[usth/MATH1.{4,5}] Calculus (not the teeth kind)

8 files changed, 4676 insertions, 7 deletions

diff --git a/README.md b/README.md
index 4ebe2ae..25b4c80 100644
--- a/README.md
+++ b/README.md
@@ -1,21 +1,23 @@
-# hsg
+# cp
 
-Bài tập luyện tập thi Olympic, học sinh giỏi Tin học, trong đó:
+This used to be my competitive programming collection, now it is more of
+a warehouse for my computer programming journey.  For historical reasons,
+this README as well as commit messages are duolingo (Anglais et Vietnamien).
 
 |    Thư mục   |                      Nguồn đề bài                      |
 | ------------ | ------------------------------------------------------ |
 | `\d{2}`      | Đề thi, kiểm tra phân theo lớp                         |
-| `COCI`       | [Giải Tin học Croatia mở rộng][0]                      |
-| `NTU`        | [Đại học Nha Trang][1]                                 |
-| `THT`        | Hội thi Tin học trẻ                                    |
+| `coci`       | [Giải Tin học Croatia mở rộng][0]                      |
 | `codechef`   | [Codechef][2]                                          |
 | `codeforces` | [Codeforces][3]                                        |
 | `cpptour`    | A Tour of C++                                          |
 | `daily`      | [/r/dailyprogrammer][4]                                |
+| `ntu`        | [Đại học Nha Trang][1]                                 |
 | `others`     | Các đề bài không rõ nguồn                              |
 | `paip`       | Paradigms of Artificial Intelligence Programming       |
 | `sicp`       | Structure and Interpretation of Computer Programs      |
 | `thinkperl6` | Think Perl 6                                           |
+| `tht`        | Hội thi Tin học trẻ                                    |
 | `toys`       | Programs that don't deserve their own repo             |
 | `usth`       | L'Université des Sciences et des Technologies de Hanoï |
 
@@ -25,7 +27,7 @@ Bài tập luyện tập thi Olympic, học sinh giỏi Tin học, trong đó:
 [3]: http://codeforces.com/
 [4]: https://www.reddit.com/r/dailyprogrammer
 
-Ở mỗi thư mục con sẽ có tệp `README.md` ghi lại đề bài. Riêng `COCI`, `NTU` và
+Ở mỗi thư mục con sẽ có tệp `README.md` ghi lại đề bài. Riêng `coci`, `ntu` và
 `codeforces` sẽ chỉ có danh sách đường dẫn tới các đề bài. Đề bài sẽ được cập
 nhật dần.
 
@@ -38,7 +40,7 @@ Phiên bản các trình dịch sử dụng test:
 | Java        | OpenJDK 11+        |
 | Lua         | Lua 5.1+           |
 | Pascal      | Free Pascal 2.6.4+ |
-| Perl 6      | Rakudo 2018.12+    |
+| Raku        | Rakudo 2018.12+    |
 | Python      | Python 3.5+        |
 | Scheme      | GNU Guile 2.0.11+  |
 
diff --git a/usth/MATH1.4/cons-sequences/README.md b/usth/MATH1.4/cons-sequences/README.md
new file mode 100644
index 0000000..493ae32
--- /dev/null
+++ b/usth/MATH1.4/cons-sequences/README.md
@@ -0,0 +1,296 @@
+# Infinite Sequences: A Case Study in Functional Python
+
+In this article, we will only consider sequences defined by a function whose
+domain is a subset of the set of all integers. Such sequences will be
+*visualized*, i.e. we will try to evaluate the first few (thousand) elements,
+using functional programming paradigm, where functions are more similar to the
+ones in math (in contrast to imperative style with side effects confusing to
+inexperenced coders). The idea is taken from
+[subsection 3.5.2 of SICP](https://mitpress.mit.edu/sites/default/files/sicp/full-text/book/book-Z-H-24.html#%_sec_3.5.2)
+and adapted to Python, which, compare to Scheme, is significantly more popular:
+Python is pre-installed on almost every modern Unix-like system, namely macOS,
+GNU/Linux and the \*BSDs; and even at MIT, the new 6.01 in Python has recently
+replaced the legendary 6.001 (SICP).
+
+One notable advantage of using Python is its huge **standard** library. For
+example the "identity sequence" (sequence defined by the identity function) can
+be imported directly from `itertools`:
+
+```python
+>>> from itertools import count
+>>> positive_integers = count(start=1)
+>>> next(positive_integers)
+1
+>>> next(positive_integers)
+2
+>>> for _ in range(4): next(positive_integers)
+... 
+3
+4
+5
+6
+```
+
+To open a Python emulator, simply lauch your terminal and run `python`. If that
+is somehow still too struggling, navigate to
+[the interactive shell on Python.org](https://www.python.org/shell/).
+
+*Let's get it started* with somethings everyone hates: recursively defined
+sequences, e.g. the famous Fibonacci (*F*<sub>*n*</sub> = *F*<sub>*n*-1</sub> +
+*F*<sub>*n*-2</sub>, *F*<sub>1</sub> = 1, *F*<sub>0</sub> = 0). Since
+[Python doesn't support](http://neopythonic.blogspot.com/2009/04/final-words-on-tail-calls.html)
+[tail recursion](https://mitpress.mit.edu/sites/default/files/sicp/full-text/book/book-Z-H-11.html#call_footnote_Temp_48),
+it's generally **not** a good idea to define anything recursively (which is,
+ironically, the only trivial *functional* solution in this case) but since we
+will only evaluate the first few terms (use the **Tab** key to indent the line
+when needed):
+
+```python
+>>> def fibonacci(n, a=0, b=1):
+... 	# To avoid making the code look complicated, n < 0 is not handled here.
+... 	return a if n == 0 else fibonacci(n - 1, b, a + b)
+... 
+>>> fibo_seq = (fibonacci(n) for n in count(start=0))
+>>> for _ in range(7): next(fibo_seq)
+... 
+0
+1
+1
+2
+3
+5
+8
+```
+
+<details><summary>Note (click to expand)</summary>
+The <code>fibo_seq</code> above is just to demonstrate how
+<code>itertools.count</code> can be use to create an infinite sequence defined
+by a function. For better performance, this should be used instead
+
+```python
+def fibonacci_sequence(a=0, b=1):
+    yield a
+    yield from fibonacci_sequence(b, a + b)
+```
+</details>
+
+It is noticable that the elements having been iterated through (using `next`)
+will disappear forever in the void (oh no!), but that is the cost we are
+willing to pay to save some memory, especially when we need to evaluate a
+member of (arbitrarily) large index to estimate the sequence's limit. One case
+in point is estimating a definite integral using
+[left Riemann sum](https://en.wikipedia.org/wiki/Riemann_sum#Left_Riemann_sum):
+
+```python
+>>> def integral(f, a, b):
+... 	def leftRiemannSum(n):
+... 		dx = (b-a) / n
+... 		def x(i): return a + i*dx
+... 		return sum(f(x(i)) for i in range(n)) * dx
+... 	return leftRiemannSum
+... 
+```
+
+The function `integral(f, a, b)` as defined above returns a function taking *n*
+as an argument. As *n* approaches ∞, its result approaches the value of the
+integral of *f* on [*a, b*]. For example, we are going to estimate π as the
+area of a semicircle whose radius is sqrt(2):
+
+```python
+>>> from math import sqrt
+>>> def semicircle(x): return sqrt(abs(2 - x*x))
+... 
+>>> pi = integral(semicircle, -sqrt(2), sqrt(2))
+>>> pi_seq = (pi(n) for n in count(start=2))
+>>> for _ in range(3): next(pi_seq) # the first few aren't quite close
+... 
+2.000000029802323
+2.514157464087051
+2.7320508224700384
+```
+
+At index around 1000, the result is somewhat acceptable:
+
+    3.1414873191059525
+    3.1414874770617427
+    3.1414876346231577
+
+Since we are comfortable with sequence of sums, let's move on to sums of a
+sequence, which are called series. For estimation, again, we are going to make
+use of infinite sequences of partial sums, which are implemented as
+`itertools.accumulate` by thoughtful Python developers.
+[Geometric](https://en.wikipedia.org/wiki/Geometric_series) and
+[*p*-series](https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/p-series.html)
+can be defined as follow:
+
+```python
+>>> from itertools import accumulate as partial_sums
+>>> def geometric_series(r, a=1): return partial_sums(a*r**n for n in count(0))
+... 
+>>> def p_series(p): return partial_sums(1 / n**p for n in count(1))
+... 
+```
+
+We can then use these to determine whether the series is convergent or
+divergent. For instance, the fact that *p*-series with *p* = 2 converges to
+π<sup>2</sup>/6 ≈ 1.6449340668482264 can be verified via
+
+```python
+>>> s = p_series(p=2)
+>>> for _ in range(11): next(s)
+... 
+1.0
+1.25
+1.3611111111111112
+1.4236111111111112
+1.4636111111111112
+1.4913888888888889
+1.511797052154195
+1.527422052154195
+1.5397677311665408
+1.5497677311665408
+1.558032193976458
+```
+
+We can observe that it takes quite a lot of steps to get the precision we would
+generally expect (*s*<sub>11</sub> is only precise to the first decimal place;
+second decimal places: *s*<sub>101</sub>; third: *s*<sub>2304</sub>). Luckily,
+many techniques for series acceleration are available.
+[Shanks transformation](https://en.wikipedia.org/wiki/Shanks_transformation),
+for instance, can be implemented as follow:
+
+```python
+>>> from itertools import islice, tee
+>>> def shanks(seq):
+... 	return map(lambda x, y, z: (x*z - y*y) / (x + z - y*2),
+... 	           *(islice(t, i, None) for i, t in enumerate(tee(seq, 3))))
+... 
+```
+
+In the code above, `lambda x, y, z: (x*z - y*y) / (x + z - y*2)` denotes the
+anonymous function (*x*, *y*, *z*) ↦ (*xz*-*y*<sup>2</sup>)/(*x*+*z*-2*y*) and
+`map` is a higher order function applying that function to respective elements
+of subsequences starting from index 1, 2, 3 of `seq`. On Python 2, one should
+import `imap` from `itertools` to get the same
+[lazy](https://en.wikipedia.org/wiki/Lazy_evaluation) behavior of `map` on
+Python 3.
+
+```python
+>>> s = shanks(p_series(2))
+>>> for _ in range(10): next(s)
+... 
+1.4500000000000002
+1.503968253968257
+1.53472222222223
+1.5545202020202133
+1.5683119658120213
+1.57846371882088
+1.5862455815659202
+1.5923993101138652
+1.5973867787856946
+1.6015104548459742
+```
+
+The result was quite satisfying, yet we can do one step futher by continuously
+applying the transformation to the sequence:
+
+```python
+>>> def compose(transform, seq):
+... 	yield next(seq)
+... 	yield from compose(transform, transform(seq))
+... 
+>>> s = compose(shanks, p_series(2))
+>>> for _ in range(10): next(s)
+... 
+1.0
+1.503968253968257
+1.5999812811165188
+1.6284732442271674
+1.6384666832276524
+1.642311342667821
+1.6425249569252578
+1.640277484549416
+1.6415443295058203
+1.642038043478661
+```
+
+Shanks transformation works on every sequence (not just sequences of partial
+sums). Back to previous example of using left Riemann sum to compute definite integral:
+
+```python
+>>> pi_seq = compose(shanks, map(pi, count(2)))
+>>> for _ in range(10): next(pi_seq)
+... 
+2.000000029802323
+2.978391111182236
+3.105916845397819
+3.1323116570377185
+3.1389379264270736
+3.140788413965646
+3.140921512857936
+3.1400282163913436
+3.1400874774021816
+3.1407097229603256
+>>> next(islice(pi_seq, 300, None))
+3.1415061302492413
+```
+
+Now having series defined, let's see if we can learn anything
+about power series. Sequence of partial sums of power series
+∑*c*<sub>*n*</sub>(*x*-*a*)<sup>*n*</sup> can be defined as
+
+```python
+>>> from operator import mul
+>>> def power_series(c, start=0, a=0):
+>>> 	return lambda x: partial_sums(map(mul, c, (x**n for n in count(start))))
+... 
+```
+
+We can use this to compute functions that can be written as
+[Taylor series](https://en.wikipedia.org/wiki/Taylor_series):
+
+```python
+>>> from math import factorial
+>>> def exp(x): return power_series(1/factorial(n) for n in count(0))(x)
+... 
+>>> def cos(x):
+... 	c = ((1 - n%2) * (1 - n%4) / factorial(n) for n in count(0))
+... 	return power_series(c)(x)
+... 
+>>> def sin(x):
+... 	c = (n%2 * (2 - n%4) / factorial(n) for n in count(1))
+... 	return power_series(c, start=1)(x)
+... 
+```
+
+Amazing! Let's test 'em!
+
+```python
+>>> e = compose(shanks, exp(1)) # this should converges to 2.718281828459045
+>>> for _ in range(4): next(e)
+... 
+1.0
+2.749999999999996
+2.718276515152136
+2.718281825486623
+```
+
+Impressive, huh? For sine and cosine, series acceleration is not even necessary:
+
+```python
+>>> from math import pi as PI
+>>> s = sin(PI/6)
+>>> for _ in range(5): next(s)
+... 
+0.5235987755982988
+0.5235987755982988
+0.49967417939436376
+0.49967417939436376
+0.5000021325887924
+>>> next(islice(cos(PI/3), 8, None))
+0.500000433432915
+```
+
+[![Creative Commons License](https://i.creativecommons.org/l/by-sa/4.0/88x31.png)](http://creativecommons.org/licenses/by-sa/4.0/)
+This work is licensed under a
+[Creative Commons Attribution-ShareAlike 4.0 International License](http://creativecommons.org/licenses/by-sa/4.0/).
diff --git a/usth/MATH1.4/homework/limits.pdf b/usth/MATH1.4/homework/limits.pdf
new file mode 100644
index 0000000..082bb88
--- /dev/null
+++ b/usth/MATH1.4/homework/limits.pdf
Binary files differdiff --git a/usth/MATH1.4/homework/limits.tex b/usth/MATH1.4/homework/limits.tex
new file mode 100644
index 0000000..c4d31e4
--- /dev/null
+++ b/usth/MATH1.4/homework/limits.tex
@@ -0,0 +1,1727 @@
+\documentclass[a4paper,12pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[english,vietnamese]{babel}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{enumerate}
+\usepackage{mathtools}
+\usepackage{pgfplots}
+\usepackage{siunitx}
+\title{Calculus Homework}
+\author{Nguyễn Gia Phong}
+\date{Winter 2018}
+\newcommand{\ud}{\,\mathrm{d}}
+\newcommand{\leibniz}[3][]{\frac{\mathrm{d} #1 #2}{\mathrm{d} #3 #1}}
+\DeclareMathOperator{\erf}{erf}
+
+\begin{document}
+\maketitle
+\setcounter{section}{1}
+\section{Limits}
+
+\setcounter{subsection}{2}
+\subsection{Limit Laws}
+Evaluate the limit:
+\[\lim_{x \to 2}\sqrt{\frac{2x^2 + 1}{3x - 2}}
+= \sqrt{\lim_{x \to 2}\frac{2x^2 + 1}{3x - 2}}
+= \sqrt{\frac{2 \cdot 2^2 + 1}{3 \cdot 2 - 2}}
+= \frac{3}{2} \tag{9}\]
+
+\[\lim_{x \to 4}\frac{x^2 - 4x}{x^2 - 3x - 4}
+= \lim_{x \to 4}\frac{x}{x + 1}
+= \frac{4}{5} \tag{12}\]
+
+\begin{align*}
+   \lim_{t \to 0}\frac{\sqrt{1 + t} - \sqrt{1 - t}}{t}
+&= \lim_{t \to 0}\frac{2t}{t(\sqrt{1 + t} + \sqrt{1 - t})} \\
+&= \lim_{t \to 0}\frac{2}{(\sqrt{1 + t} + \sqrt{1 - t})} \\
+&= \frac{2}{\sqrt{1} + \sqrt{1}} \\
+&= 1 \tag{25}
+\end{align*}
+
+\noindent\textbf{40. }Prove that
+$\lim_{x \to 0^+}\sqrt{x}e^{\sin\frac{\pi}{x}} = 0$.
+
+Given $\varepsilon > 0$, let $\delta = \frac{1}{9}\varepsilon^2$.
+If $0 < x < 0 + \delta$ then \[0 < \sqrt{x}e^{\sin\frac{\pi}{x}}
+\leq e\sqrt{\delta} < 3\sqrt{\frac{\varepsilon^2}{9}}
+\Longrightarrow |\sqrt{x}e^{\sin\frac{\pi}{x}} - 0 | < \varepsilon\]
+
+Thus, by the definition of right-hand limit,
+\[\lim_{x \to 0^+}\sqrt{x}e^{\sin\frac{\pi}{x}} = 0\]
+
+\noindent\textbf{59. }Prove that $\lim_{x \to 0}f(x) = 0$ if
+\[f(x) = \begin{cases}
+           x^2\text{ if } x\text{ is rational} \\
+           0\text{ if } x\text{ is irrational}
+         \end{cases}\]
+
+Given $\varepsilon > 0$, let $\delta = \sqrt{\varepsilon}$.
+If $0 < |x - 0| < \delta$, then $0 < x^2 < \varepsilon$
+or $|f(x) - 0| < \varepsilon$. Thus, by the definition of a limit,
+\[\lim_{x \to 0}f(x) = 0\]
+
+\noindent\textbf{61. }If $f(x) = \begin{cases}
+                                   1\text{ if } x \geq 0 \\
+                                   0\text{ if } x < 0
+                                 \end{cases}$
+and $g(x) = \begin{cases}
+              0\text{ if } x \geq 0 \\
+              1\text{ if } x < 0
+            \end{cases}$
+then $f(x)g(x) = 0$. Thus $\lim_{x \to 0}f(x)g(x) = 0$ though neither
+$\lim_{x \to 0}f(x)$ nor $\lim_{x \to 0}g(x)$ exists.
+
+\subsection{The precise definition of a limit}
+\textbf{3. }Given $f(x) = \sqrt{x}$, if $|x - 4| < 1.44$ then
+$|\sqrt{x} - 2| < 0.4$.
+
+\noindent\textbf{21. }Prove that $\lim_{x \to 2}\frac{x^2 + x - 6}{x - 2} = 5$.
+
+Given $\varepsilon > 0$, let $\delta = \varepsilon$.
+If $0 < |x - 2| < \delta$, then \[|x + 3 - 5| < \varepsilon
+\iff \left|\frac{x^2 + x - 6}{x - 2} - 5\right| < \varepsilon\]
+
+Thus, by the definition of a limit,
+$\lim_{x \to 2}\frac{x^2 + x - 6}{x - 2} = 5$.
+
+\noindent\textbf{39. }Prove that $\lim_{x \to 0}f(x)$ does not exist if
+\[f(x) = \begin{cases}
+           0\text{ if } x\text{ is rational} \\
+           1\text{ if } x\text{ is irrational}
+         \end{cases}\].
+
+Suppose $\lim_{x \to 0}f(x) = L$, hence by the definition of limit,
+for every $\varepsilon > 0$, there exists $\delta > 0$ that
+\[0 < |x - 0| < \delta \Rightarrow |f(x) - L| < \varepsilon \tag{$*$}\]
+
+For $L = 0$, consider $\varepsilon = |L - 1|$. For every $\delta$, there is
+at least one irrational $x \in (0, \delta)$, which turns $(*)$ into a false
+statement: \[0 < |x| < \delta \Rightarrow |1 - L| < |L - 1|\]
+
+For $L \neq 0$, consider $\varepsilon = |L|$. For every $\delta$, there is
+at least one rational $x \in (0, \delta)$, which turns $(*)$ into a false
+statement: \[0 < |x| < \delta \Rightarrow |L| < |L|\]
+
+Conclusion: The assumption is incorrect; in other words, $\lim_{x \to 0}f(x)$
+does not exist.
+
+\subsection{Continuity}
+\textbf{22. }Explain why the function $f$ is discontinuous at the given number
+$a = 3$.
+\begin{align*}
+f(x) &= \begin{cases}
+          \frac{2x^2 - 5x - 3}{x - 3}\text{ if } x \neq 3 \\
+          6\text{ if } x = 3
+        \end{cases}\\
+     &= \begin{cases}
+          2x + 1\text{ if } x \neq 3 \\
+          6\text{ if } x = 3
+        \end{cases}
+\end{align*}
+
+Since $\lim_{x \to 3}f(x) = \lim_{x \to 3}(2x + 1) = 7 \neq 6 = f(3)$,
+$f$ is discontinuous at $3$.
+
+\begin{tikzpicture}
+  \begin{axis}[
+    axis x line=middle, axis y line=middle,
+    xlabel={$x$}, ylabel={$f(x)$},
+    xlabel style={at=(current axis.right of origin), anchor=west},
+    ylabel style={at=(current axis.above origin), anchor=south}]
+    \addplot[domain=-1:4,blue]{2*x + 1};
+    % This part is a bit hacky but it works XD
+    \addplot[white, mark=*, only marks] coordinates {(3,7)};
+    \addplot[blue, mark=o, only marks] coordinates {(3,7)};
+    \addplot[blue, mark=*, only marks] coordinates {(3,6)};
+  \end{axis}
+\end{tikzpicture}
+
+\noindent\textbf{26. }$G(x) = \frac{x^2 + 1}{2x^2 - x - 1}$ is a rational
+function so it is continuous at every number in its domain.
+
+\noindent\textbf{38. }Since $arctan$ is an inverse trigonometric function and
+thus continuous at every number in its domain and
+$\lim_{x \to 2}\frac{x^2 - 4}{3x^2 - 6x} = \lim_{x \to 2}\frac{x + 2}{3x} =
+\frac{2}{3}$, \[\lim_{x \to 2}\arctan\frac{x^2 - 4}{3x^2 - 6x} =
+\arctan\frac{2}{3}\]
+
+\subsection{To Infinity and Beyond!}
+Find the limit:
+\[\lim_{x \to -\infty}\frac{\sqrt{9x^6 - x}}{x^3 + 1}
+= \lim_{x \to -\infty}\frac{\sqrt{9 - \frac{1}{x^5}}}{-1 - \frac{1}{x^3}}
+= -3 \tag{24}\]
+
+\subsection{Derivatives}
+\textbf{24. }If $g(x) = x^4 - 2$,
+\begin{align*}
+g'(1) &= \lim_{h \to 0}\frac{g(1 + h) - g(1)}{h}\\
+      &= \lim_{h \to 0}\frac{(1 + h)^4 - 2 - (1^4 - 2)}{h}\\
+      &= \lim_{h \to 0}\frac{h^4 + 4h^3 + 6h^2 + 4h + 1 - 1}{h}\\
+      &= \lim_{h \to 0}(h^3 + 4h^2 + 6h + 4)\\
+\end{align*}
+
+An equation of the tangent line to $g$ at $(1, -1)$:
+\[y - g(1) = g'(1)(x - 1) \iff y = 4x - 5\]
+
+\noindent Determine whether $f'(0)$ exists.
+\[f(x) = \begin{cases}
+           x\sin\frac{1}{x}\text{ if } x \neq 0 \\
+           0\text{ if } x = 0
+         \end{cases}\tag{53}\]
+\begin{align*}
+f'(0) &= \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h}\\
+      &= \lim_{h \to 0}\frac{h\sin\frac{1}{h}}{h}\\
+      &= \lim_{h \to 0}\sin\frac{1}{h}\tag{does not exist}\\
+\end{align*}
+
+\[f(x) = \begin{cases}
+           x^2\sin\frac{1}{x}\text{ if } x \neq 0 \\
+           0\text{ if } x = 0
+         \end{cases}\tag{54}\]
+\begin{align*}
+f'(0) &= \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h}\\
+      &= \lim_{h \to 0}\frac{h^2\sin\frac{1}{h}}{h}\\
+      &= \lim_{h \to 0}h\sin\frac{1}{h}
+\end{align*}
+
+Since $\forall h \neq 0, -|h| \leq h\sin\frac{1}{h} \leq |h|$
+and $\lim_{h \to 0}(-|h|) = \lim_{h \to 0}|h| = 0$,
+according to the Squeeze Theorem, $f'(0) = 0$.
+
+\section{Differentiation}
+\setcounter{subsection}{3}
+\subsection{The chain rule}
+Find the derivative of the function.
+
+\[y = \cos\sqrt{\sin(\tan{\pi x})}\tag{45}\]
+\begin{align*}
+\dot{y} &= -\sqrt{\sin(\tan{\pi x})}' \cdot \sin\sqrt{\sin(\tan{\pi x})}\\
+        &= \frac{\sin'(\tan{\pi x}) \cdot \sin\sqrt{\sin(\tan{\pi x})}}
+                {2\sqrt{\sin(\tan{\pi x})}}\\
+        &= \frac{\tan'{\pi x} \cdot \cos(\tan{\pi x})
+                 \cdot \sin\sqrt{\sin(\tan{\pi x})}}
+                {2\sqrt{\sin(\tan{\pi x})}}\\
+        &= \frac{\pi\sec^2{\pi x} \cdot \cos(\tan{\pi x})
+                 \cdot \sin\sqrt{\sin(\tan{\pi x})}}
+                {2\sqrt{\sin(\tan{\pi x})}}
+\end{align*}
+
+\[y = [x + (x + \sin^2{x})^3]^4 \tag{46}\]
+\begin{align*}
+\dot{y} &= 4[x + (x + \sin^2{x})^3]' [x + (x + \sin^2{x})^3]^3\\
+        &= 4[1 + 3(x + \sin^2{x})' (x + \sin^2{x})^2]
+            [x + (x + \sin^2{x})^3]^3\\
+        &= 4[1 + 3(1 + \sin{2x})(x + \sin^2{x})^2]
+            [x + (x + \sin^2{x})^3]^3
+\end{align*}
+
+\setcounter{subsection}{6}
+\subsection{Applications in Sciences}
+\textbf{9. }A rock is thrown vertically upward from the surface of Mars, its
+height after $t$ seconds is $h = 15t - 1.86t^2$.
+\[\frac{\ud h}{\ud t}(2)
+= (t \mapsto 15 - 3.72t)(2)
+= 7.56\text{ (m/s)}\tag{a}\]
+\[h = 25 \iff 15t - 1.86t^2 = 25
+  \iff t = \frac{375 \mp 25\sqrt{39}}{93}\tag{b}\]
+
+So at $t \approx 2.35$ s or $t \approx 5.71$ the Rock's height is 25 m. Its
+velocity at this point is
+\[v = (t \mapsto 15 - 3.72t)\left(\frac{375 \mp 25\sqrt{39}}{93}\right)
+    = \pm 6.24\text{ (m/s)}\]
+
+\pagebreak\noindent\textbf{10. }A particle moves with position function
+\[s = t^4 - 4t^3 - 20t^2 + 20t \qquad t \geq 0\]
+\[v = 20 \iff \dot{s} = 20 \iff 4t^3 - 12t^2 - 40t + 20 = 20 \tag{a}\]
+
+Since $t$ is nonnegative, the particle has a velocity of 20 m/s at $t = 0$ and
+$t = 5$ s.
+\[a = 0 \iff \dot{v} = 0 \iff 12t^2 - 24t - 40 = 0 \tag{b}\]
+
+Since $t$ is nonnegative, the acceleration is 0 at $t = \sqrt\frac{13}{3} - 1$
+s. This is when the instantaneous speed of the particle ($|v|$) reaches its
+critical value.
+
+\noindent\textbf{21. }The force $F$ acting on a body with velocity $v$ and mass
+$m = m_0 \Big/ \sqrt{1 - \frac{v^2}{c^2}}$ (where $m_0$ is the mass of the particle
+at rest and $c$ is the speed of light) is the rate of change of momentum:
+\begin{align*}
+F &= \frac{\ud(mv)}{\ud t}\\
+  &= \frac{\ud}{\ud t}\left(\frac{m_0v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\
+  &= m_0\frac{\ud v}{\ud t}\cdot
+     \frac{\ud}{\ud v}\left(\frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\
+  &= m_0a\frac{\ud}{\ud v}\left(\frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\
+  &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}}
+               - v\frac{\ud\sqrt{1 - v^2/c^2}}{\ud v}}
+              {1 - \frac{v^2}{c^2}}\\
+  &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}}
+               - \frac{v}{2}\cdot\frac{-2v}{c^2\sqrt{1 - v^2/c^2}}}
+              {1 - \frac{v^2}{c^2}}\\
+  &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}} + \frac{v^2}{c^2\sqrt{1 - v^2/c^2}}}
+              {1 - \frac{v^2}{c^2}}\\
+  &= m_0a\frac{1 - \frac{v^2}{c^2} + \frac{v^2}{c^2}}
+              {\left(1 - \frac{v^2}{c^2}\right)^\frac{3}{2}}\\
+  &= \frac{m_0a}{\left(1 - \frac{v^2}{c^2}\right)^\frac{3}{2}}
+\end{align*}
+
+\noindent\textbf{30. }The frequency of vibrations a vibrating violin string is
+given by \[f = \frac{1}{2L}\sqrt{\frac{T}{\rho}} \qquad T \geq 0, \rho > 0\]
+\begin{enumerate}[(a)]
+  \item The rate of change of the frequency with respect to
+    \begin{enumerate}[(i)]
+      \item The length:
+        $\frac{\ud f}{\ud L} = \frac{-1}{L^2}\sqrt{\frac{T}{\rho}}$.
+      \item The tension: $\frac{\ud f}{\ud T} = \frac{1}{4L\sqrt{T\rho}}$.
+      \item The density:
+        $\frac{\ud f}{\ud L} = \frac{-1}{L2}\sqrt{\frac{T}{\rho^3}}$.
+    \end{enumerate}
+  \item The pitch of a note gets higher when the string is shorter and lower
+    when the tension or density is increased.
+\end{enumerate}
+
+\noindent\textbf{35. }Applying the gas law
+\[PV = nRT \iff T = \frac{PV}{nR}\]
+
+The rate of change of temperature can be easily calculated via differentiation:
+\begin{align*}
+   \frac{\ud T}{\ud t}
+&= \frac{\ud}{\ud t}\left(\frac{PV}{nR}\right)\\
+&= \frac{1}{nR}\left(P\frac{\ud V}{\ud t} + V\frac{\ud P}{\ud t}\right)\\
+&= \frac{8.0 \cdot 0.15 + 10 \cdot 0.10}{10 \cdot 0.0821}\\
+&= \frac{1.2 + 1}{10 \cdot 0.0821}\\
+&= \frac{2}{10 \cdot 0.0821}\\
+&= 2\text{ (K/s)}
+\end{align*}
+
+(In the calculation above, significant figures are taken into consideration.)
+
+\pagebreak\subsection{Exponential Growth and Decay}
+\textbf{4. }Let $P(t)$ be the bacteria count after $t$ hours. As the bacteria
+culture grows with constant relative growth rate,
+\[\frac{\ud P}{\ud t} = kP \Longrightarrow P(t) = P(0)e^{kt}\]
+
+Since $P(2) = 400$ and $P(6) = 25600$,
+\begin{align*}
+  \begin{cases}
+    P(0)e^{2k} = 400\\
+    P(0)e^{6k} = 25600
+  \end{cases}
+  &\iff
+  \begin{cases}
+    P(0)e^{2k} = 400\\
+    e^{4k} = 64
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    P(0)e^{2k} = 400\\
+    e^{2k} = 8
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    P(0) = 50\\
+    k = \frac{\ln 8}{2} \approx 104\%
+  \end{cases}
+\end{align*}
+
+Thus (a) the relative growth rate is 104\%, (b) the initial size of the culture
+is 50 and (c) the number of bacteria after $t$ hours is $50\sqrt{8^t}$.
+
+The number of cells after 4.5 hours:
+\[P(4.5) = 50\sqrt{8^{4.5}} \approx 5382\tag{d}\]
+
+The rate of growth after 4.5 hours:
+\begin{align*}
+   \frac{\ud P}{\ud t}(4.5)
+&= 50\frac{\ud \sqrt{8}^t}{\ud t}(4.5)\\
+&= 50\left(t \mapsto \sqrt{8^t}\ln\sqrt{8}\right)(4.5)\\
+&= 25\cdot8^{2.25}\ln 8\\
+&\approx 5596\text{ (bacteria per minute)}\tag{e}
+\end{align*}
+
+The population reach 50000 when
+\[50\sqrt{8^t} = 50000 \iff 8^t = 10^6 \iff t = \log_2{100}
+\approx 6.64\text{ (days)}\tag{f}\]
+
+\noindent\textbf{8. }Given 50 mg of $^{90}$Sr which has a half-life of 28 days.
+\begin{enumerate}[(a)]
+  \item Formula of the mass remaining after $t$ days: $m(t) = 50\cdot2^{-t/28}$.
+  \item The mass remaining after 40 days:
+    $m(40) = 50\cdot\frac{1}{2}^{10/7} \approx 19\text{ (mg)}$.
+  \item To decay to a mass of 2 mg, it takes
+    $-28\log_2\frac{2}{50} \approx 130\text{ (days)}$.
+  \item The graph of the mass function:\\
+    \begin{tikzpicture}
+      \begin{axis}[
+        axis x line=middle, axis y line=middle,
+        xlabel={$t$}, ylabel={$m$},
+        xlabel style={at=(current axis.right of origin), anchor=west},
+        ylabel style={at=(current axis.above origin), anchor=south},
+        enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}]
+        \addplot[domain=0:130,magenta]{50*2^(-x/28)};
+      \end{axis}
+    \end{tikzpicture}
+\end{enumerate}
+
+\noindent\textbf{16. }Let $T(t)$ be the temperature of the coffee after $t$
+minutes. The surrounding temperature is \ang{20}C, so Newton's Law of Cooling
+states that \[\frac{\ud T}{\ud t} = k(T - 20)\]
+
+If we let $y = T - 20$, then $y(0) = T(0) - 20 = 95 - 20 = 75$, so $y$
+satisfies \[\frac{\ud y}{\ud t} = ky \iff y(t) = 75e^{kt}\]
+
+When the temperature of the coffee is \ang{70}C, its cooling rate is \ang{1}C
+per minute, i.e.
+\begin{align*}
+  \begin{cases}
+    y(t) + 20 = 70\\
+    ky(t) = -1
+  \end{cases}
+  &\iff\begin{cases}
+         y(t) = 50\\
+         k = \frac{-1}{50}
+       \end{cases}\\
+  \Longrightarrow 75e^{-t/50} = 50
+    &\iff t = 50\ln1.5 \approx 20\text{ (minutes)}
+\end{align*}
+
+\subsection{Related rates}
+\textbf{10. }A particle is moving along a hyperbola $xy = 8$
+\begin{align*}
+  \Longrightarrow \frac{\ud (xy)}{\ud t} = \frac{\ud 8}{\ud t}
+  &\iff y\frac{\ud x}{\ud t} + x\frac{\ud y}{\ud t} = 0\\
+  &\iff 2 \cdot \frac{\ud x}{\ud t} + 4 \cdot 3 = 0\\
+  &\iff \frac{\ud x}{\ud t} = -6\text{ (cm/s)}
+\end{align*}
+
+\noindent\textbf{12. }Let $D(t)$ (cm) be the diameter of the ball at minute
+$t$, its surface area is $A(D) = \pi D^2$ (cm$^2$).
+\[\frac{\ud A}{\ud t} = 1 \iff \frac{\ud A}{\ud D} \cdot \frac{\ud D}{\ud t} = 1
+  \iff 2\pi D \frac{\ud D}{\ud t} = 1 \iff \frac{\ud D}{\ud t}
+                                  = \frac{1}{2\pi D}\]
+
+Thus the decreasing rate of the diameter when it is 10 cm:
+\[\frac{\ud D}{\ud t}(10) = \frac{1}{20\pi}\text{ (cm/s)}\]
+
+\noindent\textbf{14. }
+
+\begin{tikzpicture}
+  \begin{axis}[nodes near coords align=right, xlabel={W -- E}, ylabel={S -- N}]
+    \node[label=A, shape=circle, fill, inner sep=1.5pt] at (0,0) {};
+    \addplot[->] plot coordinates {(0,0) (35,0)};
+    \node[label={180:B}, shape=circle, fill, inner sep=1.5pt] at (150,0) {};
+    \addplot[->] plot coordinates {(150,0) (150,25)};
+  \end{axis}
+\end{tikzpicture}
+\[\begin{cases}
+    \Delta x(t) = x_B(t) - x_A(t)\\
+    \Delta y(t) = y_B(t) - y_A(t)
+  \end{cases}
+\iff\begin{cases}
+      \Delta x(t) = 150 - 35t\\
+      \Delta y(t) = 25t
+  \end{cases}\]
+\[\Longrightarrow\Delta s(t) = \sqrt{1850t^2 - 10500t + 22500}
+  \Longrightarrow\frac{\ud s}{\ud t} = \frac{1850t - 5250}
+                                       {\sqrt{1850t^2 - 10500t + 22500}}\]
+\[\Longrightarrow\frac{\ud s}{\ud t}(4)
+               = \frac{1850\cdot4 - 5250}
+                      {\sqrt{1850\cdot16 - 10500\cdot4 + 22500}}
+               = \frac{2150}{\sqrt{10100}}
+               = \frac{215\sqrt{101}}{101}
+               \approx 21\text{ (km/h)}\]
+
+\noindent\textbf{27. }Let $h(t)$ (ft) be the height of the cone at minute $t$.
+Volume of the cone is
+\begin{align*}
+  V(h) = \frac{\pi h^2}{12}
+  &\Longrightarrow\frac{\ud V}{\ud t} = \frac{\pi h}{6}\cdot\frac{\ud h}{\ud t}
+                                      = 30
+  \iff\frac{\ud h}{\ud t} = \frac{180}{\pi h}\\
+  &\Longrightarrow\frac{\ud h}{\ud t}(10) = \frac{180}{10\pi} = \frac{18}{\pi}
+  \approx 5.7\text{ (ft/s)}
+\end{align*}
+
+\section{Applications of derivative}
+\subsection{Max and Min}
+Find the absolute min and max values of $f$.
+
+\[f(x) = 3x^4 - 4x^3 - 12x^2 + 1, \qquad [-2, 3]\tag{51}\]
+
+Since $f'(x) = 12x^3 - 12x^2 - 24x$, we have $f'(x) = 0$ when
+$x \in \{-1, 0, 2\}$. The values of $f$ at these critical numbers are
+\begin{align*}
+  f(-1) &= 3 + 4 - 12 + 1 = -4\\
+  f(0)  &= 1\\
+  f(2)  &= 48 - 32 - 48 + 1 = -31
+\end{align*}
+
+The values of $f$ at endpoints are
+\begin{align*}
+  f(-2) &= 48 + 32 - 48 + 1 = 33\\
+  f(3)  &= 243 - 108 - 108 + 1 = 28
+\end{align*}
+
+Comparing these five numbers and using the Closed Interval Method, we see that
+the absolute minimum value is $f(2) = -31$ and the absolute maximum value is
+$f(-2) = 33$.
+
+\[f(t) = t\sqrt{4 - t^2}, \qquad [-1, 2]\tag{55}\]
+
+Since $f'(t) = \frac{4 - 2t^2}{\sqrt{4 - t^2}}$, with $t \in [-1, 2]$, we have
+$f'(t) = 0$ when $t = \sqrt{2}$. The value of $f$ at this critical number is
+$f(\sqrt{2}) = 2$. The value of $f$ at endpoints are $f(-1) = -\sqrt{3}$ and
+$f(2) = 0$. Comparing these 3 numbers and using the Closed Interval Method, we
+see that the absolute minimum value is $f(-1) = -\sqrt{3}$ and the absolute
+maximum value is $f(\sqrt{2}) = 2$.
+
+\[f(x) = xe^{-x^2/8}, \qquad [-1, 4]\tag{59}\]
+
+With $x \in [-1, 4]$, we have $f'(x) =  \left(1-\frac{x^2}{4}\right)e^{-x^2/8}$
+when $x = 2$. Comparing values of $f$ at this critical number and endpoints,
+the minimum value is $f(-1) = -e^{-1/8}$ and
+the maximum value is $f(2) = 2e^{-1/2}$.
+
+\subsection{The Mean Theorem}
+\textbf{26. }Let $h$ be the function that $h(x) = f(x) - g(x)$. Since both $f$
+and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, $h$ inherits
+the same properties. By applying the Mean Value Theorem to $h$ on the interval
+$[a, b]$, we get a number $c \in (a, b)$ such that
+\begin{align*}
+    &h(b) - h(a) = (b - a)h'(c)\\
+\iff&f(b) - g(b) - f(a) + g(a) = (b - a)(f'(c) - g'(c))\\
+\iff&f(b) - g(b) = (b - a)(f'(c) - g'(c))
+\end{align*}
+$b - a > 0$ and $f'(c) - g'(c) < 0$ so $f(b) - g(b) < 0$ or $f(b) < g(b)$.
+
+\begin{align*}
+x > 0 &\iff x + 1 > 1\\
+      &\iff \sqrt{x + 1} > 1\\
+      &\iff \sqrt{x + 1} - 1 > 0\\
+      &\Longrightarrow \left(\sqrt{x + 1} - 1\right)^2 > 0\\
+      &\iff x + 1 - 2\sqrt{x + 1} + 1 > 0\\
+      &\iff x + 2 > 2\sqrt{x + 1}\\
+      &\iff \sqrt{1 + x} < 1 + \frac{1}{2}x\tag{27}
+\end{align*}
+
+\noindent\textbf{33. }Prove the identity
+\[\arcsin\frac{x - 1}{x + 1} = 2\arctan\sqrt{x} - \frac{\pi}{2}\]
+
+Let $\frac{-\pi}{2} \leq y = \arcsin\frac{x - 1}{x + 1} \leq \frac{\pi}{2}$ and
+$z = \arctan\sqrt{x}$, then
+\begin{align*}
+\begin{cases}
+  \sin{y} = \frac{x - 1}{x + 1}\\
+  \tan{z} = \sqrt{x}
+\end{cases}
+\Longrightarrow&
+\begin{cases}
+  \frac{\ud\sin{y}}{\ud x} = \frac{\ud}{\ud x}\left(\frac{x - 1}{x + 1}\right)\\
+  \frac{\ud\tan{z}}{\ud x} = \frac{\ud\sqrt{x}}{\ud x}
+\end{cases}\\
+\iff&
+\begin{cases}
+  \cos{y}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\
+  \left(\tan^2{z} + 1\right)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}}
+\end{cases}\\
+\iff&
+\begin{cases}
+  \sqrt{1 - \sin^2{y}}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\
+  \left(\sqrt{x}^2 + 1\right)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}}
+\end{cases}\\
+\iff&
+\begin{cases}
+  \sqrt{\frac{4x}{(x + 1)^2}}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\
+  (x + 1)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}}
+\end{cases}\\
+\iff&
+\begin{cases}
+  \frac{\ud y}{\ud x} = \frac{1}{|x + 1|\sqrt{x}}\\
+  \frac{\ud z}{\ud x} = \frac{1}{2(x + 1)\sqrt{x}}
+\end{cases}\\
+\end{align*}
+
+For all $x \geq 0$ or $x + 1 \geq 1 > 0$
+\begin{align*}
+   \frac{\ud}{\ud x}\left(2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}\right)
+&= 2\frac{\ud z}{\ud x} - \frac{\ud y}{\ud x}\\
+&= \frac{2}{2(x + 1)\sqrt{x}} - \frac{1}{(x + 1)\sqrt{x}}\\
+&= 0
+\end{align*}
+
+Thus the function
+$x \mapsto 2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}$ is constant on its
+domain $[0, \infty)$. Consequently, in $[0, \infty)$
+\begin{align*}
+   2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}
+&= \left(x \mapsto 2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}\right)(0)\\
+&= 2\arctan\sqrt{0} - \arcsin\frac{-1}{1}\\
+&= 0 - \frac{-\pi}{2}\\
+&= \frac{\pi}{2}\\
+\iff \arcsin\frac{x - 1}{x + 1} &= 2\arctan\sqrt{x} - \frac{\pi}{2}
+\end{align*}
+
+\subsection{Shape of a graph}
+\textbf{75. }Given two funtions $f$ and $g$ which are positive and concave
+upward on $I$, i.e. for all $x$ in $I$
+\[\begin{cases}
+    f(x) > 0\\
+    f''(x) > 0\\
+    g(x) > 0\\
+    g''(x) > 0
+  \end{cases}\]
+
+Second derivative of the product function $fg$:
+\[(f(x)g(x))'' = (f'(x)g(x) + f(x)g'(x))'
+  = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)\]
+
+If $f$ and $g$ are both increasing or decreasing, then $f'(x)g'(x) > 0$, which
+means $\forall x \in I, (f(x)g(x))'' > 0$, or $fg$ is concave upward on $I$.
+Otherwise, $f$ is increasing and $g$ is decreasing for instance, $fg$ may be
+either concave upward, concave downward or linear:
+
+\begin{center}
+  \begin{tikzpicture}[domain=0:8]
+    \begin{axis}[
+      axis x line = middle, axis y line = middle,
+      xlabel={$x$}, ylabel={$y$}, ymin=-8,
+      xlabel style={at=(current axis.right of origin), anchor=west},
+      ylabel style={at=(current axis.above origin), anchor=south},
+      enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1},
+      legend pos=south east]
+      \addplot[color=blue, samples=100, smooth]{-1/x};
+      \addplot[color=cyan, samples=100, smooth]{-x^(3/2)};
+      \addplot[color=red, samples=100, smooth]{sqrt(x)};
+      \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x\sqrt{x}$,
+              $f(x)g(x) = \sqrt{x}$}
+    \end{axis}
+  \end{tikzpicture}
+
+  \begin{tikzpicture}[domain=0:4]
+    \begin{axis}[
+      axis x line = middle, axis y line = middle,
+      xlabel={$x$}, ylabel={$y$}, ymin=-8,
+      xlabel style={at=(current axis.right of origin), anchor=west},
+      ylabel style={at=(current axis.above origin), anchor=south},
+      enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}]
+      \addplot[color=blue, samples=100, smooth]{-1/x};
+      \addplot[color=cyan, samples=100, smooth]{-x^3};
+      \addplot[color=red, samples=100, smooth]{x^2};
+      \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x^3$, $f(x)g(x) = x^2$}
+    \end{axis}
+  \end{tikzpicture}
+
+  \begin{tikzpicture}[domain=0:4]
+    \begin{axis}[
+      axis x line = middle, axis y line = middle,
+      xlabel={$x$}, ylabel={$y$}, ymin=-8,
+      xlabel style={at=(current axis.right of origin), anchor=west},
+      ylabel style={at=(current axis.above origin), anchor=south},
+      enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1},
+      legend pos=south east]
+      \addplot[color=blue, samples=100, smooth]{-1/x};
+      \addplot[color=cyan, samples=100, smooth]{-x^2};
+      \addplot[color=red, samples=100, smooth]{x};
+      \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x^2$, $f(x)g(x) = x$}
+    \end{axis}
+  \end{tikzpicture}
+\end{center}
+
+\pagebreak\noindent\textbf{76. }In order for $h = f(g(x))$ to be concave
+upward on $\mathbb R$
+\begin{align*}
+  h'' > 0 &\iff (f \circ g)'' > 0\\
+          &\iff ((f' \circ g) \cdot g')' > 0\\
+          &\iff (f' \circ g)' \cdot g' + (f' \circ g) \cdot g'' > 0\\
+          &\iff (f'' \circ g) \cdot (g')^2 + (f' \circ g) \cdot g'' > 0
+\end{align*}
+
+Because $f$ and $g$ are given to be concave upward on $\mathbb R$, i.e.
+$f'' > 0$ and $g'' > 0$, and $\forall x \in \mathbb R, g^2(x) \geq 0$, so if
+$f' > 0$ or $f$ is an increasing function, $h$ will be concave upward.
+
+\noindent\textbf{77. }Show that $\tan{x} > x$ for $0 < x < \frac{\pi}{2}$.
+
+Let $f$ be the function that $f(x) = \tan{x} - x$. On $(0, \frac{\pi}{2})$,
+$\sin{x}\cos{x} \neq 0$ thus $\tan{x}$ exists and is nonzero. Therefore,
+$f'(x) = \tan^2(x) > 0$ or $f$ is increasing on $[0, \frac{\pi}{2}]$, which
+means for all x in $(0, \frac{\pi}{2})$,
+\[f(x) > f(0) \iff \tan{x} - x > \tan0 - 0 \iff \tan{x} > x\]
+
+\noindent\textbf{78. } Use mathematical induction to prove that for all
+positive integer $n$,
+\[\forall x \geq 0, \qquad e^x \geq 1 + \sum_{i=1}^n\frac{x^i}{i!}\tag{$*$}\]
+
+Let $f$ be the function of domain $[0, \infty)$ that $f(x) = e^x - x$, then
+for all $x \geq 0$, $f'(x) = e^x - 1 > f'(0) = 0$ (since it is obvious that $f'$
+is an increasing function). Hence
+$\forall x \geq 0, e^x - x > e^0 - 0 = 1 \iff \forall x \geq 0, e^x > 1 + x$,
+i.e.  $(*)$ is true for $n = 1$.
+
+Suppose that $(*)$ is also true for $n = k$ ($k \in \mathbb N^*$). For all
+nonnegative $x$,
+\[e^x \geq 1 + \sum_{i=1}^k\frac{x^i}{i!}
+  \iff e^x - 1 - \sum_{i=1}^k\frac{x^i}{i!} \geq 0\]
+
+Let $g$ be the function of domain $[0, \infty)$ that
+$g(x) = e^x - \sum_{i=1}^{k+1}\frac{x^{i}}{i!}$, then for all positive $x$
+\[g'(x) = e^x - \sum_{i=1}^{k+1}\frac{ix^{i-1}}{i!}
+        = e^x - 1 - \sum_{i=2}^{k+1}\frac{x^{i-1}}{(i - 1)!}
+        = e^x - 1 - \sum_{i=1}^{k}\frac{x^{i}}{i!} \geq 0\]
+
+This means $g$ in a non-decreasing function on $[0, \infty)$
+\[e^x - \sum_{i=1}^{k+1}\frac{x^{i}}{i!}
+  \geq e^0 - \sum_{i=1}^{k+1}\frac{0^{i}}{i!} = 1\]
+
+This expression shows that $(*)$ is true for $n = k + 1$. Therefore, by
+mathematical induction, it is true for all positive integers $n$.
+
+\subsection{Rule of the Hospital}
+Find the limit
+
+\noindent\textbf{54. }Since $\lim_{x \to 0^+}\ln x = -\infty$
+and $\lim_{x \to 0^+}\frac{1}{\sqrt x} = \infty$
+\[\lim_{x \to 0^+}x^{\sqrt x}
+= \lim_{x \to 0^+}e^{\sqrt{x}\cdot\ln{x}}
+= e^{\lim_{x \to 0^+}\frac{\ln x}{1/\sqrt x}}
+= e^{\lim_{x \to 0^+}\frac{-2x\sqrt x}{x}}
+= e^{-2\lim_{x \to 0^+}\sqrt x}
+= e^0
+= 1\]
+
+\noindent\textbf{60. }Since
+$\lim_{x \to \infty}\ln{2}\ln x = \lim_{x \to \infty}(1 + \ln x) = \infty$
+\[\lim_{x \to \infty}x^\frac{\ln 2}{1 + \ln x}
+= e^{\lim_{x \to \infty}\frac{\ln{2}\ln x}{1 + \ln x}}
+= e^{\lim_{x \to \infty}\ln{2}}
+= e^{\ln 2}
+= 2\]
+
+\setcounter{subsection}{6}
+\subsection{Optimization Problems}
+\textbf{44. }Given $E(v) = \frac{aLv^3}{v - u}$
+\[\frac{\ud E}{\ud v}
+= aL\frac{3v^2(v - u) - v^3}{(v - u)^2}
+= aL\frac{2v^3 - 3uv^2}{(v - u)^2}\]
+
+Since $v > u > 0$, $E$ has only one absolute extreme value, at the only
+critical number $v = 1.5u$. Applying the First Derivative Test for Absolute
+Extreme Values, $v = 1.5u$ is shown to be the value of $v$ that minimizes $E$.
+
+\noindent\textbf{45. }Given
+$S = 6sh + \frac{3}{2}s^2(\sqrt{3}\cdot\csc\theta - \cot\theta)$.
+\[\frac{\ud S}{\ud\theta}
+= \frac{3}{2}s^2\frac{\ud}{\ud\theta}(\sqrt{3}\cdot\csc\theta - \cot\theta)
+= \frac{3s^2(1 - {\sqrt{3}\cdot\cos\theta})}{2\sin^2\theta}\]
+
+We have $\frac{\ud S}{\ud\theta} = 0$ when $\theta = \arccos\frac{\sqrt 3}{3}$.
+Applying the First Derivative Test for Absolute Extreme Values, this value of
+$\theta$ is shown to minimize $S$ to $6sh + \frac{3s^2}{\sqrt 2}$.
+
+\noindent\textbf{76. }Using Poiseuille's Law, we have the total resistance of
+the blood along the path ABC is
+\begin{multline*}
+R = R_{AB} + R_{BC}
+  = C\frac{a - b\cot\theta}{r_1^4} + C\frac{b}{r_2^4 \sin\theta}
+  = C\left(\frac{a - b\cot\theta}{r_1^4} + \frac{b\csc\theta}{r_2^4}\right)\\
+\Longrightarrow\frac{\ud R}{\ud\theta}
+= \frac{Cb}{r_1^4 \sin^2\theta} - \frac{Cb\cos\theta}{r_2^4\sin^2\theta}
+= \frac{Cb}{sin^2\theta}\left(\frac{1}{r_1^4} - \frac{\cos\theta}{r_2^4}\right)
+\end{multline*}
+
+We have $\frac{\ud R}{\ud\theta} = 0$ when $\cos\theta = (r_2/r_1)^4$. At this
+angle, the resistance is minimized (can be shown using the First Derivative
+Test for Absolute Extreme Values, but like in the two previous exercises, I'm
+too lazy to evaluate it). When $\frac{r_2}{r_1} = \frac{2}{3}$, the optimal
+branching angle is $\theta \approx \ang{79}$.
+
+\section{Integral}
+\subsection{Areas}
+\textbf{4. }Estimate the area under the graph of $f(x) = \sqrt{x}$ from $x = 0$
+to $x = 4$.
+\[\lim_{n \to \infty}R_n
+= \lim_{n \to \infty}\frac{4 - 0}{n}\sum_{i = 1}^n\sqrt\frac{4i}{n}
+= \lim_{n \to \infty}\frac{8}{n}\sum_{i = 1}^n\sqrt\frac{i}{n}\]
+\[\lim_{n \to \infty}L_n
+= \lim_{n \to \infty}\frac{4 - 0}{n}\sum_{i = 0}^{n - 1}\sqrt\frac{4i}{n}
+= \lim_{n \to \infty}\frac{8}{n}\sum_{i = 0}^{n - 1}\sqrt\frac{i}{n}\]
+
+For estimation, consider $n \to 4$:
+\[\lim_{n \to 4}R_n
+= \frac{8}{4}\sum_{i = 1}^4\sqrt\frac{i}{4}
+= \sum_{i = 1}^4\sqrt i
+= 1 + \sqrt 2 + \sqrt 3 + 2
+\approx 6.1463\]
+\[\lim_{n \to 4}L_n
+= \frac{8}{4}\sum_{i = 0}^3\sqrt\frac{i}{4}
+= \sum_{i = 0}^3\sqrt i
+= 0 + 1 + \sqrt 2 + \sqrt 3
+\approx 4.1463\]
+
+\noindent\textbf{5. }Estimate the area under the graph of $f(x) = 1 + x^2$ from
+$x = -1$ to $x = 2$.
+\begin{align*}
+  \lim_{n \to \infty}R_n
+=&\lim_{n \to \infty}\frac{2 + 1}{n}\sum_{i = 1}^n
+  f\left(-1 + i\frac{2 + 1}{n}\right)\\
+=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 1}^n
+  \left(1 + \left(\frac{3i}{n} - 1\right)^2\right)
+\end{align*}
+
+Similarly,
+\begin{align*}
+  \lim_{n \to \infty}L_n
+=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 0}^{n - 1}
+  \left(1 + \left(\frac{3i}{n} - 1\right)^2\right)\\
+  \lim_{n \to \infty}M_n
+=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 1}^n
+  \left(1 + \left(\frac{3i - 3/2}{n} - 1\right)^2\right)
+\end{align*}
+
+For $n \to 3$,
+\begin{align*}
+  \lim_{n \to 3}R_n
+=&\sum_{i = 1}^3\left(1 + (i - 1)^2\right)
+= 1 + 2 + 5
+= 8\\
+  \lim_{n \to 3}M_n
+=&\sum_{i = 1}^3\left(1 + \left(i - \frac{3}{2}\right)^2\right)
+= \frac{5}{4} + \frac{5}{4} + \frac{13}{4}
+= 5.75\\
+  \lim_{n \to 3}L_n
+=&\sum_{i = 0}^2\left(1 + (i - 1)^2\right)
+= 2 + 1 + 2
+= 5
+\end{align*}
+
+For $n \to 6$,
+\begin{align*}
+  \lim_{n \to 6}R_n
+=&\frac{1}{2}\sum_{i = 1}^6\left(1 + \left(\frac{i}{2} - 1\right)^2\right)
+= \frac{1}{2}\left(\frac{5}{4} + 1 + \frac{5}{4} + 2 + \frac{13}{4} + 5\right)
+= 6.875\\
+  \lim_{n \to 6}M_n
+=&\frac{1}{2}\sum_{i = 1}^6\left(1 + \left(\frac{2i - 5}{4}\right)^2\right)
+= \frac{1}{2}\left(\frac{25}{16} + \frac{17}{16} + \frac{17}{16} +
+                   \frac{25}{16} + \frac{41}{16} + \frac{65}{16}\right)
+= 5.9375\\
+  \lim_{n \to 6}L_n
+=&\frac{1}{2}\sum_{i = 0}^5\left(1 + \left(\frac{i}{2} - 1\right)^2\right)
+= \frac{1}{2}\left(2 + \frac{5}{4} + 1 + \frac{5}{4} + 2 + \frac{13}{4}\right)
+= 5.375
+\end{align*}
+
+\noindent\textbf{16. }The height (in feet) above the earth's surface of the
+\textit{Endeavour}, 62 seconds after liftoff, can be estimated with the assist
+of Python (which, coincidentally, has been utilized by NASA recently):
+\begin{verbatim}
+>>> time = 0, 10, 15, 20, 32, 59, 62, 125
+>>> velocity = 0, 185, 319, 447, 742, 1325, 1445, 4151
+>>> sum(map(int.__mul__, velocity,
+...         map(int.__sub__, time[1:], time[:-1])))
+122928
+\end{verbatim}
+
+\subsection{The Definite Integral}
+Evaluate the integral.
+\begin{align*}
+  \int_2^5(4 - 2x)\ud x &= \left.4x\right]_2^5 - \left.x^2\right]_2^5
+= 12 - 21 = -9\tag{21}\\
+  \int_0^2(2x - x^3)\ud x &= \left.x^3\right]_0^2 - \left.\frac{x^4}{4}\right]_0^2
+= 9 - 16 = -7\tag{24}
+\end{align*}
+
+\noindent\textbf{33. }Evaluate integral by interpreting it in terms of areas.
+\begin{align*}
+  \int_0^2 f(x)\ud x &= 4\tag{a}\\
+  \int_0^5 f(x)\ud x &= 10\tag{b}\\
+  \int_5^7 f(x)\ud x &= -3\tag{c}\\
+  \int_0^9 f(x)\ud x &= \int_0^5 f(x)\ud x + \int_5^9 f(x)\ud x
+                      = 10 - 8 = 2\tag{d}
+\end{align*}
+
+\noindent\textbf{50. }Given $f(x) = \begin{cases}
+                                      3\text{ for } x < 3\\
+                                      x\text{ for } x \geq 3
+                                    \end{cases}$
+\[\int_0^5 f(x)\ud x = \int_0^3 3\ud x + \int_3^5 x\ud x
+= \left.3x\right]_0^3 + \left.\frac{x^2}{2}\right]_3^5
+= 9 + 8 = 17\]
+
+\subsection{The Fundamental Theorem of Calculus}
+\textbf{3. }Let $g(x) = \int_0^x f(t)\ud t$.
+\begin{enumerate}[(a)]
+\item By interpreting the above integral in terms of areas, we get $g(0) = 0$,
+  $g(1) = 2$, $g(2) = 5$, $g(3) = 7$ and $g(6) = 3$.
+\item $g$ is increasing on $(0, 3)$.
+\item $g$ has a maximum value of 7 at $x = 3$.
+\item \begin{tikzpicture}
+        \begin{axis}[
+          axis x line = middle, axis y line = middle,
+          xlabel={$x$}, ylabel={$y$},
+          xlabel style={at=(current axis.right of origin), anchor=west},
+          ylabel style={at=(current axis.above origin), anchor=south},
+          enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}]
+          \addplot[color=magenta, domain=0:1]{2};
+          \addplot[color=blue, domain=0:1]{2*x};
+          \addplot[color=magenta, domain=1:2]{2*x};
+          \addplot[color=blue, domain=2:3]{-2*x^2 + 12*x -11};
+          \addplot[color=magenta, domain=2:3]{-4*x + 12};
+          \addplot[color=blue, domain=1:2]{x^2 + 1};
+          \addplot[color=magenta, domain=3:5]{-x + 3};
+          \addplot[color=blue, domain=3:5]{-x^2/2 + 3*x + 2.5};
+          \addplot[color=magenta, domain=5:6]{-2};
+          \addplot[color=blue, domain=5:6]{-2*x + 15};
+          \addplot[color=magenta, domain=6:7]{2*x - 14};
+          \addplot[color=blue, domain=6:7]{x^2 - 14*x + 51};
+          \legend{$f(x)$, $g(x)$}
+        \end{axis}
+      \end{tikzpicture}
+\end{enumerate}
+
+\noindent Find the derivative of the function.
+\begin{align*}
+   \frac{\ud}{\ud x}\int_1^{\sqrt x}\frac{z^2}{z^4 + 1}\ud z
+&= \frac{\ud}{\ud\sqrt x}\left(\int_1^{\sqrt x}\frac{z^2}{z^4 + 1}dz\right)
+   \frac{\ud\sqrt x}{\ud x}\\
+&= \frac{x}{x^2 + 1} \cdot \frac{1}{2\sqrt x}\\
+&= \frac{\sqrt x}{2x^2 + 2}\tag{14}
+\end{align*}
+
+\begin{align*}
+   \frac{\ud}{\ud x}\int_0^{\tan x}\sqrt{t + \sqrt t}\ud t
+&= \frac{\ud}{\ud\tan x}\left(\int_0^{\tan x}\sqrt{t + \sqrt t}\ud t\right)
+   \frac{\ud\tan x}{\ud x}\\
+&= \frac{\sqrt{\tan x + \sqrt{\tan x}}}{\cos^2 x}\tag{15}
+\end{align*}
+
+\noindent\textbf{64. }Given the \textbf{error function}
+\[\erf(x) = \frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\ud t
+  \Longrightarrow \erf'(x) = \frac{2e^{-x^2}}{\sqrt\pi}\]
+
+\[\int_a^b e^{-t^2}\ud t = \frac{\sqrt\pi}{2}\int_a^b \erf'(t)\ud t
+                      = \frac{\sqrt\pi}{2}[\erf(b) - \erf(a)]\tag{a}\]
+
+With $y = e^{x^2}\erf(x)$
+\begin{align*}
+y' &= \left(e^{x^2}\right)'\erf(x) + e^{x^2}\erf'(x)\\
+   &= 2xe^{x^2}\erf(x) + e^{x^2}\frac{2e^{-x^2}}{\sqrt\pi}\\
+   &= 2xe^{x^2}\erf(x) + \frac{2}{\sqrt\pi}\\
+   &= 2xy + \frac{2}{\sqrt\pi}\tag{b}
+\end{align*}
+
+\subsection{Infinite Integral}
+\textbf{56. }Let $y(x)$ be the vertical postion at a distance of $x$ miles from
+the start of the trail, then $y'(x) = f(x)$.
+Thus, $\int_3^5 f(x)\ud x = y(5) - y(3)$, which is the vertical displacement from
+3 to 5 miles.
+
+\noindent\textbf{63. }Total mass of the rod:
+\[\int_0^4 (9 + 2\sqrt x)\ud x
+= \left.9x\right]_0^4 + \left.\frac{2\sqrt{x^3}}{3}\right]_0^4
+= 36 + \frac{16}{3} = 41\frac{1}{3}\text{ (kg)}\]
+
+\noindent\textbf{64. }Amount of water flowing from the tank during the first 10
+minutes:
+\[\int_0^{10} (200 - 4t)\ud t
+= \left.200t\right]_0^{10} - \left.2t^2\right]_0^{10}
+= 2000 - 200 = 1800\text{ (l)}\]
+
+\subsection{The Substitution Rule}
+\textbf{74. }Given $f(x) = \sin\sqrt[3]x$.
+
+Since $f(-x) = \sin\sqrt[3]{-x} = \sin-\sqrt[3]x = -\sin\sqrt[3]x = -f(x)$,
+$f$ is an odd function. Hence $\int_{-2}^3 f(x)\ud x = \int_2^3 f(x)\ud x$.
+
+For $2 \leq x \leq 3$, $0 \leq \sqrt[3]2 \leq \sqrt[3]x \leq \sqrt[3]3 \pi$,
+thus $\sin\sqrt[3]x \geq 0$ and $\int_2^3 f(x)\ud x \geq 0$. Futhermore,
+$\sin\sqrt[3]x \leq 1$ so $\int_2^3 f(x)\ud x \leq \int_2^3 \ud x = 1$.
+
+\noindent Evaluate the integral.
+\begin{align*}
+   \int_{-2}^2(x + 3)\sqrt{4 - x^2}\ud x
+&= \int_{-2}^2 x\sqrt{4 - x^2}\ud x + 3\int_{-2}^2\sqrt{4 - x^2}\ud x\\
+&= 0 + 3 \cdot 2\pi\\
+&= 6\pi\tag{77}
+\end{align*}
+
+\begin{align*}
+   \int_0^{24}\left(85 - 0.18\cos\frac{\pi t}{12}\right)\ud t
+&= \left.85t\right]_0^{24}
+ - \frac{54}{25\pi}\int_0^{24}\frac{\pi t}{12}'\cos\frac{\pi t}{12}\ud t\\
+&= 2040 - \frac{54}{25\pi}\int_0^{2\pi}\cos x \ud x\\
+&= 2040 - \left.\frac{54\sin x}{25\pi}\right]_0^{2\pi}\\
+&= 2040\tag{80}
+\end{align*}
+
+\begin{align*}
+   400 + \int_0^3 450.268e^{1.12567t}\ud t
+&= 400 + 400\int_0^3 1.12567e^{1.12567t}\ud t\\
+&= 400 + \left.400e^{1.12567t}\right]_0^3\\
+&= 400e^{1.12567 \cdot 3}\\
+&\approx 11713\tag{82}
+\end{align*}
+
+\section{Applications of Integration}
+\subsection{Areas Between Curves}
+Evaluate the integral
+
+\begin{align*}
+  \int_{-1}^1\left|e^x - x^2 + 1\right|\ud x
+&=\int_{-1}^1\left(e^x - x^2 + 1\right)\ud x\\
+&=\left[e^x - \frac{x^3}{3} + x\right]_{-1}^1\\
+&=e - \frac{1}{3} + 1 - \frac{1}{e} - \frac{1}{3} + 1\\
+&=e - \frac{1}{e} + \frac{4}{3}\tag{5}
+\end{align*}
+
+\begin{align*}
+  \int_1^4\left|x^2 - 3x + 4\right|\ud x
+&=\int_1^4\left(x^2 - 3x + 4\right)\ud x\\
+&=\left[\frac{x^3}{3} - \frac{3x^2}{2} + 4x\right]_1^4\\
+&=\frac{64 - 1}{3} - \frac{48 - 3}{2} + 16 - 4\\
+&=\frac{21}{2}\tag{7}
+\end{align*}
+
+\begin{align*}
+  \int_1^2\left|\frac{1}{x} - \frac{1}{x^2}\right|\ud x
+&=\int_1^2\left(\frac{1}{x} - \frac{1}{x^2}\right)\ud x\\
+&=\left[\frac{1}{3x^3} - \frac{1}{2x^2}\right]_1^2\\
+&=\frac{1}{24} - \frac{1}{8} - \frac{1}{3} + \frac{1}{2}\\
+&=\frac{1}{12}\tag{9}
+\end{align*}
+
+\noindent\textbf{53. }Find the values of $c$ such that
+\begin{align*}
+      \int_{-|c|}^{|c|}\left|x^2 - c^2 - c^2 + x^2\right|\ud x = 576
+&\iff \int_{-|c|}^{|c|}\left(c^2 - x^2\right)\ud x = 288\\
+&\iff \left[c^2 x - \frac{x^3}{3}\right]_{-|c|}^{|c|} = 288\\
+&\iff \frac{4\left|c^3\right|}{3} = 288\\
+&\iff |c| = 6\\
+&\iff c = \pm 6
+\end{align*}
+
+\noindent\textbf{54. }Find the area of the region enclosed by the line $y = mx$
+and the curve $y = \frac{x}{x^2 + 1}$.
+
+Those two curves enclose a region if and only if the following equations has
+two unique solutions, i.e. $m \in (0, 1)$
+\[mx = \frac{x}{x^2 + 1} \iff mx^3 + (m - 1)x = 0
+  \iff x \in \left\{0, \pm\frac{1 - m}{m}\right\}\]
+
+The area of the region would then be
+\begin{align*}
+A&=\int_{\frac{m-1}{m}}^{\frac{1-m}{m}}\left|mx - \frac{x}{x^2 + 1}\right|\ud x\\
+ &=\int_{\frac{m-1}{m}}^0\left(\frac{x}{x^2 + 1} - mx\right)\ud x +
+   \int_0^{\frac{1-m}{m}}\left(mx - \frac{x}{x^2 + 1}\right)\ud x\\
+ &=\int_{\frac{m-1}{m}}^0\left(\frac{x}{x^2 + 1} - mx\right)\ud x +
+   \int_0^{\frac{1-m}{m}}\left(mx - \frac{x}{x^2 + 1}\right)\ud x\\
+ &=\int_{\frac{m-1}{m}}^0\frac{1}{x^2 + 1}\cdot\frac{\ud x^2}{\ud x}\ud x -
+   \left.\frac{mx^2}{2}\right]_{\frac{m-1}{m}}^0 +
+   \left.\frac{mx^2}{2}\right]_0^{\frac{1-m}{m}} -
+   \int_0^{\frac{1-m}{m}}\frac{1}{x^2 + 1}\cdot\frac{\ud x^2}{\ud x}\ud x\\
+ &=\frac{(m - 1)^2}{m} +
+   2\int_{\left(\frac{m-1}{m}\right)^2}^0\frac{1}{x + 1}\ud x\\
+ &=\frac{m^2 - 2m + 1}{m} +
+   2\left.\ln(|x + 1|)\right]_{\frac{m^2 - 2m + 1}{m^2}}^0\\
+ &=\frac{m^2 - 2m + 1}{m} -
+   2\ln\left(\frac{2m^2 - 2m + 1}{m^2}\right)
+\end{align*}
+
+\subsection{Volumes}
+Evaluate the integral
+
+\begin{align*}
+  \int_1^2\pi\left(2 - \frac{x}{2}\right)^2 \ud x
+&=\pi\int_1^2\left(4 - x + \frac{x^2}{4}\right)\ud x\\
+&=\pi\left[4x - x^2 + \frac{x^3}{12}\right]_1^2\\
+&=\pi\left(4 - 3 + \frac{7}{12}\right)\\
+&=\frac{19\pi}{12}\tag{1}
+\end{align*}
+
+\[\int_1^5\pi(x - 1)\ud x
+= \pi\left[\frac{x^2}{2} - x\right]_1^5
+= \pi(12 - 4)
+= 8\pi\tag{3}\]
+
+\[\int_0^9 4\pi y \ud y = \left.2\pi y^2\right]_0^9 = 162\pi\tag{5}\]
+
+\[\int_0^1\pi\left|x^2 - x^6\right|\ud x
+= \pi\left[\frac{x^3}{3} - \frac{x^7}{7}\right]_0^1
+= \frac{4\pi}{21}\tag{7}\]
+
+\begin{align*}
+   \int_{-2}^2\pi\left|\frac{x^4}{16} - 25 + 10x^2 - x^4\right|\ud x
+&= 2\pi\int_0^2\left(\frac{15x^4}{16} - 10x^2 + 25\right)\ud x\\
+&= 2\pi\left[-\frac{10x^3}{3} + 25x + \frac{3x^5}{16}\right]_0^2\\
+&= \frac{88\pi}{3}\tag{8}
+\end{align*}
+
+\begin{align*}
+   \int_0^1\pi\left|\left(\sqrt x - 1\right)^2 - \left(x^2 - 1\right)^2\right|\ud x
+&= \int_0^1\pi\left|x - 2\sqrt x - x^4 + 2x^2\right|\ud x\\
+&= \int_0^1\pi\left(x^4 - 2x^2 - x + 2\sqrt x\right)\ud x\\
+&= \pi\left[\frac{x^5}{5} - \frac{2x^3}{3}
+            - \frac{x^2}{2} + \frac{4\sqrt x^3}{3}\right]_0^1\\
+&= \pi\left(\frac{1}{5} - \frac{2}{3} - \frac{1}{2} + \frac{4}{3}\right)\\
+&= \frac{11\pi}{30}\tag{11}
+\end{align*}
+
+\begin{align*}
+   \int_0^h\left(a + \frac{x}{h}(b-a)\right)^2\ud x
+&= \int_0^h\left(a^2 - \frac{ax}{h}(b-a) + \frac{x^2}{h^2}(b-a)^2\right)\ud x\\
+&= \left[a^2 x - \frac{ax^2}{2h}(b-a) + \frac{x^3}{3h^2}(b-a)^2\right]_0^h\\
+&= ha^2 - \frac{ha}{2}(b-a) + \frac{h}{3}(b-a)^2\\
+&= ha^2 - \frac{hab}{2} + \frac{ha^2}{2}
+ + \frac{ha^2}{3} - \frac{2hab}{3} + \frac{hb^2}{3}\\
+&= \frac{11ha^2 - 7hab + 2hb^2}{6}\tag{50}
+\end{align*}
+
+\begin{align*}
+A&=\int_{-r}^r\left(\pi\left(R + \sqrt{r^2 - x^2}\right)^2 -
+                    \pi\left(R - \sqrt{r^2 - x^2}\right)^2\right)\ud x\\
+ &=\int_{-r}^r 4\pi R\sqrt{r^2 - x^2}\ud x\\
+ &=2\pi R\int_{-r}^r 2\sqrt{r^2 - x^2}\ud x\\
+ &=2\pi R \cdot \pi r^2\\
+ &=2\pi^2 R r^2\tag{61}
+\end{align*}
+
+\setcounter{subsection}{3}
+\subsection{Work}
+\textbf{7. }Spring constant: $k = F(4) / 4 = 10g / 4 = 2.5g$ (lbf/in)
+
+Work done by stretching the spring from its natural length to 6 in:
+\[\int_0^6 kx\ud x = \left.k\frac{x^2}{2}\right]_0^6 = 18k = 45g\text{ (lbf.in)}\]
+
+\noindent\textbf{9. }Suppose that 2 J of work is needed to stretch a spring from its
+natural length of 30 cm to a length of 42 cm.
+
+\begin{align*}
+      \int_0^{0.12}kx\ud x = 2
+&\iff \left.\frac{kx^2}{2}\right]_0^{0.12} = 2\\
+&\iff 0.0072k = 2\\
+&\iff k = \frac{2500}{9}\text{ (N/m)}
+\end{align*}
+
+\begin{align*}
+  \int_{0.05}^{0.1}kx\ud x
+&=\int_{0.05}^{0.1}\frac{2500x}{9}\ud x\\
+&=\left.\frac{1250x^2}{9}\right]_{0.05}^{0.1}\\
+&=\frac{1250(0.01 - 0.0025)}{9}\\
+&=\frac{25}{24}\text{ (J)}\tag{a}
+\end{align*}
+
+\[x = \frac{F}{k} = \frac{30 \cdot 9}{2500} = \frac{27}{250}\text{ (m)}\tag{b}\]
+
+\noindent Evaluate the integral
+\[\int_0^{50}mgx\ud x = \left.\frac{25gx^2}{2}\right]_0^{50} = 31250g\text{ (ft.lbf)}\tag{13}\]
+
+\begin{align*}
+W&=\lim_{n \to \infty}\sum_{i = 1}^n F_i\frac{3i}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n m_i g\frac{3i}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n A_i\frac{3}{n}\rho g\frac{3i}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n
+   3\left(1 - \frac{i}{n}\right)8\left(1 - \frac{i}{n}\right)
+   1000g\frac{3i}{n}\cdot\frac{3}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n
+   \frac{80000}{3}\left(3 - \frac{3i}{n}\right)^2\frac{3i}{n}\cdot\frac{3}{n}\\
+ &=\int_0^3 \frac{80000}{3}\left(9x - 6x^2 + x^3\right)\ud x\\
+ &=\frac{80000}{3}\left[\frac{9x^2}{2} - 2x^3 + \frac{x^4}{4}\right]_0^3\\
+ &=180000\text{ (J)}\tag{21}
+\end{align*}
+
+\begin{align*}
+W&=\lim_{n \to \infty}\sum_{i = 1}^n m_i g\frac{8i}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n A_i\frac{8}{n}\rho g\frac{8i}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n
+   \pi\left(6 - 3\frac{i}{n}\right)^2 62.5g\frac{8i}{n}\cdot\frac{8}{n}\\
+ &=\lim_{n \to \infty}\sum_{i = 1}^n
+   \frac{1125\pi g}{128}\left(16 - \frac{8i}{n}\right)^2\frac{8i}{n}\cdot\frac{8}{n}\\
+ &=\frac{1125\pi g}{128}\int_0^8\left(16 - x\right)^2 x\ud x\\
+ &=\frac{1125\pi g}{128}\left[128x^2 - \frac{32x^3}{3} + \frac{x^4}{4}\right]_0^8\\
+ &=33000\pi g\text{ (ft.lbf)}\tag{23}
+\end{align*}
+
+\subsection{Average Value of a Function}
+\textbf{9. }Given the function $f(x) = (x - 3)^2$ on $[2, 5]$.
+\[f_{ave}
+= \frac{1}{5 - 2}\int_2^5 (x - 3)^2 \ud x
+= \frac{1}{3}\left[\frac{x^3}{3} - 3x^2 + 9x\right]_2^5
+= 1\tag{a}\]
+
+Since $x \in [2, 5]$,
+\[f(c) = f_{ave} \iff (c - 3)^2 = 1 \iff c = 4\tag{b}\]
+
+\begin{tikzpicture}
+  \begin{axis}[
+    axis x line=middle, axis y line=middle,
+    xlabel={$x$}, ylabel={$y$}, area style,
+    xlabel style={at=(current axis.right of origin), anchor=west},
+    ylabel style={at=(current axis.above origin), anchor=south}]
+    \addplot[domain=2:5,blue]{x^2 - 6*x + 9};
+    \addplot[domain=2:5]{1};
+  \end{axis}
+\end{tikzpicture}
+
+\noindent\textbf{12. }Given $f(x) = 2\sin x - \sin 2x$ on $[0, \pi]$.
+\begin{align*}
+f_{ave} &= \frac{1}{\pi}\int_0^\pi\left(2\sin x - \sin 2x\right)\ud x\\
+        &= \frac{1}{\pi}\left[\frac{\cos 2x}{2} - 2\cos x\right]_0^\pi\\
+        &= \frac{4}{\pi}
+\end{align*}
+
+$f(c) = f_{ave} \iff 2\sin x - \sin 2x = \frac{4}{\pi}$, i.e.
+$x \approx 1.24$ or $x \approx 2.81$ on $[0, \pi]$.
+
+\noindent\textbf{13. }Since $f$ is continuous, apply Mean Value Theorem on
+$[1, 3]$,
+
+\[\exists c \in [1, 3], f(c) = \frac{1}{3 - 1}\int_1^3 f(x)\ud x
+= \frac{8}{2} = 4\]
+
+\section{Techniques of Integration}
+\subsection{Integration by Parts}
+Evaluate the integral
+\begin{align*}
+   \int x \cos 5x \ud x
+&= \int\frac{x}{5} \ud\sin 5x\\
+&= \frac{x\sin 5x}{5} - \int\sin 5x \ud\frac{x}{5}\\
+&= \frac{x\sin 5x}{5} + \frac{x\cos 5x}{25} + C\tag{3}
+\end{align*}
+
+\begin{align*}
+   \int e^{2\theta}\sin 3\theta \ud\theta
+&= \int\frac{\sin 3\theta}{2} \ud e^{2\theta}\\
+&= \frac{e^{2\theta}\sin 3\theta}{2}
+ - \int\frac{3}{2}e^{2\theta}\cos 3\theta \ud\theta\\
+&= \frac{e^{2\theta}\sin 3\theta}{2}
+ - \frac{3}{2}\int\frac{\cos 3\theta}{2} \ud e^{2\theta}\\
+&= \frac{e^{2\theta}\sin 3\theta}{2}
+ - \frac{3e^{2\theta}\cos 3\theta}{4}
+ - \frac{9}{4}\int e^{2\theta}\sin 3\theta \ud\theta\\
+&= \frac{4}{13}e^{2\theta}
+   \left(\frac{\sin 3\theta}{2} - \frac{\cos 3\theta}{4}\right) + C\\
+&= \frac{e^{2\theta}}{13}(2\sin 3\theta - 3\cos 3\theta) + C\tag{17}
+\end{align*}
+
+\setcounter{section}{8}
+\section{Differential Equations}
+\setcounter{subsection}{2}
+\subsection{Separable Equations}
+Solve the equation.
+\begin{align*}
+  \leibniz{y}{x} = xy^2
+  &\iff y^{-2}\ud y = x\ud x\\
+  &\;\Longrightarrow \int\frac{\ud y}{y^2} = \int x\ud x
+  \qquad\text{(for $y\neq 0$)}\\
+  &\iff C_y - \frac{1}{y} = C_x + \frac{x^2}{2}\\
+  &\iff \frac{C - x^2}{2} = \frac{1}{y}\\
+  &\iff y = \frac{2}{C - x^2}\tag{1}
+\end{align*}
+\begin{align*}
+  (y + \sin y)\leibniz{y}{x} = x + x^3
+  &\iff \int(y + \sin y)\ud y = \int(x + x^3)\ud x\\
+  &\iff \frac{y^2}{2} - \cos y = \frac{x^2}{2} + \frac{x^4}{4} + C\tag{5}\\
+\end{align*}
+\begin{align*}
+  \leibniz{y}{t} = \frac{t}{y\exp(y + t^2)}
+  &\iff \int ye^y\ud y = \int te^{-t^2}\ud t\\
+  &\iff (y - 1)e^y = C - \frac{1}{2e^{t^2}}\tag{7}
+\end{align*}
+\begin{align*}
+  \leibniz{u}{t} = \frac{2t + \sec^2 t}{2u}
+  &\iff \int 2u\ud u = \int(2t + \sec^2 t)\ud t\\
+  &\iff u^2 = t^2 + \tan t + C\tag{13}
+\end{align*}
+Since $u(0) = -5$, $u = -\sqrt{t^2 + \tan t + 25}$.
+\begin{align*}
+  \leibniz{y}{x} = xy
+  &\iff \int\frac{\ud y}{y} = \int x\ud x\qquad\text{(since $y\neq 0$)}\\
+  &\iff \ln|y| = \frac{x^2}{2} + C\\
+  &\iff |y| = \exp\left(\frac{x^2}{2} + C\right)\tag{19}
+\end{align*}
+Since $y(0) = 1$, $y = \exp(x^2/2)$.\pagebreak
+\begin{align*}
+  y(x) = 2 + \int_2^x[t - ty(t)]\ud t
+  &\;\Longrightarrow y - 2 = \int(x - xy)\ud x\\
+  &\iff \leibniz{(y - 2)}{x} = x - xy\\
+  &\iff \int\frac{\ud y}{1 - y} = \int x\ud x\\
+  &\iff C - \frac{x^2}{2} = \ln|1 - y|\\
+  &\iff y = 1 \pm \exp\left(C - \frac{x^2}{2}\right)\tag{33}
+\end{align*}
+Since $y(2) = 2$ (which can be trivially obtained from the original condition),
+$y = 1 + \exp(2 - x^2/2)$.
+
+\subsection{Models for Population Growth}
+\textbf{3.} The Pacific halibut fishery has been modeled
+by the differential equation
+\begin{align*}
+  \leibniz{y}{t} = ky\left(1 - \frac{y}{M}\right)
+  &\;\Longrightarrow \int\left(\frac{1}{y} + \frac{1}{M-y}\right)\ud y
+  = \int k\ud t\\
+  &\iff \ln|y| - \ln|M - y| = kt + C\\
+  &\iff \left|\frac{M}{y} - 1\right| = e^{-kt-C}\\
+  &\iff \frac{M}{y} = 1 \pm e^{-kt-C}\\
+  &\iff y = \frac{M}{1 \pm e^{-kt-C}}
+  \tag{$*$}
+\end{align*}
+
+As $M = 8\times 10^7$, $k = 0.71$ and $y(0) = 2\times 10^7$,
+from $(*)$ we get $\pm e^{-C} = 3$ and thus
+\[y = \frac{M}{1 + 3e^{-kt}}\]
+
+For $t = 1$, $y \approx 3.2\times 10^7$.
+For $y = 4\times 10^7$, $t = (\ln 3)/0.71$.
+
+\noindent\textbf{5.} Suppose a population grows according to a logistic model
+\[\leibniz{P}{t} = kP\left(1 - \frac{P}{M}\right)
+\iff P(t) = \frac{M}{1 \pm e^{-kt-C}}\]
+with initial population $P(0) = 1000$ and carrying capacity $M = 10000$.
+
+Suppose $P(1) = 2500$,
+\[\begin{dcases}
+  \frac{10000}{1\pm e^{-C}} &= 1000\\
+  \frac{10000}{1\pm e^{-k-C}} &= 2500
+\end{dcases}
+\iff\begin{cases}
+  \pm e^{-C} &= 9\\
+  \pm e^{-k-C} &= 3
+\end{cases}
+\iff\begin{cases}
+  \pm := +\\
+  C = -\ln 9\\
+  k = \ln 3
+\end{cases}\]
+
+After another 3 years, the population will be
+\[P(3) = \left(t\mapsto\frac{10000}{1+3^{2-t}}\right)(1+3) = 9000\]
+
+\subsection{Linear Equations}
+Solve the differential equation.
+\begin{align*}
+  \leibniz{y}{x} + y = x
+  &\iff e^x\leibniz{y}{x} + y\leibniz{e^x}{x} = xe^x\\
+  &\iff \int\ud ye^x = \int x\ud e^x\\
+  &\iff ye^x = e^x(x - 1) + C\\
+  &\iff y = x - 1 + Ce^{-x}\tag{7}
+\end{align*}
+\begin{align*}
+  x\leibniz{y}{x} + y = \sqrt x
+  &\iff \int\ud xy = \int\sqrt x\ud x\\
+  &\iff xy = \frac{2x\sqrt x}{3} + C\\
+  &\iff y = \frac{2\sqrt x}{3} + \frac{C}{x}\tag{9}
+\end{align*}
+\begin{align*}
+  x^2\leibniz{y}{x} + 2xy = \ln x
+  &\iff \int\ud yx^2 = \int\ln x\ud x\\
+  &\iff yx^2 = x(\ln x - 1) + C\\
+  &\iff y = \frac{\ln x - 1}{x} + \frac{C}{x^2}
+\end{align*}
+Since $y(1) = 2$, $C = 3$.
+\begin{align*}
+  L\leibniz{I}{t} + RI = \mathcal E
+  &\iff e^{Rt/L}\left(\leibniz{I}{t} + \frac{R}{L}I\right)
+  = \frac{\mathcal E}{L}e^{Rt/L}\\
+  &\iff \int\ud Ie^{Rt/L} = \frac{\mathcal E}{L}\int e^{Rt/L}\ud t\\
+  &\iff Ie^{Rt/L} = \frac{\mathcal E}{R}e^{Rt/L} + C\\
+  &\iff I = \frac{\mathcal E}{R} + \frac{C}{\exp(Rt/L)}
+\end{align*}
+Since $\mathcal E = 40$ V, $L = 2$ H, $R = 10\,\Omega$ and $I(0) = 0$,
+$I(t) = 4 - 4/\exp 5t$ and $I(0.1) = 4 - 4/\sqrt e$.
+
+\allowdisplaybreaks
+\setcounter{section}{10}
+\section{Lazy Evaluation}
+\setcounter{subsection}{2}
+\subsection{The Integral Test and Estimates of Sums}
+\textbf{34. }Using Leonhard Euler's calculation of the exact sum of
+the $p$-series with $p = 2$:
+\[\zeta(2) = \sum_{n=1}^\infty\frac{1}{n^2}
+= \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} = \frac{\pi^2}{6}\]
+\[\sum_{n=2}^\infty\frac{1}{n^2} = \lim_{n\to\infty}\sum_{i=2}^n\frac{1}{i^2}
+= \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} - \frac{1}{1^2}
+= \frac{\pi^2}{6} - 1\tag{a}\]
+\[\sum_{n=3}^\infty\frac{1}{(n + 1)^2}
+= \lim_{n\to\infty}\sum_{i=4}^{n+1}\frac{1}{i^2}
+= \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} - \sum_{i=1}^3\frac{1}{i^2}
+= \frac{\pi^2}{6} - \frac{49}{36}\tag{b}\]
+\[\sum_{n=1}^\infty\frac{1}{(2n)^2}
+= \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{4i^2}
+= \frac{1}{4}\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2}
+= \frac{\pi^2}{24}\tag{c}\]
+
+\noindent Determine if the series is convergent or divergent using
+the Integral Test.
+
+\[\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}\tag{22}\]
+
+\begin{align*}
+  \int_2^\infty\frac{1}{x(\ln x)^2}
+&= \lim_{t\to\infty}\int_2^t\frac{1}{x(\ln x)^2}\ud x\\
+&= \lim_{t\to\infty}\int_2^t\frac{1}{(\ln x)^2}\ud\ln x\\
+&= \lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{1}{x^2}\ud x\\
+&= \lim_{t\to\infty}\left.\frac{-1}{x}\right]_{\ln 2}^{\ln t}\\
+&= \lim_{t\to\infty}\left(\frac{1}{\ln{2}} - \frac{1}{\ln t}\right)\\
+&= \frac{1}{\ln 2}
+\end{align*}
+
+Thus by the Integral Test, the given series is convergent.
+
+\[\sum_{n=3}^\infty\frac{n^2}{e^n}\tag{24}\]
+
+\begin{align*}
+  \int_3^\infty\frac{x^2}{e^x}
+&= \lim_{t\to\infty}\int_3^t\frac{x^2}{e^x}\ud x\\
+&= \lim_{t\to\infty}\int_3^t -x^2 \ud e^{-x}\\
+&= \lim_{t\to\infty}\left(\int_3^t e^{-x}\ud x^2
+ - \left.x^2 e^{-x}\right]_3^t\right)\\
+&= \lim_{t\to\infty}\left(-\int_3^t 2x\ud e^{-x}
+ + \left.\frac{x^2}{e^x}\right]_t^3\right)\\
+&= \lim_{t\to\infty}\left(\int_3^t e^{-x}\ud 2x
+ + \left[\frac{2x}{e^x} + \frac{x^2}{e^x}\right]_t^3\right)\\
+&= \lim_{t\to\infty}\left(2\int_3^t e^{-x}\ud x
+ + \left.\frac{2x + x^2}{e^x}\right]_t^3\right)\\
+&= \lim_{t\to\infty}\left.\frac{2 + 2x + x^2}{e^x}\right]_t^3\\
+&= \lim_{t\to\infty}\left(\frac{17}{e^3} - \frac{2 + 2t + t^2}{e^t}\right)\\
+&= \frac{17}{e^3}
+\end{align*}
+
+Thus by the Integral Test, the given series is convergent.\pagebreak
+
+\[\sum_{n=1}^\infty\frac{\cos(\pi n)}{\sqrt n}\tag{27}\]
+
+Since $x \mapsto \cos(\pi x)/\sqrt n$ is neither positive
+(e.g. $\cos 3\pi/\sqrt 3 = -1$) nor ultimately decreasing, the Integral Test
+cannot be used to determine whether the series is convergent.
+
+\subsection{The Comparison Test}
+Determine whether the series is convergent or divergent.
+\[\sum_{n=1}^\infty\frac{\sqrt{n^4 + 1}}{n^3 + n^2}\tag{25}\]
+
+We use the Limit Comparison Test with
+\[a_n = \frac{\sqrt{n^4 + 1}}{n^3 + n^2} \qquad b_n = \frac{1}{n}\]
+and obtain
+\[\lim_{n\to\infty}\frac{a_n}{b_n}
+= \lim_{n\to\infty}\frac{\sqrt{n^4 + 1}}{n^2 + n}
+= \lim_{n\to\infty}\frac{\sqrt{1 + \frac{1}{n^4}}}{1 + \frac{1}{n}}
+= 1 > 0\]
+
+Since this limit exists and $\sum\frac{1}{n}$ is divergent ($p$-series with
+$p = 1$), the given series diverges by the Limit Comparison Test.
+
+\[\sum_{n=1}^\infty\frac{1}{n!}\tag{29}\]
+\[\lim_{n\to\infty}\frac{1/(n+1)!}{1/n!}
+= \lim_{n\to\infty}\frac{1}{n + 1} = 0 < 1\]
+
+Thus by the Ratio Test, the given series is absolutely convergent.
+
+\[\sum_{n=1}^\infty\frac{n!}{n^n}\tag{30}\]
+\[\frac{n!}{n^n} = \frac{2}{n^2}\cdot\frac{n!}{2n^{n-2}} \leq \frac{2}{n^2}\]
+
+Since both $\sum n!/n^n$ and $\sum 2/n^2$ are series with positive terms and
+$\sum 2/n^2$ converges because it is a constant time of $p$-series with
+$p = 2$, by the Comparison Test, $\sum n!/n^n$ is convergent.
+
+\[\sum_{n=1}^\infty\frac{1}{n\sqrt[n]n}\tag{32}\]
+
+We use the Limit Comparison Test with
+\[a_n = \frac{1}{n\sqrt[n]n} \qquad b_n = \frac{1}{n}\]
+and obtain
+\[\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{1}{\sqrt[n]n}\]
+
+Since $\frac{1}{\sqrt[n]n}' = \frac{1 - \ln n}{n^2\sqrt[n]n}$ is negative on
+$(e, \infty)$, $n \mapsto \frac{1}{\sqrt[n]n}$ is ultimately decreasing.
+Additionally, $\frac{1}{\sqrt[n]n} \geq \frac{1}{\sqrt[n]1} = 1$ on this
+interval, thus
+\[\lim_{n\to\infty}\frac{1}{\sqrt[n]n}
+= \inf\left\{\frac{1}{\sqrt[n]n}: n \in \mathbb N_3\right\} = 1 > 0\]
+
+Therefore, the given series diverges by the Limit Comparison Test,
+as $\sum\frac{1}{n}$ is divergent ($p$-series with $p = 1$).
+
+\subsection{Alternating Series}
+Test the series for convergence or divergence.
+
+\[\sum_{n=1}^\infty (-1)^n\frac{n^n}{n!}\tag{19}\]
+
+Since $\frac{(n + 1)^{n + 1}}{(n + 1)!} > \frac{n^n}{n!}$, the given
+alternating series diverges.
+
+\[\sum_{n=1}^\infty (-1)^n\left(\sqrt{n + 1} - \sqrt n\right)\tag{20}\]
+
+For all $n$,
+\begin{align*}
+      n^2 + 2n < n^2 + 2n + 1
+&\iff \sqrt{n(n + 2)} < n + 1\\
+&\iff n + \sqrt{n(n + 2)} + n + 2 < 4n + 4\\
+&\iff \sqrt{n + 2} + \sqrt n < 2\sqrt{n + 1}\\
+&\iff \sqrt{n + 2} - \sqrt{n + 1} < \sqrt{n + 1} - \sqrt n\tag{i}
+\end{align*}
+\[\lim_{n\to\infty}\left(\sqrt{n + 1} - \sqrt n\right)
+= \lim_{n\to\infty}\frac{1}{\sqrt{n + 1} + \sqrt n}
+= \lim_{n\to\infty}\frac{1/\sqrt n}{\sqrt{1 + 1/\sqrt{n}} + 1}
+= 0\tag{ii}\]
+
+Thus, by the Alternating Series Test, the given series is convergent.
+
+\subsection{Absolute Convergence}
+Determine whether the series is absolutely convergent,
+conditionally convergent, or divergent.
+
+\[\sum_{n=2}^\infty\left(\frac{-2n}{n + 1}\right)^{5n}\tag{22}\]
+\[\lim_{n\to\infty}\sqrt[n]{\left(\frac{2n}{n + 1}\right)^{5n}}
+= \lim_{n\to\infty}\left(\frac{2n}{n + 1}\right)^5\\
+= \left(\lim_{n\to\infty}\frac{2}{1 + 1/n}\right)^5\\
+= 32 > 1\]
+
+Thus the given series diverges by the Root Test.
+
+\[\sum_{n=1}^\infty\prod_{i=1}^n\frac{2i}{3i + 2}\tag{30}\]
+\[\lim_{n\to\infty}\frac{\prod_{i=1}^{n+1}\frac{2i}{3i + 2}}
+                        {\prod_{i=1}^n\frac{2i}{3i + 2}}
+= \lim_{n\to\infty}\frac{2n + 2}{3n + 5}
+= \lim_{n\to\infty}\frac{2 + 2/n}{3 + 5/n}
+= \frac{2}{3} < 1\]
+
+Thus by the Ratio Test, the given series is absolutely convergent.
+
+\setcounter{subsection}{7}
+\subsection{Power Series}
+Find the radius of convergence and the interval of convergence of the series.
+
+\[\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n + 1)!}\tag{14}\]
+
+Let $a_n = (-1)^n x^{2n+1}/(2n + 1)!$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\left|\frac{x^2}{4n^2 + 10n + 6}\right|
+= 0 < 1\]
+
+Thus by the Ratio Test, the series is convergent for all $x$ and the radius of
+convergence is $R = \infty$.
+
+\[\sum_{n=1}^\infty\frac{3^n (x+4)^n}{\sqrt n}\tag{17}\]
+
+Let $a_n = 3^n (x+4)^n / \sqrt n$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\left|\frac{3(x + 4)\sqrt n}{\sqrt{n + 1}}\right|
+= 3|x + 4|\]
+
+Using the Ratio Test, we see that the series converges if $|x + 4| < 1/3$ and
+it diverges if $|x + 4| > 1/3$, thus the radius of convergence is $R = 1/3$.
+
+When $|x + 4| = 1/3$, the series is either $\sum (-3)^n / \sqrt n$ or
+$\sum 3^n / \sqrt n$, both of which diverge by the Test for Divergence.
+Therefore the interval of convergence is $(-13/3, -11/3)$.
+
+\[\sum_{n=1}^\infty\frac{b^n}{\ln n}(x - a)^n,\qquad b > 0\tag{22}\]
+
+Let $a_n = b^n (x - a)^n / \ln n$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\left|\frac{b(x - a)\ln n}{\ln(n + 1)}\right|
+= b|x - a|\]
+
+Using the Ratio Test, we see that the series converges if $|x - a| < b^{-1}$
+and it diverges if $|x - a| > b^{-1}$, thus the radius of convergence is
+$R = b^{-1}$.
+
+When $|x - a| = b^{-1}$, the series is $\sum (\pm b)^n / \ln n$, which diverges
+by the Test for Divergence. Therefore the interval of convergence is
+$(a - b^{-1}, a + b^{-1})$.
+
+\[\sum_{n=2}^\infty\frac{x^{2n}}{n(\ln n)^2}\tag{26}\]
+
+Let $a_n = x^{2n} / n / (\ln n)^2$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\left|\frac{x^2 n \ln^2 n}{(n + 1)\ln^2(n + 1)}\right|
+= x^2\]
+
+Using the Ratio Test, we see that the series converges if $|x| < 1$ and it
+diverges if $|x| > 1$, therefore the radius of convergence is $R = 1$.
+
+When $x = \pm 1$, $a_n = n^{-1}/(\ln n)^2$, which is defined by a continuous,
+positive and decreasing function $x \mapsto x^{-1}/(\ln x)^2$ on $[2, \infty)$.
+\begin{align*}
+   \int_2^\infty\frac{1}{x(\ln x)^2}\ud x
+&= \lim_{t\to\infty}\int_2^t\frac{1}{(\ln x)^2}\ud\ln x\\
+&= \lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{1}{x^2}\ud x\\
+&= \lim_{t\to\infty}\left.\frac{1}{3x^3}\right]_{\ln t}^{\ln 2}\\
+&= \frac{1}{3(\ln 2)^3}
+\end{align*}
+
+By the Integral Test, $\sum_{n=2}^\infty n^{-1}/(\ln n)^2$ converges, and thus
+the interval if of convergence of the given power series is $[-1, 1]$.
+
+\subsection{Representations of Functions as Power Series}
+Find a power series representation for the function and determine the interval
+of convergence.
+
+\[f(x) = \frac{5}{1 - 4x^2}
+       = 5\sum_{n=0}^\infty\left(4x^2\right)^n
+       = \sum_{n=0}^\infty 5 \cdot 2^{2n} x^{2n}\tag{4}\]
+
+Interval of convergence is $(-1, 1)$.
+
+\[f(x) = \frac{x^2}{a^3 - x^3}
+       = \frac{x^2}{a^3}\sum_{n=0}^\infty\left(\frac{x}{a}\right)^{3n}
+       = \sum_{n=0}^\infty\frac{x^{3n + 2}}{a^{3n + 3}}\tag{10}\]
+
+Interval of convergence is $(-a, a)$.
+
+\begin{align*}
+f(x) &= \frac{x + 2}{2x^2 - x - 1}\\
+     &= \frac{1}{x - 1} - \frac{1}{2x + 1}\\
+     &= -\sum_{n=0}^\infty x^n - \sum_{n=0}^\infty (-2x)^n\\
+     &= \sum_{n=0}^\infty (-1 - (-2)^n)x^n\tag{12}
+\end{align*}
+
+Interval of convergence is $(-1, 1) \cap (-1/2, 1/2) = (-1/2, 1/2)$.
+
+\noindent\textbf{40. }Find the sum of the series when $|x| < 1$.
+\begin{align*}
+   \sum_{n=1}^\infty nx^{n - 1}
+&= \sum_{n=1}^\infty x^{n - 1} + \sum_{n=1}^\infty (n - 1)x^{n - 1}\\
+&= \sum_{n=0}^\infty x^n + \sum_{n=0}^\infty nx^n\\
+&= \frac{1}{1 - x} + x\sum_{n=1}^\infty nx^{n - 1}\\
+&= \frac{1}{(1 - x)^2}\tag{a}
+\end{align*}
+
+\[\sum_{n=1}^\infty nx^n
+= x\sum_{n=1}^\infty nx^{n - 1}
+= \frac{x}{(1 - x)^2}\tag{b.i}\]
+
+\[\sum_{n=1}^\infty \frac{n}{2^n}
+= \left(x \mapsto \frac{x}{(1 - x)^2}\right)\left(\frac{1}{2}\right)
+= 2\tag{b.ii}\]
+
+\begin{align*}
+   \sum_{n=2}^\infty n(n - 1)x^n
+&= \sum_{n=2}^\infty 2(n - 1)x^n + \sum_{n=2}^\infty (n - 1)(n - 2)x^n\\
+&= 2\sum_{n=1}^\infty (n - 1)x^n + x\sum_{n=1}^\infty n(n - 1)x^n\\
+&= 2\left(\sum_{n=1}^\infty nx^n + 1 - \sum_{n=0}^\infty x^n\right) : (1 - x)\\
+&= 2\left(\frac{x}{(1 - x)^2} + 1 - \frac{1}{1 - x}\right) : (1 - x)\\
+&= \frac{2x^2}{(1 - x)^3}\tag{c.i}
+\end{align*}
+
+\[\sum_{n=2}^\infty \frac{n^2 - n}{2^n}
+= \left(x \mapsto \frac{2x^2}{(1 - x)^3}\right)\left(\frac{1}{2}\right)
+= 4\tag{c.ii}\]
+
+\[\sum_{n=1}^\infty \frac{n^2}{2^n}
+= \sum_{n=2}^\infty \frac{n^2 - n}{2^n} + \sum_{n=1}^\infty nx^n
+= 4 + 2 = 6\tag{c.iii}\]
+
+\subsection{Taylor and Maclaurin Series}
+Find the Taylor series for $f$ centered at the given value of $a$ and the
+associative radius of convergence.
+
+\[f(x) = \ln x,\qquad a = 2\tag{15}\]
+\begin{align*}
+   f(x)
+&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\
+&= \ln 2 + \sum_{n=1}^\infty\left(x\mapsto\binom{-1}{n-1}\frac{1}{nx^n}\right)
+                            (2)\cdot(x - 2)^n\\
+  &= \ln 2 + \sum_{n=1}^\infty(-1)^{n-1}\frac{(x - 2)^n}{n2^n}
+\end{align*}
+
+Let $a_n = (-1)^{n-1}(x - 2)^n/(n2^n)$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\frac{n}{2n + 2}|x - 2|
+= \frac{|x - 2|}{2}\]
+
+Using the Ratio Test, we see $f(x) = \ln 2 + \sum a_n$ converges if
+$|x - 2| < 2$ and it diverges if $|x - 2| > 2$, therefore the associative
+radius of convergence is $R = 2$.
+
+\[f(x) = \frac{1}{x},\qquad a = -3\tag{16}\]
+\begin{align*}
+   f(x)
+&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\
+&= \sum_{n=0}^\infty
+   \left(x\mapsto\binom{-1}{n}\frac{1}{x^{n+1}}\right)(-3)\cdot(x + 3)^n\\
+&= \sum_{n=0}^\infty(-1)^n\frac{(x + 3)^n}{(-3)^{n+1}}\\
+&= \sum_{n=0}^\infty\frac{-(x + 3)^n}{3^{n+1}}
+\end{align*}
+
+Let $a_n = (x + 3)^n/3^{n+1}$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\frac{|x + 3|}{3}
+= \frac{|x + 3|}{3}\]
+
+Using the Ratio Test, we see that the series converges if $|x + 3| < 3$ and it
+diverges if $|x + 3| > 3$, therefore the associative radius of convergence is
+$R = 3$.
+
+\[f(x) = \sin x,\qquad a = \frac{\pi}{2}\tag{18}\]
+\begin{align*}
+f(x) &= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\
+     &= \sum_{n=0}^\infty\frac{\cos\frac{n\pi}{2}}{n!}
+                         \left(x - \frac{\pi}{2}\right)^n\\
+     &= \sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}
+                         \left(x - \frac{\pi}{2}\right)^{2n}
+\end{align*}
+
+Let $a_n = (-1)^n(x - \pi/2)^{2n}/(2n)!$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\frac{(x - \pi/2)^2}{(2n + 2)(2n + 1)}
+= 0 < 1\]
+
+Using the Ratio Test, we see that the series converges for all $x$, thus the
+associative radius of convergence is $R = \infty$.
+
+\[f(x) = \sqrt x,\qquad x = 16\tag{20}\]
+\begin{align*}
+   f(x)
+&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\
+&= \sum_{n=0}^\infty\left(x\mapsto\binom{\frac{1}{2}}{n}x^{1/2-n}\right)(16)
+                    \cdot (x - 16)^n\\
+&= \sum_{n=0}^\infty\binom{\frac{1}{2}}{n}\frac{4(x - 16)^n}{16^n}
+\end{align*}
+
+Let $a_n = 4\binom{1/2}{n}(x - 16)^n/16^n$,
+\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
+= \lim_{n\to\infty}\left|\frac{1/2 - n}{n + 1}\right|\frac{|x - 16|}{16}
+= \frac{|x - 16|}{16}\]
+
+Using the Ratio Test, we see that the series converges if $|x - 16| < 16$ and
+it diverges if $|x - 16| > 16$, therefore the associative radius of convergence
+is $R = 16$.
+
+\noindent\textbf{55. }Use series to evaluate the limit.
+\begin{align*}
+   \lim_{x \to 0}\frac{x - \ln(1 + x)}{x^2}
+&= \lim_{x \to 0}\frac{x - \sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}}{x^2}\\
+&= \lim_{x \to 0}\frac{\sum_{n=2}^\infty (-1)^n\frac{x^n}{n}}{x^2}\\
+&= \lim_{x \to 0}\sum_{n=0}^\infty (-1)^n\frac{x^n}{n + 2}\\
+&= \lim_{x \to 0}\frac{1}{2}
+ + \lim_{x \to 0}\sum_{n=1}^\infty (-1)^n\frac{x^n}{n + 2}\\
+&= \frac{1}{2}
+\end{align*}
+
+\newpage\noindent Find the sum of the series.
+\begin{align*}
+   \sum_{n=0}^\infty\frac{(-\ln 2)^n}{n!}
+&= \left(x \mapsto \sum_{n=0}^\infty\frac{x^n}{n!}\right)(-\ln 2)\\
+&= (x \mapsto e^x)\left(\ln\frac{1}{2}\right)\\
+&= \exp\left(\ln\frac{1}{2}\right)\\
+&= \frac{1}{2}\tag{68}
+\end{align*}
+
+\begin{align*}
+\sum_{n=0}^\infty\frac{(-1)^n}{(2n + 1)2^{2n + 1}}
+&= \left(x \mapsto \sum_{n=0}^\infty(-1)^n\frac{2^{2n + 1}}{2n + 1}\right)
+   \left(\frac{1}{2}\right)\\
+&= \left(x \mapsto \tan^{-1}x\right)\left(\frac{1}{2}\right)\\
+  &= \tan^{-1}\frac{1}{2}\tag{70}
+\end{align*}
+
+\noindent\textbf{72. }If $f(x) = \left(1 + x^3\right)^{30}$, what is
+$f^{(58)}(0)$?
+\begin{align*}
+  f(x) = \sum_{n=0}^{30}\binom{30}{n}x^{3n}
+&\Longrightarrow f'(x) = \sum_{n=0}^{30}\binom{30}{n}x^{3n - 1}3n\\
+&\Longrightarrow f''(x) = \sum_{n=1}^{30}\binom{30}{n}x^{3n - 2}3n(3n - 1)\\
+&\Longrightarrow f^{(58)}(x) = \sum_{n=20}^{30}\binom{30}{n}x^{3n - 58}
+                                               \prod_{i=0}^{57}(3n - i)\\
+&\Longrightarrow f^{(58)}(0) = \sum_{n=20}^{30}\binom{30}{n}0^{3n - 58}
+                                               \prod_{i=0}^{57}(3n - i) = 0
+\end{align*}
+\end{document}
diff --git a/usth/MATH1.5/homework/cursived.pdf b/usth/MATH1.5/homework/cursived.pdf
new file mode 100644
index 0000000..c7a2337
--- /dev/null
+++ b/usth/MATH1.5/homework/cursived.pdf
Binary files differdiff --git a/usth/MATH1.5/homework/cursived.tex b/usth/MATH1.5/homework/cursived.tex
new file mode 100644
index 0000000..59a3ac7
--- /dev/null
+++ b/usth/MATH1.5/homework/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
+\[\begin{cases}
+  \nabla f = \lambda\nabla\sum_{i=1}^n x_i\\
+  \sum_{i=1}^n x_i = c
+\end{cases}
+\iff
+\begin{cases}
+  \left<\frac{x_1^{1-2/n}}{n}, \ldots, \frac{x_i^{1-2/n}}{n}\right>f
+  = \lambda\left<x_1, x_2, \ldots, x_n\right>\\
+  \sum_{i=1}^n x_i = c
+\end{cases}\]
+
+\[\Longrightarrow
+\begin{cases}
+  x_1 = x_2 = \ldots = x_n\\
+  \sum_{i=1}^n x_i = c
+\end{cases}
+\iff x_1 = x_2 = \ldots = x_n = \frac{c}{n}\]
+
+At $x_1 = x_2 = \ldots = x_n = c/n$, $f(x_1, x_2, \ldots, x_n) = c/n$.
+As $c/n > 0 = f(c, 0, \ldots, 0)$, $c/n$ is the maximum value of $f$
+on the given constraint.
+
+\exercise{48} By AM-GM inequality,
+as $\sum_{i=1}^n x_i^2 = \sum_{i=1}^n y_i^2 = 1$,
+
+\[\sum_{i=1}^n x_i y_i \leq \sum_{i=1}^n\frac{x_i^2 + y_i^2}{2} = 1\]
+with equality when $\sum_{i=1}^n(x_i - y_i)^2 = 0$.
+
+\subsection*{Problem Plus}
+\exercise{1} A rectangle with length L and width W is cut into four smaller
+rectangles by two lines parallel to the sides.
+
+Let $x, y$ be two nonnegative numbers satisfying $x \leq L$ and $y \leq W$.
+The sum of the squares of the areas of the smaller rectangles would then be
+\begin{align*}
+  f(x, y) &= x^2y^2 + x^2(W-y)^2 + (L-x)^2y^2 + (L-x)^2(W-y)^2\\
+  &= (x^2 + (L-x)^2)(y^2 + (W-y)^2)\\
+\end{align*}
+
+By AM-GM inequality, $f(x, y) \geq 4x(L-x)y(W-y)$ with the equality
+$f(x, y) = L^2W^2/4$ if and only if $x = L - x = L/2$ and $y = W - y = y/2$.
+
+On the other hand,
+\begin{align*}
+  \begin{cases}
+    0 \leq x \leq L\\
+    0 \leq y \leq W
+  \end{cases}
+  &\Longrightarrow
+  \begin{cases}
+    2x(L - x) \geq 0\\
+    2y(W - y) \geq 0
+  \end{cases}
+  \iff
+  \begin{cases}
+    L^2 \geq x^2 + (L-x)^2\\
+    W^2 \geq y^2 + (W-y)^2
+  \end{cases}\\
+  &\Longrightarrow
+  f(x, y) \leq L^2W^2
+\end{align*}
+with equality when $(x, y) \in \{(0, 0), (0, W), (L, W), (L, 0)\}$.
+
+\exercise{3} A long piece of galvanized sheet metal with width $w$ is to be
+bent into a symmetric form with three straight sides to make a rain gutter.
+
+Cross-section area, with $0 \leq x \leq w/2$
+and $0 \leq \theta \leq \max\left(\arccos\frac{2x-w}{2x}, \pi\right)$
+\begin{align*}
+  A(x, \theta) &= (w - 2x + x\cos\theta)x\sin\theta\\
+  &= wx\sin\theta - x^2\left(2\sin\theta - \frac{\sin2\theta}{2}\right)
+\end{align*}
+
+First derivatives:
+\begin{align*}
+  A_x &= w\sin\theta - 2x\left(2\sin\theta - \frac{\sin2\theta}{2}\right)\\
+  A_\theta &= wx\cos\theta - x^2(2\cos\theta - \cos2\theta)
+\end{align*}
+
+Critical points occur when
+\[A_x = A_\theta = 0 \iff
+\begin{cases}
+  w\sin\theta = 2x\left(2\sin\theta - \dfrac{\sin2\theta}{2}\right)\\
+  wx\cos\theta = x^2(2\cos\theta - \cos2\theta)
+\end{cases}\tag{$*$}\]
+
+\begin{tikzpicture}
+  \begin{axis}[
+    axis x line=middle, axis y line=middle,
+    xmin=-0.15, xmax=0.75, xlabel={$\frac{x}{w}$},
+    ymin=-0.7, ymax=3.8, ylabel={$\theta$},
+    xlabel style={at=(current axis.right of origin), anchor=west},
+    ylabel style={at=(current axis.above origin), anchor=south}]
+    \addplot[domain=0.25:0.5, color=red]{acos(1 - 0.5/x)/57.3};
+    \addplot[magenta] plot coordinates {(0.5,1.57) (0.5,0)};
+    \addplot[blue] plot coordinates {(0,0) (0.5,0)};
+    \addplot[cyan] plot coordinates {(0,0) (0,3.14)};
+    \addplot[green] plot coordinates {(0,3.14) (0.25,3.14)};
+    \legend{$C$, $L_0$, $L_1$, $L_2$, $L_3$}
+  \end{axis}
+\end{tikzpicture}
+
+For $x = 0$ (along $L_2$), it is obvious that the area is 0. For $x \neq 0$,
+\begin{align*}
+  (*) &\iff
+  \begin{cases}
+    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
+    w\sin\theta(2\cos\theta-\cos2\theta) = w\cos\theta(4\sin\theta-\sin2\theta)
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
+    2\cos\theta - \cos2\theta = \cos\theta(4 - 2\cos\theta)
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
+    -\cos2\theta = 2\cos\theta - 2\cos^2\theta
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    x = \frac{w\cos\theta}{2\cos\theta - \cos2\theta}\\
+    1 = 2\cos\theta
+  \end{cases}\\
+  &\iff
+  \begin{cases}
+    x = \frac{w}{3}\\
+    \theta = \frac{\pi}{3}
+  \end{cases}
+\end{align*}
+At this point, $A(x, \theta) = w^2/4\sqrt3$.
+
+Along $C$, $A\left(x, \arccos\frac{2x-w}{2x}\right)
+= \frac{1}{4}\sqrt{w(4x-w)(w-2x)^2} \in \left[0, \frac{w^2}{12\sqrt3}\right]$.
+
+Along $L_0$, $A(w/2, \theta)
+= \frac{w^2}{8}\sin(\pi - 2\theta) \in [0, w^2/8]$.
+
+Along $L_1$ and $L_3$, $A(x, \theta) = A(x, 0) = A(x, \pi) = 0$.
+
+In conclusion, the maximum cross-section is $\frac{w^2}{4\sqrt3}$
+at $(x, \theta) = (w/3, \pi/3)$.
+
+\exercise{4} For what values of $r$ is the function
+\[f(x, y, z) =
+\begin{cases}
+  \dfrac{(x + y + z)^r}{x^2 + y^2 + z^2}&\text{if }(x, y, z) \neq (0, 0, 0)\\
+  0&\text{if }(x, y, z) = (0, 0, 0)\\
+\end{cases}\]
+continuous on $\mathbb{R}^3$?
+
+Along $y = z = 0$, as $x \to 0$, $f(x, 0, 0) = x^{r-2} \to \infty$
+(or the limit might not exist at all) for $r < 2$
+and $f(x, 0, 0) = 1$ for $r = 2$.
+Therefore for $r \leq 2$, $f$ is discontinuous at $(0, 0, 0)$.
+
+It is not difficult to show that for $r > 2$, $f$ is continuous.
+For every positive number $\varepsilon$,
+let $\delta = (\varepsilon/3^r)^{1/(2r-2)}$, then from
+\begin{align*}
+  &0 < \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2} < \delta\\
+  \iff &0 < \sqrt{x^2 + y^2 + z^2}
+          < \left(\frac{\varepsilon}{3^r}\right)^\frac{1}{2r-2}\\
+  \iff &0 < \frac{3^r(x^2 + y^2 + z^2)^r}{x^2 + y^2 + z^2} < \varepsilon
+\end{align*}
+and
+\[(x + y + z)^2 \leq 3(x^2 + y^2 + z^2)
+\iff |x + y + z|^r \leq 3^r(x^2 + y^2 + z^2)^r\]
+we get
+\[0 < \frac{|x + y + z|^r}{x^2 + y^2 + z^2} < \varepsilon
+\iff |f(x, y, z) - 0| < \varepsilon\]
+
+Thus by definition, for $r > 2$, $f(x, y, z) \to 0$ as $(x, y, z)\to(0, 0, 0)$,
+hence $f$ is continuous on $\mathbb{R}^3$.
+
+\exercise{5} Suppose $f$ is a differentiable function of one variable.
+Show that all tangent planes to the surface $z = xf(y/x)$
+intersect in a common point.
+
+Let $t = y/x$,
+\begin{align*}
+\tho{z}{x} &= f(t) + x\tho{f(t)}{x}
+            = f(t) + x\leibniz{f}{t}\tho{(y/x)}{x}
+            = f(t) - t\leibniz{f}{t}\\
+\tho{z}{y} &= x\tho{f(t)}{y}
+            = x\leibniz{f}{t}\tho{(y/x)}{y}
+            = \leibniz{f}{t}
+\end{align*}
+
+Equation of the tangent plane to the given surface at $P(a, b, af(b/a))$ is
+\begin{align*}
+     &z - af\left(\frac{b}{a}\right) = \left(f\left(\frac{b}{a}\right)
+        - \frac{b}{a}\cdot\leibniz{f}{t}\left(\frac{b}{a}\right)\right)(x - a)
+        + \leibniz{f}{t}\left(\frac{b}{a}\right)(y - b)\\
+\iff &z = xf\left(\frac{b}{a}\right) + \leibniz{f}{t}\left(\frac{b}{a}\right)
+          \left(y - \frac{bx}{a}\right)\\
+\iff &\left(f\left(\frac{b}{a}\right)
+            - \frac{b}{a}\cdot\leibniz{f}{t}\left(\frac{b}{a}\right)\right)x
+      + \leibniz{f}{t}\left(\frac{b}{a}\right)y - z = 0
+\end{align*}
+
+Since the equation is homogenous, the tangent plane always goes through origin
+$O(0, 0, 0)$.
+
+\section{Multiple Integrals}
+\subsection{Double Integrals over Rectangles}
+\exercise{1} Use a Riemann sum with $m=3$ and $n=2$ to estimate the volume
+of the solid that lies below the surface $z = xy$ and above the rectangle
+$R = [0, 6] \times [0, 4]$.
+
+Take the sample point to be the upper right corner of each square,
+\[V \approx \sum_{i=1}^3\sum_{j=1}^2 ij \cdot 4 = 288\tag{a}\]
+
+Take the sample point to be the center of each square,
+\[V \approx \sum_{i=1}^3\sum_{j=1}^2 (2i-1)(2j-1)4 = 144\tag{b}\]
+
+\exercise{13} Evaluate the double integral by first identifying it
+as the volume of a solid.
+\[\iint_{[-2,2]\times[1,6]}(4 - 2y)\ud A = 0\]
+
+\subsection{Integrated Integrals}
+Calculate the integrated integrals.
+\[\int_1^4\int_0^2(6x^2 - 2x)\ud y\ud x
+= \int_1^4(12x^2 - 4x)\ud x = 222\tag{3}\]
+\[\int_{-3}^3\int_0^{\pi/2}(y + y^2\cos x)\ud x\ud y
+= \int_{-3}^3 y^2\ud y = 0\tag{7}\]
+\[\iint_{[0,\pi/2]^2}\sin(x - y)\ud A
+= \int_0^{\pi/2}(\cos y - \sin y)\ud y = 0\tag{15}\]
+\begin{align*}
+  \iint_{[0,1]\times[-3,3]}\frac{xy^2}{x^2 + 1}\ud A
+  &= \int_0^1\frac{x}{x^2 + 1}\ud x \cdot \int_{-3}^3 y^2\ud y\\
+  &= \frac{1}{2}\int_0^1\frac{\ud x}{x+1}
+     \cdot \left[\frac{y^3}{3}\right]_{-3}^3\\
+  &= 9\ln(x + 1)\big]_0^1\\
+  &= 9\ln 2\tag{17}
+\end{align*}
+\begin{align*}
+  \iint_{[0,2]\times[0,3]}ye^{-xy}\ud A
+  &= \int_0^3\int_0^2 ye^{-xy}\ud x\ud y\\
+  &= \int_0^3(1 - e^{-2y})\ud y\\
+  &= \left[y + \frac{e^{-2y}}{2}\right]_0^3\\
+  &= \frac{1}{2e^6} + \frac{5}{2}\tag{21}
+\end{align*}
+\[\iint_{[-1,1]\times[-2,2]}\left(1-\frac{x^2}{4}-\frac{y^2}{9}\right)\ud A
+= \int_{-1}^1\left(\frac{92}{27} - x^2\right)\ud x = \frac{166}{27}\tag{27}\]
+\[\iint_{[0,4]\times[0,5]}(16 - x^2)\ud A
+= \int_0^4(80 - 5x^2)\ud x = \frac{640}{3}\tag{30}\]
+
+\exercise{40} Fubini's and Clairaut's theorems are similar in the way that
+for continuous functions, order of variables are interchangeable in integration
+and differentiation. By the Fundamental Theorem and these two theorems,
+if $f(x, y)$ is continuous on $[a, b]\times[c, d]$ and
+\[g(x, y) = \int_a^x\int_c^y g(s, t)\ud t\ud s\]
+for $a < x < b$ and $c < y < d$, then $g_{xy} = g_{yx} = f(x, y)$.
+
+\subsection{Double Integrals over General Regions}
+Evaluate the iterated integral.
+\[\int_0^1\int_0^{s^2}\cos s^3\ud t\ud s
+= \int_0^1 s^2\cos s^3\ud s
+= \left[\frac{\sin s^3}{3}\right]_0^1
+= \frac{\sin 1}{3}\tag{5}\]
+\[\int_0^\pi\int_0^{\sin x}x\ud y\ud x
+= \int_0^\pi x\sin x\ud x
+= [\sin x - x\cos x]_0^\pi
+= \pi\tag{9}\]
+\[\int_{-1}^2\int_{y^2}^{y+2}y\ud x\ud y
+= \int_{-1}^2(2y + y^2 - y^3)\ud y
+= \left[y^2 + \frac{y^3}{3} - \frac{y^4}{4}\right]_{-1}^2
+= \frac{9}{4}\tag{15}\]
+\[\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(2x - y)\ud y\ud x
+= \int_{-2}^2 4x\sqrt{4 - x^2}\ud x
+= 0\tag{21}\]
+\begin{align*}
+  \int_1^2\int_1^{7-3y}xy\ud x\ud y
+  &= \int_1^2\left(\frac{9y^3}{2} - 21y^2 + 24y\right)\ud y\\
+  &= \left[\frac{9y^4}{8} - 7y^3 + 12y^2\right]_1^2\\
+  &= \frac{31}{8}\tag{25}
+\end{align*}
+\[\int_1^2\int_0^{\ln x} f(x, y)\ud y\ud x
+= \int_0^{\ln 2}\int_{e^y}^2 f(x, y)\ud x\ud y\tag{47}\]
+\[\int_0^1\int_{3y}^3 e^{x^2}\ud x\ud y
+= \int_0^3\int_0^{x/3} e^{x^2}\ud y\ud x
+= \int_0^3\frac{xe^{x^2}}{3}\ud x
+= \left.\frac{e^{x^2}}{6}\right]_0^3
+= \frac{e^9 - 1}{6}\tag{49}\]
+
+\subsection{Double Integrals in Polar Coordinates}
+Evaluate the given integral.
+\[\int_0^{3\pi/2}\int_0^4 f(r\cos\theta, r\sin\theta)r\ud r\ud\theta\tag{1}\]
+\begin{align*}
+  \int_{\pi/4}^{\pi/2}\int_0^2(2\cos\theta - \sin\theta)r^2\ud r\ud\theta
+  &= \int_{\pi/2}^{\pi/4}\frac{8}{3}(2\cos\theta - \sin\theta)\ud\theta\\
+  &= \frac{8}{3}\left[2\sin\theta + \cos\theta\right]_{\pi/4}^{\pi/2}\\
+  &= \frac{16}{3} - 4\sqrt 2\tag{8}
+\end{align*}
+\begin{align*}
+  \int_{-\pi/2}^{\pi/2}\int_0^2 re^{-r^2}\ud r\ud\theta
+  &= \int_{-\pi/2}^{\pi/2}\frac{1 - e^{-4}}{2}\ud\theta\\
+  &= \pi\frac{1 - e^{-4}}{2}\tag{11}
+\end{align*}
+\begin{align*}
+  \int_0^{2\pi}\int_0^{\sqrt{1/2}}\left(\sqrt{1 - r^2} - r\right)r\ud r\ud\theta
+  &= \pi\int_0^{\sqrt{1/2}}\left(\sqrt{1 - r^2} - r\right)\ud r^2\\
+  &= \pi\int_0^{1/2}(\sqrt{1 - x} - \sqrt x)\ud x\\
+  &= \frac{\pi}{3}(2 - \sqrt 2)\tag{25}
+\end{align*}
+\begin{align*}
+  \int_0^\pi\int_0^3 r\sin r^2\ud r\ud\theta
+  &= \int_0^9\frac{\pi\sin x}{2}\ud x\\
+  &= \left.\frac{\pi\cos x}{-2}\right]_0^9\\
+  &= \frac{\pi}{2}(1 - \cos 9)\tag{29}
+\end{align*}
+
+\exercise{40} We define the improper integral
+(over the entire plane $\mathbb{R}^2$)
+\begin{align*}
+  I &= \iint_{\mathbb{R}^2}\exp(-x^2-y^2)\ud A\\
+  &= \int_{-\infty}^\infty\int_{-\infty}^\infty\exp(-x^2-y^2)\ud x\ud y\\
+  &= \lim_{a\to\infty}\iint_{D_a}\exp(-x^2-y^2)\ud A
+\end{align*}
+where $D_a$ is the disk with radius $a$ and center the origin.
+
+By changing to polar coordinates,
+\begin{align*}
+  I &= \lim_{a\to\infty}\int_0^{2\pi}\int_0^a\exp(-a^2)a\ud a\ud\theta\\
+  &= \lim_{a\to\infty}\int_0^a-\pi\exp(-a^2)\ud-a^2\\
+  &= -\pi\lim_{a\to\infty}\int_0^{-a^2}e^b\ud b\\
+  &= -\pi\lim_{a\to\infty}\left.e^b\right]_0^{-a^2}\\
+  &= \pi\lim_{a\to\infty}(1 - \exp(-a^2))\\
+  &= \pi\tag{a}
+\end{align*}
+
+As $\exp(-x^2-y^2)$ is continuous on $\mathbb{R}^2$,
+\[\int_{-\infty}^\infty\exp(-x^2)\ud x\int_{-\infty}^\infty\exp(-y^2)\ud y
+= I = \pi\tag{b}\]
+
+Thus $\int_{-\infty}^\infty\exp(-x^2)\ud x = \sqrt I = \sqrt\pi$ and
+$\int_{-\infty}^\infty\exp(-x^2/2)\ud x = \sqrt{2\pi}$.
+
+\subsection{Applications of Double Integrals}
+\exercise{2} The total charge on the disk is
+\[\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{x^2 + y^2}\ud y\ud x
+= \int_0^{2\pi}\int_0^1 r^2\ud r\ud\theta
+= \left.2\pi\frac{r^3}{3}\right]_0^1
+= \frac{2\pi}{3}\]
+
+\noindent Find the mass and center of mass of the lamina that occupies the
+regions $D$ and has the given density function $\rho$.
+\[D = [1, 3]\times[1, 4];\qquad\rho(x, y) = ky^2\tag{3}\]
+\[m = \int_1^3\ud x \cdot \int_1^4 ky^2\ud y = 42k\]
+\begin{align*}
+  \bar x &= \frac{k}{m}\int_1^3\int_1^4 xy^2\ud y\ud x
+= \frac{21k}{m}\int_1^3 x\ud x
+= \frac{84k}{m}
+= 2\\
+  \bar y &= \frac{k}{m}\int_1^3\int_1^4 y^3\ud y\ud x
+= \frac{2k}{m}\int_1^4 y^3\ud y
+= \frac{255k}{m}
+= \frac{85}{28}
+\end{align*}
+
+\[D = \{(x, y)\,|\,-1 \leq x \leq 1,\,0 \leq y \leq 1 - x^2\},\qquad
+\rho(x, y) = ky\tag{7}\]
+\[m = \int_{-1}^1\int_0^{1-x^2} ky\ud y\ud x
+= \frac{k}{2}\int_{-1}^1 (x^4 - 2x^2 + 1)\ud x
+= \frac{8k}{15}\]
+\begin{align*}
+  \bar x &= \frac{k}{m}\int_{-1}^1\int_0^{1-x^2} xy\ud y\ud x
+  = \frac{15}{8}\int_{-1}^1 (x^5 - 2x^3 + x)\ud x
+  = 0\\
+  \bar y &= \frac{k}{m}\int_{-1}^1\int_0^{1-x^2} y^2\ud y\ud x
+  = \frac{8}{45}\int_{-1}^1 (1 - x^2)^3\ud x
+  = \frac{4}{7}
+\end{align*}
+\pagebreak
+
+\[D = \left\{(x, y)\,\Big|\,0\leq y\leq\sin\frac{\pi x}{L},\,
+0\leq x\leq L\right\},\qquad\rho(x, y) = y\tag{9}\]
+\[m = \int_0^L\int_0^{\sin(\pi x/L)}y\ud y\ud x
+= \int_0^L\frac{\sin^2(\pi x/L)}{2}\ud x
+= \left[\frac{x}{4} - \frac{L}{8\pi}\sin\frac{2\pi x}{L}\right]_0^L
+= \frac{L}{4}\]
+\begin{align*}
+  \bar x &= \int_0^L\int_0^{\sin(\pi x/L)}\frac{xy}{m}\ud y\ud x
+= \int_0^L\frac{2x\sin^2(\pi x/L)}{L}\ud x
+= \frac{L}{2}\\
+  \bar y &= \int_0^L\int_0^{\sin(\pi x/L)}\frac{y^2}{m}\ud y\ud x
+= \int_0^L\frac{4\sin^3(\pi x/L)}{3L}\ud x
+= \frac{16}{9\pi}
+\end{align*}
+
+\[D = \{(x, y)\,|\,0\leq x\leq 1,\,0\leq y\leq\sqrt{1-x^2}\},\qquad
+\rho(x, y) = ky\tag{11}\]
+\[m = \int_0^1\int_0^{\sqrt{1-x^2}}ky\ud y\ud x
+= \int_0^{\pi/2}\sin\theta\ud\theta\cdot\int_0^1 kr^2\ud r
+= \frac{k}{3}\]
+\begin{align*}
+  \bar x &= \int_0^1\int_0^{\sqrt{1-x^2}}3xy\ud y\ud x
+= \int_0^{\pi/2}\cos\theta\sin\theta\ud\theta\cdot\int_0^1 3r^3\ud r
+  = \frac{3}{8}\\
+  \bar y &= \int_0^1\int_0^{\sqrt{1-x^2}}3y^2\ud y\ud x
+= \int_0^{\pi/2}\sin^2\theta\ud\theta\cdot\int_0^1 3r^3\ud r
+= \frac{3\pi}{16}
+\end{align*}
+
+\subsection{Surface area}
+Find the area of the surface.
+
+\exercise{3} The part of the plane $3x + 2y + z = 6$
+that lies in the first octant.
+\begin{align*}
+  A &= \int_0^2\int_0^{3-1.5x}\sqrt{1 + \left(\tho{z}{x}\right)^2
+     + \left(\tho{z}{y}\right)^2}\ud y\ud x\\
+    &= \int_0^2\int_0^{3-1.5x}\sqrt{14}\ud y\ud x\\
+    &= \int_0^2\left(3 - \frac{3}{2}x\right)\sqrt{14}\ud x\\
+    &= \left[3x\sqrt{14} - \frac{3x^2\sqrt{14}}{4}\right]_0^2\\
+    &= 3\sqrt{14}
+\end{align*}
+
+\exercise{9} The part of the surface $z = xy$
+that lies within the cylinder $x^2 + y^2 = 1$.
+\begin{align*}
+  A &= \iint_D\sqrt{1 + \left(\tho{xy}{x}\right)^2
+     + \left(\tho{xy}{y}\right)^2}\ud A\\
+    &= \int_0^{2\pi}\int_0^1 r\sqrt{1 + r^2}\ud r\ud\theta\\
+    &= \pi\int_0^1\sqrt{1 + t}\ud t\\
+    &= \left.\frac{2\pi\sqrt{(1 - t)^3}}{3}\right]_0^1\\
+    &= \frac{2\pi}{3}\left(2\sqrt{2} - 1\right)
+\end{align*}
+
+\exercise{12} The part of the sphere $x^2 + y^2 + z^2 = 4z$
+that lies inside the paraboloid $z = x^2 + y^2$,
+in which it has the equation $z = 2 + \sqrt{4 - x^2 - y^2}$.
+\begin{align*}
+  A &= \iint_D\sqrt{1 + \left(\tho{}{x}\left(2 + \sqrt{4 - x^2 - y^2}\right)\right)^2
+     + \left(\tho{}{y}\left(2 + \sqrt{4 - x^2 - y^2}\right)\right)^2}\ud A\\
+    &= \iint_D\sqrt\frac{4}{4 - x^2 - y^2}\ud A\\
+    &= \int_0^{2\pi}\int_0^{\sqrt 3}r\sqrt\frac{4}{4 - r^2}\ud r\ud\theta\\
+    &= 2\pi\int_0^3\sqrt\frac{1}{4 - t}\ud t\\
+    &= \left.-4\pi\sqrt{4 - t}\right]_0^3\\
+    &= 4\pi
+\end{align*}
+
+\subsection{Triple Integrals}
+Evaluate the integral.
+\[\int_0^1\int_0^3\int_{-1}^2 xyz^2\ud y\ud z\ud x
+= \int_0^1\int_0^3\frac{3xz^2}{2}\ud z\ud x
+= \int_0^1\frac{27x}{2}\ud x
+= \frac{27}{4}\tag{1}\]
+\begin{align*}
+  \int_0^2\int_0^{z^2}\int_0^{y-z}(2x - y)\ud x\ud y\ud z
+  &= \int_0^2\int_0^{z^2}(z^2 - yz)\ud y\ud z\\
+  &= \int_0^2\left(z^4 - \frac{z^5}{2}\right)\ud z\\
+  &= \frac{16}{15}\tag{3}
+\end{align*}
+\[\int_0^3\int_0^x\int_{x-y}^{x+y}y\ud z\ud y\ud x
+= \int_0^3\int_0^x 2y^2\ud y\ud x
+= \int_0^3\frac{2x^3}{3}\ud x
+= \frac{27}{2}\tag{9}\]
+\begin{align*}
+  \int_0^\pi\int_0^{\pi-x}\int_0^x\sin y\ud z\ud y\ud x
+  &= \int_0^\pi\int_0^{\pi-x}x\sin y\ud y\ud x\\
+  &= \int_0^\pi(x + x\cos y)\ud x\\
+  &= \frac{\pi^2}{2} - 2\tag{12}
+\end{align*}
+\begin{align*}
+  \int_0^1\int_0^{3x}\int_0^{\sqrt{9-y^2}}z\ud z\ud y\ud x
+  &= \int_0^1\int_0^{3x}\frac{9 - y^2}{2}\ud y\ud x\\
+  &= \int_0^1\frac{27x - 9x^3}{2}\ud x\\
+  &= \frac{45}{8}\tag{18}
+\end{align*}
+\begin{align*}
+  \int_0^2\int_0^{4-2x}\int_0^{4-2x-y}\ud z\ud y\ud x
+  &= \int_0^2\int_0^{4-2x}(4 - 2x - y)\ud y\ud x\\
+  &= \int_0^2\frac{(4 - 2x)^2}{2}\ud x\\
+  &= \frac{16}{3}\tag{19}
+\end{align*}
+\begin{align*}
+  \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{-1}^{4-z}\ud y\ud z\ud x
+  &= \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(5 - z)\ud z\ud x\\
+  &= \int_{-2}^2 10\sqrt{4 - x^2}\ud x\\
+  &= 20\pi\tag{22}
+\end{align*}
+
+\subsection{Triple Integrals in Cylindrical Coordinates}
+\exercise{1} Change from cylindrical coordinates to rectangular coordinates.
+\begin{enumerate}[(a)]
+  \item $\left(4, \frac{\pi}{3}, -2\right)
+    \rightarrow \left(2, 2\sqrt 3, -2\right)$
+  \item $\left(2, \frac{-\pi}{2}, 1\right) \rightarrow \left(0, -2, 1\right)$
+\end{enumerate}
+
+\exercise{3} Change from rectangular coordinates to cylindrical coordinates.
+\begin{enumerate}[(a)]
+  \item $\left(-1, 1, 1\right)
+    \rightarrow \left(\sqrt 2, \frac{3\pi}{4}, 1\right)$
+  \item $\left(-2, 2\sqrt 3, 3\right)
+    \rightarrow \left(4, \frac{2\pi}{3}, 3\right)$
+\end{enumerate}
+
+\exercise{7} In cylindrical coordinates $(r, \theta, z)$, $z = 4 - r^2$
+is the paraboloid $z = 4 - x^2 - y^2$ in Cartesian coordinates.
+
+\exercise{15\&17\&21} Evaluate the integral.
+\[\int_{-\pi/2}^{\pi/2}\int_0^2\int_0^{r^2}r\ud z\ud r\ud\theta
+= \pi\int_0^2 r^3\ud r
+= 4\pi\tag{15}\]
+\[\iiint_E\sqrt{x^2 + y^2}\ud V
+= \int_0^{2\pi}\int_0^4\int_{-5}^4 r^2\ud z\ud r\ud\theta
+= 18\pi\left.\frac{r^3}{3}\right]_0^4
+= 384\pi\tag{17}\]
+\begin{align*}
+  \iiint_E x^2\ud V
+  &= \int_0^{2\pi}\int_0^2\int_{z/2}^1 r^3\cos^2\theta\ud r\ud z\ud\theta\\
+  &= \int_0^{2\pi}\cos^2\theta\ud\theta\int_0^2\int_{z/2}^1 r^3\ud r\ud z\\
+  &= \left[\frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2}\right]_0^{2\pi}
+     \int_0^2\left(\frac{1}{4} - \frac{z^4}{64}\right)\ud z\\
+  &= \frac{2\pi}{5}\tag{21}
+\end{align*}
+
+\section{Vector Calculus}
+\setcounter{subsection}{1}
+\subsection{Line Integrals}
+Evaluate the integral.
+\begin{align*}
+&\int_{-\pi/2}^{\pi/2}4\cos t(4\sin t)^4
+ \sqrt{\left(\leibniz{4\cos t}{t}\right)^2
+     + \left(\leibniz{4\sin t}{t}\right)^2}\ud t\\
+=\,&4096\int_{-\pi/2}^{\pi/2}\sin^4 t\ud\sin t
+= 4096\int_{-1}^1 w^4\ud w
+= \frac{8192}{5}\tag{3}
+\end{align*}
+\begin{align*}
+  \int_{\left\{(x, y)\in[1,4]\times[1,2]\,|\,y=\sqrt x\right\}}
+  \left(x^2 y^3 - \sqrt x\right)\ud y
+  &= \int_1^2(t^7 - t)\leibniz{t}{t}\ud t\\
+  &= \left[\frac{t^8}{8} - \frac{t^2}{2}\right]_1^2\\
+  &= \frac{243}{8}\tag{5}
+\end{align*}
+\begin{align*}
+& \int_0^2(x + x)\ud x + \int_2^3(x + 6 - 2x)\ud x
++ \int_0^1(2y)^2\ud y + \int_1^0(3-x)^2\ud y\\
+=\,&4 + \frac72 + \frac43 - \frac{19}{3}
+=\frac{5}{2}\tag{7}
+\end{align*}
+\begin{align*}
+  &\int_2^0 x^2\ud x + \int_0^4 x^2\ud x + \int_0^2 y^2\ud y + \int_2^3\ud y\\
+= &\int_2^4 x^2\ud x + \int_0^3 y^2\ud y
+= \left.\frac{x^3}{3}\right]_2^4 + \left.\frac{y^3}{3}\right]_0^3
+= 13\tag{8}
+\end{align*}
+\begin{align*}
+   \int_0^1(11y^7\unit\i + 3t^6\unit\j)\ud(11t^4\unit\i + t^3\unit\j)
+&= \int_0^1(11y^7\unit\i + 3t^6\unit\j)\cdot(44t^3\unit\i + 3t^2\unit\j)\ud t\\
+  &= \int_0^1(484t^{10} + 9t^8)\ud t\\
+&= \left[44t^11 + t^9\right]_0^1\\
+&= 45\tag{19}
+\end{align*}
+\begin{align*}
+ &\int_0^1(\sin t^3\unit\i + \cos t^2\unit\j + t^4\unit k)
+  \ud(t^3\unit\i + t^2\unit\j + t\unit k)\\
+=&\int_0^1\sin x\ud x + \int_0^1\cos y\ud y + \int_0^1 z^4\ud z\\
+=&\,\frac{6}{5} - \cos 1 - \sin 1\tag{21}
+\end{align*}
+\begin{align*}
+  &\int_0^{2\pi}(t - \sin t)\ud(t - \sin t) + (3 - \cos t)\ud(1 - \cos t)\\
+= &\int_0^{2\pi}((t - \sin t)(1 - \cos t) + (3 - \cos t)\sin t)\ud t\\
+= &\int_0^{2\pi}(t - t\cos t + 2\sin t)\ud t\\
+=\,&\left[\frac{t^2}{2} - t\sin t - 3\cos t\right]_0^{2\pi}
+= 2\pi^2\tag{39}
+\end{align*}
+\begin{align*}
+  &\,2\int_0^{2\pi}\left(4 + \frac{x^2 - y^2}{100}\right)
+  \sqrt{(-10\sin t)^2 + (10\cos t)^2}\ud t\\
+= &\int_0^{2\pi}\left(800 + (10\cos t)^2 - (10\sin t)^2\right)\ud t\\
+= &\,100\int_0^{2\pi}(8 + \cos 2t)\ud t\\
+= &\,\left[8t - \frac{\sin 2t}{2}\right]_0^{2\pi}
+= 16\pi\tag{48}
+\end{align*}
+
+\subsection{The Fundamental Theorem for Line Integrals}
+Evaluate the integrals.
+\begin{align*}
+   &\int_C(x^2\unit\i + y^2\unit\j)\cdot\ud(x\unit\i + 2x^2\unit\j)\\
+=\,&(f\mapsto f(2, 8) - f(-1, 2))\left((x, y)\mapsto\frac{x^3 + y^3}{3}\right)
+= 513\tag{12}
+\end{align*}
+\begin{align*}
+   &\int_C(xy^2\unit\i + x^2y\unit\j)\cdot\ud\mathbf{r}\\
+=\,&(f\mapsto f(2, 1) - f(0, 1))\left((x, y)\mapsto\frac{x^2y^2}{2}\right)
+= 2\tag{13}
+\end{align*}
+
+\subsection{Green's Theorem}
+Evaluate the integrals.
+\begin{align*}
+  \int_C\left(y+e^{\sqrt x}\right)\ud x + (2x + \cos y^2)\ud y
+  &= \int_0^1\int_{y^2}^{\sqrt y}\ud x\ud y\\
+  &= \int_0^1(\sqrt y - y^2)\ud y\\
+  &= \left[\frac{2\sqrt{y^3}}{3} - \frac{y^3}{3}\right]_0^1\\
+  &= \frac{1}{3}\tag{7}
+\end{align*}
+\begin{align*}
+  \int_{x^2+y^2=4}y^3\ud x - x^3\ud y
+  &= \iint_{x^2+y^2=4}(-3x^2-3y^2)\ud A\\
+  &= -3\int_0^{2\pi}\int_0^2 r^3\ud r\ud\theta\\
+  &= -6\pi\left.\frac{r^4}{4}\right]_0^2\\
+  &= -24\pi\tag{9}
+\end{align*}
+\begin{align*}
+  \int_C(1-y^3)\ud x + (x^3+\exp y^2)\ud y
+  &= \iint_D(3x^2 + 3y^2)\ud A\\
+  &= 3\int_0^{2\pi}\int_2^3 r^3\ud r\ud\theta\\
+  &= 6\pi\left.\frac{r^4}{4}\right]_2^3\\
+  &= \frac{195}{8}\pi\tag{10}
+\end{align*}
+\begin{align*}
+  &\int_C(y\cos x - xy\sin x)\ud x + (xy + x\cos x)\ud y\\
+= &-\iint_D(y + \cos x - x\sin x - \cos x + x\sin x)\ud A\\
+= &-\int_0^2\int_0^{4-2x}y\ud y\ud x = \frac{16}{-3}\tag{11}
+\end{align*}
+\begin{align*}
+  &\int_C(\exp-x + y^2)\ud x + (\exp-y + x^2)\ud y\\
+  =&-\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}(2x - 2y)\ud y\ud x\\
+  =&-\int_{-\pi/2}^{\pi/2}(2x\cos x - \cos^2 x)\ud x
+  =\frac\pi 2\tag{12}
+\end{align*}
+\[\int_0^1\int_0^{1-x}(y^2 - x)\ud y\ud x
+= \int_0^1\left(\frac{(1-x)^3}{3} + x^2 - x\right)\ud x
+= \frac{-1}{12}\tag{17}\]
+\begin{align*}
+  \int_\text{cycloid}y\ud x + \int_\text{segment}y\ud x
+  &= \int_{2\pi}^0(1-\cos t)\ud(t-\sin t) + 0\\
+  &= \int_{2\pi}^0\left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right)\ud t\\
+  &= \left[\frac{3t}{2} - 2\sin t + \frac{\sin 2t}{4}\right]_{2\pi}^0
+  = 3\pi\tag{19}
+\end{align*}
+
+\subsection{Curl and Divergence}
+\exercise{19} Since the divergence of curl of $\mathbf G$ is $1 \neq 0$,
+there does not exist a vector field $\mathbf G$ satisfying the given condition.
+
+\subsection{Parametric Surfaces and Their Areas}
+\exercise{19} One parametric representation for the surface $x + y + z = 0$ is
+$\mathbf{r}(u, v) = \langle u, v, -u-v\rangle$.
+
+\exercise{23} One parametric representation for the sphere $x^2 + y^2 + z^2 = 4$
+above the cone $\sqrt{x^2 + y^2}$ is $\mathbf{r}(u, v) =
+\langle 2\cos u\cos v, 2\cos u\sin v, 2\sin u\rangle$.
+
+\exercise{39} The plane intersects with $Ox$ at $A(2, 0, 0)$, with $Oy$
+at $B(0, 3, 0)$ and with $Oz$ at $C(0, 0, 6)$. The area of the triangle $ABC$
+is $|\mathbf{AB}\times\mathbf{AC}|/2 = 3\sqrt{14}$.
+
+\exercise{42} Surface of the cone $\sqrt{x^2 + y^2}$:
+\[\iint_D\sqrt{1 + \left(\frac{x}{\sqrt{x^2+y^2}}\right)^2
++ \left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2}\ud A
+= \iint_D\sqrt 2\ud A\]
+
+For the part lying between $y = x$ and $y = x^2$, the area is
+\[\int_0^1\int_{x^2}^x\sqrt{2}\ud y\ud x
+= \sqrt 2\int_0^1(x - x^2)\ud x
+= \frac{\sqrt 2}{6}\]
+
+\exercise{43} Area of the surface:
+\[\int_0^1\int_0^1\sqrt{1 + x + y}\ud y\ud x
+= \frac{4 - 32\sqrt 2}{15} + \frac{12\sqrt 3}{5}\]
+
+\exercise{45} Area of $z = xy$ within $x^2 + y^2 = 1$:
+\[\iint_D\sqrt{1 + x^2 + y^2}\ud A
+= \int_0^{2\pi}\int_0^1\sqrt{1 + r^2}r\ud r\ud\theta
+= \pi\int_1^2\sqrt t\ud t
+= \frac{2\pi}{3}\left(\sqrt 8 - 1\right)\]
+
+\exercise{49} Area of the surface with given parametric equation
+$\mathbf{r}(u, v) = \langle u^2, uv, v^2/2\rangle$ within $0 \leq u \leq 1$ and
+$0 \leq v \leq 2$:
+\[\iint_D|\mathbf{r}_u\times\mathbf{r}_v|\ud A
+= \int_0^2\int_0^1(2u^2 + v^2)\ud u\ud v
+= \int_0^2\left(\frac{2}{3} + v^2\right)\ud v = 4\]
+
+\subsection{Surface Integrals}
+Evaluate the surface integrals.
+\begin{align*}
+   \iint_S(x + y + z)\ud S
+&= \int_0^2\int_0^1(4u + v + 1)\sqrt{14}\ud v\ud u\\
+&= \int_0^2\left(4u + \frac{3}{2}\right)\sqrt{14}\ud u\\
+&= 11\sqrt{14} \tag{5}
+\end{align*}
+\begin{align*}
+   \int_0^2\int_0^3 x^2y(1+2x+3y)\sqrt{1 + 4 + 9}\ud x\ud y
+&= \int_0^2\left(27y^2 + \frac{99}{2}y\right)\sqrt{14}\ud y\\
+&= 171\sqrt{14}\tag{9}
+\end{align*}
+\begin{align*}
+  &\int_0^1\int_0^1\left(xy\unit\i + yz\unit\j + zx\unit k\right)
+  \cdot\left(\unit\i + 0\unit\j - 2x\unit k\right)
+  \times\left(0\unit\i + \unit\j - 2y\unit k\right)\ud y\ud x\\
+= &\int_0^1\int_0^1\left(xz + 2y^2z + 2x^2y\right)\ud y\ud x\\
+= &\int_0^1\int_0^1((x + 2y^2)(4 - x^2 - y^2) + 2x^2y)\ud y\ud x\\
+= &\int_0^1\int_0^1(4x - x^3 - xy^2 + 8y^2 - 2x^2y^2 - 2y^4 + 2x^2y)\ud y\ud x\\
+= &\int_0^1\left(4x - x^3 - \frac{x}{3} + \frac{8}{3}
+  - \frac{2x^2}{3} - \frac{2}{5} + x^2\right)\ud x\\
+= &\,2 - \frac{1}{4} - \frac{1}{6} + \frac{8}{3}
+  - \frac{2}{9} - \frac{2}{5} + \frac{1}{3}
+= \frac{713}{180}\tag{23}
+\end{align*}
+
+\section{Second-Order Linear Equations}
+\subsection{Homogeneous Linear Equations}
+Solve the differential equation.
+
+\[y'' - y' - 6y = 0\tag{1}\]
+
+The auxiliary equation is $r^2 - r - 6 = 0$ whose roots are $r = -2, 3$.
+Therefore, the general solution of the given differential equation is
+\[y = \frac{c_1}{e^{2x}} + c_2 e^{3x}\]
+
+\[y'' + 16y = 0\tag{3}\]
+
+The auxiliary equation is $r^2 + 16 = 0$ whose roots are $r = \pm 4i$.
+Therefore, the general solution of the given differential equation is
+\[y = c_1\cos 4x + c_2\sin 4x\]
+
+\[9y'' - 12y' + 4y = 0\tag{5}\]
+
+The auxiliary equation is $9r^2 - 12r + 4 = 0$
+whose roots are $r_1 = r_2 = 2/3$.
+Therefore, the general solution of the given differential equation is
+\[y = (c_1 + c_2 x)e^{2x/3}\]
+
+\[2y'' = y'\tag{7}\]
+
+The auxiliary equation is $2r^2 = r$ whose roots are $r = 0, 1/2$.
+Therefore, the general solution of the given differential equation is
+$y = c_1 + c_2\sqrt{e^x}$.
+
+\[y'' - 6y' + 8y = 0,\qquad y(0) = 2,\qquad y'(0) = 2\tag{17}\]
+
+The auxiliary equation is $r^2 - 6r + 8 = 0$ whose roots are $r = 2, 4$.
+Therefore, the general solution of the given differential equation is
+\[y = c_1 e^{2x} + c_2 e^{4x} \Longrightarrow y' = 2c_1 e^{2x} + 4c_2 e^{4x}\]
+
+Since $y(0) = y'(0) = 2$,
+\[c_1 + c_2 = 2c_1 + 4c_2 = 2 \iff (c_1, c_2) = (3, -1)
+\iff y = 3e^{2x} - e^{4x}\]
+
+\[9y'' + 12y' + 4y = 0,\qquad y(0) = 1,\qquad y'(0) = 0\tag{19}\]
+
+The auxiliary equation is $9r^2 + 12r + 4 = 0$
+whose roots are $r_1 = r_2 = -2/3$.
+Therefore, the general solution of the given differential equation is
+\[y = \frac{c_1 + c_2 x}{e^{2x/3}}
+\Longrightarrow y' = \frac{c_2 - 2c_2 x/3 - 2c_1/3}{e^{2x/3}}\]
+
+As $y(0) = 1$, $c_1 = 1$ and as $y'(0) = 0$, $c_2 = 2/3$, thus
+\[y = \left(1 + \frac{2x}{3}\right)e^{-2x/3}\]
+
+\subsection{Nonhomogeneous Linear Equations}
+Solve the differential equation.
+
+\[y'' - 2y' - 3y = \cos 2x\tag{1}\]
+
+The auxiliary equation of $y'' - 2y' - 3y = 0$ is $r^2 - 2r - 3 = 0$
+with roots $r = -1, 3$. So the solution of the complementary equation is
+\[y_c = \frac{c_1}{e^x} + c_2 e^{3x}\]
+
+Since $G(x) = \cos 2x$ is cosine function, we seek a particular solution
+of the form $y_p = A\sin 2x + B\cos 2x$. Then $y_p' = 2A\cos 2x - 2B\sin 2x$
+and $y_p'' = -4y$ so, substituting into the given differential equation,
+we have
+\begin{multline*}
+  (4A - 7B)\cos 2x - (7A + 4B)\sin 2x = \cos 2x\\
+\iff\begin{cases}
+  4A - 7B = 1\\
+  7A + 4B = 0
+\end{cases}
+\iff\begin{dcases}
+  A = \frac{4}{65}\\
+  B = \frac{-7}{65}
+\end{dcases}
+\end{multline*}
+
+Thus the general solution of the given differential equation is
+\[y = y_c + y_p
+= \frac{c_1}{e^x} + c_2 e^{3x} + \frac{4\sin 2x}{65} - \frac{7\cos 2x}{65}\]
+
+\[y'' + 9y = \frac{1}{e^{2x}}\tag{3}\]
+
+The auxiliary equation of $y'' + 9y = 0$ is $r^2 + 9 = 0$
+whose roots are $r = \pm 3i$.
+Therefore, the general solution of the given differential equation is
+\[y_c = c_1\cos 3x + c_2\sin 3x\]
+
+Since $G(x) = e^{-2x}$ is an exponential function, we seek
+a particular solution of an exponential function as well:
+\[y_p = Ae^{-2x}
+\Longrightarrow y_p' = -2Ae^{-2x}
+\Longrightarrow y_p'' = 4Ae^{-2x}\]
+
+Substituting these into the differential equation, we get
+\[\frac{13A}{e^{2x}} = \frac{1}{e^{2x}}
+\iff A = \frac{1}{13}
+\iff y_p = \frac{1}{13e^{2x}}\]
+
+Thus the general solution of the given differential equation is
+\[y = y_c + y_p = c_1\cos 3x + c_2\sin 3x + \frac{1}{13e^{2x}}\]
+
+\[y'' - 4y = e^x\cos x,\qquad y(0) = 1,\qquad y'(0) = 2\tag{8}\]
+
+The auxiliary equation of $y'' + 4y = 0$ is $r^2 + 4 = 0$
+whose roots are $r = \pm 2i$.
+Therefore, the general solution of the given differential equation is
+\[y_c = c_1\cos 2x + c_2\sin 2x\]
+
+We seek a particular solution of the form $y_p = e^x(A\sin x + B\cos x)$.
+Substituting this into the given differential equation we get
+\begin{multline*}
+  2e^x(A\cos x - B\sin x) + 4e^x(A\sin x + B\cos x) = e^x\cos x\\
+  \iff\begin{cases}
+    2A + 4B = 1\\
+    4A - 2B = 0
+  \end{cases}
+  \iff\begin{cases}
+    A = 0.1\\
+    B = 0.2
+  \end{cases}
+\end{multline*}
+
+Thus the general solution of the given differential equation is
+\begin{align*}
+  y &= y_c + y_p = c_1\cos 2x + c_2\sin 2x + e^x(0.1\sin x + 0.2\cos x)\\
+  \Longrightarrow y' &= 2c_2\cos 2x - 2c_1\sin 2x + e^x(0.3\cos x - 0.1\sin x)
+\end{align*}
+
+From $y(0) = 1$ we obtain $c_1 = 0.8$ and from $y'(0) = 2$ we have $c_2 = 0.85$.
+Thus the solution of the initial-value problem is
+\[y = 0.8\cos 2x + 0.85\sin 2x + e^x(0.1\sin x + 0.2\cos x)\]
+
+\[y'' - y' = xe^x,\qquad y(0) = 2,\qquad y'(0) = 1\tag{9}\]
+
+The auxiliary equation of $y'' - y' = 0$ is $r^2 - r = 0$ with roots $r = 0, 1$.
+So the solution of the complementary equation is
+\[y_c = c_1 + c_2 e^x\]
+
+Base on instinct, we seek a particular solution of the form $y_p = (A + x)e^x$.
+Substituting this into the given differential equation we get
+\[(2 + A + x)e^x + (1 + A + x)e^x = xe^x
+\iff 3 + 2A = 0
+\iff A = \frac{-3}{2}\]
+
+Thus the general solution of the given differential equation is
+\begin{align*}
+y &= y_c + y_p
+   = c_1 + c_2 e^x + \left(x - \frac{3}{2}\right)e^x
+   = c_1 + (x + C)e^x\\
+\Longrightarrow y' &= (x + C + 1)e^x
+\end{align*}
+
+From $y'(0) = 1$ we get $C = 0$ and from $y(0) = 2$ we get $c_1 = 2$.
+Hence the solution of the initial-value problem is $y = c_1 + (x + C)e^x$.
+\end{document}
diff --git a/usth/MATH1.5/homework/review.pdf b/usth/MATH1.5/homework/review.pdf
new file mode 100644
index 0000000..10f4b74
--- /dev/null
+++ b/usth/MATH1.5/homework/review.pdf
Binary files differdiff --git a/usth/MATH1.5/homework/review.tex b/usth/MATH1.5/homework/review.tex
new file mode 100644
index 0000000..55ad0a0
--- /dev/null
+++ b/usth/MATH1.5/homework/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}