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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 +``` + +[](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} |