p_X(2) += \binom{2}{1}\frac{3}{4}\left(1 - \frac{3}{4}\right) ++ 2\binom{2}{2}\left(\frac{3}{4}\right)^2 += \frac{3}{2}\] + +\exercise 6 Let $X$ be the number of red traffic lights Alyssa encounters, +$X$ is a binomial random variable whose PMF is +\[p_X(x) = \binom{4}{x}0.5^4\] + +The mean of $X$ is +\[\E[X] = \frac{1}{16}\sum_{x=0}^4 x\binom{4}{x} = 2\] + +The variance of $X$ is +\[\var(X) = \E\left[X^2\right] - (\E[X])^2 += \frac{1}{16}\sum_{x=0}^4 x^2\binom{4}{x} - 4 = 1\] + +\exercise 7 Let $X_i$ be 1 if the $i$th person gets his or her own hat and 0 +otherwise, then for all positive integer $i \le n$ +\[\E[X_i] = p_{X_i}(1) = \frac{(n - 1)!}{n!} = \frac{1}{n}\] +since if we fix one hat to its owner, there are $(n - 1)!$ arrangements +for the rest. Due to the linearity property of expectation, +\[\E[X] = \E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n\E[X_i] = 1\] + +\exercise 8 Consider four independent rolls of a six-sided die. +Let $X$ and $Y$ be the number of ones and twos obtained respectively, +both are binomial random variables: +\[p_X(k) = p_Y(k) += \binom{4}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4 - k} += \binom{4}{k}\frac{5^{4 - k}}{1296}\] + +Given $Y = y$, $X$ is the number of ones in the remaining $4 - y$ rolls, +each of which can take the values other than two equally likely: +\[p_{X|Y}(x|y) += \binom{4 - y}{x}\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{4 - y - x} += \binom{4 - y}{k}\frac{4^{4 - y - x}}{625}\] + +Thus the joint PMF of $X$ and $Y$ is +\[p_{X,Y}(x, y) = p_Y(y)p_{X|Y}(x|y) += \binom{4}{x}\binom{4 - y}{k}\frac{5^{4 - x}4^{4 - y - x}}{810\,000}\] + +\exercise 9 Given the joint PMF of two discrete random variables $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }(x, y)\in\{0, 1, 2\}\times\{0, 1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item Consider all cases: + \begin{align*} + \sum_x\sum_y p_{X,Y}(x, y) = 1 + &\iff \sum_{x=0}^2\sum_{y=0}^3 c(2x + y) = 1\\ + &\iff \sum_{x=0}^2 c(8x + 6) = 1\\ + &\iff c(24 + 18) = 1\\ + &\iff c = \frac{1}{42} + \end{align*} + \item $P(X = 2, Y = 1) = (2\cdot 2 + 1)/42 = 5/42$. + \item Similarly, $P(X \ge 1, Y \le 2) = 4/7$. + \item The marginal PMF of $X$: + \[p_X(x) = \sum_y p_{X,Y}(x, y) + = \sum_{y=0}^3\frac{2x + y}{42} = \frac{4x + 3}{21}\] + \item The marginal PMF of $Y$: + \[p_Y(y) = \sum_x p_{X,Y}(x, y) + = \sum_{x=0}^2\frac{2x + y}{42} = \frac{2 + y}{14}\] + \item Since $p_X(2)p_Y(1) \neq p_{X,Y}(2, 1)$, + the two variables are dependent. + \item Given $X = 2$, + \[p_{Y|X}(y|2) = \frac{p_{X,Y}(2, y)}{p_X(2)} = \frac{4 + y}{22} + \Longrightarrow p_{Y|X}(1|2) = \frac{5}{22}\] + \item Given $Y = 2$, + \[p_{X|Y}(x|2) = \frac{p_{X,Y}(x, 2)}{p_Y(2)} = \frac{x + 1}{6} + \Longrightarrow p_{X|Y}(3|2) = \frac{2}{3}\] +\end{enumerate} + +\exercise{10} Given the joint PMF of two discrete random variables $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + cxy &\text{where }(x, y)\in\{1, 2, 3\}\times\{1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item Consider all cases: + \begin{align*} + \sum_x\sum_y p_{X,Y}(x, y) = 1 + &\iff \sum_{x=1}^3\sum_{y=1}^3 cxy = 1\\ + &\iff 36c = 1\\ + &\iff c = \frac{1}{36} + \end{align*} + \item $P(X = 2, Y = 3) = 1/6$. + \item Similarly, $P(1\le X\le 2, Y\le 2) = 1/4$. + \item By the result of (e), $P(X\ge 2) = 5/6$, + $P(Y < 2) = P(Y = 1) = P(X = 1) = 1/6$ and $P(Y = 3) = 1/2$. + \item The marginal PMF of $X$: + \[p_X(x) = \sum_y p_{X,Y}(x, y) + = \sum_{y=1}^3\frac{xy}{36} = \frac{x}{6}\] + The marginal PMF of $Y$: + \[p_Y(y) = \sum_x p_{X,Y}(x, y) + = \sum_{x=1}^3\frac{xy}{36} = \frac{y}{6}\] + \item Since $p_{X,Y}(x, y) = p_X(x)p_Y(y)$, $X$ and $Y$ are independent. +\end{enumerate} +\pagebreak + +\section{Continuous Random Variable 1} +\subsection{PDF and CDF} +\exercise 1 Given a PDF such that +\[f_X(x) = \begin{cases} + cx^2 &\text{if }0 < x < 3\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item By the normalization property, + \[1 = \int_{-\infty}^{\infty}f_X(x)\ud x = \int_0^3 cx^2\ud x = 9c + \Longrightarrow c = \frac{1}{9}\] + \item $\displaystyle P(1 < X < 2) = \int_1^2\frac{x^2}{9}\ud x = \frac{7}{27}$ + \item The CDF of $X$: + \[F_X(x) = \begin{dcases} + 0 &\text{if }x \le 0\\ + 1 &\text{if }x \ge 3\\ + \int_0^x\frac{t^2}{9}\ud t = \frac{x^3}{27} &\text{otherwise} + \end{dcases}\] + \item $P(1 < X \le 2) = F_X(2) - F_X(1) + = \dfrac{8}{27} - \dfrac{1}{27} = \dfrac{7}{27}$ +\end{enumerate} + +\exercise 2 Denote the event that the day is sunny as $A$, $P(A) = 2/3$. +\begin{align*} + f_{X|A}(x) &= \begin{cases} + b &\text{if }15 \le x \le 20\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{15}^{20}b\ud x = 1 + \iff b = \frac{1}{5}\\ + f_{X|A^\C}(x) &= \begin{cases} + c &\text{if }20 \le x \le 25\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{20}^{25}c\ud x = 1 + \iff c = \frac{1}{5} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{2}{15} &\text{if }15 \le x < 20\\ + \frac{1}{15} &\text{if }20 \le x < 25\\ + 0 &\text{otherwise} + \end{dcases}\] +\pagebreak + +\exercise 3 Given a random variable $X$ with the PDF +$f_X(x) = \dfrac{c}{x^2 + 1}$. +\begin{enumerate}[(a)] + \item By the normalization property, + \[\int_{-\infty}^\infty\frac{c}{x^2 + 1}\ud x = 1 + \iff \left.c\arctan x\right|_{-\infty}^\infty = 1 + \iff c\pi = 1 \iff c = \frac{1}{\pi}\] + \item The probability that $X^2$ to lie between 1/3 and 1 is + \begin{align*} + P\left(\frac{1}{3} < X^2 < 1\right) + &= P\left(-1 < X < \frac{-1}{\sqrt 3}\right) + + P\left(\frac{1}{\sqrt 3} < X < 1\right)\\ + &= \left.\frac{\arctan x}{\pi}\right|_{-1}^{\frac{-1}{\sqrt 3}} + + \left.\frac{\arctan x}{\pi}\right|_{\frac{1}{\sqrt 3}}^{1} + = \frac{1}{6} + \end{align*} + \item The CDF of $X$: + \[F_X(x) = \int_{-\infty}^{x}\frac{1}{\pi\left(x^2 + 1\right)}\ud t + = \frac{\arctan x}{\pi} + \frac{1}{2}\] +\end{enumerate} + +\exercise 4 Given a random variable $X$ with the CDF +\[F_X(x) = \begin{cases} + 1 - \exp(-2x) &\text{if } x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\] +\begin{enumerate}[(a)] + \item Let $g$ be the antiderivative of $f_X$, $g$ is a constant function + in $(-\infty, 0)$ and for all positive $x$ + \[g(x) = g(0) + 1 - \exp(-2x) + \Longrightarrow f_X(x) = 2\exp(-2x)\] + \item The probability that $X > 2$ is + \[P(X > 2) = P(\Omega) - P(X \le 2) = 1 - F_X(2) = \frac{1}{e^4}\] + \item The probability that $-3 < X \le 4$ is + \[P(-3 < X \le 4) = F_X(4) - F_X(-3) = 1 - \frac{1}{e^8}\] +\end{enumerate} + +\exercise 5 Given random variable with the following PDF: +\[f_X(x) = \begin{dcases} + \frac{10}{x^2} &\text{if }x > 10\\ + 0 &\text{otherwise} +\end{dcases} +\Longrightarrow F_X(x) = \begin{dcases} + 1 - \frac{10}{x} &\text{if }x > 10\\ + 0 &\text{otherwise} +\end{dcases}\tag{b}\] + +\[P(X > 20) = P(\Omega) - F_X(20) = 1 - 1 + \frac{10}{20} = \frac{1}{2}\tag{a}\] + +Let $Y$ be the number out of six devices that will function for at least +15 hours, $Y$ is a binomial random variable whose PMF is +\begin{align*} + p_Y(y) &= \binom{6}{y} P^y(X \ge 15) P^{6-y}(X < 15)\\ + &= \binom{6}{y} \left(1 - F_X(15)\right)^y F_X^{6-y}(15)\\ + &= \binom{6}{y} \left(\frac{10}{15}\right)^y + \left(1 - \frac{10}{15}\right)^{6-y}\\ + &= \binom{6}{y} \frac{2^y}{3^6} +\end{align*} + +Denote $A$ as the event that at least three out of six devices will function +for at least 15 hours, +\[P(A) = \sum_{y=3}^6 p_Y(y) = \frac{656}{729}\tag{c}\] + +\subsection{Expectation, Variance and STD\protect\footnote{No, not that STD.}} +\exercise 6 Given a random variable $X$ with the PDF +$f_X(x) = \frac{\lambda}{2}\exp(-\lambda|x|)$. +\begin{align*} + \int_{-\infty}^\infty f_X(x)\ud x + &= \int_{-\infty}^0 \frac{\lambda}{2}\exp(\lambda x)\ud x + + \int_0^\infty \frac{\lambda}{2}\exp(-\lambda x)\ud x\\ + &= \frac{1}{2}\left(\int_{-\infty}^0\ud\exp(\lambda x) + - \int_0^\infty\ud\exp(-\lambda x)\right)\\ + &= \frac{1 - (-1)}{2} = 1 +\end{align*} + +Let $g(x) = x f_X(x)$, the mean of $X$ is +$\E[X] = \displaystyle\int_{-\infty}^\infty g(x)\ud x$. Since $g(-x) = -g(x)$ +for all $x$, $\E[X] = 0$. + +The variance of $X$ can be calculated as +$\var(X) = \E\left[X^2\right] - \E^2[X] = \E\left[X^2\right]$ +or $\var(X) = \displaystyle\int_{-\infty}^\infty x^2 f_X(x)\ud x$. + +Let $h(x) = x^2 f_X(x)$, we have $h(-x) = h(x)$ for all $x$ and thus +\begin{align*} + \E\left[X^2\right] &= 2\int_{-\infty}^\infty h(x)\ud x\\ + &= 2\int_0^\infty\frac{\lambda x^2}{2}\exp(-\lambda x)\ud x\\ + &= 2\int_0^\infty\frac{x^2}{-2}\ud\exp(-\lambda x)\\ + &= 2\int_0^\infty\exp(-\lambda x)\ud\frac{x^2}{2} + - \int_0^\infty\ud x^2\exp(-\lambda x)\\ + &= 2\int_0^\infty x\exp(-\lambda x)\ud x\\ + &= \frac{-2}{\lambda}\int_0^\infty x\ud\exp(-\lambda x)\\ + &= \frac{2}{\lambda}\left(\int_0^\infty\exp(-\lambda x)\ud x + - \int_0^\infty\ud x\exp(-\lambda x)\right)\\ + &= \frac{-2}{\lambda^2}\int_0^\infty\ud\exp(-\lambda x) + = \frac{2}{\lambda^2} +\end{align*} + +\exercise 7 Given a random variable $X$ with PDF +\[f_X(x) = \begin{cases} + 2\exp(-2 x) &\text{if }x > 0\\ + 0 &\text{otherwise} +\end{cases}\] + +Since $X$ is exponentially distributed with the parameter $\lambda = 2$, +\[\E[X] = \sigma_X = \frac{1}{\lambda} = \frac{1}{2},\, +\var(X) = \frac{1}{\lambda^2} = \frac{1}{4}\] + +Thus $\E\left[X^2\right] = \var(X) + \E^2[X] = 1/2$. + +\exercise 8 Given a random variable $X$ with PDF +\[f_X(x) = \begin{cases} + a + bx^2 &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization probability, +\[\int_0^1\left(a + bx^2\right)\ud x = 1 \iff a + \frac{b}{3} = 1\] + +Since $\E[X] = 3/5$, +$\displaystyle\int_0^1\left(ax + 3x^3\right)\ud x = \frac{3}{5}$ +or $a/2 + b/4 = 0.6$ +and thus $a = 0.6$ and $b = 1.2$. + +\exercise 9 The expectation is +\begin{align*} + \E[X] &= \int_0^\infty\frac{x^2}{e^x}\ud x\\ + &= -\int_0^\infty x^2\ud e^{-x}\\ + &= \int_0^\infty e^{-x}\ud x^2 - \int_0^\infty\ud\frac{x^2}{e^x}\\ + &= \int_0^\infty\frac{2x}{e^x}\ud x\\ + &= -2\int_0^\infty x\ud e^{-x}\\ + &= 2\int_0^\infty e^{-x}\ud x - 2\int_0^\infty\ud\frac{x}{e^x}\\ + &= -2\int_0^\infty\ud e^{-x} = 2 +\end{align*} + +\exercise{10} The $X$ is exponentially distributed by the PDF +$f_X(x) = \dfrac{1}{3}\exp\dfrac{-x}{3}$. +\begin{enumerate}[(a)] + \item $\E[X] = \sigma_X = 3$ and $\var(X) = 9$. + \item The CDF of $X$ is $F_X(x) = 1 - \exp(-x/3)$ and thus + \[P(2 < X \le 4) = F_X(4) - F_X(2) + = \exp\frac{-2}{3} - \exp\frac{-4}{3}\] +\end{enumerate} +\pagebreak + +\section{Continuous Random Variable 2} +\exercise 1 Let $X$ be the time to repair the machine, +\begin{align*} + f_X(x) &= \begin{dcases} + \frac{1}{2}\exp\frac{-x}{2} &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases}\\ + \Longrightarrow F_X(x) &= \begin{dcases} + 1 - \exp\frac{-x}{2} &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} +\begin{enumerate}[(a)] + \item $P(X > 2) = 1 - F_X(2) = \exp\dfrac{-2}{2} = \dfrac{1}{e}$ + \item Let $A = \{X > 10\}$ and $B = \{X > 8\}$, we have + $P(B) = \exp\dfrac{-8}{2} = \dfrac{1}{e^4}$ and + $P(A\cap B) = P(A) = \exp\dfrac{-10}{2} = \dfrac{1}{e^5}$. + Hence the conditional probability that a repair exceeding eight hours + takes at least 10 hours is + \[P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{1}{e}\] +\end{enumerate} + +\exercise 2 Denote the event that the day is sunny as $A$, $P(A) = 2/3$. +\begin{align*} + f_{X|A}(x) &= \begin{cases} + b &\text{if }15 \le x \le 23\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{15}^{20}b\ud x = 1 + \iff b = \frac{1}{8}\\ + f_{X|A^\C}(x) &= \begin{cases} + c &\text{if }20 \le x \le 25\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{20}^{25}c\ud x = 1 + \iff c = \frac{1}{5} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{1}{12} &\text{if }15 \le x < 20\\ + \frac{3}{20} &\text{if }20 \le x < 23\\ + \frac{1}{15} &\text{if }23 \le x < 25\\ + 0 &\text{otherwise} + \end{dcases}\] + +\exercise 3 Let $X$ be the waiting time +and $A$ be the event that one arrives at the station before 7:15, +we have $P(A) = 0.25$, $P\left(A^\C\right) = 0.75$. + +\begin{align*} + f_{X|A}(x) &= \begin{dcases} + \frac{1}{5} &\text{if }0 \le x < 5\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_{X|A^\C}(x) &= \begin{dcases} + \frac{1}{15} &\text{if }5 \le x \le 15\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{1}{10} &\text{if }0 \le x < 5\\ + \frac{1}{20} &\text{if }5 \le x < 15\\ + 0 &\text{otherwise} + \end{dcases}\] + +\exercise 4 Let $X$ be a random variable with PDF +\[f_X(x) = \begin{dcases} + \frac{x}{4} &\text{if }1 < x \le 3\\ + 0 &\text{otherwise} +\end{dcases}\] +and $A = \{X \ge 2\}$. +\begin{enumerate}[(a)] + \item $X$ has the mean of + $\displaystyle\E[X] = \int_1^3\frac{x^2}{4}\ud x = \frac{13}{6}$. + The CDF of $X$ is + \[F_X(x) = \begin{dcases} + 0 &\text{if }x \le 1\\ + \frac{x^2 - 1}{8} &\text{if }1 < x \le 3\\ + 1 &\text{otherwise} + \end{dcases}\] + thus $P(A) = P(\Omega) - P(X < 2) = 1 - F_X(2) = \dfrac{5}{8}$. + \[f_{X|A}(x) = \frac{P(\{X = x\}\cap A)}{P(A)} + = \frac{8}{5}P(\{X = x\}\cap A)\] + It is trivial that $\{X = x\}\cap A = \varnothing$ if $x < 2$ + and $P(\{X = x\}\cap A) = f_X(x)$ otherwise, so + \[f_{X|A}(x) = \begin{dcases} + \frac{2x}{5} &\text{if }2 \le x \le 3\\ + 0 &\text{otherwise} + \end{dcases} + \Longrightarrow \E[X|A] = \int_2^3\frac{2x^2}{5}\ud x = \frac{38}{15}\] + \item Let $Y = X^2$, the $Y$ has the expectation of + \[\E[Y] = \E\left[X^2\right] = \int_1^3\frac{x^3}{4}\ud x = 5\] + + The variance of $Y$ is + \[\var(Y) = \E\left[Y^2\right] - \E^2[Y] = \E\left[X^4\right] - 5^2 + = \int_1^3\frac{x^5}{4}\ud x - 25 = \frac{16}{3}\] +\end{enumerate} + +\exercise 5 Let $X$ be Alyssa's waiting time and $A$ be the event there is +a customer ahead, then $P(A) = P\left(A^\C\right) = 0.5$ and +\begin{align*} + f_{X|A}(x) &= \begin{cases} + \lambda\exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\\ + \Longrightarrow F_{X|A}(x) &= \begin{cases} + 1 - \exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\\ + p_{X|A^\C}(x) &= \begin{cases} + 1 &\text{if }x = 0\\ + 0 &\text{otherwise} + \end{cases}\\ + \Longrightarrow F_{X|A^\C}(x) &= \begin{cases} + 1 &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases} +\end{align*} + +Therefore +\begin{align*} + F_X(x) &= P(A)F_{X|A}(x) + P\left(A^\C\right)F_{X^\C}(x)\\ + &= \begin{dcases} + 1 - 0.5\exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +\exercise 6 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + cxy &\text{if }0 < x < 4\text{ and }1 < y < 5\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item By the normalization probability + \[\int_0^4\int_1^5 cxy\ud y\ud x = 1 \iff 96c = 1 \iff c = \frac{1}{96}\] + \item $\displaystyle P(1 < X < 2, 2 < Y < 3) + = \int_1^2\int_2^3\frac{xy}{96}\ud y\ud x = \frac{5}{128}$ + \item $\displaystyle P(X \ge 3, Y \le 2) + = \int_3^4\int_1^2\frac{xy}{96}\ud y\ud x = \frac{7}{128}$ + \item The marginal PDF of $X$ is $\displaystyle f_X(x) + = \int_1^5\frac{xy}{96}\ud y = \frac{x}{8}$ and that of $Y$ is + $\displaystyle f_Y(y) = \int_0^4\frac{xy}{96}\ud x = \frac{y}{12}$. + \item The region with nonzero probability where $X + Y < 3$ is + $\{(x, y) \in \mathbb R^2 \mid 0 < x < 2, 1 < y < 3 - x\}$, thus + \[P(X + Y < 3) = \int_0^2\int_1^{3-x}\frac{xy}{96}\ud y\ud x + = \int_0^2\frac{x^3 - 6x^2 + 8x}{192}\ud x = \frac{1}{48}\] + \item Let $C_u$ be the line $X + 2Y = u$, then the PDF of $U = X + 2Y$ is + \[f_U(u) = P(X + 2Y = u) = \int_{C_u}f_{X,Y}(x, y)\ud s\] + where $\ud s$ is the infinitesimal length of $C_u$. + + Let $t$ satisfy $x = u - 2t$ and $y = t$ we get + \begin{align*} + f_U(u) &= \int_{-\infty}^\infty f_{X,Y}(u - 2t, t) + \sqrt{\left(\frac{\ud x}{\ud t}\right)^2 + + \left(\frac{\ud y}{\ud t}\right)^2}\ud t\\ + &= \int_{-\infty}^\infty f_{X,Y}(u - 2t, t)\sqrt 5\ud t + \end{align*} + + For $u < 2$, with $x \in (0, 4)$, + \[0 < u - 2t < 4 \Longrightarrow 2t < u < 2 \iff y = t < 1\] + and thus $f_{X,Y}(u - 2t, t) = 0$. Similarly, the joint PDF of $X$ and $Y$ + also equals zero when $u > 15$, so $f_U(u) = 0$ + for $u \in \mathbb R \setminus [2, 14]$. + + For $u \in [2, 6]$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff \begin{cases} + 0 < u - 2t < 4\\ + 1 < t < 5 + \end{cases}\\ + &\iff \frac{u}{2} - 2 \le 1 < t < \frac{u}{2} < 5\\ + &\iff 1 < t < \frac{u}{2} + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96}\int_1^{u/2}(ut - 2t^2)\ud t + = \frac{\sqrt 5}{2304}(u^3 - 12u + 16)$. + + For $u \in (6, 10)$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff 1 < \frac{u}{2} - 2 < t < \frac{u}{2} < 5\\ + &\iff \frac{u}{2} - 2 < t < \frac{u}{2} + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96} + \int_{u/2-2}^{u/2}(ut - 2t^2)\ud t + = \frac{\sqrt 5}{288}(u^2 - 2u)$. + + For $u \in [10, 14]$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff 1 < \frac{u}{2} - 2 < t < 5 \le \frac{u}{2}\\ + &\iff \frac{u}{2} - 2 < t < 5 + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96}\int_{u/2-2}^5(ut - 2t^2)\ud t + = \frac{\sqrt 5}{2304}(348u - u^3 - 2128)$. +\end{enumerate} + +\exercise 7 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + 8xy &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item The marginal PDFs are + \begin{align*} + f_X(x) &= \begin{dcases} + \int_0^x 8xy\ud y = 4x^3 &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \begin{dcases} + \int_y^1 8xy\ud x = 4y - 4y^3 &\text{if }0 \le y \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item The conditional PDFs are + \begin{align*} + f_{X|Y}(x|y) = \frac{f_{X,Y}(x, y)}{f_Y(y)} &= \begin{dcases} + \frac{2x}{1 - y^2} &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_{Y|X}(y|x) = \frac{f_{X,Y}(x, y)}{f_X(x)} &= \begin{dcases} + \frac{2y}{x^2} &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item The conditional expectations are + \begin{align*} + \E[X|Y=y] &= \int_{-\infty}^\infty xf_{X|Y}(x|y)\ud x\\ + &= \begin{dcases} + \int_y^1\frac{2x^2}{1-y^2}\ud x = \frac{2x^2 + 2x + 2}{3x + 3} + &\text{if }0 \le y \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + \E[Y|X=x] &= \int_{-\infty}^\infty yf_{Y|X}(y|x)\ud y\\ + &= \begin{dcases} + \int_0^x\frac{2y^2}{x^2}\ud y = \frac{2x}{3} &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item It is trivial that given $X = x \in \mathbb R \setminus [0, 1]$, + $\var(Y|X=x) = 0$. Otherwise, + \begin{align*} + \var(Y|X=x) &= \E\left[Y^2|X=x\right] - \E^2[Y|X=x]\\ + &= \int_0^x\frac{2y^3}{x^2}\ud y - \left(\frac{2x}{3}\right)^2 + = \frac{x^2}{2} - \frac{4x^2}{9} = \frac{x^2}{16} + \end{align*} +\end{enumerate} + +\exercise 8 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + \exp(-x-y) &\text{if }x \ge 0\text{ and }y \ge 0\\ + 0 &\text{otherwise} +\end{cases}\] + +The marginal PDFs are +\begin{align*} + f_X(x) &= \begin{dcases} + \int_0^\infty\exp(-x-y)\ud y = \exp(-x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \begin{dcases} + \int_0^\infty\exp(-x-y)\ud x = \exp(-y) &\text{if }y \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +It is noticeable that $X$ and $Y$ are independent, +thus $f_{X|Y}(x|y) = f_X(x)$ and $f_{Y|X}(y|x) = f_Y(y)$. +\pagebreak + +\section{Continuous Random Variable 3} +\subsection{PDF and CDF} +\exercise 1 Given $Z \sim \N(0, 1)$. +\begin{enumerate}[(a)] + \item $P(Z > 1.2) = 1 - \Phi(1.2) = 1 - 0.8849 = 0.1101$ + \item $P(-2 < Z < 2) = \Phi(2) - \Phi(-2) = 2\Phi(2) - 1 + = 2\cdot 0.9772 - 1 = 0.9544$ + \item $P(-1.2 < Z < 1) = \Phi(1) + \Phi(1.2) - 1 + = 0.8413 + 0.8849 - 1 = 0.7262$ +\end{enumerate} + +\exercise 2 Given $X \sim \N(4, 9)$. Let $Y = \dfrac{X - 4}{3}$, +$Y$ is a standard normal random variable +and $F_X(x) = \Phi\left(\dfrac{x - 4}{3}\right)$. +\begin{enumerate}[(a)] + \item $P(X > 6) = 1 - F_X(6) = 1 - \Phi\left(\dfrac{6 - 4}{3}\right) + \approx 1 - \Phi(0.67) = 0.2514$ + \item $P(X > 1) = 1 - F_X(6) = 1 - \Phi\left(\dfrac{1 - 4}{3}\right) + = 1 - \Phi(-1) = \Phi(1) = 0.8413$ +\end{enumerate} + +\exercise 3 Let $X$ be the annual snowfall in inches +and $Y = \dfrac{X - 60}{20}$, we have $Y \sim \N(0, 1)$ +and $F_X(x) = \Phi\left(\dfrac{x - 60}{20}\right)$. +The probability that snowfall will be at least 80 inches is +\[P(X \ge 80) = 1 - F_X(80) = 1 - \Phi\left(\frac{80 - 60}{20}\right) += 1 - \Phi(1) = 1 - 0.8413 = 0.1587\] + +\exercise 4 Let $X$ be the number of customers arriving during an one-hour +period, $X$ is a Poisson random variable whose PMF is +\begin{multline*} + p_X(x) = \frac{24^x}{e^{24}x!},\qquad x \in \mathbb N\\ + \Longrightarrow P(X < 15) = \sum_{x=0}^{14}p_X(x) + = \sum_{x=0}^{14}\frac{24^x}{e^{24}x!} \approx 0.019825332823463673 +\end{multline*} + +\subsection{Covariance and Correlation Coefficient} +\exercise 5 Given the joint PMF of $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }x \in \{0, 1, 2\}\text{ and }y \in \{0, 1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization property, +\[\sum_{x=0}^2\sum_{y=0}^3 c(2x + y) = 1 \iff c = \frac{1}{42}\] +\begin{enumerate}[(a)] + \item The maginal PMFs are + \begin{align*} + p_X(x) &= \sum_{y=0}^3 p_{X,Y}(x, y) = \begin{dcases} + \frac{4x + 3}{21} &\text{if }x \in \{0, 1, 2\}\\ + 0 &\text{otherwise} + \end{dcases}\\ + p_Y(y) &= \sum_{x=0}^2 p_{X,Y}(x, y) = \begin{dcases} + \frac{y + 2}{14} &\text{if }y \in \{0, 1, 2, 3\}\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + + Therefore we can compute these expectations: + \begin{align*} + \E[X] &= \sum_{x=0}^2 x p_X(x) + = \sum_{x=0}^2\frac{4x^2 + 3x}{21} = \frac{29}{21}\\ + \E[Y] &= \sum_{y=0}^3 y p_Y(y) + = \sum_{y=0}^3\frac{y^2 + 2y}{14} = \frac{13}{7}\\ + \E[XY] &= \sum_{x=0}^2\sum_{y=0}^3 xy p_{X,Y}(x,y) + = \sum_{x=0}^2\sum_{y=0}^3\frac{2x^2 y + xy^2}{42} = \frac{17}{7} + \end{align*} + \item The variances of these variables can be calculated as + \begin{align*} + \var(X) &= \E\left[X^2\right] - \E^2[X] = \frac{230}{21}\\ + \var(Y) &= \E\left[Y^2\right] - \E^2[Y] = \frac{25}{7} + \end{align*} + where + \begin{align*} + \E\left[X^2\right] &= \sum_{x=0}^2 x^2 p_X(x) + = \sum_{x=0}^2\frac{4x^3 + 3x^2}{21} = \frac{17}{7}\\ + \E\left[Y^2\right] &= \sum_{y=0}^3 y^2 p_Y(y) + = \sum_{y=0}^3\frac{y^3 + 2y^2}{14} = \frac{32}{7} + \end{align*} + \item $\cov(X, Y) = \E[XY] - \E[X]\E[Y] = \dfrac{-20}{147}$ so + \[\rho(X, Y) = \dfrac{\cov(X, Y)}{\sqrt{\var(X)\var(Y)}} \approx -0.027\] +\end{enumerate} + +\exercise 6 Given the joint PDF of $X$ and $Y$ as followed +\[f_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }(x, y) \in (2, 6) \times (0, 5)\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization property, +\[\int_2^6\int_0^5 c(2x + y)\ud y\ud x = 1 \iff c = \frac{1}{210}\] +\begin{enumerate}[(a)] + \item The maginal PMFs are + \begin{align*} + f_X(x) &= \int_0^5 f_{X,Y}(x, y)\ud y = \begin{dcases} + \frac{4x + 5}{84} &\text{if }2 < x < 6\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \int_2^6 f_{X,Y}(x, y)\ud x = \begin{dcases} + \frac{2y + 16}{105} &\text{if }0 < y < 5\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + + Therefore we can compute these expectations: + \begin{align*} + \E[X] &= \int_2^6 x f_X(x)\ud x + = \int_1^6\frac{4x^2 + 5x}{84}\ud x = \frac{268}{63}\\ + \E[Y] &= \int_0^5 y f_Y(y)\ud y + = \int_0^5\frac{2y^2 + 16y}{105}\ud y = \frac{170}{63}\\ + \E[XY] &= \int_2^6\int_0^5 xy f_{X,Y}(x,y)\ud y\ud x + = \int_2^6\int_0^5\frac{2x^2 y + xy^2}{210}\ud y\ud x = \frac{80}{7} + \end{align*} + \item The variances of these variables can be calculated as + \begin{align*} + \var(X) &= \E\left[X^2\right] - \E^2[X] = \frac{5036}{3969}\\ + \var(Y) &= \E\left[Y^2\right] - \E^2[Y] = \frac{16225}{7938} + \end{align*} + where + \begin{align*} + \E\left[X^2\right] &= \int_2^6 x^2 f_X(x) + = \int_2^6\frac{4x^3 + 5x^2}{84} = \frac{1220}{63}\\ + \E\left[Y^2\right] &= \int_0^5 y^2 f_Y(y) + = \int_0^5\frac{2y^3 + 16y^2}{105} = \frac{1175}{126} + \end{align*} + \item $\cov(X, Y) = \E[XY] - \E[X]\E[Y] = \dfrac{-200}{3969}$ so + \[\rho(X, Y) = \dfrac{\cov(X, Y)}{\sqrt{\var(X)\var(Y)}} \approx -0.0313\] +\end{enumerate} + +\subsection{Derived Distribution} +\exercise 7 With $X$ being uniform on $[0, 1]$, by the normalization property, +$f_X(x) = 1$ $\Longrightarrow F_X(x) = x$ on this interval. Given $Y = \sqrt X$, +\[F_Y(y) = P(Y \le y) = P\left(\sqrt X \le y\right) = F_X\left(y^2\right) = y^2 +\Longrightarrow f_y(y) = \frac{\ud F_Y}{\ud y} = 2y\] +if $0 < Y < 1$, otherwise $f_Y(y) = 0$. + +\exercise 8 Let $X$ be the speed in miles per hour, +\[f_X(x) = \begin{dcases} + \frac{1}{30} &\text{if }30 \le x \le 60\\ + 0 &\text{otherwise} +\end{dcases} +\Longrightarrow F_X(x) = \begin{dcases} + 0 &\text{if }x < 30\\ + \frac{x}{30} - 1 &\text{if }30 \le x < 60\\ + 1 &\text{otherwise} +\end{dcases}\] +then the duration of the trip is $Y = 180/X$. Where $3 \le Y \le 6$, +\begin{multline*} + F_Y(y) = P\left(\frac{180}{X} \le y\right) + = P\left(X \ge \frac{180}{y}\right) + = 1 - F_X\left(\frac{180}{y}\right) = 2 - \frac{6}{y}\\ + \Longrightarrow f_Y(y) = \frac{\ud F_Y}{\ud y} = \frac{6}{y^2} +\end{multline*} + +Otherwise, it is obvious that $f_Y(y) = 0$. + +\exercise{9}\footnote{IMHO this is a really poor example to demonstrate +the usefulness of this method.} Let $A$ be the event that one arrives at +the station before 7:15, we have $P(A) = 0.25$ and $P\left(A^\C\right) = 0.75$. +The probability that $X = x$ is 0.05 for all $x \in [0, 20)$ +and is 0 otherwise, thus +\begin{align*} + F_{X|A}(x) &= \begin{dcases} + 0 &\text{if }x < 0\\ + \frac{\int_0^x 0.05\ud t}{0.25} = \frac{0.05x}{0.25} + &\text{if }0 \le x < 5\\ + 1 &\text{otherwise} + \end{dcases}\\ + F_{X|A^\C}(x) &= \begin{dcases} + 0 &\text{if }x < 5\\ + \frac{\int_5^x 0.05\ud t}{0.25} = \frac{0.05x - 0.25}{0.75} + &\text{if }5 \le x < 20\\ + 1 &\text{otherwise} + \end{dcases} +\end{align*} + +With $Y = 5 - X$ if $A$ and $Y = 20 - X$ otherwise, +\begin{multline*} + \begin{aligned} + F_Y(y) &= P(Y \le y)\\ + &= 1 - P(Y > y)\\ + &= P(A)P(5 - X > y|A) + P\left(A^\C\right)P\left(20 - X > y|A^\C\right)\\ + &= 1 - 0.25P(X < 5 - y|A) - 0.75P\left(X < 20 - y|A^\C\right)\\ + &= 1 - 0.25F_{X|A}(5 - y) - 0.75F_{X|A^\C}(20 - y)\\ + &= \begin{dcases} + 0 &\text{if }y < 0\\ + 1 - 0.05(5 - y) - 0.05(20 - y) + 0.25 = 0.1y &\text{if }0 \le y < 5\\ + 1 - 0.05(20 - y) + 0.25 = 0.05y + 0.25 &\text{if }5 \le y < 15\\ + 1 &\text{otherwise} + \end{dcases} + \end{aligned}\\ + \Longrightarrow f_Y(y) = \frac{\ud F_Y}{\ud y} = \begin{dcases} + 0.1 &\text{if }0 \le y < 5\\ + 0.05 &\text{if }5 \le y < 15\\ + 0 &\text{otherwise} + \end{dcases} +\end{multline*} +\pagebreak + +\setcounter{section}{8} +\section{Limit Theorem} +\exercise 1 Let $S_{100} = \sum_{i=1}^{100}X_i$, $M_{100} = S_{100}/100$ +is the sample mean. We have +\[Z_{100} = \frac{S_{100} - 100\cdot 10}{4\sqrt{100}} = 2.5M_{100} - 25\] + +Since 100 is large, we can use the approximation +$P(Z_{100} \le z) \approx \Phi(z)$: +\begin{multline*} + P(S_{100} \le 900) = P(M_{100} \le 9) = P(2.5M_{100} - 25 \le 22.5 - 25)\\ + = P(Z_n \le -2.5) \approx \Phi(-2.5) = 1 - \Phi(2.5) = 0.0062 +\end{multline*} + +\exercise 2 Let $X$ be the weight of a box in lbs, and denote +\[Z_{49} = \frac{\sum_{i=1}^{49}X_i - 49\cdot 205}{15\sqrt{49}} += \frac{S_{49}}{105} - \frac{287}{3}\] + +Since 49 is large, we can use the following approximation +to compute the probability that all 49 boxes can be +safely loaded onto the freight elevator and transported +\[P(S_{49} \le 9800) = P\left(Z_{49} \le \frac{-7}{3}\right) +\approx 1 - \Phi\left(\frac 7 3\right) = 0.0099\] + +\exercise 3 Let $X$ be the number of tickets to be purchased by a student, +and denote +\[Z_{100} = \frac{\sum_{i=1}^{100}X_i - 100\cdot 2.4}{2\sqrt{100}} += \frac{S_{100}}{20} - 12\] + +Since 100 is large, we can use the following approximation +to compute the probability that all 100 students will be able to purchase +the tickets they desire from the 250 that is left: +\[P(S_{100} \le 250) = P(Z_{100} \le 0.5) +\approx \Phi(0.5) = 0.6915\] + +\exercise 4 Let $X$ be the time in minutes to complete one problem, +and denote +\[Z_{40} = \frac{\sum_{i=1}^{40}X_i - 40\cdot 5}{2\sqrt{40}} += \frac{S_{40} - 200}{4\sqrt{10}}\] + +Since 40 is large, we can use the following approximation +to compute the probability that all 40 problems within 3 hours +\[P(S_{40} \le 180) = P\left(Z_{49} \le -\sqrt\frac{5}{2}\right) +\approx 1 - \Phi(1.58) = 0.0571\] + +\exercise 5 Let $X$ be the size in MB of an image and denote +\[Z_{80} = \frac{\sum_{i=1}^{80}X_i - 80\cdot 0.6}{0.4\sqrt{80}} += \frac{5S_{40} - 240}{8\sqrt{5}}\] + +Since 80 is large, we can use the following formula to approximate +the probability that the total size is between 47 and 56 MB +\[P(47 \le S_{80} \le 56) += P\left(\frac{\sqrt 5}{-8} \le Z_{49} \le \sqrt 5\right) +\approx \Phi(2.23) - 1 + \Phi(0.28) = 0.5974\] + +\exercise 6 After 11 weeks, the station is supplied +\[74\,000 + 47\,000 \cdot 11 = 591\,000\text{ (gallons)}\] + +Let $X$ be the gasoline consumption in gallons a week and denote +\[Z_{11} = \frac{\sum_{i=1}^{11}X_i - 11\cdot 50\,000}{10\,000\sqrt{11}} += \frac{S_{11} - 550\,000}{10\,000\sqrt{11}}\] + +While 11 is not exactly large, for the ease of calculation, +we still use the approximation $P(Z_{11} \le z) \approx \Phi(z)$. + +\begin{enumerate}[(a)] + \item The probability that the remain will be below 20\,000 gallons is + \begin{multline*} + P(591\,000 - S_{11} < 20\,000) + = P(S_{11} > 571\,000) + = P\left(Z_{11} > \frac{21}{10\sqrt{11}}\right)\\ + \approx 1 - \Phi(0.63) = 0.2643 + \end{multline*} + \item Let $w$ be the weekly delivery satisfying the probability + that below 20\,000 gallons will be remained is 0.5\%, we have + \begin{align*} + &P(74\,000 + 11w - S_{11} < 20\,000) = 0.005\\ + \iff &P(S_{11} > 54\,000 + 11w) = 0.005\\ + \iff &P\left(Z_{11}>\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.005\\ + \iff &1 - \Phi\left(\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.005\\ + \iff &\Phi\left(\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.995\\ + \iff &\frac{11w-496\,000}{10\,000\sqrt{11}} = 2.57\\ + \iff &w = 52840 + \end{align*} +\end{enumerate} +\end{document} -- cgit 1.4.1