├── .gitignore ├── README.md ├── code └── ps2a │ └── q1.py ├── images ├── ps2a │ ├── q1_even.png │ └── q1_odd.png ├── ps2b │ ├── q1a.jpg │ └── q2.jpg ├── ps3 │ ├── k_eq_0.png │ └── k_eq_1.png ├── ps4 │ └── q6b.jpg ├── ps5 │ ├── q1a.jpg │ ├── q1b.jpg │ ├── q2a.jpg │ └── q3b.jpg ├── ps6 │ ├── q3a.jpg │ └── q3b.jpg ├── ps7 │ └── q2.jpg └── ps8a │ ├── 1_keq1.jpg │ ├── 1_keq2.jpg │ ├── 1_keq3.jpg │ ├── 2c.jpg │ ├── 3a.jpg │ ├── p1_4e8.jpg │ ├── p1_4h2.jpg │ ├── p1_4h3.jpg │ ├── p1_4i2-integration.jpg │ └── p1_4i2.jpg ├── ps2a.pdf ├── ps2a.tex ├── ps2b.pdf ├── ps2b.tex ├── ps3.pdf ├── ps3.tex ├── ps4.pdf ├── ps4.tex ├── ps5.pdf ├── ps5.tex ├── ps6.pdf ├── ps6.tex ├── ps7.pdf ├── ps7.tex ├── ps8a.pdf └── ps8a.tex /.gitignore: -------------------------------------------------------------------------------- 1 | *.aux 2 | *.log 3 | *.out 4 | *.zip 5 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # About 2 | 3 | Unofficial solutions to Problem Sets for [MIT 18.01 Single Variable Calculus](https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/) taught in Fall 2006 by Professor David Jerison. 4 | 5 | Time is needed for me to organize my answers and type in what I've done into Latex. See the [Progress section](#progress) for what has / hasn't been done. 6 | 7 | Feel free to open an issue if you see any mistakes 8 | 9 | 10 | ## Requesting help / solutions 11 | 12 | These are questions that I just can't seem to solve no matter what. Any help here will be appreciated. 13 | 14 | - PS2A Q4b 15 | - PS2B Q1d 16 | 17 | 18 | ## Progress 19 | 20 | - PS1 (pending) 21 | - PS2 (done) 22 | - PS3 (done) 23 | - PS4 (done) 24 | - PS5 (done) 25 | - PS6 (done) 26 | - PS7 (done) 27 | - PS8A (done) 28 | - PS8B (pending) 29 | 30 | 31 | ## Copyright 32 | 33 | All questions in the problem sets are copyright to MIT OpenCourseWare. 34 | 35 |

All solutions provided in this repo are licensed under the Creative Commons Attribution 4.0 International License. 36 | -------------------------------------------------------------------------------- /code/ps2a/q1.py: -------------------------------------------------------------------------------- 1 | from __future__ import division 2 | from __future__ import print_function 3 | 4 | import os.path 5 | 6 | import matplotlib.patches as mpatches 7 | import matplotlib.pyplot as plt 8 | import numpy as np 9 | 10 | def _target(x): 11 | return (x * x + 1) / (x * x - 1) 12 | 13 | def _gradient(x): 14 | return -4 * x / (x * x - 1)**2 15 | 16 | def _odd_fn(x): 17 | return -2 * x / (x * x - 1) 18 | 19 | def _odd_gradient(x): 20 | return 2 * (x * x + 1) / (x * x - 1)**2 21 | 22 | _NR_POINTS = 1000 23 | _MAIN_PLOT_COLOR = "blue" 24 | _GRADIENT_PLOT_COLOR = "green" 25 | 26 | def _main(): 27 | x_left_limit = -5 28 | x_right_limit = 5 29 | y_lower_limit = -5 30 | y_upper_limit = 5 31 | vec_target = np.vectorize(_target) 32 | vec_gradient = np.vectorize(_gradient) 33 | 34 | plt.figure(figsize=(480/96, 360/96,), dpi=96,) 35 | axes = plt.gca() 36 | axes.set_xlim([x_left_limit, x_right_limit]) 37 | axes.set_ylim([y_lower_limit, y_upper_limit]) 38 | first_x_part = np.linspace(x_left_limit, -1.000001, _NR_POINTS) 39 | second_x_part = np.linspace(-0.999999, 0.999999, _NR_POINTS) 40 | third_x_part = np.linspace(1.000001, x_right_limit, _NR_POINTS) 41 | all_x_parts = (first_x_part, second_x_part, third_x_part,) 42 | for x_part in all_x_parts: 43 | plt.plot(x_part, vec_target(x_part), color=_MAIN_PLOT_COLOR) 44 | plt.plot(x_part, vec_gradient(x_part), color=_GRADIENT_PLOT_COLOR) 45 | 46 | plt.axhline(y=0, color="black") 47 | plt.axvline(x=-1, color="gray", linestyle="dashed",) 48 | plt.axvline(x=1, color="gray", linestyle="dashed",) 49 | plt.title("Graph of y = (x^2 + 1) / (x^2 - 1) and its derivative") 50 | plt.xlabel("x") 51 | plt.ylabel("y") 52 | main_plot_patch = mpatches.Patch(color=_MAIN_PLOT_COLOR, label="main plot",) 53 | derivative_plot_patch = mpatches.Patch(color=_GRADIENT_PLOT_COLOR, label="derivative",) 54 | plt.legend(handles=[main_plot_patch, derivative_plot_patch,]) 55 | plt.savefig("q1_even.png", dpi=96,) 56 | 57 | vec_odd_fn = np.vectorize(_odd_fn) 58 | vec_odd_gradient = np.vectorize(_odd_gradient) 59 | plt.clf() 60 | axes = plt.gca() 61 | axes.set_xlim([x_left_limit, x_right_limit]) 62 | axes.set_ylim([y_lower_limit, y_upper_limit]) 63 | for x_part in all_x_parts: 64 | plt.plot(x_part, vec_odd_fn(x_part), color=_MAIN_PLOT_COLOR,) 65 | plt.plot(x_part, vec_odd_gradient(x_part), color=_GRADIENT_PLOT_COLOR,) 66 | plt.axhline(y=0, color="black") 67 | plt.axvline(x=-1, color="gray", linestyle="dashed",) 68 | plt.axvline(x=1, color="gray", linestyle="dashed",) 69 | plt.title("Graph of y = (-2x) / (x^2 - 1) and its derivative") 70 | plt.xlabel("x") 71 | plt.ylabel("y") 72 | main_plot_patch = mpatches.Patch(color=_MAIN_PLOT_COLOR, 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https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{graphicx} 12 | \graphicspath{{images/ps2a/}} 13 | 14 | \usepackage{hyperref} 15 | 16 | \leftmargin=0.25in 17 | \oddsidemargin=0.25in 18 | \textwidth=6.0in 19 | \topmargin=-0.25in 20 | \textheight=9.25in 21 | 22 | \definecolor{Gray}{gray}{0.9} 23 | 24 | \begin{document} 25 | 26 | \begin{center} 27 | \large\textbf{MIT 18.01 Problem Set 2A Unofficial Solutions} 28 | \end{center} 29 | 30 | \begin{tcolorbox} 31 | \textbf{Q1)} Graph the even and odd functions you found in Problem 1, Part II of PS1. Directly below, graph their derivatives. Do this qualitatively using your estimation of the slope. Do not use the formulas for the derivatives (except to check your work if you want). You can use a graphing calculator to check your answer, provided that you mention it in Problem 0. (Note, however, that you may not use books, notes or calculators during tests, so it is unwise to rely on a graphing calculator here.) 32 | \end{tcolorbox} 33 | 34 | \begin{align*} 35 | \frac{x - 1}{x + 1} &= \frac{x - 1}{x + 1} \cdot \frac{x - 1}{x - 1}\\ 36 | &= \frac{x^2 - 2x + 1}{x^2 - 1}\\ 37 | &= \frac{x^2 + 1}{x^2 - 1} - \frac{2x}{x^2 - 1} 38 | \end{align*} 39 | 40 | so the even function is $\frac{x^2 + 1}{x^2 - 1}$ and the odd function is -$\frac{2x}{x^2 - 1}$ 41 | 42 | \begin{center} 43 | \includegraphics[scale=0.8]{q1_even.png} 44 | \end{center} 45 | 46 | \begin{center} 47 | \includegraphics[scale=0.8]{q1_odd.png} 48 | \end{center} 49 | 50 | 51 | \begin{tcolorbox} 52 | \textbf{Q2a)} Compute $(d/dx) tan^3(x^4)$ 53 | \end{tcolorbox} 54 | 55 | \begin{align*} 56 | \frac{d}{dx} tan^3 (x^4) &= \frac{d}{dx} (tan(x^4))^3\\ 57 | &= 3 \cdot (tan(x^4))^2 \cdot \frac{d}{dx} tan(x^4)\\ 58 | &= 3(tan(x^4))^2 \cdot 4x^3 sec^2(x^4)\\ 59 | &= 12x^3 (sec(x^4)tan(x^4))^2 60 | \end{align*} 61 | 62 | 63 | \begin{tcolorbox} 64 | \textbf{Q2b)} Compute $(d/dy)(sin^2y\ cos^2y)$ 65 | \end{tcolorbox} 66 | 67 | \begin{align*} 68 | \frac{d}{dy} (sin^2y\ cos^2y) &= 2sin(y)cos(y) \cdot cos^2y + sin^2y \cdot 2cos(y) \cdot -sin(y)\\ 69 | &= 2sin(y)cos^3(y) - 2sin^3(y)cos(y)\\ 70 | &= 2sin(y)cos(y)(cos^2(y) - sin^2(y)) 71 | \end{align*} 72 | 73 | 74 | \begin{tcolorbox} 75 | \textbf{Q3a)} If $y = uv$, show that $y'' = u''v + 2u'v' + uv''$ 76 | \end{tcolorbox} 77 | 78 | \begin{align*} 79 | y' &= u'v + uv'\\ 80 | y'' &= u''v + u'v' + u'v' + uv''\\ 81 | &= u''v + 2u'v' + uv'' 82 | \end{align*} 83 | 84 | 85 | \begin{tcolorbox} 86 | \textbf{Q3b)} Find $y'''$. 87 | \end{tcolorbox} 88 | 89 | \begin{align*} 90 | y''' &= u'''v + u''v' + 2u''v' + 2u'v'' + u'v'' + uv'''\\ 91 | &= u'''v + 3u''v' + 3u'v'' + uv''' 92 | \end{align*} 93 | 94 | 95 | \begin{tcolorbox} 96 | \textbf{Q4a)} The function $cos^{-1}x$ is the inverse of the $cos\theta$ on $0 \leq \theta \leq \pi$. Use implicit differentiation to derive the formula for $(d/dx)cos^{-1}x$. Pay particular attention to the sign of the square root. (See the book or lecture for the case of the inverse of sine.) 97 | \end{tcolorbox} 98 | 99 | Let $y = cos^{-1}x$. Then $cos(y) = x$. Differentiating implicitly with respect to $x$: 100 | 101 | \begin{align*} 102 | -sin(y) \cdot \frac{dy}{dx} &= 1\\ 103 | \frac{dy}{dx} &= -\frac{1}{sin(y)} 104 | \end{align*} 105 | 106 | $cos(y) = x$ represents a triangle whose angle is $y$, adjacent side has length $x$ and hypotenuse has length $1$. Hence its opposite side has length $\sqrt{1 - x^2}$ and $sin(y) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$. We take the positive square root because for $0 \leq x \leq \pi$, $sin(x) \geq 0$. 107 | 108 | \begin{align*} 109 | \frac{dy}{dx} &= -\frac{1}{sin(y)}\\ 110 | &= -\frac{1}{\sqrt{1 - x^2}} 111 | \end{align*} 112 | 113 | 114 | \begin{tcolorbox} 115 | \textbf{Q4b)} Without calculation, explain why $(d/dx)cos^{-1}x + (d/dx)sin^{-1}x = 0$. 116 | \end{tcolorbox} 117 | 118 | 119 | \begin{tcolorbox} 120 | \textbf{Q5)} Do 8.2/8ac, 10, 11; 8.4/18, 19a. 121 | \end{tcolorbox} 122 | 123 | I do not have the textbook. Skipped. 124 | 125 | 126 | \begin{tcolorbox} 127 | \textbf{Q6)} Derive the formula for $D(u_1u_2 \cdot \cdot \cdot u_n)$ from PS1, Part II, 7b, using logarithmic differentiation. 128 | \end{tcolorbox} 129 | 130 | \begin{align*} 131 | D(u_1u_2 \cdot \cdot \cdot u_n) &= D(e^{ln(u_1u_2 \cdot \cdot \cdot u_n)})\\ 132 | &= D(ln(u_1u_2 \cdot \cdot \cdot u_n)) \cdot e^{ln(u_1u_2 \cdot \cdot \cdot u_n)}\\ 133 | &= D(ln(u_1) + ln(u_2) + \cdot \cdot \cdot ln(u_n)) \cdot (u_1 u_2 \cdot \cdot \cdot u_n)\\ 134 | &= (D(ln(u_1)) + D(ln(u_2)) + \cdot \cdot \cdot + D(ln(u_n))) \cdot (u_1 u_2 \cdot \cdot \cdot u_n)\\ 135 | &= (\frac{u_1'}{u_1} + \frac{u_2'}{u_2} + \cdot \cdot \cdot + \frac{u_n'}{u_n}) \cdot (u_1 u_2 \cdot \cdot \cdot u_n)\\ 136 | &= u_1' u_2 \cdot \cdot \cdot u_n + u_1 u_2' \cdot \cdot \cdot u_n + \cdot \cdot \cdot + u_1 u_2 \cdot \cdot \cdot u_n' 137 | \end{align*} 138 | 139 | \end{document} 140 | -------------------------------------------------------------------------------- /ps2b.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps2b.pdf -------------------------------------------------------------------------------- /ps2b.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{hyperref} 12 | \graphicspath{{images/ps2b/}} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 2B Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \renewcommand{\thefootnote}{\roman{footnote}} 30 | \textbf{Q1) Golf balls} The area of a section of a sphere of radius $R$ between two parallel planes that are a distance $h$ apart is \footnote{This formula will be derived in Unit 4. Two examples may convince you that it is reasonable. For $h = R$, it gives the area of the hemisphere, $2 \pi R^2.$ For $h = 2R$ it gives $4 \pi R^2$, the area of the whole sphere.}\\ 31 | \begin{center} 32 | area of a spherical section = $2 \pi h R$ 33 | \end{center} 34 | \ \\ 35 | Slice the sphere of radius $R$ by a horizontal plane. The portion of the plane inside the sphere is a disk of radius $r \leq R$. The portion of the spherical surface above the plane is called a \emph{spherical cap}. For example, if the plane passes through the center, then the disk has radius $r = R$, its circumference is the equator, and the spherical cap is the Northern Hemisphere. More generally, a spherical cap is the portion of surface of the Earth north of a latitude line. The formula above applies to regions between two latitude lines, and, in particular, to spherical caps.\\ 36 | \\ 37 | a) Consider a spherical cap which is the portion of the surface of the sphere above horizontal plane that slices the sphere at or above its center. Find the area of the cap as a function of $R$ and $r$. Do this by finding first the formula for the height $h$ of the spherical cap in terms of $r$ and $R$. (This height is the vertical distance from the horizontal slicing plane to the North Pole.) Then use your formula for $h$ and the formula above for the area of spherical sections. 38 | \end{tcolorbox} 39 | 40 | \begin{center} 41 | \includegraphics[scale=0.8]{q1a.jpg} 42 | \end{center} 43 | 44 | In the above diagram, we see that the distance between the 2 horizontal slicing planes is $R - h$, where $h$ is the height of the spherical cap. The insight is that, the distance from the center of the sphere to any point on the circumference is $R$. We form a right triangle using this line and the line with the radius $r$ of the spherical cap. 45 | 46 | By the Pythagorean formula, $(R - h)^2 = R^2 - r^2$. Hence 47 | 48 | \begin{align*} 49 | R - h &= \sqrt{R^2 - r^2}\\ 50 | h &= R - \sqrt{R^2 - r^2} 51 | \end{align*} 52 | 53 | Then the area of the spherical cap is $2 \pi h R = 2 \pi (R - \sqrt{R^2 - r^2}) R = 2 \pi R (R - \sqrt{R^2 - r^2})$ 54 | 55 | 56 | \begin{tcolorbox} 57 | \textbf{Q1b)} Express the formula for the area of a spherical cap in terms of $R^2$ and $r / R$. (This is natural because the proportional scaling $cr$ and $cR$ changes the area by the factor $c^2$.) Then use the linear and quadratic approximations to $(1 + x)^{1/2}$ near $x = 0$ to find a good and an even better approximation to the area of the spherical cap, appropriate when the ratio $r / R$ is small. (Hint: What is $x$?) Simplify your answers as far as possible: the approximation corresponding to the linear approximation to $(1 + x)^{1/2}$ should be very familiar. 58 | \end{tcolorbox} 59 | 60 | In part (a), we see that the area of the spherical cap is $2 \pi R (R - \sqrt{R^2 - r^2})$. This can also be expressed as 61 | 62 | \begin{align*} 63 | 2 \pi R (R - \sqrt{R^2 - r^2}) &= 2 \pi R (R - \sqrt{R^2 (1 - \frac{r^2}{R^2})}) \\ 64 | &= 2 \pi R (R - R \sqrt{1 - \frac{r^2}{R^2}}) \\ 65 | &= 2 \pi R^2 (1 - \sqrt{1 - \frac{r^2}{R^2}}) \\ 66 | \end{align*} 67 | 68 | When $\frac{r}{R}$ is small, $\frac{r^2}{R^2} \approx 0$. Let $x = -\frac{r^2}{R^2}$. We are going to use the linear approximation of $(1 + x)^{1/2}$. Let $f(x) = (1 + x)^{1/2}$. Then $f(0) = (1 + 0)^{1/2} = 1$ and $f'(x) = \frac{1}{2} (1 + x)^{-1/2}$ and $f'(0) = \frac{1}{2}(1 + 0)^{-1/2} = \frac{1}{2}$. 69 | 70 | Hence the linear approximation to $(1 + x)^{1/2} = f(0) + f'(0)(x - 0) = 1 + \frac{1}{2}x$ and the linear approximation to $\sqrt{1 - \frac{r^2}{R^2}}$ is thus $1 + \frac{1}{2}(-\frac{r^2}{R^2}) = 1 - \frac{r^2}{2R^2}$ 71 | 72 | Linear approximation to area of spherical cap is thus $2 \pi R^2 (1 - (1 - \frac{r^2}{2R^2})) = 2 \pi R^2 (\frac{r^2}{2R^2}) = \pi r^2$ 73 | 74 | For quadratic approximation to $\sqrt{1 - \frac{r^2}{R^2}}$, we need to compute the 2nd derivative of $f(x) = (1 + x)^{1/2}$, which is $f''(x) = -\frac{1}{4}(1 + x)^{-3/2}$. Now, $f''(0) = -\frac{1}{4}(1 + 0)^{-3/2} = -\frac{1}{4}$ 75 | 76 | Hence, quadratic approximation to $\sqrt{1 - \frac{r^2}{R^2}}$ is: 77 | 78 | \begin{align*} 79 | 1 + \frac{1}{2}(-\frac{r^2}{R^2} - 0) - \frac{\frac{1}{4}}{2}(-\frac{r^2}{R^2} - 0)^2 &= 1 - \frac{r^2}{2R^2} - \frac{r^4}{8R^4} 80 | \end{align*} 81 | 82 | Quadratic approximation to area of spherical cap is 83 | 84 | \begin{align*} 85 | 2 \pi R^2 (1 - (1 - \frac{r^2}{2R^2} - \frac{r^4}{8R^4})) &= 2 \pi R^2 (\frac{r^2}{2R^2} + \frac{r^4}{8R^4}) \\ 86 | &= \pi r^2 + \pi \frac{r^4}{4R^2} 87 | \end{align*} 88 | 89 | 90 | \begin{tcolorbox} 91 | \textbf{Q1c)} The following problem appeared on a middle school math contest exam. The numbers have been changed to protect the innocent. Consider a golf ball that is 3 centimeters in diameter with 100 hemispherical dimples of diameter 3 millimeters. (Note that this is not a realistic golf ball because the dimples are too deep). Find the area of the golf ball rounded to the nearest $1/100$ of a square centimeter using the approximation $\pi \approx 3.14$. (The students were given three minutes. We are spending more time on it.) \\ 92 | 93 | Under the rules of the contest, an incorrectly rounded answer was counted as wrong with no partial credit, so correct numerical approximations were crucial. Some students objected that they could not figure out the area of portion of the large sphere that is removed when a dimple is inserted. A careless examiner had assumed that the students would use the approximation that the area removed for each dimple was nearly the same as the area of a flat disk. We are going to figure out whether this approximation is adequate or gives the wrong answer according to the rules.\\ 94 | 95 | Write down formulas for the surface area of the golf ball in the three cases listed below. (Put in 100 dimples, but leave $r, R$, and $\pi$ as letters.)\\ 96 | \\ 97 | i) the approximation pretending that the removed surface is flat (what is the relationship between this and the approximations of part (b)?)\\ 98 | \\ 99 | ii) the higher order approximation you derived in part (b)\\ 100 | \\ 101 | iii) the exact formula\\ 102 | 103 | Finally, evaluate each of the answers for the given values $r = .15$ and $R = 1.5$ centimeters, and find the accuracy of the approximations. 104 | \end{tcolorbox} 105 | 106 | Area of the golf ball rounded to nearest $1 / 100$ of a square centimeter is $4 \pi (3 / 2)^2 \approx 28.27 cm^2$ 107 | 108 | \textbf{Part (i)} 109 | 110 | Each dimple removes an area of approximately $\pi r^2$ 111 | 112 | Hence removing 100 dimples leaves $4 \pi R^2 - 100 \pi r^2$ surface area remaining 113 | 114 | \textbf{Part (ii)} 115 | 116 | Each dimple removes an area of approximately $\pi r^2 + \pi \frac{r^4}{4R^2}$ 117 | 118 | Hence removing 100 dimples leaves $4 \pi R^2 - 100 \pi r^2 - 25 \pi \frac{r^4}{R^2}$ surface area remaining. 119 | 120 | \textbf{Part (iii)} 121 | 122 | Exact area removed per dimple is $2 \pi R^2 (1 - \sqrt{1 - \frac{r^2}{R^2}})$ 123 | 124 | Hence removing 100 dimples leaves $4 \pi R^2 - 200 \pi R^2 (1 - \sqrt{1 - \frac{r^2}{R^2}})$ surface area remaining. 125 | 126 | \textbf{Calculations using $r = .15$ and R = $1.5$ centimeters} 127 | 128 | For part (i), we get $28.27 - 100 \pi (0.15)^2 \approx 21.20 cm^2$ 129 | 130 | For part (ii), we get $28.27 - 100 \pi (0.15)^2 - 25 \pi \frac{(0.15)^4}{1.5^2} \approx 21.18 cm^2$ 131 | 132 | For part (iii), we get $28.27 - 200 \pi (1.5)^2 (1 - \sqrt{1 - \frac{0.15^2}{1.5^2}}) \approx 21.18cm^2$ 133 | 134 | Hence quadratic approximation is more accurate. 135 | 136 | 137 | \begin{tcolorbox} 138 | \textbf{Q1d)} Although nobody noticed it at the time, the examiner who created this problem made a much bigger mistake. With the diameters actually given, it would have been impossible for the number of dimples given to be placed on the golf ball without overlap. Give a (reasonable) estimate for the largest number of dimples that can fit on our golf ball. 139 | \end{tcolorbox} 140 | 141 | \textbf{Skipped. (Attempted but got very weird figures that contradicted the question)} 142 | 143 | 144 | \begin{tcolorbox} 145 | \textbf{Q2)} Draw the graph of $f(x) = 1/(1 + x^2)$ and, directly underneath, it with the graphs of $f'(x)$ and $f''(x)$. Label critical points and inflection points on the graph of $f$ with their coordinates. Draw vertical lines joining these special points of the graph of $f$ to the corresponding points on the graphs below. 146 | \end{tcolorbox} 147 | 148 | \begin{center} 149 | \includegraphics[]{q2.jpg} 150 | \end{center} 151 | 152 | Now for the hard work. Let's compute $f'(x), f''(x), f'''(x)$ and $f''''(x)$. 153 | 154 | \begin{align*} 155 | f'(x) = \frac{d}{dx} \frac{1}{1 + x^2} &= \frac{d}{dx}(1 + x^2)^{-1} \\ 156 | &= -1 \cdot (1 + x^2)^{-2} \cdot 2x \\ 157 | &= -\frac{2x}{(1 + x^2)^2} \\ 158 | \\ 159 | f''(x) &= \frac{d}{dx} (-\frac{2x}{(1 + x^2)^2}) \\ 160 | &= \frac{d}{dx} \frac{-2x}{(1 + x^2)^2} \\ 161 | &= \frac{-2(1 + x^2)^2 - (-2x)\cdot 2(1 + x^2)(2x)}{((1 + x^2)^2)^2} \\ 162 | &= \frac{-2(1 + x^2)^2 + 8x^2 (1 + x^2)}{(1 + x^2)^4} \\ 163 | &= \frac{-2(1 + x^2) + 8x^2}{(1 + x^2)^3} \\ 164 | &= \frac{-2 - 2x^2 + 8x^2}{(1 + x^2)^3} \\ 165 | &= \frac{6x^2 - 2}{(1 + x^2)^3} \\ 166 | \\ 167 | f'''(x) &= \frac{d}{dx} \frac{6x^2 - 2}{(1 + x^2)^3} \\ 168 | &= \frac{12x (1 + x^2)^3 - (6x^2 - 2)3(1 + x^2)^2 (2x)}{((1 + x^2)^3)^2} \\ 169 | &= \frac{12x (1 + x^2)^3 - (36x^3 - 12x)(1 + x^2)^2}{(1 + x^2)^6} \\ 170 | &= \frac{12x (1 + x^2) - 36x^3 + 12x}{(1 + x^2)^4} \\ 171 | &= \frac{12x + 12x^3 - 36x^3 + 12x}{(1 + x^2)^4} \\ 172 | &= \frac{-24x^3 + 24x}{(1 + x^2)^4} \\ 173 | \\ 174 | f''''(x) &= \frac{d}{dx} \frac{-24x^3 + 24x}{(1 + x^2)^4} \\ 175 | &= \frac{(-72x^2 + 24) (1 + x^2)^4 - (-24x^3 + 24x)4(1 + x^2)^3 2x}{((1 + x^2)^4)^2} \\ 176 | &= \frac{(-72x^2 + 24) (1 + x^2)^4 - (-192x^4 + 192x^2)(1 + x^2)^3}{(1 + x^2)^8} \\ 177 | &= \frac{(-72x^2 + 24) (1 + x^2) + 192x^4 - 192x^2}{(1 + x^2)^5} \\ 178 | &= \frac{-72x^2 - 72x^4 + 24 + 24x^2 + 192x^4 - 192x^2}{(1 + x^2)^5} \\ 179 | &= \frac{120x^4 - 240x^2 + 24}{(1 + x^2)^5} 180 | \end{align*} 181 | 182 | \subsection*{For the graph of $f(x) = \frac{1}{1 + x^2}$} 183 | 184 | There are no discontinuities for $f(x) = \frac{1}{1 + x^2}$. There are also no x-intercepts. In fact, this graph is always positive because the numerator $1$ is positive and the denominator $1 + x^2$ is always positive so dividing the numerator by the denominator is also always positive. 185 | 186 | For the y-intercept, $x = 0$. In this case, $y = \frac{1}{1 + 0^2} = 1$ so $(0, 1)$ is the y-intercept. 187 | 188 | Now for the critical points of $f(x) = \frac{1}{1 + x^2}$. These occur when $f'(x) = -\frac{2x}{(1 + x^2)^2} = 0$, or equivalently when $-2x = 0$ or $x = 0$. At $x = 0$, $f(x) = \frac{1}{1 + 0^2} = 1$ so $(0, 1)$ is a critical point (this point also happens to be the y-intercept). Computing the second derivative, we see that $f''(0) = \frac{6(0)^2 - 2)}{(1 + 0^2)^3} = -2 < 0$ so $(0, 1)$ is a maximum point. 189 | 190 | For the inflection points of $f(x) = \frac{1}{1 + x^2}$. These occur when $f''(x) = \frac{6x^2 - 2}{(1 + x^2)^3} = 0$. Or equivalently, when $6x^2 - 2 = 0$ and $6x^2 = 2$ and $3x^2 = 1$ and $x^2 = \frac{1}{2}$ and $x = \pm \frac{1}{\sqrt{3}}$. At both these points, $f(x) = \frac{1}{1 + 1/3} = \frac{1}{4/3} = \frac{3}{4}$. Hence $(-\frac{1}{\sqrt{3}}, \frac{3}{4})$ and $(\frac{1}{\sqrt{3}}, \frac{3}{4})$ are the inflection points. 191 | 192 | Now for the limits. $\lim\limits_{x \rightarrow \pm\infty} \frac{1}{1 + x^2} = 0$ 193 | 194 | With the above information, we can construct the graph of $f(x) = \frac{1}{1 + x^2}$ 195 | 196 | \subsection*{For the graph of $f'(x) = -\frac{2x}{(1 + x^2)^2}$} 197 | 198 | The x-intercept occurs at $f'(x) = -\frac{2x}{(1 + x^2)^2} = 0$ which is when $x = 0$. So $(0, 0)$ is the x-intercept. For the y-intercept, $f'(0) = -\frac{2(0)}{(1 + 0^2)^2} = 0$ so $(0, 0)$ is also the y-intercept. 199 | 200 | The inflection points of $f(x)$ are at $x = \pm\frac{1}{\sqrt{3}}$. These now become the critical points of $f'(x)$. At $x = -\frac{1}{\sqrt{3}}, f'(x) = -\frac{2 (- 1 / \sqrt{3})}{(1 + (- 1 / \sqrt{3})^2)^2} = \frac{2 / \sqrt{3}}{(1 + 1/3)^2} = \frac{2}{\sqrt{3} \cdot \frac{16}{9}} = \frac{18}{16 \sqrt{3}} = \frac{9}{8 \sqrt{3}}$ so $(-\frac{1}{\sqrt{3}}, \frac{9}{8 \sqrt{3}})$ is a critical point. At $x = \frac{1}{\sqrt{3}}$, $f'(x) = -\frac{2 (1 / \sqrt{3})}{(1 + (1 / \sqrt{3})^2)^2} = -\frac{2}{\sqrt{3} (1 + 1/3)^2} = -\frac{2}{\sqrt{3} \cdot \frac{16}{9}} = -\frac{2 * 9}{\sqrt{3} \cdot 16} = -\frac{9}{8 \sqrt{3}}$ so $(\frac{1}{\sqrt{3}}, -\frac{9}{8 \sqrt{3}})$ is the other critical point. Now, $f'''(-\frac{1}{\sqrt{3}}) \approx -2.9228357377724787 < 0$ so $(-\frac{1}{\sqrt{3}}, \frac{9}{8 \sqrt{3}})$ is a maximum point. Similarly $f'''(\frac{1}{\sqrt{3}}) \approx 2.9228357377724787 > 0$ so $(\frac{1}{\sqrt{3}}, -\frac{9}{8 \sqrt{3}})$ is a minimum point. 201 | 202 | $\lim\limits_{x \rightarrow \pm\infty} -\frac{2x}{(1 + x^2)^2} = \lim\limits_{x \rightarrow \pm\infty} -\frac{2}{2(1 + x^2)2x} = 0$. 203 | 204 | Notice that at $x < 0$, $-2x > 0$ and $(1 + x^2)^2 > 0$ so $f'(x) = -\frac{2x}{(1 + x^2)^2} > 0$. At $x > 0, -2x < 0$ and $(1 + x^2)^2 > 0$ so $f'(x) = -\frac{2x}{(1 + x^2)^2} < 0$. The crossover point comes at $(0, 0)$. 205 | 206 | With the above information, we can construct the graph of $f'(x) = -\frac{2x}{(1 + x^2)^2}$ 207 | 208 | \subsection*{For the graph of $f''(x) = \frac{6x^2 - 2}{(1 + x^2)^3}$} 209 | 210 | The inflection points of $f(x)$ are now the x-intercept of $f''(x)$. At $x = \pm \frac{1}{\sqrt{3}}$, $f''(x) = 0$. 211 | 212 | For the y-intercept, $f''(0) = \frac{6(0)^2 - 2}{(1 + 0^2)^3} = -2$ so $(0, -2)$ is the y-intercept. 213 | 214 | Notice also that the only $x$ terms are $x^2$ which have even powers. So this is an even function. Let us then consider the branch where $x > 0$ and we can construct the $x < 0$ case from there. 215 | 216 | For the critical points, let $f'''(x) = \frac{-24x^3 + 24x}{(1 + x^2)^4} = 0$. Then $-24x^3 + 24x = 0$ and $24x(- x^2 + 1) = 0$ so $x = 0$ or $-x^2 + 1 = 0$ which is equivalent to $x^2 - 1 = 0$ or $x = \pm 1$. At $x = 1$, $f''(1) = \frac{6(1)^2 - 2}{(1 + 1^2)^3} = \frac{4}{2^3} = \frac{1}{2}$ so $(1, \frac{1}{2})$ is a critical point. Using the second derivative of $f''(x)$ which is $f''''(x)$, we see that $f''''(\frac{1}{2}) = -9.33888 < 0$ so $(1, \frac{1}{2})$ is a maximum point. Similarly, $f''''(0) = 24 > 0$ so $(0, -2)$ is a minimum point. 217 | 218 | Now for the limits. $\lim\limits_{x \rightarrow \infty}\frac{6x^2 - 2}{(1 + x^2)^3} = \lim\limits_{x \rightarrow \infty}\frac{12x}{3(1 + x^2)^2 2x} = \lim\limits_{x \rightarrow \infty}\frac{2}{(1 + x^2)^2} = 0$. 219 | 220 | With the above information, we can construct the graph of $f''(x) = \frac{6x^2 - 2}{(1 + x^2)^3}$ 221 | 222 | \end{document} 223 | -------------------------------------------------------------------------------- /ps3.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps3.pdf -------------------------------------------------------------------------------- /ps3.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{graphicx} 12 | \graphicspath{{images/ps3/}} 13 | 14 | \usepackage{hyperref} 15 | 16 | \leftmargin=0.25in 17 | \oddsidemargin=0.25in 18 | \textwidth=6.0in 19 | \topmargin=-0.25in 20 | \textheight=9.25in 21 | 22 | \definecolor{Gray}{gray}{0.9} 23 | 24 | \begin{document} 25 | 26 | \begin{center} 27 | \large\textbf{MIT 18.01 Problem Set 3 Unofficial Solutions} 28 | \end{center} 29 | 30 | \textbf{NOTE:} This might be the hardest Problem Set in the entire course. Some of the problems here may seem straightforward after you see the solutions - that is not the case when I was solving them and they are anything but straightforward, especially Q3. 31 | 32 | I spent a lot amount of time solving the problems in this Problem Set (and I skipped some parts) while I was taking the course, then many weeks after I completed the course, I extracted and refined the relevant stuff from the very messy original work I did on paper, correcting any mistakes found in the process. 33 | 34 | If you are in the process of taking this class, don't get too discouraged if you can't solve everything on this Problem Set. 35 | 36 | \begin{tcolorbox} 37 | \textbf{Q2. Hypocycloid.} Show that every tangent line to the curve $ x^{2/3} + y^{2/3} = 1 $ in the first quadrant has the property that portion of the line in the first quadrant has length 1. (Use implicit differentiation; this is the same as problem 45 page 114 of text.) 38 | \end{tcolorbox} 39 | 40 | \begin{align*} 41 | x^{2/3} + y^{2/3} = 1 42 | \end{align*} 43 | 44 | Using implicit differentiation with respect to $x$, 45 | 46 | \begin{align*} 47 | \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = 0 \\ 48 | \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \\ 49 | \end{align*} 50 | 51 | Assuming $x > 0$, $y > 0$, 52 | \begin{align*} 53 | \frac{dy}{dx} &= \frac{-\frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3}} \\ 54 | &= -\frac{x^{-1/3}}{y^{-1/3}} \\ 55 | &= - (\frac{y}{x})^{1/3} 56 | \end{align*} 57 | 58 | The tangent line has equation 59 | 60 | \begin{align*} 61 | y - y_0 = -(\frac{y_0}{x_0})^{1/3} (x - x_0) 62 | \end{align*} 63 | 64 | for $x_0 > 0, y_0 > 0$ \ . To find the length of the tangent line in the first quadrant, we need to find its x-intercept and y-intercept. 65 | 66 | First, the y-intercept. When $x = 0$, 67 | 68 | \begin{align*} 69 | y - y_0 &= -(\frac{y_0}{x_0})^{1/3} (0 - x_0) \\ 70 | y &= y_0 -(\frac{y_0}{x_0})^{1/3} (- x_0) \\ 71 | y &= y_0 + (\frac{y_0}{x_0})^{1/3} x_0 \\ 72 | y &= y_0 + x_0^{2/3} y_0^{1/3} 73 | \end{align*} 74 | 75 | Onto the x-intercept. When $y = 0$, 76 | 77 | \begin{align*} 78 | 0 - y_0 &= -(\frac{y_0}{x_0})^{1/3}(x - x_0) \\ 79 | y_0 (\frac{x_0}{y_0})^{1/3} &= x - x_0 \\ 80 | x &= x_0 + y_0(\frac{x_0}{y_0})^{1/3} \\ 81 | x &= x_0 + x_0^{1/3} y_0^{2/3} 82 | \end{align*} 83 | 84 | So the tangent line cuts the y-axis at $y = y_0 + x_0^{2/3} y_0^{1/3}$ and cuts the x-axis at $x = x_0 + x_0^{1/3} y_0^{2/3}$. We assumed $x_0 > 0, y_0 > 0$ so these 2 coordinates are positive and the length of the tangent line in the first quadrant is simply $\sqrt{x^2 + y^2}$, which is 85 | 86 | \begin{align*} 87 | \sqrt{x^2 + y^2} &= \sqrt{(x_0 + x_0^{1/3} y_0^{2/3})^2 + (y_0 + x_0^{2/3} y_0^{1/3})^2} \\ 88 | &= \sqrt{x_0^2 + 2x_0^{4/3}y_0^{2/3} + x_0^{2/3} y_0^{4/3} + y_0^2 + 2x_0^{2/3}y_0^{4/3} + x_0^{4/3}y_0^{2/3}} \\ 89 | &= \sqrt{x_0^2 + y_0^2 + 3x_0^{4/3}y_0^{2/3} + 3x_0^{2/3}y_0^{4/3}} \\ 90 | &= \sqrt{x_0^{4/3} (x_0^{2/3} + 3y_0^{2/3}) + y_0^{4/3} (y_0^{2/3} + 3x_0^{2/3})} \\ 91 | &= \sqrt{x_0^{4/3} (x_0^{2/3} + y_0^{2/3} + 2y_0^{2/3}) + y_0^{4/3} (y_0^{2/3} + x_0^{2/3} + 2x_0^{2/3})} \\ 92 | &= \sqrt{x_0^{4/3} (1 + 2y_0^{2/3}) + y_0^{4/3} (1 + 2x_0^{2/3})} \tag*{(Using $x_0^{2/3} + y_0^{2/3} = 1$)} \\ 93 | &= \sqrt{x_0^{4/3} + 2x_0^{4/3}y_0^{2/3} + y_0^{4/3} + 2x_0^{2/3}y_0^{4/3}} \\ 94 | &= \sqrt{2x_0^{2/3}y_0^{2/3} (x_0^{2/3} + y_0^{2/3}) + x_0^{4/3} + y_0^{4/3}} \\ 95 | &= \sqrt{2x_0^{2/3}y_0^{2/3} + x_0^{4/3} + y_0^{4/3}} \\ 96 | &= \sqrt{x_0^{2/3} (2y_0^{2/3} + x_0^{2/3}) + y_0^{4/3}} \\ 97 | &= \sqrt{x_0^{2/3} (x_0^{2/3} + y_0^{2/3} + y_0^{2/3}) + y_0^{4/3}} \\ 98 | &= \sqrt{x_0^{2/3} (1 + y_0^{2/3}) + y_0^{4/3}} \\ 99 | &= \sqrt{x_0^{2/3} + x_0^{2/3}y_0^{2/3} + y_0^{4/3}} \\ 100 | &= \sqrt{y_0^{2/3} (x_0^{2/3} + y_0^{2/3}) + x_0^{2/3}} \\ 101 | &= \sqrt{y_0^{2/3} + x_0^{2/3}} \\ 102 | &= \sqrt{1} \\ 103 | &= 1 104 | \end{align*} 105 | 106 | And hence every tangent line to the curve $x_0^{2/3} + y_0^{2/3} = 1$ in the first quadrant has the property that the portion of the line in the first quadrant has length $1$. 107 | 108 | \begin{tcolorbox} 109 | \textbf{Q3. Sensitivity of measurement, revisited.} 110 | 111 | a) Recall that in problem 2, PS1/Part II, $L^2 + 20000^2 = h^2$. Use implicit differentiation to calculate $dL/dh$. Compare the linear approximation $dL/dh$ to the error $\Delta L / \Delta h$ computed in examples on PS1. Explain why $ \Delta L / \Delta h <= dL/dh $ if the derivative is evaluated at the left endpoint of the interval of uncertainty (or, in other words, $\Delta h < 0$). In what range of values of $h$ is it true that $|\Delta L| <= 2|\Delta h|$ ? 112 | \end{tcolorbox} 113 | 114 | \begin{tcolorbox} 115 | Use implicit differentiation to find $\frac{dL}{dh}$ 116 | \end{tcolorbox} 117 | \begin{align*} 118 | L^2 + 20000^2 &= h^2 \\ 119 | 2L\frac{dL}{dh} &= 2h \\ 120 | \frac{dL}{dh} &= \frac{h}{L} 121 | \end{align*} 122 | 123 | \begin{tcolorbox} 124 | Compare the linear approximation $dL/dh$ to the error $\Delta L / \Delta h$ computed in examples on PS1: 125 | \end{tcolorbox} 126 | 127 | This is for $h = 25000$ with altitude $20000$. Then $L = \sqrt{25000^2 - 20000^2} = 15000$ and $\frac{dL}{dh} = \frac{h}{L} = 25000 / 15000 = 5 / 3$. 128 | \begin{center} 129 | \begin{tabular}{|c|c|c|} 130 | \hline 131 | \rowcolor{Gray} 132 | $\Delta h$ & $\Delta L / \Delta h$ & $\Delta L / \Delta h <= \frac{dL}{dh}$ \\ \hline 133 | $1$ & $1.666607$ & True \\ \hline 134 | $10^{-1}$ & $1.666661$ & True \\ \hline 135 | $10^{-2}$ & $1.666666$ & True \\ \hline 136 | $-1$ & $1.666726$ & False \\ \hline 137 | $-10^{-1}$ & $1.666673$ & False \\ \hline 138 | $-10^{-2}$ & $1.666667$ & False \\ \hline 139 | \end{tabular} 140 | \end{center} 141 | 142 | Now for $h = 20001$ with altitude $20000$. $L = \sqrt{20001^2 - 20000^2} \approx 200.0024999843752$ and $\frac{dL}{dh} = \frac{h}{L} \approx 100.00374996093818$. 143 | \begin{center} 144 | \begin{tabular}{|c|c|c|} 145 | \hline 146 | \rowcolor{Gray} 147 | $\Delta h$ & $\Delta L / \Delta h$ & $\Delta L / \Delta h <= \frac{dL}{dh}$ \\ \hline 148 | $1$ & $82.847283$ & True \\ \hline 149 | $10^{-1}$ & $97.621539$ & True \\ \hline 150 | $10^{-2}$ & $99.755002$ & True \\ \hline 151 | $-1$ & $200.002500$ & False \\ \hline 152 | $-10^{-1}$ & $102.637058$ & False \\ \hline 153 | $-10^{-2}$ & $100.254998$ & False \\ \hline 154 | \end{tabular} 155 | \end{center} 156 | 157 | \begin{tcolorbox} 158 | Explain why $\Delta L / \Delta h <= dL / dh$ if the derivative is evaluated at the right endpoint of the interval of uncertainty (or, in other words, $\Delta h > 0$). 159 | \end{tcolorbox} 160 | 161 | \textbf{NOTE: There is a typo with this part of the question.} It uses the phrase "left endpoint of the interval of uncertainty" in conjunction with $\Delta h > 0$. It is only after completing the part above where we have more evidence that $\Delta L / \Delta h <= \frac{dL}{dh}$ should be correct and is what the question should be asking for. \\ 162 | 163 | We have that $L^2 + 20000^2 = h^2$. $L >= 0$, and it's pretty obvious that $h >= 20000$ since $L >= 0$, and that $h > L$. To see how $\frac{h}{L}$ changes according to a change in $h$, we take limits on both ends of $h$. 164 | 165 | \begin{equation*} 166 | \lim_{h\to20000^+} \frac{h}{L} = \lim_{h\to20000^+} \frac{h}{\sqrt{h^2 - 20000^2}} 167 | = \lim_{h\to20000^+} \sqrt{\frac{h^2}{h^2 - 20000^2}} 168 | = \lim_{h\to20000^+} \sqrt{\frac{h^2 / h^2}{h^2 / h^2 - 20000^2 / h^2}} \\ 169 | \end{equation*} 170 | \begin{equation*} 171 | = \lim_{h\to20000^+} \sqrt{\frac{1}{1 - (\frac{20000}{h})^2}} = \infty 172 | \end{equation*} 173 | \begin{equation*} 174 | \lim_{h\to\infty} \frac{h}{L} = \lim_{h\to\infty} \sqrt{\frac{1}{1 - (\frac{20000}{h})^2}} = \sqrt{\frac{1}{1 - 0}} = 1 175 | \end{equation*} 176 | 177 | So we can conclude that $\frac{h}{L} >= 1$. Now, going back to $L^2 + 20000^2 = h^2$, suppose we add $\Delta h > 0$ to $h$. Then the resulting $L$ will be $L + \Delta L$ for some $\Delta L > 0$. 178 | 179 | \begin{align*} 180 | (h + \Delta h)^2 &= (L + \Delta L)^2 + 20000^2 \\ 181 | h + \Delta h &= \sqrt{(L + \Delta L)^2 + 20000^2} \\ 182 | \Delta h &= \sqrt{(L + \Delta L)^2 + 20000^2} - h \\ 183 | &= \sqrt{L^2 + 2L\Delta L + (\Delta L)^2 + 20000^2} - h \\ 184 | &= \sqrt{(L^2 + 20000^2) + 2L\Delta L + (\Delta L)^2} - h \\ 185 | &= \sqrt{h^2 + 2L\Delta L + (\Delta L)^2} - h \\ 186 | &<= \sqrt{h^2 + 2h\Delta L + (\Delta L)^2} - h \tag*{(Since $h >= L$ and $\Delta L > 0$)} \\ 187 | &= \sqrt{(h + \Delta L)^2} - h \\ 188 | &= h + \Delta L - h \\ 189 | &= \Delta L 190 | \end{align*} 191 | 192 | And $\Delta L >= \Delta h$. Since both $\Delta h > 0$ and $\Delta L > 0$, we have that $\frac{\Delta L}{\Delta h} >= 1$. Earlier we proved that $\frac{h}{L} >= 1$. Unfortunately, these 2 facts combined \textbf{cannot} help us prove that $\frac{\Delta L}{\Delta h} <= \frac{dL}{dh} = \frac{h}{L}$ for $\Delta h > 0$. In fact, you can try adding $\Delta h < 0$ (which means $\Delta L < 0$) and reach the same conclusion that $\frac{\Delta L}{\Delta h} >= 1$. That is also what we observe when computing the values for $\frac{\Delta L}{\Delta h}$ in the tables above. \\ 193 | 194 | We shall instead offer a more hand-wavy explanation for the phenomenon (which may be wrong). I have tried extremely hard to reason about this problem for a long time and this is the only sufficiently reasonable explanation I can come up with.\\ 195 | 196 | Earlier, we proved that $\lim_{h \rightarrow 20000^+} \frac{h}{L} = \infty$ and $\lim_{h \rightarrow \infty} \frac{h}{L} = 1$. Since $\frac{dL}{dh} = \frac{h}{L}$, the rate of change of $L$ with respect to $h$ is decreasing (but positive) as $h$ increases. As such, for $\Delta h > 0$, $\frac{\Delta L}{\Delta h}$, which is the actual change in $L$ over the actual change in $h$, decreases as $\Delta h$ increases. In fact, due to the decreasing $\frac{dL}{dh}$, $\Delta L <= \frac{dL}{dh} * \Delta h$ for $\Delta h > 0$, which gives us $\frac{\Delta L}{\Delta h} <= \frac{dL}{dh} = \frac{h}{L}$. 197 | 198 | \begin{tcolorbox} 199 | In what range of values of $h$ is it true that $|\Delta L| <= 2|\Delta h|$ ? 200 | \end{tcolorbox} 201 | 202 | $|\Delta L| <= 2|\Delta h|$ implies that $\frac{|\Delta L|}{|\Delta h|} <= 2$, which implies that $|\frac{\Delta L}{\Delta h}| <= 2$. We know that $\Delta L$ and $\Delta h$ have the same sign, so this simplifies to $\frac{\Delta L}{\Delta h} <= 2$. 203 | 204 | For $\Delta h > 0$, we have $\frac{\Delta L}{\Delta h} <= \frac{dL}{dh} = \frac{h}{L}$. Therefore, we need to find $h$ such that $\frac{h}{L} <= 2$. 205 | 206 | \begin{align*} 207 | \frac{h}{L} &<= 2 \\ 208 | h &<= 2L \\ 209 | &= 2\sqrt{h^2 - 20000^2} 210 | \end{align*} 211 | 212 | Squaring both sides, 213 | 214 | \begin{align*} 215 | h^2 &<= 4(h^2 - 20000^2) \\ 216 | 3h^2 &>= 4 * 20000^2 \\ 217 | h^2 &>= \frac{4 * 20000^2}{3} \\ 218 | h &>= \sqrt{\frac{4 * 20000^2}{3}} \approx 23094.010760368154 219 | \end{align*} 220 | 221 | But that is just for $\Delta h > 0$. How about for $\Delta h < 0$? We also require that $\frac{\Delta L}{\Delta h} <= 2$. Since $\Delta h$ and $\Delta L$ have the same sign, we want to find $\Delta h < 0$ that will result in the highest magnitude $\Delta L$. 222 | 223 | Earlier we proved that $\frac{dL}{dh} = \frac{h}{L}$ and $\lim_{h \rightarrow 20000^+} \frac{h}{L} = \infty$ and $\lim_{h \rightarrow \infty} \frac{h}{L} = 1$. Hence $\frac{dL}{dh}$ is decreasing (but positive) as $h$ increases. Hence for $\Delta h < 0$, it makes sense that the maximum $|\Delta L|$ arises from a change in $h$ that brings it all the way back to its minimum allowed value of $20000$. In other words, $\Delta h = -(h - 20000)$. The corresponding value of $\Delta L$ is then $-L$, because if $h$ goes back to $20000$, the constraint $L^2 + 20000^2 = h^2$ causes $L$ to go to $0$. Then $\frac{\Delta L}{\Delta h} = \frac{-L}{-(h - 20000)} = \frac{L}{h - 20000}$ and we require this to be less than or equal to $2$. 224 | 225 | \begin{align*} 226 | \frac{L}{h - 20000} &<= 2 \\ 227 | L &<= 2(h - 20000) \\ 228 | \sqrt{h^2 - 20000^2} &<= 2h - 40000 \\ 229 | h^2 - 20000^2 &<= (2h - 40000)^2 \\ 230 | h^2 &<= 4h^2 - 160000h + 40000^2 + 20000^2 \\ 231 | 3h^2 - 160000h + 2*10^9 &>= 0 232 | \end{align*} 233 | 234 | Let $y = 3h^2 - 160000h + 2*10^9$. We solve for $y = 0$, getting $h = \frac{-(-160000) + \sqrt{(-160000)^2 - 4 \cdot 3 \cdot 2 * 10^9}}{2 * 3}$, getting $h = 33333.333333333336$ and $h = 20000$. $\frac{dy}{dh} = 6h - 160000$. At the critical points, $6h - 160000 = 0$ so $h = \frac{160000}{6} = 26666 \frac{2}{3}$. $\frac{d^2 y}{dh^2} = 6 > 0$ so $h = 26666 \frac{2}{3}$ is a minimum point. At $h = 26666 \frac{2}{3}, y = -133333332 < 0$. 235 | 236 | This is a pretty "standard" quadratic function, so $y >= 0$ for $h <= 20000$ and $y <= 0$ for $20000 <= h <= 33333 \frac{1}{3}$ and $y >= 0$ for $h >= 33333 \frac{1}{3}$. By the constraints of the problem, $h$ must be at least $20000$, so we need to take $h >= 33333 \frac{1}{3}$. Then $\frac{\Delta L}{\Delta h} <= \frac{L}{h - 20000} <= 2$ for $\Delta h < 0$. Since $33333 \frac{1}{3} > 23094.010760368154$, for $h >= 33333 \frac{1}{3}$, we will also have $\frac{\Delta L}{\Delta h} <= 2$ for $h > 0$. 237 | 238 | Hence for $|\Delta L| <= 2 |\Delta h|$ or equivalently $\frac{\Delta L}{\Delta h} <= 2$, we need $h >= 33333 \frac{1}{3}$. 239 | 240 | \begin{tcolorbox} 241 | \textbf{Q3. Sensitivity of measurement, revisited.} 242 | 243 | b) Suppose that the Planet Quirk is a not only flat, but one-dimensional (a straight line). There are several satellites at height $20,000$ kilometres and you get readings saying that satellite 1 is directly above the point $x_1 \pm 10^{-10}$ and is at a distance $h_1 = 21,000 \pm 10^{-2}$ from you, satellite 2 is directly above $x_2 \pm 10^{-10}$ and at a distance $h_2 = 52,000 \pm 10^{-2}$. Where are you and to what accuracy? Hint: Consider separately the cases $x1 < x2$ and $x2 > x1$. 244 | \end{tcolorbox} 245 | 246 | \textbf{NOTE:} I am only doing 1 part of this question and I'm not sure whether my solution is correct. This question also covers propagation of error, which is not covered in the lectures. 247 | 248 | Suppose we are at point $q$ of Planet Quirk. We'll follow the hint given in this question. 249 | 250 | \textbf{Case 1:} $x_1 < x_2$ 251 | 252 | \textbf{Case 1a:} $q < x_1 < x_2$ 253 | 254 | We can safely ignore $q = x_1$, because if $q = x_1$, $h_1 = 20,000$ but that is not the case. 255 | 256 | Let's ignore the uncertainties in measurement. Then we have the following equations: 257 | 258 | \begin{align} 259 | (x_1 - q)^2 &= h_1^2 - 20000^2 \\ 260 | (x_2 - q)^2 &= h_2^2 - 20000^2 261 | \end{align} 262 | 263 | From the above 2 equations we see that 264 | 265 | \begin{align} 266 | q = x_1 - \sqrt{h_1^2 - 20000^2} \\ 267 | q = x_2 - \sqrt{h_2^2 - 20000^2} 268 | \end{align} 269 | 270 | In general, we see that given an $x$ and $h$, we have $q = x - \sqrt{h^2 - 20000^2}$ 271 | 272 | If these YouTube videos are correct and I've interpreted their content correctly: 273 | 274 | \url{https://www.youtube.com/watch?v=N0OYaG6a51w} \\ 275 | \url{https://www.youtube.com/watch?v=V0ZRvvHfF0E} 276 | 277 | we can treat $x$ and $h$ as variables, get their partial derivatives, and compute the uncertainty in $q$ as follows: 278 | 279 | \begin{align*} 280 | \Delta q = \sqrt{(\frac{\partial q}{\partial x} \cdot \Delta x)^2 + (\frac{\partial q}{\partial h} \cdot \Delta h)^2} 281 | \end{align*} 282 | 283 | Now, $\frac{\partial q}{\partial x} = 1$ and $\frac{\partial q}{\partial h} = -\frac{1}{2} \cdot 2h \cdot (h^2 - 20000^2)^{-1/2} = -\frac{h}{\sqrt{h^2 - 20000^2}}$. 284 | 285 | Based on $x_1$, $\Delta x_1 = \pm 10^-10$, $h_1 = 21000$, $\Delta h_1 = 10^{-2}$, 286 | 287 | \begin{align*} 288 | \Delta q &= \sqrt{(1 \cdot \pm 10^{-10})^2 + (-\frac{21000}{21000^2 - 20000^2} \cdot \pm 10^{-2})^2} \\ 289 | &= \sqrt{10^{-20} + \frac{21000^2}{21000^2 - 20000^2} \cdot 10^{-4}} \\ 290 | &\approx 0.03162277660168379 291 | \end{align*} 292 | 293 | Then 294 | 295 | \begin{align*} 296 | q &= x_1 - \sqrt{h_1^2 - 20000^2} \pm \Delta q \\ 297 | &= x_1 - \sqrt{21000^2 - 20000^2} \pm \Delta q \\ 298 | &\approx x_1 - 6403.1242374328485 \pm 0.03162277660168379 299 | \end{align*} 300 | 301 | Based on $x_2$, $\Delta x_2 = \pm 10^-10$, $h_2 = 52000$, $\Delta h_2 = 10^{-2}$, 302 | 303 | \begin{align*} 304 | \Delta q &= \sqrt{(1 \cdot \pm 10^{-10})^2 + (-\frac{52000}{52000^2 - 20000^2} \cdot \pm 10^{-2})^2} \\ 305 | &= \sqrt{10^{-20} + \frac{52000^2}{52000^2 - 20000^2} \cdot 10^{-4}} \\ 306 | &\approx 0.01 307 | \end{align*} 308 | 309 | Then 310 | 311 | \begin{align*} 312 | q &= x_2 - \sqrt{h_2^2 - 20000^2} \pm \Delta q \\ 313 | &= x_2 - \sqrt{52000^2 - 20000^2} \pm \Delta q \\ 314 | &\approx x_2 - 48000 \pm 0.01 315 | \end{align*} 316 | 317 | Now, I believe that these 2 measurements of $q$ along with uncertainty $\Delta q$ should partially overlap and the interval of $q$ should be the intersection of these 2 intervals. I also believe that it is a contradiction if the intervals are non overlapping. However, I am not able to prove either of that. 318 | 319 | This is also the only case in this part of the question I will tackle. I have spent waaaaay too much time on this Problem Set and this is the final part of the Problem Set I worked on in my writeup. Apologies to readers for the incompleteness. 320 | 321 | \begin{tcolorbox} 322 | \textbf{Q3. Sensitivity of measurement, revisited.} 323 | 324 | c) Express $dL/dh$ in terms of the angle between the line of sight to the satellite and the horizontal from the person on the ground. (When expressed using the line-of-sight angle, the formula also works for a curved planet like Earth.) 325 | \end{tcolorbox} 326 | 327 | \textbf{Note to reader:} I'm not sure whether my solution to this part is correct. 328 | 329 | \begin{align*} 330 | cos \theta &= \frac{L}{h} 331 | \end{align*} 332 | 333 | Differentiating implicitly wrt $h$, 334 | 335 | \begin{align*} 336 | -sin\theta \cdot \frac{d\theta}{dh} &= \frac{dL}{dh} \cdot \frac{1}{h} + L \cdot (-\frac{1}{h^2}) \\ 337 | \frac{dL}{dh} &= (-sin\theta \cdot \frac{d\theta}{dh} + \frac{L}{h^2}) * h \\ 338 | &= \frac{L}{h} - h \cdot sin\theta \cdot \frac{d\theta}{dh} 339 | \end{align*} 340 | 341 | 342 | 343 | \begin{tcolorbox} 344 | \textbf{Q4. More sensitivity of measurement.} 345 | 346 | Consder a parabolic mirror with equation $y = -1/4 + x^2$ and focus at the origin. (See Problem Set 1.) A ray of light traveling down vertically along the line $x = a$ hits the mirror at the point $(a, b)$ where $b = -1/4 + a^2$ and goes to the origin along a ray at angle $\theta$ measured from the positive x-axis. 347 | 348 | a) Find the formula for $tan\theta$ in terms of $a$ and $b$, and calculate $d\theta / da$ using implicit differentiation. (Express your answer in terms of $a$ and $\theta$.) 349 | \end{tcolorbox} 350 | 351 | Assuming $a \neq 0$, $tan \theta = \frac{b}{a} = \frac{-1/4 + a^2}{a} = a - \frac{1}{4a}$ 352 | 353 | Differentiating implicitly with respect to $a$, 354 | 355 | \begin{align*} 356 | (sec^2 \theta) \cdot \frac{d \theta}{da} &= 1 + \frac{1}{4a^2} \\ 357 | \frac{d \theta}{da} &= cos^2 \theta (1 + \frac{1}{4a^2}) \\ 358 | &= cos^2 \theta + \frac{cos^2 \theta}{4a^2} 359 | \end{align*} 360 | 361 | \begin{tcolorbox} 362 | \textbf{Q4. More sensitivity of measurement.} 363 | 364 | b) If the telescope records a star at $\theta = -\pi / 6$ and the measurement is accurate to $10^{-3}$ radians, use part (a) to give an estimate as to the location of the star in the variable $a$. 365 | \end{tcolorbox} 366 | 367 | \textbf{NOTE:} This makes use of propagation of error, which isn't really covered in the lectures. 368 | 369 | Since we are given $\theta$ and its accuracy, we actually want $\frac{da}{d \theta}$. 370 | 371 | \begin{align*} 372 | tan \theta &= a - \frac{1}{4a} \\ 373 | sec^2 \theta &= \frac{da}{d \theta} + \frac{1}{4a^2} \cdot \frac{da}{d \theta} \\ 374 | sec^2 \theta &= \frac{da}{d \theta} (1 + \frac{1}{4a^2}) \\ 375 | da &= \frac{sec^2 \theta}{(1 + \frac{1}{4a^2})} \cdot d \theta \\ 376 | &= \frac{sec^2 \theta}{\frac{4a^2 + 1}{4a^2}} \cdot d \theta \\ 377 | &= \frac{4a^2 sec^2 \theta}{4a^2 + 1} \cdot d \theta \\ 378 | &= \frac{4a^2}{cos^2 \theta (4a^2 + 1)} \cdot d \theta 379 | \end{align*} 380 | 381 | which turns out to be exactly the same thing as if we started from what we have in part (a): 382 | 383 | \begin{align*} 384 | \frac{d \theta}{da} &= cos^2 \theta + \frac{cos^2 \theta}{4a^2} \\ 385 | da &= \frac{1}{cos^2 \theta + \frac{cos^2 \theta}{4a^2}} \cdot d \theta \\ 386 | &= \frac{1}{\frac{4a^2 cos^2 \theta + cos^2 \theta}{4a^2}} \cdot d \theta \\ 387 | &= \frac{4a^2}{4a^2 cos^2 \theta + cos^2 \theta} \cdot d \theta \\ 388 | &= \frac{4a^2}{cos^2 \theta (4a^2 + 1)} \cdot d \theta 389 | \end{align*} 390 | 391 | At $\theta = -\frac{\pi}{6}$, $tan \theta = tan (-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$. But $tan \theta = a - \frac{1}{4a}$. Hence 392 | 393 | \begin{align*} 394 | -\frac{1}{\sqrt{3}} &= a - \frac{1}{4a} \\ 395 | (-\frac{1}{\sqrt{3}})^2 &= (\frac{a \cdot 4a - 1}{4a})^2 \\ 396 | \frac{1}{3} &= (\frac{4a^2 - 1}{4a})^2 \\ 397 | \frac{1}{3} &= \frac{16a^4 - 8a^2 + 1}{16a^2} \\ 398 | 16a^2 &= 3(16a^4 - 8a^2 + 1) \\ 399 | 48a^4 - 24a^2 + 3 - 16a^2 &= 0 \\ 400 | 48a^4 - 40a^2 + 3 &= 0 \\ 401 | a^2 &= \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 48 \cdot 3}}{2 \cdot 48} \\ 402 | &= \frac{40 \pm \sqrt{1024}}{96} \\ 403 | &= \frac{40 \pm 32}{96} 404 | \end{align*} 405 | 406 | So $a^2 = \frac{40 + 32}{96} = \frac{72}{96} = \frac{3}{4}$ or $a^2 = \frac{40 - 32}{96} = \frac{8}{96} = \frac{1}{12}$. 407 | 408 | Then $a = \pm \frac{\sqrt{3}}{2}$ or $a = \pm \frac{1}{\sqrt{12}} = \pm \frac{1}{\sqrt{3 * 4}} = \pm \frac{1}{2 \sqrt{3}} = \pm \frac{\sqrt{3}}{6}$. 409 | 410 | When $a = \frac{\sqrt{3}}{2}$, $da = \frac{4a^2}{cos^2 \theta (4a^2 + 1)} \cdot d \theta = \frac{4 (\frac{\sqrt{3}}{2})^2}{cos^2 (-\frac{\pi}{6}) (4 \cdot (\frac{\sqrt{3}}{2})^2 + 1)} \cdot d \theta = \frac{4 \cdot \frac{3}{4}}{(\frac{\sqrt{3}}{2})^2 (4 \cdot \frac{3}{4} + 1)} \cdot d \theta = \frac{3}{\frac{3}{4} (3 + 1)} \cdot d \theta = \frac{3}{3} \cdot d \theta = d \theta$. This is also the same for $a = -\frac{\sqrt{3}}{2}$. Hence for $a = \pm \frac{\sqrt{3}}{2}, da = d \theta = 10^{-3}$. 411 | 412 | When $a = \frac{\sqrt{3}}{6}, da = \frac{4a^2}{cos^2 \theta (4a^2 + 1)} \cdot d \theta = \frac{4 (\frac{\sqrt{3}}{6})^2}{cos^2 (-\frac{\pi}{6}) (4 \cdot (\frac{\sqrt{3}}{6})^2 + 1)} \cdot d \theta = \frac{4 \cdot \frac{3}{36}}{(\frac{\sqrt{3}}{2})^2 (4 \cdot \frac{3}{36} + 1)} \cdot d \theta = \frac{\frac{3}{9}}{\frac{3}{4}(\frac{3}{9} + 1)} \cdot d \theta = \frac{\frac{3}{9}}{\frac{3}{4} \cdot \frac{12}{9}} \cdot d \theta = \frac{1}{3} d \theta$. This is also the same for $a = -\frac{\sqrt{3}}{6}$. Hence for $a = \pm \frac{\sqrt{3}}{6}, da = \frac{1}{3} d \theta = \frac{1}{3} \cdot 10^{-3}$. 413 | 414 | \begin{tcolorbox} 415 | \textbf{Q4. More sensitivity of measurement.} 416 | 417 | c) (optional; no credit) Solve for $a$ as a function of $\theta$ alone and doublecheck your answers to parts (a) and (b). 418 | \end{tcolorbox} 419 | 420 | \begin{align*} 421 | tan \theta &= a - \frac{1}{4a} \\ 422 | tan \theta &= \frac{4a^2 - 1}{4a} \\ 423 | 4a \cdot tan \theta &= 4a^2 - 1 \\ 424 | 4a^2 - 4 (tan \theta) \cdot a - 1 &= 0 \\ 425 | a &= \frac{-(-4 \cdot tan \theta) \pm \sqrt{(-4 \cdot tan \theta)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \\ 426 | &= \frac{4 \cdot tan \theta \pm \sqrt{16 \cdot tan^2 \theta + 16}}{8} \\ 427 | &= \frac{4 \cdot tan \theta \pm \sqrt{16 sec^2 \theta}}{8} \\ 428 | &= \frac{4 \cdot tan \theta \pm 4 \cdot sec \theta}{8} \\ 429 | &= \frac{tan \theta \pm sec \theta}{2} 430 | \end{align*} 431 | 432 | \textbf{Case 1:} $a = \frac{tan \theta + sec \theta}{2} = \frac{tan (-\frac{\pi}{6}) + sec (-\frac{\pi}{6})}{2} = \frac{-\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}{2} = \frac{\frac{1}{\sqrt{3}}}{2} = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$ 433 | 434 | \begin{align*} 435 | \frac{da}{d \theta} &= \frac{sec^2 \theta + sec \theta tan \theta}{2} \\ 436 | da &= \frac{1}{2} (sec^2 \theta + sec \theta tan \theta) \cdot d \theta \\ 437 | &= \frac{1}{2} (sec^2 (-\frac{\pi}{6}) + sec(-\frac{\pi}{6}) tan(-\frac{\pi}{6})) \cdot d \theta \\ 438 | &= \frac{1}{2} ((\frac{2}{\sqrt{3}})^2 + \frac{2}{\sqrt{3}}(-\frac{1}{\sqrt{3}})) \cdot d \theta \\ 439 | &= \frac{1}{2} (\frac{4}{3} - \frac{2}{3}) \cdot d \theta \\ 440 | &= \frac{1}{2} \cdot \frac{2}{3} \cdot d \theta \\ 441 | &= \frac{1}{3} d \theta 442 | \end{align*} 443 | 444 | which agrees with what we've got in part (b). 445 | 446 | \textbf{Case 2:} $a = \frac{tan \theta - sec \theta}{2} = \frac{tan (-\frac{\pi}{6}) - sec (-\frac{\pi}{6})}{2} = \frac{-\frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{2} = \frac{-\frac{3}{\sqrt{3}}}{2} = -\frac{3}{2 \sqrt{3}} = -\frac{3 \sqrt{3}}{6} = -\frac{\sqrt{3}}{2}$ 447 | 448 | \begin{align*} 449 | \frac{da}{d \theta} &= \frac{sec^2 \theta - sec \theta tan \theta}{2} \\ 450 | da &= \frac{1}{2} (sec^2 \theta - sec \theta tan \theta) \cdot d \theta \\ 451 | &= \frac{1}{2} (sec^2 (-\frac{\pi}{6}) - sec(-\frac{\pi}{6}) tan(-\frac{\pi}{6})) \cdot d \theta \\ 452 | &= \frac{1}{2} ((\frac{2}{\sqrt{3}})^2 - \frac{2}{\sqrt{3}}(-\frac{1}{\sqrt{3}})) \cdot d \theta \\ 453 | &= \frac{1}{2} (\frac{4}{3} + \frac{2}{3}) \cdot d \theta \\ 454 | &= \frac{1}{2} \cdot 2 \cdot d \theta \\ 455 | &= d \theta 456 | \end{align*} 457 | 458 | which agrees with what we've got in part (b). 459 | 460 | \begin{tcolorbox} 461 | \textbf{Q5. Newton's method} \\ 462 | a) Compute the cube root of 9 to 6 significant figures using Newton's method. Give the general formula, and list numerical values, starting with $x_0 = 2$. At what iteration $k$ does the method surpass the accuracy of your calculator or computer? (Display your answers to the accuracy of your calculator or computer.) 463 | \end{tcolorbox} 464 | 465 | Let $x = \sqrt[3]{9}$. This is our quantity of interest. But we need to convert this to something we can solve using Newton's method. Cubing both sides, we have $x^3 = 9$ and hence $x^3 - 9 = 0$. 466 | 467 | Recall that the formula for Newton's method is given by $x_{n+1} = x_n - \frac{y_n}{f'(x_n)} = x_n - \frac{f(x_n)}{f'(x_n)}$, where $x_n$ is the $n$th guess. In this case, $f(x) = x^3 - 9$ and $f'(x) = \frac{d}{dx} x^3 - 9 = 3x^2$. Since $2^3 = 8$ is close to 9 and is a cube that we know, let's use use $x_0 = 2$ as our initial guess. 468 | 469 | \begin{center} 470 | \begin{tabular}{|c|c|c|c|} 471 | \hline 472 | \rowcolor{Gray} 473 | $n$ & $x_n$ & $|f(x_n) - 9|$ & $|x_n - \sqrt[3]9|$ \\ \hline 474 | 0 & 2 & $|2^3 - 9| = 1$ & $|2 - \sqrt[3]9| \approx 0.080084$ \\ \hline 475 | 1 & $2 - \frac{2^3 - 9}{3\cdot2^2} = 2 + \frac{1}{12} = \frac{25}{12}$ & $|(\frac{25}{12})^3 - 9| \approx 0.042245$ & $|\frac{25}{12} - \sqrt[3]9| \approx 0.003250$ \\ \hline 476 | 2 & $\frac{25}{12} - \frac{(25/12)^3 - 9}{3\cdot (25/12)^2} \approx 2.080088888888889$ & $6.575597 * 10^{-5}$ & $\approx 5.065837*10^{-06}$ \\ \hline 477 | 3 & $2.0800838230642413$ & $1.601403* 10^{-10}$ & $\approx 1.233724*10^{-11}$ \\ \hline 478 | 4 & $2.080083823051904$ & $0.0$ & $0.0$ \\ \hline 479 | \end{tabular} 480 | \end{center} 481 | 482 | Calculations done using Python 2.7 . I've truncated the numbers for the final 2 columns to 6 decimal places. So we see that by iteration 5 (at $n = 4$), both ${x_n}^3 - 9$ and $x_n - \sqrt[3]9$ exceed Python 2.7's accuracy. Which is pretty fast. In fact, this happens by iteration 3 for a handheld calculator I'm using. 483 | 484 | \begin{tcolorbox} 485 | \textbf{Q5. Newton's method} \\ 486 | b) For each step $x_k, \ k = 0, 1, ... ,$ say whether the value is i) larger or smaller than $9^{1/3}$; ii) larger or smaller than the perceding value $x_{k-1}$. Illustrate on the graph $x^3 - 9$ why this is so. 487 | \end{tcolorbox} 488 | 489 | \begin{center} 490 | \includegraphics[scale=0.8]{k_eq_0.png} 491 | \end{center} 492 | 493 | The graph in blue is $x^3 - 9$. The point in magenta is $x_0 = 2$ plugged into $y = x^3 - 9$. The point where the red line $y = 0$ cuts the curve $y = x^3 - 9$ represents the number $\sqrt[3]9$. We see that $x_0 < \sqrt[3]9$. It should be obvious that the gradient of $y = x^3 - 9$ at $x_0 = 2$ is positive and the $y_0 = 2^3 - 9 = -1$ is negative. Hence $x_1 = x_0 - \frac{x_0}{f'(x_0)}$ is a positive value and $x_1 > x_0$. 494 | 495 | \begin{center} 496 | \includegraphics[scale=0.8]{k_eq_1.png} 497 | \end{center} 498 | 499 | For $k = 1$, we see that $f(x_1) = x_1^3 - 9 > 0$ and $f'(x_1) > 0$ and hence $\frac{f(x_1)}{f'(x_1)}$ is a positive quantity, so $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} < x_1$. 500 | 501 | It is not very meaningful to carry on for $k > 1$ because Newton's method converges very quickly in this case and I believe you get the idea from these 2 iterations. 502 | 503 | \begin{tcolorbox} 504 | \textbf{Q5. Newton's method} \\ 505 | c) Find a quadratic approximation to $9^{1/3}$, and estimate the difference between the quadratic approximation and the exact answer. (Hint: To get a reasonable quadratic approximation, use $9 = 8(1 + 1/8)$.) 506 | \end{tcolorbox} 507 | 508 | \begin{align*} 509 | \sqrt[3]9 = (8 + 1)^{1/3} = (8(1 + \frac{1}{8}))^{1/3} = 2\cdot(1 + \frac{1}{8})^{1/3} 510 | \end{align*} 511 | 512 | Since $\frac{1}{8}$ is kind of small, we will use the quadratic approximation to $2(1 + x)^{1/3}$ where $x = \frac{1}{8}$ to get an approximation to $\sqrt[3]9$. 513 | 514 | Let's first derive the quadratic approximation to $(1+x)^r$ for $x$ near 0. 515 | 516 | \begin{align*} 517 | (1 + x)^r &\approx (1 + 0)^r + r(1 + 0)^{r-1}x + \frac{r(r-1)(1+0)^{r-2}}{2!}x^2 \\ 518 | &= 1 + rx + \frac{r(r-1)}{2}x^2 519 | \end{align*} 520 | 521 | Substitute $x = \frac{1}{8}, r = \frac{1}{3}$ into $(1 + x)^r$, we get: 522 | 523 | \begin{align*} 524 | (1 + \frac{1}{8})^{1/3} &= 1 + \frac{1}{3} \cdot \frac{1}{8} + \frac{\frac{1}{3}(-\frac{2}{3})}{2}(\frac{1}{8})^2 \\ 525 | &= 1 + \frac{1}{24} - \frac{1}{9} \cdot \frac{1}{64} \\ 526 | &= 1 + \frac{24 - 1}{576} = \frac{576 + 23}{576} = \frac{599}{576} 527 | \end{align*} 528 | 529 | But we are interested in $\sqrt[3]9$ 530 | 531 | \begin{align*} 532 | \sqrt[3]9 &= 2(1 + \frac{1}{8})^{1/3} \approx 2 \cdot \frac{599}{576} = \frac{599}{288} \approx 2.079861111111111 533 | \end{align*} 534 | 535 | The difference between this quadratic approximation and the exact answer is approximately 536 | 537 | \begin{align*} 538 | |\sqrt[3]9 - \frac{599}{288}| \approx 0.0002227119407929301 539 | \end{align*} 540 | 541 | 542 | \begin{tcolorbox} 543 | \textbf{Q6. Hypocycloid, again} \\ 544 | Here we derive the equation for the hypocycloid of Problem 2 from the sweeping out property directly. This takes quite a bit longer. We will look at the hypocycloid from yet another (easier) point of view later on. \\ 545 | \\ 546 | Think of the first quadrant of the $xy$-plane as representing the region to the right of a wall with the ground as the positive $x$-axis and the wall as the positive $y$-axis. A unit length ladder is placed vertically against the wall. The bottom of the ladder is at $x = 0$ and slides to the right along the $x$-axis until the ladder is horizontal. At the same time, the top of the aldder is dragged down the $y$-axis ending at the origin $(0, 0)$. We are going to describe the region swept out by this motion, in other words, the blurry region formed in a photograph of the motion if the eye of the camera is open the whole time.\\ 547 | \\ 548 | a) Suppose that $L_1$ is the line segment from $(0,y_1)$ to $(x_1,0)$ and $L_2$ is the line segment from $(0,y_2)$ to $(x_2,0)$. Find the formula for the point of intersection $(x_3,y_3)$ of the two line segments. Don't expect the formula to be simple: It must involve all four parameters $x_1, x_2, y_1,$ and $y_2$. But simplify as much as possible! \\ 549 | \\ 550 | It's important to make sure you have the right formula before proceeding further. You can doublecheck your formulas in several ways. (This is optional.)\\ 551 | \\ 552 | i) If $y_2 = 0$, then $x_3 = x_1$.\\ 553 | \\ 554 | ii) When the $x's$ and $y's$ are interchanged the formulas should be the same. What transformation of the plane does the exchange of $x$ and $y$ represent?\\ 555 | \\ 556 | iii) It is impossible to find $x_3$ and $y_3$ if the lines are parallel, so the denominator in the formula must be zero when $L_1$ and $L_2$ have the same slope.\\ 557 | \\ 558 | iv) Rescaling all variables by a factor $c$ leaves the formula unchanged, so the numerator of the formula for $x_3$ and $y_3$ should have degree (in all variables) one greater than the denominator. 559 | \end{tcolorbox} 560 | 561 | $L_1$ has equation 562 | 563 | \begin{align*} 564 | y - y_1 &= \frac{0 - y_1}{x_1 - 0} \cdot (x - x_1) \\ 565 | y - y_1 &= -\frac{y_1}{x_1} \cdot x \\ 566 | y &= -\frac{y_1}{x_1} \cdot x + y_1 567 | \end{align*} 568 | 569 | $L_2$ has equation 570 | 571 | \begin{align*} 572 | y - y_2 &= \frac{0 - y_2}{x_2 - 0} \cdot (x - x_2) \\ 573 | y - y_2 &= -\frac{y_2}{x_2} \cdot x \\ 574 | y &= -\frac{y_2}{x_2} \cdot x + y_2 575 | \end{align*} 576 | 577 | When $L_1$ and $L_2$ intersect at $(x_3, y_3)$, these 2 equations have the same value at $y_3$, so 578 | 579 | \begin{align*} 580 | y_3 = -\frac{y_1}{x_1} \cdot x_3 + y_1 &= -\frac{y_2}{x_2} \cdot x_3 + y_2 \\ 581 | \frac{y_2}{x_2} \cdot x_3 - \frac{y_1}{x_1} \cdot x_3 &= y_2 - y_1 \\ 582 | (\frac{y_2}{x_2} - \frac{y_1}{x_1}) \cdot x_3 &= y_2 - y_1 \\ 583 | \frac{x_1 y_2 - x_2 y_1}{x_1 x_2} \cdot x_3 &= y_2 - y_1 \\ 584 | x_3 &= \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} \tag*{(Assuming $x_1 y_2 \neq x_2 y_1$)} 585 | \end{align*} 586 | 587 | Substitute $x_3 = \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1}$ into equation of $L_1$ to get $y_3$: 588 | 589 | \begin{align*} 590 | y_3 &= -\frac{y_1}{x_1} \cdot \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} + y_1 \\ 591 | &= \frac{-y_1 (y_2 - y_1) x_2}{x_1 y_2 - x_2 y_1} + \frac{x_1 y_1 y_2 - x_2 y_1^2}{x_1 y_2 - x_2 y_1} \\ 592 | &= \frac{-x_2 y_1 y_2 + x_2 y_1^2 + x_1 y_1 y_2 - x_2 y_1^2}{x_1 y_2 - x_2 y_1} \\ 593 | &= \frac{y_1 y_2 (x_1 - x_2)}{x_1 y_2 - x_2 y_1} 594 | \end{align*} 595 | 596 | \begin{tcolorbox} 597 | Check (i) If $y_2 = 0$, then $x_3 = x_1$. 598 | \end{tcolorbox} 599 | 600 | \begin{align*} 601 | x_3 &= \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} \\ 602 | &= \frac{(0 - y_1) x_1 x_2}{x_1 \cdot 0 - x_2 y_1} \\ 603 | &= \frac{- x_1 x_2 y_1}{- x_2 y_1} \\ 604 | &= x_1 605 | \end{align*} 606 | 607 | \begin{tcolorbox} 608 | Check (ii) When the $x's$ and $y's$ are interchanged the formulas should be the same. What transformation of the plane does the exchange of $x$ and $y$ represent?\\ 609 | \end{tcolorbox} 610 | 611 | Original equations: 612 | 613 | \begin{align*} 614 | x_3 &= \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} \\ 615 | y_3 &= \frac{y_1 y_2 (x_1 - x_2)}{x_1 y_2 - x_2 y_1} 616 | \end{align*} 617 | 618 | Swap $x's$ and $y's$ in $x_3 = \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1}$: 619 | 620 | \begin{align*} 621 | y_3 &= \frac{(x_2 - x_1) y_1 y_2}{y_1 x_2 - y_2 x_1} \\ 622 | &= \frac{- y_1 y_2 (x_1 - x_2)}{-(x_1 y_2 - x_2 y_1)} \\ 623 | &= \frac{y_1 y_2 (x_1 - x_2)}{x_1 y_2 - x_2 y_1} 624 | \end{align*} 625 | 626 | Swap $x's$ and $y's$ in $y_3 = \frac{y_1 y_2 (x_1 - x_2)}{x_1 y_2 - x_2 y_1}$: 627 | 628 | \begin{align*} 629 | x_3 &= \frac{x_1 x_2 (y_1 - y_2)}{y_1 x_2 - y_2 x_1} \\ 630 | &= \frac{-(y_2 - y_1) x_1 x_2}{-(x_1 y_2 - x_2 y_1)} \\ 631 | &= \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} 632 | \end{align*} 633 | 634 | This exchange of $x$ and $y$ should represent reflection about the line $y = x$. 635 | 636 | \begin{tcolorbox} 637 | Check (iii) It is impossible to find $x_3$ and $y_3$ if the lines are parallel, so the denominator in the formula must be zero when $L_1$ and $L_2$ have the same slope.\\ 638 | \end{tcolorbox} 639 | 640 | Suppose the lines are parallel. Then the slopes of $L_1$ and $L_2$ must be the same. Hence 641 | 642 | \begin{align*} 643 | -\frac{y_1}{x_1} &= -\frac{y_2}{x_2} \\ 644 | -x2 y_1 &= -x_1 y_2 \\ 645 | x_1 y_2 - x_2 y_1 &= 0 646 | \end{align*} 647 | 648 | So the denominator $x_1 y_2 - x_2 y_1$ in the formula for $x_3$ and $y_3$ must be zero. 649 | 650 | \begin{tcolorbox} 651 | Check (iv) Rescaling all variables by a factor $c$ leaves the formula unchanged, so the numerator of the formula for $x_3$ and $y_3$ should have degree (in all variables) one greater than the denominator. 652 | \end{tcolorbox} 653 | 654 | Let's scale all $x_1, x_2, y_1, y_2$ by some factor $c \neq 0$. So $x_1$ is now $c x_1$, $x_2$ is now $c x_2$, $y_1$ is now $c y_1$, $y_2$ is now $c y_2$. 655 | 656 | \begin{align*} 657 | x_3 &= \frac{(c y_2 - c y_1) c x_1 c x_2}{c x_1 c y_2 - c x_2 c y_1} \\ 658 | &= \frac{c (y_2 - y_1) c^2 x_1 x_2}{c^2(x_1 y_2 - x_2 y_1)} \\ 659 | &= \frac{c^3 (y_2 - y_1) x_1 x_2}{c^2(x_1 y_2 - x_2 y_1)} \\ 660 | &= \frac{c (y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} 661 | \end{align*} 662 | 663 | which is $c$ times the original $x_3$. 664 | 665 | \begin{align*} 666 | y_3 &= \frac{c y_1 c y_2 (c x_1 - c x_2)}{c x_1 c y_2 - c x_2 c y_1} \\ 667 | &= \frac{c^2 y_1 y_2 \cdot c (x_1 - x_2)}{c^2 (x_1 y_2 - x_2 y_1)} \\ 668 | &= \frac{c^3 y_1 y_2 (x_1 - x_2)}{c^2 (x_1 y_2 - x_2 y_1)} \\ 669 | &= \frac{c y_1 y_2 (x_1 - x_2)}{x_1 y_2 - x_2 y_1} 670 | \end{align*} 671 | 672 | which is $c$ times the original $y_3$. 673 | 674 | 675 | \begin{tcolorbox} 676 | \textbf{Q6. Hypocycloid, again} \\ 677 | b) Write the equation involving $x_2$ and $y_2$ that expresses the property that ladder $L_2$ has length one. We will suppose that $L_1$ represents the ladder at a fixed position, and $L_2$ tends to $L_1$. Thus 678 | 679 | \begin{align*} 680 | x_2 = x_1 + \Delta x; y_2 = y_1 + \Delta y 681 | \end{align*} 682 | 683 | Use implicit differentiaton (related rates) to find 684 | 685 | \begin{align*} 686 | \lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} 687 | \end{align*} 688 | 689 | (Express the limit as a function of the fixed values $x_1$ and $y_1$.) 690 | \end{tcolorbox} 691 | 692 | The equation involving $x_2$ and $y_2$ that expresses the property that ladder $L_2$ has length one is $x_2^2 + y_2^2 = 1$. In fact, since $L_1$ and $L_2$ are the same ladder but in different positions, the general equation is $x^2 + y^2 = 1$. 693 | 694 | For a fixed $x = x_1$ and $y = y_1$, we are given that $x_2 = x_1 + \Delta x$ and $y_2 = y_1 + \Delta y$. Then $\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} = \frac{dy}{dx}$ evaluated at $x = x_1, y = y_1$. Using implicit differentiation on $x^2 + y^2 = 1$ 695 | 696 | \begin{align*} 697 | x^2 + y^2 &= 1 \\ 698 | 2x + 2y \cdot \frac{dy}{dx} &= 0 \\ 699 | \frac{dy}{dx} &= -\frac{2x}{2y} \\ 700 | &= -\frac{x}{y} 701 | \end{align*} 702 | 703 | And at $x = x_1, y = y_1$, this is $-\frac{x_1}{y_1}$. So $\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} = -\frac{x_1}{y_1}$ 704 | 705 | \begin{tcolorbox} 706 | \textbf{Q6. Hypocycloid, again} \\ 707 | c) Substitute $x_2 = x_1 + \Delta x$ and $y_2 = y_1 + \Delta y$ into the formula in part (a) for $x_3$ and use part (b) to compute 708 | 709 | \begin{align*} 710 | X = \lim_{x_2 \rightarrow x_1} x_3 = \lim_{\Delta x \rightarrow 0} x_3 711 | \end{align*} 712 | 713 | Simplify as much as possible. Deduce, by symmetry alone, the formula for 714 | 715 | \begin{align*} 716 | Y = \lim_{x_2 \rightarrow x_1} y_3 717 | \end{align*} 718 | \end{tcolorbox} 719 | 720 | \begin{align*} 721 | x_3 &= \frac{(y_2 - y_1) x_1 x_2}{x_1 y_2 - x_2 y_1} \\ 722 | &= \frac{(y_1 + \Delta y - y_1) x_1 (x_1 + \Delta x)}{x_1 (y_1 + \Delta y) - (x_1 + \Delta x) y_1} \\ 723 | &= \frac{(\Delta y) x_1 (x_1 + \Delta x)}{x_1 y_1 + x_1 \Delta y - x_1 y_1 - (\Delta x) y} \\ 724 | &= \frac{(\Delta y) x_1 (x_1 + \Delta x)}{x_1 \Delta y - (\Delta x) y_1} 725 | \end{align*} 726 | 727 | We will be making use of $\lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = -\frac{x_1}{y_1}$ below: 728 | 729 | \begin{align*} 730 | X = \lim_{x_2 \rightarrow x_1} x_3 = \lim_{\Delta x \rightarrow 0} x_3 &= \lim_{\Delta x \rightarrow 0} \frac{(\Delta y) x_1 (x_1 + \Delta x)}{x_1 \Delta y - (\Delta x) y_1} \\ 731 | &= \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \cdot \frac{x_1 (x_1 + \Delta x)}{x_1 \cdot \frac{\Delta y}{\Delta x} - y_1} \\ 732 | &= (\lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}) \cdot (\lim_{\Delta x \rightarrow 0}\frac{x_1 (x_1 + \Delta x)}{x_1 \cdot \frac{\Delta y}{\Delta x} - y_1}) \\ 733 | &= -\frac{x_1}{y_1} \cdot \frac{\lim_{\Delta x \rightarrow 0}(x_1 (x_1 + \Delta x))}{\lim_{\Delta x \rightarrow 0}(x_1 \cdot \frac{\Delta y}{\Delta x} - y_1)} \\ 734 | &= -\frac{x_1}{y_1} \cdot \frac{\lim_{\Delta x \rightarrow 0} (x_1^2 + x_1 \Delta x)}{\lim_{\Delta x \rightarrow 0} (x_1 \cdot \frac{\Delta y}{\Delta x}) - \lim_{\Delta x \rightarrow 0} y_1} \\ 735 | &= -\frac{x_1}{y_1} \cdot \frac{\lim_{\Delta x \rightarrow 0} x_1^2 - \lim_{\Delta x \rightarrow 0} x_1 \Delta x}{x_1 \cdot \lim_{\Delta x \rightarrow 0} (\frac{\Delta y}{\Delta x}) - y_1} \\ 736 | &= -\frac{x_1}{y_1} \cdot \frac{x_1^2 - x_1 \lim_{\Delta x \rightarrow 0} \Delta x}{x_1 \cdot (-\frac{x_1}{y_1}) - y_1} \\ 737 | &= -\frac{x_1}{y_1} \cdot \frac{x_1^2 - x_1 \cdot 0}{\frac{-x_1^2 - y_1^2}{y_1}} \\ 738 | &= -\frac{x_1}{y_1} \cdot \frac{x_1^2 y_1}{-(x_1^2 + y_1^2)} \\ 739 | &= \frac{x_1^3}{x_1^2 + y_1^2} \\ 740 | &= x_1^3 \tag*{(Ladder $L_1$ has length $x_1^2 + y_1^2 = 1$)} 741 | \end{align*} 742 | 743 | In Q6(a) check (ii) we deduced that swapping the $x's$ and $y's$ in the formulae for $x_3$ and $y_3$ will not change the equations. Hence by symmetry, 744 | 745 | \begin{align*} 746 | Y = \lim_{\Delta x \rightarrow 0} y_3 = \lim_{\Delta x \rightarrow 0} y_1^3 = y_1^3 747 | \end{align*} 748 | 749 | \begin{tcolorbox} 750 | \textbf{Q6. Hypocycloid, again} \\ 751 | d) Show that $X^{2/3} + Y^{2/3} = 1$. (The limit point $(X,Y)$ that you found in part (c) is expressed as a function of $x_1$ and $y_1$. This is the unique point of the ladder $L_1$ that is also part of the boundary curve of the region swept out by the family of ladders.) 752 | \end{tcolorbox} 753 | 754 | By part (c), we have $X = x_1^3, Y = y_1^3$. Then $X^{2/3} + Y^{2/3} = (x_1^3)^{2/3} + (y_1^3)^{2/3} = x_1^2 + y_1^2 = 1$ 755 | 756 | \end{document} 757 | -------------------------------------------------------------------------------- /ps4.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps4.pdf -------------------------------------------------------------------------------- /ps4.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | \graphicspath{{images/ps4/}} 11 | 12 | \usepackage{hyperref} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 4 Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \textbf{Q1a)} Use the mean value property to show that if $f(0) = 0$ and $f'(x) \geq 0$, then $f(x) \geq 0$ for all $x \geq 0$. 30 | \end{tcolorbox} 31 | 32 | For any $x > 0$, $f(x) = f(0) + f'(c)(x - 0)$, for some $0 < c < x$ 33 | 34 | Since $f'(x) \geq 0$ for all $x$, then for all $x > 0$, $f(x) \geq f(0)$. 35 | 36 | Since $f(x) \geq f(0) \geq 0$ for all $x > 0$, then $f(x) \geq 0$ for all $x > 0$. 37 | 38 | Since $f(0) = 0$ and $f(x) \geq 0$ for all $x > 0$, then $f(x) \geq 0$ for all $x \geq 0$. 39 | 40 | 41 | \begin{tcolorbox} 42 | \textbf{Q1b)} Deduce from part (a) that $ln(1 + x) \leq x$ for $x \geq 0$. Hint: Use $f(x) = x - ln(1 + x)$. 43 | \end{tcolorbox} 44 | 45 | Let $f(x) = x - ln(1 + x)$ 46 | 47 | Then $f(0) = 0 - ln(1 + 0) = 0 - 0 = 0$ 48 | 49 | $f'(x) = 1 - \frac{1}{1 + x}$ 50 | 51 | For all $x \geq 0$, $1 + x \geq 1$ and $\frac{1}{1 + x} \leq 1$ and $1 - \frac{1}{1 + x} \geq 0$ 52 | 53 | Since $f'(x) = 1 - \frac{1}{1 + x} \geq 0$, therefore $f'(x) \geq 0$ for $x \geq 0$. 54 | 55 | By 1(a), $f(x) = x - ln(1 + x) \geq 0$ for $x \geq 0$ 56 | 57 | Then $x \geq ln(1 + x)$ for $x \geq 0$ 58 | 59 | 60 | \begin{tcolorbox} 61 | \textbf{Q1c)} Use the same method as in (b) to show $ln(1 + x) \geq x - x^2 / 2$ and $ln(1 + x) \leq x - x^2 / 2 + x^3 / 3$ for $x \geq 0$. 62 | \end{tcolorbox} 63 | 64 | Let $f_1(x) = ln(1 + x) - x + \frac{x^2}{2}$ 65 | 66 | $f_1(0) = ln(1 + 0) - 0 + \frac{0^2}{2} = 0$ 67 | 68 | $f_1'(x) = \frac{1}{1 + x} - 1 + x = \frac{1 - (1 + x) + x(1 + x)}{1 + x} = \frac{1 - 1 - x + x + x^2}{1 + x} = \frac{x^2}{1 + x}$ 69 | 70 | Since $1 + x \geq 1$ for all $x \geq 0$ and $x^2 \geq 0$ for all $x \geq 0$, then $f_1'(x) = \frac{x^2}{1 + x} \geq 0$ for all $x \geq 0$. 71 | 72 | By 1(a), $f_1(x) \geq 0$ for all $x \geq 0$, which is equivalent to $ln(1 + x) - x - \frac{x^2}{2} \geq 0$ for all $x \geq 0$. Hence $ln(1 + x) \geq x - \frac{x^2}{2}$ for all $x \geq 0$. 73 | \\ 74 | 75 | Let $f_2(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ln(1 + x)$ 76 | 77 | $f_2(0) = 0 - \frac{0^2}{2} + \frac{0^3}{3} - ln(1 + 0) = 0$ 78 | 79 | $f_2'(x) = 1 - x + x^2 - \frac{1}{1 + x} = \frac{1(1 + x) - x(1 + x) + x^2 (1 + x) - 1}{1 + x} = \frac{1 + x - x - x^2 + x^2 + x^3}{1 + x} = \frac{x^3}{1 + x}$ 80 | 81 | Since $1 + x \geq 1$ for all $x \geq 0$ and $x^3 \geq 0$ for all $x \geq 0$, then $f_2'(x) = \frac{x^3}{1 + x} \geq 0$ for all $x \geq 0$. 82 | 83 | By 1(a), $f_2(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ln(1 + x) \geq 0$ for all $x \geq 0$. Hence $x - \frac{x^2}{2} + \frac{x^3}{3} \geq ln(1 + x)$ for all $x \geq 0$. 84 | 85 | 86 | \begin{tcolorbox} 87 | \textbf{Q1d)} Find the pattern in (b) and (c) and make a general conjecture. 88 | \end{tcolorbox} 89 | 90 | General conjecture: 91 | 92 | $ln(1 + x) \leq x - \frac{x^2}{2} + \frac{x^3}{3} + ... + \frac{x^{2n + 1}}{2n + 1} = \sum\limits_{k=1}^{2n + 1} (-1)^{k+1} \frac{x^k}{k}$ for all $x \geq 0, n \geq 0$ 93 | 94 | $ln(1 + x) \geq x - \frac{x^2}{2} + \frac{x^3}{3} + ... + \frac{x^{2n}}{2n} = \sum\limits_{k=1}^{2n} (-1)^{k+1} \frac{x^k}{k}$ for all $x \geq 0, n \geq 1$ 95 | 96 | Let $f_1(x) = (\sum\limits_{k=1}^{2n + 1} (-1)^{k+1} \frac{x^k}{k}) - ln(x + 1)$ for any $n \geq 1$. 97 | 98 | $f_1(0) = (\sum\limits_{k=1}^{2n + 1} (-1)^{k+1} \frac{0^k}{k}) - ln(0 + 1) = 0$ 99 | 100 | \begin{align*} 101 | f_1'(x) &= (\sum\limits_{k=1}^{2n + 1} (-1)^{k+1} x^{k-1}) - \frac{1}{x + 1} \\ 102 | &= \frac{(x+1)(\sum\limits_{k=1}^{2n + 1} (-1)^{k+1} x^{k-1}) - 1}{x + 1} \\ 103 | &= \frac{(\sum\limits_{k=1}^{2n + 1} (-1)^{k+1} (x^k + x^{k - 1})) - 1}{x + 1} \\ 104 | &= \frac{(x^1 + x^0) - (x^2 + x^1) + ... + (-1)^{2n + 1 + 1}(x^{2n + 1} + x^{2n}) - 1}{x + 1} \\ 105 | &= \frac{x^0 + x^{2n + 1} - 1}{x + 1} \\ 106 | &= \frac{x^{2n + 1}}{x + 1} 107 | \end{align*} 108 | 109 | Since $x + 1 \geq 1$ for $x \geq 0$ and $x^{2n + 1} \geq 0$ for $n \geq 0, x \geq 0$, then $f_1'(x) = \frac{x^{2n + 1}}{x + 1} \geq 0$ for all $x \geq 0$. 110 | 111 | By 1(a), $f_1(x) = (\sum\limits_{k=1}^{2n+1} (-1)^{k+1} \frac{x^k}{k}) - ln(x + 1) \geq 0$ for all $x \geq 0, n \geq 0$. Hence $ln(x + 1) \leq \sum\limits_{k=1}^{2n+1} (-1)^{k+1} \frac{x^k}{k}$ for all $x \geq 0, n \geq 0$. 112 | \\ 113 | \\ 114 | 115 | Let $f_2(x) = ln(x + 1) - \sum\limits_{k=1}^{2n} (-1)^{k+1} \frac{x^k}{k}$ for any $n \geq 0$. 116 | 117 | $f_2(0) = ln(0 + 1) - \sum\limits_{k=1}^{2n} (-1)^{k+1} \frac{0^k}{k} = 0$ 118 | 119 | \begin{align*} 120 | f_2'(x) &= \frac{1}{x + 1} - \sum\limits_{k=1}^{2n} (-1)^{k+1} x^{k-1} \\ 121 | &= \frac{1 - (x + 1)\sum\limits_{k=1}^{2n} x^{k - 1}}{x + 1} \\ 122 | &= \frac{1 - \sum\limits_{k=1}^{2n}(x^k + x^{k - 1})}{x + 1} \\ 123 | &= \frac{1 - ((x^1 + x^0) - (x^2 + x^1) + ... + (-1)^{2n + 1}(x^{2n} + x^{2n - 1}))}{x + 1} \\ 124 | &= \frac{1 - (x^0 - x^{2n})}{x + 1} \\ 125 | &= \frac{x^{2n}}{x + 1} 126 | \end{align*} 127 | 128 | For $x \geq 0$, $x + 1 \geq 1$ and $x^{2n} \geq 0$ for $x \geq 0, n \geq 1$. Hence $f_2'(x) = \frac{x^{2n}}{x + 1} \geq 0$ for $n \geq 1, x \geq 0$. 129 | 130 | By 1(a), $f_2(x) = ln(x + 1) - \sum\limits_{k=1}^{2n} (-1)^{k+1} x^{k-1} \geq 0$ for all $n \geq 1, x \geq 0$. 131 | 132 | Then $ln(x + 1) \geq \sum\limits_{k=1}^{2n} (-1)^{k+1} x^{k-1}$ for all $n \geq 1, x \geq 0$. 133 | 134 | 135 | \begin{tcolorbox} 136 | \textbf{Q1e)} Show that $ln(1 + x) \leq x$ for $-1 < x \leq 0$. (Use the change of variable $u = -x$.) 137 | \end{tcolorbox} 138 | 139 | Let $u = -x$. Then $x = -u$. 140 | 141 | We want to show that $ln(1 - u) \leq -u$ for $0 \leq u < 1$. 142 | 143 | Let $f(u) = -u - ln(1 - u)$ 144 | 145 | $f(0) = -0 - ln(1 - 0) = 0$ 146 | 147 | $f'(u) = -1 - \frac{-1}{1 - u} = -1 + \frac{1}{1-u} = \frac{-1(1-u) + 1}{1 - u} = \frac{u - 1 + 1}{1 - u} = \frac{u}{1 - u}$ 148 | 149 | Since $0 \leq u < 1$, then $1 - u > 0$. Hence $f'(u) = \frac{u}{1 - u} \geq 0$ for $0 \leq u < 1$ 150 | 151 | By 1(a), $f(u) = -u - ln(1 - u) \geq 0$ for $0 \leq u < 1$. 152 | 153 | Then $f(x) = -(-x) - ln(1 - (-x)) = x - ln(1 + x) \geq 0$ for $0 \leq -x < 1$ or $-1 < x \leq 0$. 154 | 155 | Hence $ln(1 + x) \leq x$ for $-1 < x \leq 0$ 156 | 157 | 158 | \begin{tcolorbox} 159 | \textbf{Q2a)} Do 5.3/68 160 | \end{tcolorbox} 161 | 162 | I do not have the textbook. Skipped. 163 | 164 | 165 | \begin{tcolorbox} 166 | \textbf{Q2b)} Show that both of the following integrals are correct, and explain. 167 | 168 | \begin{align*} 169 | \int tan x\ sec^2 x\ dx = (1/2) tan^2 x; \int tan x\ sec^2 x\ dx = (1/2) sec^2 x 170 | \end{align*} 171 | \end{tcolorbox} 172 | 173 | Let $u = tan\ x$. We have 174 | 175 | \begin{align*} 176 | \int tan x\ sec^2 dx &= \int (sec^2 x)\ tan x\ dx \\ 177 | &= \int u'\ u\ du \\ 178 | &= \frac{1}{2} u^2 + c \\ 179 | &= \frac{1}{2} tan^2 x + c 180 | \end{align*} 181 | 182 | Now we want to prove that $\int tan x\ sec^2 x\ dx = (1/2) sec^2 x$. Let $u = sec\ x$. Then $\int u' \ u\ du = \frac{1}{2}u^2 + c = \frac{1}{2} sec^2 x + c$ 183 | 184 | 185 | \begin{tcolorbox} 186 | \textbf{Q3)} (Lec 16, 6 pts: 3 + 3)\\ 187 | a) Do 8.6/5 (answer in back of book)\\ 188 | b) Do 8.6/6 (optional?) 189 | \end{tcolorbox} 190 | 191 | I do not have the textbook. Skipped. 192 | 193 | 194 | \begin{tcolorbox} 195 | \textbf{Q4)} (Lec 16, 7 pts: 2 + 3 + 2) Do 3F-5abc\\ 196 | \\ 197 | 3F-5 Air pressure satisfies the differential equation $dp/dh = -(.13)p$, where $h$ is the altitude from sea level measured in kilometres.\\ 198 | a) At sea level the pressure is $1kg/cm^2$. Solve the equation and find the pressure at the top of Mt. Everest (10 km). 199 | \end{tcolorbox} 200 | 201 | \begin{align*} 202 | \frac{dp}{dh} &= -(.13)p\\ 203 | \frac{1}{p} dp &= -.13 dh\\ 204 | \int \frac{1}{p} dp &= \int -.13 dh\\ 205 | ln\ p &= -.13h + c\\ 206 | p &= e^{-.13h + c}\\ 207 | p &= e^{-.13h} \cdot e^c\\ 208 | p &= A e^{-.13h} 209 | \end{align*} 210 | 211 | where $A = e^c$. When $h = 0, p = 1$ (sea level pressure is $1$). Then $p = 1 = A e^{-.13(0)} = A$. Hence $p = A e^{-.13h} = 1 \cdot e^{-.13h} = e^{-.13h}$ 212 | 213 | At the top of Mt Everest, $p = e^{-.13(10)} \approx 0.2725\ kg/cm^2$ 214 | 215 | \begin{tcolorbox} 216 | \textbf{Q4)} (Lec 16, 7 pts: 2 + 3 + 2) Do 3F-5abc\\ 217 | \\ 218 | 3F-5 b) Find the difference in pressure between the top and bottom of the Green building. (Pretend it's 100 meters tall starting at sea level.) Compute the numerical value using a calculator. Then use instead the linear approximation to $e^x$ near $x = 0$ to estimate the percentage drop in pressure from the bottom to the top of the Green Building. 219 | \end{tcolorbox} 220 | 221 | Difference = $e^{-.13(0.1)} - e^{-13(0)} = e^{-.013} - 1 \approx -0.012915865\ kg/cm^2$ 222 | 223 | Linear approximation to $e^x$ near $x = 0$ is $1 + x$. But because we are dealing with $e^{-.13h}$, for $h \approx 0$, $e^{-.13h} \approx e^{-.13(0)} + -.13e^{-.13(0)}h = 1 - .13h$ 224 | 225 | Pressure at top - Pressure at bottom $\approx 1 - .13(0.1) - 1 \approx -0.013\ kg/cm^2$ 226 | \begin{tcolorbox} 227 | \textbf{Q4)} (Lec 16, 7 pts: 2 + 3 + 2) Do 3F-5abc\\ 228 | \\ 229 | 3F-5 c) Use the linear approximation $\Delta p \approx p'(0) \Delta h$ and compute $p'(0)$ directly from the differential equation to find the drop in pressure from the bottom to top of the Green Building. Notice that this gives an answer without even knowing the solution to the differential equation. Compare with the approximation in part (b). What does the linear approximation $p'(0)\Delta h$ give for the pressure at the top of Mt. Everest? 230 | \end{tcolorbox} 231 | 232 | \begin{align*} 233 | \Delta p &\approx p'(0)\Delta h\\ 234 | \frac{\Delta p}{\Delta h} &\approx p'(0)\\ 235 | \lim_{\Delta h \rightarrow 0} \frac{dp}{dh} &= \frac{dp}{dh}\\ 236 | \frac{\Delta p}{\Delta h} &\approx \frac{dp}{dh} = -.13p 237 | \end{align*} 238 | 239 | We will use $p'(0) \approx \frac{\Delta p}{\Delta h} \approx -.13p$. When $h = 0, p = 1$. Hence $1 = p'(0) \approx -.13p = -.13(1) = -.13$, which implies that every kilometre increase in altitude causes pressure to decrease by $-.13\ kg/cm^2$. 240 | 241 | Drop in pressure from bottom to top of green building = $-(-.13(0.1)) = 0.013\ kg/cm^2$ which was exactly the same as what we got using the linear approximation in part (b). 242 | 243 | For pressure at top of Mt Everest, we have $\Delta h = 10$. Hence 244 | 245 | \begin{align*} 246 | p(0) + p'(0)\Delta h &= 1 + -.13(10)\\ 247 | &= 1 - 1.3\\ 248 | &= -0.3\ kg/cm^2 249 | \end{align*} 250 | 251 | which is totally incorrect because $\Delta h = 10$ is much bigger than $0$ and causes the approximation to be wildly inaccurate. 252 | 253 | 254 | \begin{tcolorbox} 255 | \textbf{Q5)} Calculate $\int_{0}^{1} e^x dx$ using lower Riemann sums. (You will need to sum a geometric series to get a usable formula for the Riemann sum. To take the limit of Riemann sums, you will need to evaluate $\lim\limits_{n \to \infty} n(e^{1/n} - 1)$, which can be done using the standard linear approximation to the exponential function.) 256 | \end{tcolorbox} 257 | 258 | \begin{align*} 259 | \int_{0}^{1} e^x dx &= \lim\limits_{n \to \infty} \sum_{k=0}^{n-1} e^{\frac{k}{n}} \frac{1}{n}\\ 260 | &= \lim\limits_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} e^{\frac{k}{n}}\\ 261 | &= \lim\limits_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} (e^{\frac{1}{n}})^k\\ 262 | &= \lim\limits_{n \to \infty} \frac{1}{n} \cdot \frac{1 - (e^{\frac{1}{n}})^n}{1 - e^{\frac{1}{n}}}\\ 263 | &= \lim\limits_{n \to \infty} \frac{1}{n} \cdot \frac{1 - e}{1 - e^\frac{1}{n}}\\ 264 | &= \lim\limits_{n \to \infty} \frac{1 - e}{n(1 - e^{\frac{1}{n}})} 265 | \end{align*} 266 | 267 | The linear approximation to $e^x$ near $x = 0$ is $e^0 + e^0 x = 1 + x$. Since $n \rightarrow \infty, \frac{1}{n} \rightarrow 0$ and $e^{\frac{1}{n}} \approx 1 + \frac{1}{n}$. Then 268 | 269 | \begin{align*} 270 | \int_{0}^{1} e^x dx &= \lim\limits_{n \to \infty} \frac{1 - e}{n(1 - e^{\frac{1}{n}})}\\ 271 | &\approx \lim\limits_{n \to \infty} \frac{1 - e}{n(1 - (1 + \frac{1}{n}))}\\ 272 | &= \lim\limits_{n \to \infty} \frac{1 - e}{n(-\frac{1}{n})}\\ 273 | &= \lim\limits_{n \to \infty} \frac{1 - e}{-1}\\ 274 | &= \lim\limits_{n \to \infty} e - 1\\ 275 | &= e - 1 276 | \end{align*} 277 | 278 | 279 | \begin{tcolorbox} 280 | \textbf{Q6) More about the hypocycloid} We use differential equations to find the curve with the property that the potrion of its tangent line in the first quadrant has fixed length.\\ 281 | \\ 282 | a) Suppose that a line through the point $(x_0, y_0)$ has slope $m_0$ and that the point is in the first quadrant. Let $L$ denote the length of the portion of the line in the first quadrant. Calculate $L^2$ in terms of $x_0, y_0$ and $m_0$. (Do not expand or simplify.) 283 | \end{tcolorbox} 284 | 285 | Equation of tangent line is given by $y - y_0 = m_0(x - x_0)$. When $x = 0$ 286 | 287 | \begin{align*} 288 | y - y_0 &= m_0(0 - x_0)\\ 289 | y &= -m_0 x_0 + y_0 290 | \end{align*} 291 | 292 | When $y = 0$, 293 | 294 | \begin{align*} 295 | 0 - y_0 &= m_0(x - x_0)\\ 296 | -\frac{y_0}{m_0} &= x - x_0\\ 297 | x &= x_0 - \frac{y_0}{m_0} 298 | \end{align*} 299 | 300 | Substituting the above equations for $x$ and $y$ into $L^2 = x^2 + y^2$, we have 301 | 302 | \begin{align*} 303 | L^2 &= x^2 + y^2\\ 304 | &= (x_0 - \frac{y_0}{m_0})^2 + (-m_0 x_0 + y_0)^2 305 | \end{align*} 306 | 307 | 308 | \begin{tcolorbox} 309 | \textbf{Q6b)} Suppose that $y = f(x)$ is a graph on $0 \leq x \leq L$ satisfying $f(0) = L$ and $f(L) = 0$ and such that the portion of each tangent line to the graph in the first quadrant has the same length $L$. Find the differential equation that $f$ satisfies. Express it in terms of $L, x, y$ and $y' = dy/dx$. (Hints: This requires only thought, not computation. Note that $y = f(x), y' = f'(x)$. Don't take square roots, the expression using $L^2$ is much easier to use. Don't expand or simplify; that would make things harder in the next step.) 310 | \end{tcolorbox} 311 | 312 | \begin{center} 313 | \includegraphics[]{q6b.jpg} 314 | \end{center} 315 | 316 | Apologies for the poor quality image. 317 | 318 | Let the x-intercept of the tangent line be point $d$. Given a point $(x, y)$ on the tangent line, we have 319 | 320 | \begin{align*} 321 | \frac{dy}{dx} &= \frac{0 - y}{d - x}\\ 322 | d - x &= \frac{-y}{\frac{dy}{dx}}\\ 323 | d &= x - \frac{y}{\frac{dy}{dx}} 324 | \end{align*} 325 | 326 | We also see from the diagram that $(\Delta L)^2 = y^2 + (d - x)^2 = y^2 + (x - \frac{y}{\frac{dy}{dx}} - x)^2 = y^2 + (-\frac{y}{\frac{dy}{dx}})^2 = y^2 + (\frac{y}{\frac{dy}{dx}})^2$ 327 | 328 | The y-intercept of the graph is $y - x \cdot \frac{dy}{dx}$. This is also the height of the triangle in the diagram. Using similar triangles, we have 329 | 330 | \begin{align*} 331 | \frac{y}{\Delta L} &= \frac{y - x \cdot \frac{dy}{dx}}{L}\\ 332 | (\frac{y}{\Delta L})^2 &= (\frac{y - x \cdot \frac{dy}{dx}}{L})^2 333 | \end{align*} 334 | 335 | Substituting $(\Delta L)^2 = y^2 + (\frac{y}{\frac{dy}{dx}})^2$ which we got above: 336 | 337 | \begin{align*} 338 | (\frac{y}{\Delta L})^2 &= (\frac{y - x \cdot \frac{dy}{dx}}{L})^2\\ 339 | \frac{y^2}{y^2 + \frac{y^2}{(\frac{dy}{dx})^2}} &= (\frac{y - x \cdot \frac{dy}{dx}}{L})^2\\ 340 | L^2 y^2 &= (y - x \cdot \frac{dy}{dx})^2 (y^2 + \frac{y^2}{(\frac{dy}{dx})^2}) 341 | \end{align*} 342 | 343 | 344 | \begin{tcolorbox} 345 | \textbf{Q6c)} Differentiate the equation in part (b) with respect to $x$. Simplify and write in the form\\ 346 | \begin{center} 347 | (something)$(xy' - y)y'' = 0$ 348 | \end{center} 349 | 350 | \ \\ 351 | (This starts out looking horrendous, but simplifies considerably.) 352 | \end{tcolorbox} 353 | 354 | \begin{align*} 355 | L^2 y^2 &= (y - x \cdot \frac{dy}{dx})^2 (y^2 + \frac{y^2}{(\frac{dy}{dx})^2})\\ 356 | L^2 y^2 &= (y^2 - 2xy \cdot \frac{dy}{dx} + x^2 \cdot (\frac{dy}{dx})^2) (y^2 + \frac{y^2}{(\frac{dy}{dx})^2})\\ 357 | &= y^4 + \frac{y^4}{(\frac{dy}{dx})^2} - 2xy^3 \cdot \frac{dy}{dx} - \frac{2xy^3}{\frac{dy}{dx}} + x^2y^2 \cdot (\frac{dy}{dx})^2 + x^2 y^2 358 | \end{align*} 359 | 360 | Differentiate implicitly with respect to $x$, 361 | 362 | \begin{align*} 363 | L^2 \cdot 2y \cdot y' &= 4y^3 \cdot y' + \frac{4y^3(y')(y')^2 - y^4 \cdot 2y'(y'')}{(y')^4} - 2y^3y' - 6xy^2(y')^2 - 2xy^3y'' - \frac{2y^3y' + 6xy^2(y')^2 - 2xy^3y''}{(y')^2}\\ 364 | &\ \ \ + 2xy^2(y')^2 + 2x^2y(y')^3 + 2x^2y^2y'y'' + 2xy^2 + 2x^2yy'\\ 365 | &= 2y^3y' + \frac{4y^3(y')^3 - 2y^4y'y''}{(y')^4} - 4xy^2(y')^2 - 2xy^3y'' - \frac{2y^3y' + 6xy^2(y')^2 - 2xy^3y''}{(y')^2} + 2x^2y(y')^3\\ 366 | &\ \ \ + 2x^2y^2y'y'' + 2xy^2 + 2x^2yy'\\ 367 | &= \frac{1}{(y')^4}(2y^3(y')^5 + 4y^3(y')^3 - 2y^4y'y'' - 4xy^2(y')^6 - 2xy^3(y')^4y'' - 2y^3(y')^3 - 6xy^2(y')^4 + 2xy^3(y')^2y''\\ 368 | &\ \ \ + 2x^2y(y')^7 + 2x^2y^2(y')^5y'' + 2xy^2(y')^4 + 2x^2y(y')^5) 369 | \end{align*} 370 | \begin{align*} 371 | L^2 \cdot 2y \cdot y' \cdot (y')^4 &= 2y^3(y')^5 + 4y^3(y')^3 - 2y^4y'y'' - 4xy^2(y')^6 - 2xy^3(y')^4y'' - 2y^3(y')^3 - 6xy^2(y')^4 + 2xy^3(y')^2y''\\ 372 | &\ \ \ + 2x^2y(y')^7 + 2x^2y^2(y')^5y'' + 2xy^2(y')^4 + 2x^2y(y')^5 373 | \end{align*} 374 | 375 | In part (a), we got the result $L^2 = (x_0 - \frac{y_0}{m_0})^2 + (-m_0 x_0 + y_0)^2$ for any point $(x_0, y_0)$ on the tangent. Rewriting $m_0$ as $y'$, this becomes $L^2 = (x_0 - \frac{y_0}{y'})^2 + (-y' x_0 + y_0)^2$. Substituting the point $x_0 = x, y_0 = y$ for some point $(x, y)$ on the tangent line, this becomes $L^2 = (x - \frac{y}{y'})^2 + (-y' x + y)^2 = (y - xy')^2 + (x - \frac{y}{y'})^2$. Substituting that into $L^2 \cdot 2y \cdot y' \cdot (y')^4$, we have 376 | 377 | \begin{align*} 378 | L^2 \cdot 2y \cdot y' \cdot (y')^4 &= ((y - xy')^2 + (x - \frac{y}{y'})^2) \cdot 2y \cdot y' \cdot (y')^4\\ 379 | &= 2(y^2 - 2xyy' + x^2(y')^2 + x^2 - \frac{2xy}{y'} + \frac{y^2}{(y')^2}) \cdot y \cdot (y')^5\\ 380 | &= 2y^3(y')^5 - 4xy^2(y')^6 + 2x^2y(y')^7 + 2x^2y(y')^5 - 4xy^2(y')^4 + 2y^3(y')^3 381 | \end{align*} 382 | 383 | Then 384 | \begin{align*} 385 | L^2 \cdot 2y \cdot y' \cdot (y')^4 &= 2y^3(y')^5 + 4y^3(y')^3 - 2y^4y'y'' - 4xy^2(y')^6 - 2xy^3(y')^4y'' - 2y^3(y')^3 - 6xy^2(y')^4 + 2xy^3(y')^2y''\\ 386 | &\ \ \ + 2x^2y(y')^7 + 2x^2y^2(y')^5y'' + 2xy^2(y')^4 + 2x^2y(y')^5 387 | \end{align*} 388 | 389 | becomes 390 | \begin{align*} 391 | &2y^3(y')^5 - 4xy^2(y')^6 + 2x^2y(y')^7 + 2x^2y(y')^5 - 4xy^2(y')^4 + 2y^3(y')^3 \\ 392 | &= 2y^3(y')^5 + 4y^3(y')^3 - 2y^4y'y'' - 4xy^2(y')^6 - 2xy^3(y')^4y'' - 2y^3(y')^3 - 6xy^2(y')^4 + 2xy^3(y')^2y''\\ 393 | &\ \ \ + 2x^2y(y')^7 + 2x^2y^2(y')^5y'' + 2xy^2(y')^4 + 2x^2y(y')^5 394 | \end{align*} 395 | 396 | which simplifies to 397 | 398 | \begin{align*} 399 | 0 = -2y^4y'y'' - 2xy^3(y')^4y'' + 2xy^3(y')^2y'' + 2x^2y^2(y')^5y''\\ 400 | (y^3y' + xy^2(y')^4)(xy' - y)y'' = 0 401 | \end{align*} 402 | 403 | 404 | \begin{tcolorbox} 405 | \textbf{Q6d)} Show that one solution to the equation in part (c) is $x^{2/3} + y^{2/3} = L^{2/3}$. What about two other possibilities, namely, those solving $y'' = 0$ and $xy' - y = 0$? 406 | \end{tcolorbox} 407 | 408 | In part (c), we derived $(y^3y' + xy^2(y')^4)(xy' - y)y'' = 0$. Let's solve for $y^3y' + xy^2(y')^4 = 0$ 409 | 410 | \begin{align*} 411 | y^3y' + xy^2(y')^4 &= 0\\ 412 | y^3y' &= -xy^2(y')^4\\ 413 | y^3 &= -xy^2(y')^3\\ 414 | -y^3 &= xy^2(y')^3\\ 415 | -y &= x^{1/3}y^{2/3}y'\\ 416 | -y &= x^{1/3}y^{2/3}\frac{dy}{dx}\\ 417 | x^{-1/3} &= -y^{-1/3}\frac{dy}{dx}\\ 418 | x^{-1/3} dx &= -y^{-1/3} dy\\ 419 | \int x^{-1/3} dx &= \int -y^{-1/3} dy\\ 420 | \frac{3}{2} x^{2/3} &= -\frac{3}{2} y^{2/3} + c\\ 421 | \frac{3}{2} x^{2/3} + \frac{3}{2} y^{2/3} &= c\\ 422 | x^{2/3} + y^{2/3} &= \frac{2}{3}c 423 | \end{align*} 424 | 425 | Now we have to prove that $c = \frac{3}{2}L^{2/3}$ for $x^{2/3} + y^{2/3} = \frac{2}{3}c = \frac{2}{3} \cdot \frac{3}{2}L^{2/3} = L^{2/3}$. 426 | 427 | In solving for $y^3y' + xy^2(y')^4 = 0$, we had a step $x^{-1/3} = -y^{-1/3}\frac{dy}{dx}$. From that, we derive $\frac{dy}{dx} = \frac{x^{-1/3}}{-y^{-1/3}} = -x^{-1/3}y^{1/3}$. Or equivalently, $y' = -x^{-1/3}y^{1/3}$. 428 | 429 | In part (c), we also showed that $L^2 = (y - xy')^2 + (x - \frac{y}{y'})^2$. We will expand that and then substitute $y' = -x^{-1/3}y^{1/3}$: 430 | 431 | \begin{align*} 432 | L^2 &= (y - xy')^2 + (x - \frac{y}{y'})^2\\ 433 | &= x^2 - 2xy\frac{1}{y'} + \frac{y^2}{(y')^2} + y^2 - 2xyy' + x^2(y')^2\\ 434 | &= x^2 - 2xy(-x^{1/3}y^{-1/3}) + y^2(x^{2/3}y^{-2/3}) + y^2 - 2xy(-x^{-1/3}y^{1/3}) + x^2(x^{-2/3}y^{2/3})\\ 435 | &= x^2 + 2x^{4/3}y^{2/3} + x^{2/3}y^{4/3} + y^2 + 2x^{2/3}y^{4/3} + x^{4/3}y^{2/3}\\ 436 | &= x^2 + y^2 + 3x^{4/3}y^{2/3} + 3x^{2/3}y^{4/3}\\ 437 | &= (x^{2/3} + y^{2/3})(x^{4/3} + y^{4/3}) + 2x^{4/3}y^{2/3} + 2x^{2/3}y^{4/3}\\ 438 | &= (x^{2/3} + y^{2/3})(x^{4/3} + y^{4/3}) + 2(x^{2/3}y^{2/3})(x^{2/3} + y^{2/3})\\ 439 | &= (x^{2/3} + y^{2/3})(x^{4/3} + y^{4/3} + 2x^{2/3}y^{2/3})\\ 440 | &= (x^{2/3} + y^{2/3})(x^{2/3} + y^{2/3})^2\\ 441 | &= (x^{2/3} + y^{2/3})^3 442 | \end{align*} 443 | 444 | Hence $x^{2/3} + y^{2/3} = L^{2/3} = \frac{2}{3}c$ and $c = \frac{3}{2} L^{2/3}$ as desired. And we also get $x^{2/3} + y^{2/3} = L^{2/3}$\\ 445 | 446 | Another possibility in the equation $(y^3y' + xy^2(y')^4)(xy' - y)y'' = 0$ is that $y'' = 0$. 447 | 448 | \begin{align*} 449 | y'' &= 0\\ 450 | \int y'' dy &= \int 0 dx\\ 451 | y' &= c\\ 452 | \int y' dy &= \int c dx\\ 453 | y &= cx + d 454 | \end{align*} 455 | 456 | for some constants $c, d$. 457 | 458 | For the problem to be meaningful, the length tangent line, $L$, must be positive. Hence some part of the tangent line has to lie in the first quadrant. We know that for a line of the form $y = cx + d$, the tangent line to any point on the line must be the line itself. Hence the part of the tangent line that lies in the is precisely the part of the $y = cx + d$ itself that lies in the first quadrant, and it must be of length $L$. 459 | 460 | Let the y-intercept of the line be $d$. We assume $L$ to be finite, so the line $y = cx + d$ must be decreasing and hence its y-intercept $d > 0$ and its gradient $c < 0$. Its x-intercept can be obtained by equating $y = 0$ and solving: 461 | 462 | \begin{align*} 463 | 0 &= cx + d\\ 464 | x &= -\frac{d}{c} 465 | \end{align*} 466 | 467 | We must have that $L^2 = d^2 + (-\frac{d}{c})^2$. Hence 468 | 469 | \begin{align*} 470 | L^2 &= d^2 + (-\frac{d}{c})^2\\ 471 | L^2 &= d^2 + + \frac{d^2}{c^2}\\ 472 | d^2 - L^2 &= -\frac{d^2}{c^2}\\ 473 | c^2 &= -\frac{d^2}{d^2 - L^2}\\ 474 | c^2 &= \frac{d^2}{L^2 - d^2}\\ 475 | c^2 &= \pm \sqrt{\frac{d^2}{L^2 - d^2}} 476 | \end{align*} 477 | 478 | Since $c < 0$, we must have $c^2 = -\sqrt{\frac{d^2}{L^2 - d^2}}$. Then $y = -\sqrt{\frac{d^2}{L^2 - d^2}}x + d$ gives the equation for all such lines for a fixed $L$ and any $d > 0$.\\ 479 | 480 | Another possibility in the equation $(y^3y' + xy^2(y')^4)(xy' - y)y'' = 0$ is that $xy' - y = 0$. 481 | 482 | \begin{align*} 483 | xy' - y &= 0\\ 484 | x\frac{dy}{dx} - y &= 0\\ 485 | x\frac{dy}{dx} &= y\\ 486 | \frac{1}{y}dy &= \frac{1}{x}dx\\ 487 | \int \frac{1}{y}dy &= \int \frac{1}{x}dx\\ 488 | ln|y| &= ln|x| + c 489 | \end{align*} 490 | 491 | for some constant $c$. Since $x, y > 0$ (because we are only concerned about the part of the graph in the first quadrant), $ln|y| = ln|x| + c = ln(y) + ln(x) + c$ and we have $y = e^{ln(x) + c} = e^c \cdot x$ for some constant $c$. 492 | 493 | We see that lines of the form $y = e^c \cdot x$ pass through the origin. Furthermore, for any $c$, we have $e^c > 0$. Hence $y = e^c \cdot x$ is an increasing function that is linear in $x$. Since the tangent line to any point on the line is the line itself and it is an increasing function that starts from $y = 0$ at $x = 0$, its length in the first quadrant is unbounded. Hence this is not a meaningful case. 494 | 495 | \end{document} 496 | -------------------------------------------------------------------------------- /ps5.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps5.pdf -------------------------------------------------------------------------------- /ps5.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{hyperref} 12 | \graphicspath{{images/ps5/}} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 5 Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \textbf{Q1a)} Suppose that at the beginning of day 0, some time last summer, the temperature in Boston was $y(0) = 65^{\circ}$ Fahrenheit and that over a 50-day period, the temperature increased according to the rule $y'(t) = y(t) / 100$, with time $t$ measured in days. Find the formula for $y$, and draw a graph of temperature on days 3 and 4, $3 <= t <= 5$, and label with the correct day and shade in the regions whose areas represent the average temperature each of the two days.\footnote{The continuous average of a function is $\frac{1}{b - 1} \int_{a}^{b} f(x) dx $ . In this case $b - a = 1$, so the average is the same as the integral. For more, see Notes, AV and Lecture 23.} 30 | \end{tcolorbox} 31 | 32 | Based on the rule given, we calculate the temperatures from $t = 0$ to $t = 5$: 33 | 34 | \begin{center} 35 | \begin{tabular}{|c|c|} 36 | \hline 37 | \rowcolor{Gray} 38 | $t$ & $y(t)$ \\ \hline 39 | 0 & 65 \\ \hline 40 | 1 & 65 + 65 / 100 = 65.65 \\ \hline 41 | 2 & 65.65 + 65.65 / 100 = 66.3065 \\ \hline 42 | 3 & 66.3065 + 66.3065 / 100 = 66.969565 \\ \hline 43 | 4 & 66.969565 + 66.969565 / 100 = 67.63926065 \\ \hline 44 | 5 & 67.63926065 + 67.63926065 / 100 = 68.3156532565 \\ \hline 45 | \end{tabular} 46 | \end{center} 47 | 48 | From the formula $y'(t) = y(t) / 100$, we get $\frac{dy}{dt} = \frac{y}{100}$. Then 49 | 50 | \begin{align*} 51 | \frac{dy}{dt} &= \frac{y}{100} \\ 52 | \frac{1}{y} dy &= \frac{1}{100} dt \\ 53 | ln(|y|) &= \frac{1}{100} t + C \\ 54 | \end{align*} 55 | 56 | Since $y > 0$ and is increasing, 57 | 58 | \begin{align*} 59 | ln(y) &= \frac{1}{100}t + C \\ 60 | y &= e^{\frac{1}{100}t + C} \\ 61 | &= e^{C} \cdot e^{\frac{1}{100}t} \\ 62 | &= Ae^{\frac{1}{100}t} 63 | \end{align*} 64 | 65 | At $t = 0, y = 65$. Hence $65 = Ae^{\frac{1}{100} \cdot 0} = A$. Then $ln(y) = 65e^{\frac{1}{100}t}$ \\ 66 | 67 | Graph: 68 | 69 | \begin{center} 70 | \includegraphics{q1a.jpg} 71 | \end{center} 72 | 73 | \begin{tcolorbox} 74 | \textbf{Q1b)} The number of \emph{cooling degree days} is the sum of reach day of the difference between the average temeperature for that day and $65^{\circ}$. The number is used to estimate the demand for electricity for air conditioning. Draw a second graph of $y$ for the whole 50 days and shade in the region whose area represents the total number of degree days. Write a formula for this total area as the difference between $65 \cdot 50$ and a definite integral. Evaluate the definite integral using the fundamental theorem of calculus. (Alternatively, write the whole quantity as an integral expressing the area between curves as in Lecture 21, and 7.2.) 75 | \end{tcolorbox} 76 | 77 | \begin{center} 78 | \includegraphics{q1b.jpg} 79 | \end{center} 80 | 81 | \begin{align*} 82 | A &= \int_0^{50} 65 e^{\frac{1}{100}t} dt - 65 \cdot 50 \\ 83 | &= 6500e^{\frac{1}{100}t}\bigg]_0^{50} - 65 \cdot 50 \\ 84 | &= 6500 (e^{\frac{50}{100}} - e^{0}) - \frac{1}{2} \cdot 6500 \\ 85 | &= 6500 (e^{\frac{1}{2}} - \frac{3}{2}) \\ 86 | &\approx 966.688 87 | \end{align*} 88 | 89 | 90 | \begin{tcolorbox} 91 | \textbf{Q1c)} (extra credit) Compute the definite integral in part (b) directly by evaluating a lower Riemann sum and taking a limit. Follow the procedure in Problem 5, PS4, but with different scale factors. This rather elaborate calculation shows how much time and effort we save every time we use the fundamental theorem and the change of variable formula in integrals. 92 | \end{tcolorbox} 93 | 94 | \begin{center} 95 | $\int_0^{50} 65 e^{\frac{1}{100}t} dt = 65 \int_0^{50} e^{\frac{1}{100}t} dt$ 96 | \end{center} 97 | 98 | Since $\frac{d}{dt} e^{\frac{1}{100}t} = \frac{1}{100} e^{\frac{1}{100}t} > 0$, $e^{\frac{1}{100}t}$ is an increasing function. Hence the left Riemann sum is the lower Riemann sum. 99 | 100 | Left Riemann sum of $\int_0^{50} e^{\frac{1}{100}t} dt = \sum\limits_{i=0}^{n-1} e^{\frac{1}{100}t_i} \Delta t$ where $\Delta t = \frac{50 - 0}{n} = \frac{50}{n}$ 101 | 102 | \begin{align*} 103 | \sum\limits_{i=0}^{n-1} e^{\frac{1}{100}(\frac{50i}{n})} \cdot \frac{50}{n} &= \frac{50}{n} \sum\limits_{i=0}^{n-1} e^{\frac{50i}{100n}} \\ 104 | &= \frac{50}{n} \sum\limits_{i=0}^{n-1} e^{\frac{i}{2n}} \\ 105 | &= \frac{50}{n} \cdot \frac{1(1 - (e^{\frac{1}{2n}})^n)}{1 - e^{\frac{1}{2n}}} \\ 106 | &= \frac{50}{n} \cdot \frac{1 - e^{\frac{1}{2}}}{1 - e ^{\frac{1}{2n}}} 107 | \end{align*} 108 | 109 | Use linear approximation for $1 - e^{\frac{1}{2n}}$. 110 | 111 | The linear approximation for $e^x = 1 + x$ for $x \approx 0$. Since $\frac{1}{2n} \rightarrow 0$ as $n \rightarrow +\infty$, then $1 - e^{\frac{1}{2n}} \approx 1 - (1 + \frac{1}{2n}) = -\frac{1}{2n}$ 112 | 113 | Then $n(1 - e^{\frac{1}{2n}}) \approx n \cdot (- \frac{1}{2n}) \approx -\frac{1}{2}$. As $n \rightarrow +\infty$, 114 | 115 | \begin{align*} 116 | \sum\limits_{i=0}^{n-1} e^{\frac{1}{100}(\frac{50i}{n})} \cdot \frac{50}{n} &\approx \lim_{n \rightarrow +\infty} \frac{50}{n} \cdot \frac{1 - e^{\frac{1}{2}}}{1 - e ^{\frac{1}{2n}}} \\ 117 | &\approx \lim_{n \rightarrow +\infty} \frac{50 \cdot (1 - e^{\frac{1}{2}})}{-\frac{1}{2}} \\ 118 | &= -100(1 - e^{\frac{1}{2}}) \\ 119 | &= 100e^{\frac{1}{2}} - 100 120 | \end{align*} 121 | 122 | Thus $\int_0^{50} e^{\frac{1}{100}t} dt \approx 100e^{\frac{1}{2}} - 100$ and $65 \int_0^{50} e^{\frac{1}{100}t} dt \approx 65(100e^{\frac{1}{2}} - 100) = 6500e^{\frac{1}{2}} - 6500$ 123 | 124 | 125 | \begin{tcolorbox} 126 | \textbf{Q2} Consider the function $f(x) = \int_0^{x} cos(t^2) dt$. There is no expression for $f(x)$ in terms of standard elementary functions. It is known as a Fresnel integral, along with the corresponding sine integral. \\ 127 | a) Draw a rough sketch of $cos(t^2)$, showing the first positive and negative zeros. What does the curve look like at $t = 0$? Is the function even or odd? 128 | \end{tcolorbox} 129 | 130 | \begin{center} 131 | \includegraphics{q2a.jpg} 132 | \end{center} 133 | 134 | It is an even function. 135 | 136 | \begin{tcolorbox} 137 | \textbf{Q2b)} List the critical points of $f(x)$ in the entire range $-\infty < x < \infty$. Which critical points are local maxima and which ones are local minima? 138 | \end{tcolorbox} 139 | 140 | $\frac{d}{dx}cos(x^2) = 2x(-sin(x^2)) = -2x\ sin(x^2)$ 141 | 142 | At critical points, $-2x\ sin(x^2) = 0$. Either $x = 0$ or $sin(x^2) = 0$ which implies $x^2 = n\pi$ for some integer $n >= 0$, which means $x = \pm\sqrt{n\pi}$ for some integer $n >= 0$ 143 | 144 | $\frac{d}{dx}-2x\ sin(x^2) = -2sin(x^2) - 4x^2cos(x^2)$ 145 | 146 | At $x = 0$, $-2sin(x^2) = 4x^2cos(x^2)$. Either $x = 0$ or $-\frac{1}{2}tan(x^2) = x^2$. 147 | 148 | At $x = \pm\sqrt{n\pi}$ for odd $n > 0$, $-4(\pm\sqrt{n\pi})^2cos((\pm\sqrt{n\pi})^2) = -4n\pi cos(n\pi) = -4n\pi (-1) = 4n\pi > 0$. So minimum points are at $\pm\sqrt{n\pi}$ for odd $n > 0$. 149 | 150 | At $x = \pm\sqrt{n\pi}$ for even $n$, $-4(\pm\sqrt{n\pi})^2cos((\pm\sqrt{n\pi})^2) = -4n\pi cos(n\pi) = -4n \pi < 0$. So maximum points are at $x = \pm\sqrt{n\pi}$ for even $n > 0$. 151 | 152 | When $-4x^2cos(x^2) = 0$, either $x^2 = 0 (x = 0)$ or $cos(x^2) = 0$, which implies $x^2 = \frac{\pi}{2} + n\pi$ for any integer $n >= 0$. That is $\frac{\pi + 2n\pi}{2} = \pi(\frac{2n + 1}{2})$. Hence $x = \pm\sqrt{\pi(\frac{2n + 1}{2})}$ for any integer $n >= 0$. 153 | 154 | In summary: 155 | 156 | $x = \pm\sqrt{n\pi}$ for odd integer $n > 0$ are minimum points. \\ 157 | $x = \pm\sqrt{n\pi}$ for even integer $n > 0$ are maximum points. \\ 158 | $x = 0$ is inflection point. \\ 159 | $x = \pm\sqrt{\pi(\frac{2n+1}{2})}$ for any integer $n >= 0$ are inflection points. 160 | 161 | 162 | \begin{tcolorbox} 163 | \textbf{Q2c)} Sketch the graph of $f$ on the interval $-2 <= x <= 2$, with labels for the critical points \textbf{and inflection points}. (The drawing should be qualitatively correct, but just estimate the values of $f$ at the labelled points.) 164 | \end{tcolorbox} 165 | 166 | Skipped. 167 | 168 | 169 | \begin{tcolorbox} 170 | \textbf{Q2d)} Estimate $f(0.1)$ to six decimal places. 171 | \end{tcolorbox} 172 | 173 | $f(0.1) = \int_0^{0.1} cos(t^2) dt$ 174 | 175 | Compute fourth derivative of $cos(t^2)$ 176 | 177 | \begin{align*} 178 | \frac{d}{dt}cos(t^2) &= 2t(-sin(t^2)) = -2tsin(t^2) \\ 179 | \frac{d^2}{dt^2}cos(t^2) &= \frac{d}{dt}(-2tsin(t^2)) \\ 180 | &= -2sin(t^2) - 2t(2tcos(t^2)) \\ 181 | &= -2sin(t^2) - 4t^2cos(t^2) \\ 182 | \frac{d^3}{dt^3}cos(t^2) &= \frac{d}{dt}(-2sin(t^2) - 4t^2cos(t^2)) \\ 183 | &= -2(2t)cos(t^2) - 8tcos(t^2) - 4t^2(2t \cdot (-sin(t^2))) \\ 184 | &= -4tcos(t^2) - 8tcos(t^2) + 8t^3sin(t^2) \\ 185 | &= -12tcos(t^2) + 8t^3sin(t^2) \\ 186 | \frac{d^4}{dt^4}cos(t^2) &= \frac{d}{dt}(-12tcos(t^2) + 8t^3sin(t^2)) \\ 187 | &= -12cos(t^2) - 12t(2t \cdot (-sin(t^2))) + 24t^2 sin(t^2) + 8t^3 \cdot 2tcos(t^2) \\ 188 | &= -12cos(t^2) + 24t^2sin(t^2) + 24t^2sin(t^2) + 16t^4cos(t^2) \\ 189 | &= -12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) 190 | \end{align*} 191 | 192 | The error to $\int_0^{0.1} cos(t^2) dt$ is bounded from above by 193 | 194 | \begin{equation*} 195 | \max_{0 \leq t \leq 0.1} \Big| -12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) \Big| \cdot \frac{(0.1 - 0)h^4}{180} 196 | \end{equation*} 197 | 198 | where $h = \frac{b - a}{N}$ 199 | 200 | First, we try to find this: 201 | 202 | \begin{equation*} 203 | \max_{0 \leq t \leq 0.1} \Big| -12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) \Big| 204 | \end{equation*} 205 | 206 | Since $0 \leq t \leq 0.1$ and both $t^2$ and $sin(t^2)$ are increasing on that range, $48t^2sin(t^2)$ is maximized at $t = 0.1$, with a value of approximately $0.0479992$. Which is very close to 0. 207 | 208 | Now, we look at $-12cos(t^2) + 16t^4cos(t^2) = (16t^4 - 12)cos(t^2)$ on that same range. 209 | 210 | On $0 \leq t \leq 0.1$, $16t^4 - 12$ is increasing. On that range, $-12 \leq 16t^4 - 12 \leq -11.99999984$ which is around $-12$. $cos(t^2)$ is maximized at $t = 0$ and decreases as $t \rightarrow 0.1$. However, at $t = 0.1$, $cos(t^2) = cos(0.1^2) \approx 0.99995 \approx 1$. Hence $16t^4cos(t^2) - 12cos(t^2) \approx -12$ on $0 \leq t \leq 0.1$ and it makes very little difference which value we pick. 211 | 212 | Hence on $0 \leq t \leq 0.1$, $-12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) \approx 0 - 12 = -12$. 213 | 214 | Then $\Big|-12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) \Big| \approx \Big|-12\Big| = 12$ 215 | 216 | Then 217 | 218 | \begin{equation*} 219 | \max_{0 \leq t \leq 0.1} \Big| -12cos(t^2) + 48t^2sin(t^2) + 16t^4cos(t^2) \Big| \cdot \frac{(0.1 - 0)h^4}{180} 220 | \end{equation*} 221 | 222 | is just $12 \cdot \frac{(0.1 - 0)h^4}{180}$ and the error of $\int_0^{0.1} cos(t^2) dt$ is $\leq 12 \cdot \frac{(0.1 - 0)h^4}{180}$ 223 | 224 | Since we want to estimate $f(0.1)$ to 6 decimal places, we choose that this error is less than $0.0000005$ 225 | 226 | Which means 227 | 228 | \begin{align*} 229 | 12 \cdot \frac{(0.1 - 0)h^4}{180} &= \frac{0.1h^4}{15} < 0.0000005 \\ 230 | 0.1 h^4 &< 0.0000075 \\ 231 | h^4 &<= 0.000075 \\ 232 | \end{align*} 233 | 234 | Substitute $h = \frac{0.1}{N}$ 235 | 236 | \begin{align*} 237 | (\frac{0.1}{N})^4 &< 0.000075 \\ 238 | N^4 &> \frac{0.1^4}{0.000075} \\ 239 | N &> (\frac{0.1^4}{0.000075})^{\frac{1}{4}} \approx 1.0745699 240 | \end{align*} 241 | 242 | Let's use $N = 4$ 243 | 244 | \begin{align*} 245 | \int_{0}^{0.1} cos(t^2) dt &\approx \frac{1}{3} \cdot \frac{0.1}{4} (cos(0^2) + 4cos(0.025^2) + 2cos(0.05^2) + 4cos(0.075^2) + cos(0.1^2)) \\ 246 | &\approx 0.09999899740072335 247 | \end{align*} 248 | 249 | Wolfram Alpha gives us $0.099999$ so this is good enough. 250 | 251 | 252 | \begin{tcolorbox} 253 | \textbf{Q2e)} Fresnel integrals are sometimes expressed using different scaling of the variables.\\ 254 | \textbf{Q2e(i)} Let $g(x) = \int_{0}^{x} cos((\pi/2)u^2) du$. Make a change of variables to show that $f(x) = c_1g(c_2 x)$ for some constants $c_1$ and $c_2$. Why did we choose the factor $\pi/2$? 255 | \end{tcolorbox} 256 | 257 | $g(x) = \int_{0}^{x} cos(\frac{\pi}{2}u^2) du$ 258 | 259 | Let $f(x) = \int_0^{x} cos(t^2) dt$. We want to show that $f(x) = c_1g(c_2x)$ 260 | 261 | Let $t = \sqrt{\frac{\pi}{2}} u$. Then $t^2 = \frac{\pi}{2}u^2$ and $dt = \sqrt{\frac{\pi}{2}} du$ 262 | 263 | \begin{align*} 264 | f(x) = \int_0^{x / \sqrt{\pi/2}} cos(\frac{\pi}{2}u^2) \sqrt{\frac{\pi}{2}} du &= \sqrt{\frac{\pi}{2}} \int_0^{x / \sqrt{\pi / 2}} cos(\frac{\pi}{2} u^2) du \\ 265 | = \sqrt{\frac{\pi}{2}} g(\frac{1}{\sqrt{\pi / 2}} x) 266 | \end{align*} 267 | 268 | so $c_1 = \sqrt{\frac{\pi}{2}}, c_2 = \frac{1}{\sqrt{\pi / 2}}$ 269 | 270 | Why choose the factor $\frac{\pi}{2}$? 271 | 272 | For $cos(t^2)$, $y = 1$ at $t^2 = 0, 2\pi$; $y = 0$ at $t^2 = \frac{\pi}{2}, \frac{3\pi}{2}, ...$ and $y = -1$ at $t^2 = 2(n+1)\pi$. All these $t$ happen to be multiples of $\frac{\pi}{2}$. By using a factor of $\frac{\pi}{2}$, we get $f(x) = \pi g(\frac{1}{\sqrt{\pi / 2}} x)$ which factors out a $\frac{1}{\sqrt{\pi / 2}}$. Hence, for $g$, instead of dealing with $t^2$ which are multiples of $\frac{\pi}{2}$, we get to deal with $t^2 = 1, \frac{3}{2}, 2$, etc which are easier numbers to work with. 273 | 274 | This is not a great explanation but it was what I could come up with. 275 | 276 | 277 | \begin{tcolorbox} 278 | \textbf{Q2e(ii)} Let $h(x) = \int_0^x \frac{cos v}{\sqrt{v}} dv$. (This integral is called \emph{improper} because $1 / \sqrt{v}$ is infinite \footnote{Although the integrand is infinite, the area under the curve is finite. The function $h$ is continuous, $h(0) = 0$, but its graph has infinite slope at $x = 0$} at $v = 0$.) Make a different change of variable to show that $f(x) = ch(x^2)$ for some constant $c$ (assume that $x > 0$). 279 | \end{tcolorbox} 280 | 281 | Let $t^2 = v$. Then $2t dt = dv$ and $dt = \frac{dv}{2t} = \frac{dv}{2\sqrt{v}}$ or $\frac{dv}{-2\sqrt{v}}$ (reject this because $t$ starts from 0, the lower bound of the integral). 282 | 283 | Lower bound: $t = 0, v = t^2 = 0$ 284 | 285 | Upper bound: $t = x, v = x^2$ 286 | 287 | Then 288 | 289 | \begin{align*} 290 | f(x) &= \int_0^x cos(t^2) dt \\ 291 | &= \int_0^{x^2} cos(v) \frac{dv}{2\sqrt{v}} \\ 292 | &= \frac{1}{2} \int_0^{x^2} \frac{cos(v)}{\sqrt{v}} dv \\ 293 | &= \frac{1}{2} h(x^2) 294 | \end{align*} 295 | 296 | 297 | \begin{tcolorbox} 298 | \textbf{Q2e(iii)} Let $k(x) = \sqrt{x} \int_0^1 cos(xt^2)dt$, $x > 0$. Use the change of variable $z = xt^2$ and part (ii) to find the relationship between the functions $k$ and $f$. Hint: Which quantities are variable and which are constant? (Not assigned: you can also work out this relationship using a change of variables like the one in part (i).) 299 | \end{tcolorbox} 300 | 301 | $k(x) = \sqrt{x} \int_0^1 cos(xt^2)dt$, $x > 0$. $x$ is constant and $t$ is the variable. 302 | 303 | Let $z = xt^2$. Then $dz = 2xt dt$ and $dt = \frac{dz}{2xt} = \frac{dz}{2x\sqrt{z/x}}$ or $\frac{dz}{2x(-\sqrt{z/x})}$ (rejected because lower bound for $t$ is $0$ so the overall value has to be non negative) 304 | 305 | Lower bound: $t = 0, z = x(0)^2 = 0$ 306 | 307 | Upper bound: $t = 1, z = x(1)^2 = x$ 308 | 309 | \begin{align*} 310 | k(x) &= \sqrt{x} \int_0^1 cos(xt^2) dt \\ 311 | &= \sqrt{x} \int_0^x cos(z) \frac{dz}{2x\sqrt{z/x}} \\ 312 | &= \sqrt{x} \int_0^x cos(z) \frac{dz}{2\sqrt{x}\sqrt{z}} \\ 313 | &= \frac{1}{2} \int_0^x cos(z) \frac{dz}{\sqrt{z}} \\ 314 | &= \frac{1}{2} h(x) 315 | \end{align*} 316 | 317 | So $c = \frac{1}{2}$. 318 | 319 | 320 | \begin{tcolorbox} 321 | \textbf{Q3a)} Do 7.3/22 322 | \end{tcolorbox} 323 | 324 | \textbf{Skipped}. We do not have the textbook. 325 | 326 | 327 | \begin{tcolorbox} 328 | \textbf{Q3b)} Find the volume of the region in 3-space with $x > 0$, $y > 0$ and $z > 0$ given by 329 | \begin{align*} 330 | z^2 < x + y < z 331 | \end{align*} 332 | Hint: First find the area of the horizontal cross-sections. 333 | \end{tcolorbox} 334 | 335 | For $z^2 / 2 < z$, we need $z / 2 < 1$ (Division by $z$ is ok since $z > 0$ and there is no change of sign) 336 | 337 | Then $z < 2$. Hence $0 < z < 2$. 338 | 339 | Area in terms of $z$ for one slice = $\frac{1}{2} * z * z - \frac{1}{2} * \frac{z^2}{2} * \frac{z^2}{2} = \frac{z^2}{2} - \frac{z^4}{8}$ 340 | 341 | Pictorially, it looks like the following for a fixed $z$: 342 | 343 | \begin{center} 344 | \includegraphics[scale=0.5]{q3b.jpg} 345 | \end{center} 346 | 347 | Volume is 348 | 349 | \begin{align*} 350 | \int_0^2 \frac{z^2}{2} - \frac{z^4}{8} dz &= \frac{z^3}{6} - \frac{z^5}{40}\bigg]_0^2 \\ 351 | &= \frac{8}{6} - \frac{32}{40} \\ 352 | &= \frac{8 * 20 - 32 * 3}{120} \\ 353 | &= \frac{64}{120} \\ 354 | &= \frac{8}{15} 355 | \end{align*} 356 | 357 | \end{document} 358 | -------------------------------------------------------------------------------- /ps6.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps6.pdf -------------------------------------------------------------------------------- /ps6.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{hyperref} 12 | \graphicspath{{images/ps6/}} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 6 Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \textbf{Q1)} Do 7.4/12 and 13. 30 | \end{tcolorbox} 31 | 32 | \textbf{Skipped.} We do not have the textbook. 33 | 34 | 35 | \begin{tcolorbox} 36 | \textbf{Q2)} The voltage $V$ of the house current is given by\\ 37 | \begin{center} 38 | $V(t) = Csin(120\pi t)$ 39 | \end{center} 40 | where $t$ is time, in seconds and $C$ is a constant amplitude. The square root of the average value of $V^2$ over one period of $V(t)$ (or cycle) is called the \emph{root-mean-square} voltage, abbreviated RMS. This is what the voltage meter on a house records. For house current, find the RMS in terms of the constant $C$. (The peak voltage delivered to the house is $\pm C$. The units of $V^2$ are square volts; when we take the square root again after averaging, the units become volts again.) 41 | \end{tcolorbox} 42 | 43 | Every cycle of the $sin$ function corresponds to $2 \pi$. This happens every $t = \frac{2 \pi}{120 \pi} = \frac{1}{60}$ seconds. 44 | 45 | Since $V(t) = Csin(120\pi t)$, $V^2(t) = C^2 sin^2(120\pi t)$. The average value of $V^2$ over 1 cycle of $V(t)$ is: 46 | 47 | \begin{align*} 48 | \frac{1}{\frac{1}{60} - 0} \int_0^{\frac{1}{60}} C^2 sin^2(120\pi t) dt &= 60C^2 \int_0^{\frac{1}{60}} sin^2(120\pi t) dt \\ 49 | &= 60C^2(\frac{1}{2}t - \frac{1}{240\pi} sin(120\pi t) cos(120\pi t)) \bigg]_0^{1/60} \\ 50 | &= 60C^2(\frac{1}{2} \cdot \frac{1}{60} - \frac{1}{240\pi} sin(2\pi) cos(2\pi) - (\frac{1}{2} \cdot 0 - \frac{1}{240\pi} sin(0) cos(0))) \\ 51 | &= 60C^2 \cdot \frac{1}{120} \\ 52 | &= \frac{C^2}{2} 53 | \end{align*} 54 | 55 | Then RMS $= \sqrt{\frac{C^2}{2}} = \frac{C}{\sqrt{2}} = \frac{\sqrt{2}C}{2}$ 56 | 57 | 58 | \begin{tcolorbox} 59 | \textbf{Q3a)} What is the probability that $x^2 < y$ if $(x, y)$ is chosen from the unit square $0 \leq x \leq 1$, $0 \leq y \leq 1$ with probability equal to the area. 60 | \end{tcolorbox} 61 | 62 | \begin{center} 63 | \includegraphics[scale=0.4]{q3a.jpg} 64 | \end{center} 65 | 66 | \begin{align*} 67 | P(x^2 < y) &= \frac{\int_0^1 1 dx - \int_0^1 x^2 dx}{\int_0^1 1 dx} \\ 68 | &= \frac{x \bigg]_0^1 - \frac{x^3}{3}\bigg]_0^1}{x \bigg]_0^1} \\ 69 | &= \frac{1 - \frac{1}{3}}{1} \\ 70 | &= \frac{2}{3} 71 | \end{align*} 72 | 73 | 74 | \begin{tcolorbox} 75 | \textbf{Q3b)} What is the probability that $x^2 < y$ if $(x, y)$ is chosen from the square $0 \leq x \leq 2$, $0 \leq y \leq 2$ with probability \textbf{proportional} to the area. (Probability = Part/Whole). 76 | \end{tcolorbox} 77 | 78 | \begin{center} 79 | \includegraphics[scale=0.5]{q3b.jpg} 80 | \end{center} 81 | 82 | For the line $y = x^2$ $y = 2$, $x = \sqrt{2}$. Once $x \ge \sqrt{2}$, $y > 2$. Hence $x \leq \sqrt{2}$. 83 | 84 | \begin{align*} 85 | P(x^2 < y) &= \frac{\int_0^2 2 dx - \int_0^{\sqrt{2}} x^2 dx - \int_{\sqrt{2}}^2 2 dx}{\int_0^2 2 dx} \\ 86 | &= \frac{2x \bigg]_0^2 - \frac{x^3}{3} \bigg]_0^{\sqrt{2}} - 2x \bigg]_{\sqrt{2}}^2}{2x \bigg]_0^2} \\ 87 | &= \frac{4 - \frac{(\sqrt{2})^3}{3} - (4 - 2\sqrt{2})}{4} \\ 88 | &= \frac{4 - \frac{2\sqrt{2}}{3} - 4 + 2\sqrt{2}}{4} \\ 89 | &= \frac{2\sqrt{2} - \frac{2\sqrt{2}}{3}}{4} \\ 90 | &= \frac{\frac{6 \sqrt{2} - 2\sqrt{2}}{3}}{4} \\ 91 | &= \frac{\frac{4 \sqrt{2}}{3}}{4} \\ 92 | &= \frac{\sqrt{2}}{3} 93 | \end{align*} 94 | 95 | 96 | \begin{tcolorbox} 97 | \textbf{Q3c)} Evaluate \\ 98 | \begin{center} 99 | $W = \int_0^{\infty} e^{-at} dt = \lim\limits_{N \rightarrow \infty} \int_0^N e^{-at} dt$ 100 | \end{center} 101 | This is known as an improper integral because it represents the area of an unbounded region. We are using the letter $W$ to signify "whole". \\ 102 | The probability that a radioactive particle will decay some time in the interval $0 \leq t \leq T$ is \\ 103 | \begin{center} 104 | $P([0, T]) = \frac{\text{PART}}{\text{WHOLE}} = \frac{1}{W} \int_0^T e^{-at} dt$ 105 | \end{center} 106 | \begin{center} 107 | \end{center} 108 | Note that $P([0, \infty)) = 1 = 100\%$ 109 | \end{tcolorbox} 110 | 111 | \begin{align*} 112 | \int_0^{\infty} e^{-at} dt &= -\frac{1}{a} e^{-at} \bigg]_0^{\infty} \\ 113 | &= -\frac{1}{a} (e^{-a \cdot \infty} - e^{-a \cdot 0}) \\ 114 | &= -\frac{1}{a} (\frac{1}{e^{a \cdot \infty}} - \frac{1}{e^{a \cdot 0}}) \\ 115 | &= -\frac{1}{a} (0 - \frac{1}{e^0}) \\ 116 | &= -\frac{1}{a} (-1) \\ 117 | &= \frac{1}{a} 118 | \end{align*} 119 | 120 | 121 | \begin{tcolorbox} 122 | \textbf{Q3d)} The half-life is the time $T$ for which $P([0, T]) = 1/2$. Find the value of $a$ and $W$ for which the half-life is $T = 1$. Suppose that a radioactive particle has a half-life of 1 second. What is the probability that it survives to time $t = 1$, but decays some time during the interval $1 \leq t \leq 2$? (Give an integral formula, and use a calculator to get an approximate numerical answer.) 123 | \end{tcolorbox} 124 | 125 | We saw in part (c) that $W = \frac{1}{a}$. We want to find $a$ and $W$ such that $P([0, 1]) = \frac{1}{2}$ 126 | 127 | \begin{align*} 128 | P([0, 1]) &= \frac{\int_0^1 e^{-at} dt}{W} \\ 129 | &= a \int_0^1 e^{-at} dt \\ 130 | &= a(-\frac{1}{a} e^{-at} \bigg]_0^1) \\ 131 | &= a \cdot (-\frac{1}{a}) e^{-at} \bigg]_0^1 \\ 132 | &= -1(e^{-a} - e^0) \\ 133 | &= -1(\frac{1}{e^a} - 1) \\ 134 | &= 1 - \frac{1}{e^a} \\ 135 | &= \frac{1}{2} 136 | \end{align*} 137 | 138 | So $1 - \frac{1}{e^a} = \frac{1}{2}$. Then $\frac{1}{2} = \frac{1}{e^a}$ and $e^a = 2$ so $a = ln(2)$ and $W = \frac{1}{a} = \frac{1}{ln(2)}$ 139 | 140 | From $0 \leq t \leq 1$, the particle decays with probability $\frac{1}{2}$ and survives with probability $\frac{1}{2}$. These 2 cases are mutually exclusivge and from $1 \leq t \leq 2$, the particle decays with probability $\frac{1}{W} \int_1^2 e^{-at} dt$ regardless of what happens from $0 \leq t \leq 1$. 141 | 142 | Hence probability of particule surviving to $t = 1$ and decaying some time in $1 \leq t \leq 2$ is 143 | 144 | \begin{align*} 145 | \frac{1}{2} \cdot \frac{1}{W} \int_1^2 e^{-at} dt &= \frac{1}{2} * a(-\frac{1}{a} e^{-at} \bigg]_1^2) \\ 146 | &= -\frac{1}{2} (e^{-at} \bigg]_1^2) \\ 147 | &= -\frac{1}{2} (e^{-2a} - e^{-a}) \\ 148 | &= \frac{1}{2} e^{-a} - \frac{1}{2} e^{-2a} \\ 149 | &= \frac{1}{2} e^{-ln(2)} - \frac{1}{2} e^{-2ln(2)} \\ 150 | &= \frac{1}{2} e^{ln(2^{-1})} - \frac{1}{2} e^{ln(2^{-2})} \\ 151 | &= \frac{1}{2} e^{ln(\frac{1}{2})} - \frac{1}{2} e^{ln(\frac{1}{4})} \\ 152 | &= \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{4} \\ 153 | &= \frac{1}{4} - \frac{1}{8} \\ 154 | &= \frac{1}{8} 155 | \end{align*} 156 | 157 | 158 | \begin{tcolorbox} 159 | \textbf{Q4)} The basis for Simson's rule is the following formula. Let $x_1$ be the midpoint of the interval $[x_0, x_2]$, and denote its length by $2h$. Consider any three points $(x_0, y_0)$, $(x_1, y_1)$, $(x_2, y_2)$. There is a unique quadratic function (parabola) \\ 160 | \begin{center} 161 | $y = Ax^2 + Bx + C$ 162 | \end{center} 163 | \begin{center} 164 | \end{center} 165 | whose graph passes through the three points. Simpson's rule says that the area under the parabola above $[x_0, x_2]$ is 166 | \begin{center} 167 | $\frac{h}{3}(y_0 + 4y_1 + y_2)$ 168 | \end{center} 169 | \begin{center} 170 | \end{center} 171 | This problem is devoted to proving this formula. It is significant because it illustrates how calculations can be simplified by using symmetry, and by looking ahead to see what you need.\\ 172 | \\ 173 | Since the area will be the same if the parabola is translated to the left or right, we may assume that $x_0 = -h$, $x_1 = 0$, and $x_2 = h$. Then in terms of the rest of the data (i.e., $h$ and the $y_i$) \\ 174 | \\ 175 | make a sketch and determine $C$;\\ 176 | \\ 177 | show, by integrating, that to find the area we need only determine $A$ (or better, $2Ah^2$);\\ 178 | \\ 179 | determine $2Ah^2$ using the data;\\ 180 | \\ 181 | put the results together to establish the formula for area. 182 | \end{tcolorbox} 183 | 184 | Since we are assuming that $x_1 = 0$, the easiest way to get $C$ is to substitute $x = x_1 = 0$ and $y = y_1$ into $y = Ax^2 + Bx + C$. From that, we get $y_1 = A(0)^2 + B(0) + C$ or $y_1 = C$. 185 | 186 | Again, substituting $(x_2, y_2)$ into $y = Ax^2 + Bx + C$, we get $f(x_2) = f(h) = A(h^2) + B(h) + C = Ah^2 + Bh + y_1 = y_2$. Then $Bh = -Ah^2 - y_1 + y_2$ and $B = -Ah - \frac{y_1}{h} - \frac{y_2}{h}$ assuming $h \neq 0$. 187 | 188 | \begin{align*} 189 | \int_{-h}^h Ax^2 + Bx + C dx &= \int_{-h}^h Ax^2 + (-Ah - \frac{y_1}{h} + \frac{y_2}{h})x + y_1 dx \\ 190 | &= \bigg[ \frac{Ax^3}{3} - \frac{Ahx^2}{2} - \frac{y_1x^2}{2h} + \frac{y_2x^2}{2h} + y_1x \bigg]_{-h}^h \\ 191 | &= \frac{Ah^3}{3} - \frac{Ahh^2}{2} - \frac{y_1h^2}{2h} + \frac{y_2h^2}{2h} + y_1h - (\frac{A(-h)^3}{3} - \frac{Ah(-h)^2}{2} - \frac{y_1(-h)^2}{2h} + \frac{y_2(-h)^2}{2h} + y_1(-h)) \\ 192 | &= \frac{Ah^3}{3} - \frac{Ah^3}{2} - \frac{y_1h}{2} + \frac{y_2h}{2} + y_1h - (-\frac{Ah^3}{3} - \frac{Ah^3}{2} - \frac{y_1h}{2} + \frac{y_2h}{2} - y_1h) \\ 193 | &= \frac{2Ah^3}{3} + 2y_1h \\ 194 | &= \frac{h}{3}(2Ah^2 + 6y_1) 195 | \end{align*} 196 | 197 | If we know $x_0$ and $x_2$, then we will know $h$. We will also know $x_1$ and hence $y_1$. Then we only need to determine $A$ to find the area. 198 | 199 | Now we will determine $2Ah^2$ from the data. Earlier, we have that $y_2 = Ah^2 + Bh + y_1$. From this, we get $Ah^2 = y_2 - Bh - y_1$. 200 | 201 | If we substitute $(x_0 = -h, y_0)$ into $y = Ax^2 + Bx + C$, we get $y_0 = A(-h)^2 + B(-h) + C = Ah^2 - Bh + C$ and $Ah^2 = y_0 + Bh - C$ and since $C = y_1$, this is $Ah^2 = y_0 + Bh - y_1$. 202 | 203 | Now we have 204 | 205 | \begin{align*} 206 | Ah^2 &= y_2 - Bh - y_1 \\ 207 | Ah^2 &= y_0 + Bh - y_1 208 | \end{align*} 209 | 210 | Adding them together gives us $2Ah^2 = y_2 - Bh - y_1 + y_0 + Bh - y_1 = y_0 - 2y_1 + y_2$. Then $2Ah^2 + 6y_1 = y_0 - 2y_1 + y_2 + 6y_1 = y_0 + 4y_1 + y_2$. 211 | 212 | Then 213 | 214 | \begin{align*} 215 | \int_{-h}^h Ax^2 + Bx + C dx &= \frac{h}{3}(2Ah^2 + 6y_1) \\ 216 | &= \frac{h}{3}(y_0 + 4y_1 + y_2) 217 | \end{align*} 218 | 219 | which is exactly what we want. 220 | 221 | 222 | \begin{tcolorbox} 223 | \textbf{Q5)} Use a calculator to make a table of values of the integrand and find approximations to the Fresnel integral $\int_0^a cos(t^2) dt$ for $a = \sqrt{\pi / 2}$, using Simpson's rule with four and eight intervals. (The exact answer to five decimal places is $1.22505$. Record your approximations to six decimal places to compare.) 224 | \end{tcolorbox} 225 | 226 | With 4 intervals, each interval is of size $\frac{1}{4} \sqrt{\frac{\pi}{2}}$. 227 | 228 | \begin{center} 229 | \begin{tabular}{|c|c|} 230 | \hline 231 | \rowcolor{Gray} 232 | $x$ & $cos(x^2)$ \\ \hline 233 | $0$ & $cos(0^2) = 1$ \\ \hline 234 | $\frac{1}{4} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{1}{16} \cdot \frac{\pi}{2}) = cos(\frac{\pi}{32})$ \\ \hline 235 | $\frac{1}{2} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{1}{4} \cdot \frac{\pi}{2}) = cos(\frac{\pi}{8})$ \\ \hline 236 | $\frac{3}{4} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{9}{16} \cdot \frac{\pi}{2}) = cos(\frac{9\pi}{32})$ \\ \hline 237 | $\sqrt{\frac{\pi}{2}}$ & $cos(\frac{\pi}{2}) = 0$ \\ \hline 238 | \end{tabular} 239 | \end{center} 240 | 241 | \begin{align*} 242 | \int_0^a cos(t^2) dt &= \frac{\frac{1}{4} \sqrt{\frac{\pi}{2}}}{3} (cos(0^2 ) + 4cos(\frac{\pi}{32}) + 2cos(\frac{\pi}{8}) + 4cos(\frac{9\pi}{32}) + cos(\frac{\pi}{2})) \\ 243 | &\approx 0.978219 244 | \end{align*} 245 | 246 | With 8 intervals, each interval is of size $\frac{1}{8} \sqrt{\frac{\pi}{2}}$ 247 | 248 | \begin{center} 249 | \begin{tabular}{|c|c|} 250 | \hline 251 | \rowcolor{Gray} 252 | $x$ & $cos(x^2)$ \\ \hline 253 | $0$ & $cos(0^2) = 1$ \\ \hline 254 | $\frac{1}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{1}{64} \cdot \frac{\pi}{2}) = cos(\frac{\pi}{128})$ \\ \hline 255 | $\frac{2}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{4}{64} \cdot \frac{\pi}{2}) = cos(\frac{\pi}{32})$ \\ \hline 256 | $\frac{3}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{9}{64} \cdot \frac{\pi}{2}) = cos(\frac{9\pi}{128})$ \\ \hline 257 | $\frac{4}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{16}{64} \cdot \frac{\pi}{2}) = cos(\frac{\pi}{8})$ \\ \hline 258 | $\frac{5}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{25}{64} \cdot \frac{\pi}{2}) = cos(\frac{25\pi}{128})$ \\ \hline 259 | $\frac{6}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{36}{64} \cdot \frac{\pi}{2}) = cos(\frac{9\pi}{32})$ \\ \hline 260 | $\frac{7}{8} \sqrt{\frac{\pi}{2}}$ & $cos(\frac{49}{64} \cdot \frac{\pi}{2}) = cos(\frac{49\pi}{128})$ \\ \hline 261 | $\sqrt{\frac{\pi}{2}}$ & $cos(\frac{\pi}{2}) = 0$ \\ \hline 262 | \end{tabular} 263 | \end{center} 264 | 265 | \begin{align*} 266 | \int_0^a cos(t^2) dt &= \frac{\frac{1}{8} \sqrt{\frac{\pi}{2}}}{3} (cos(0^2) + 4cos(\frac{\pi}{128}) + 2cos(\frac{\pi}{32}) + 4cos(\frac{9\pi}{128}) + 2cos(\frac{\pi}{8}) + 4cos(\frac{25\pi}{128}) + 2cos(\frac{9\pi}{32}) \\ 267 | & \ \ \ + 4cos(\frac{49\pi}{128}) + cos(\frac{\pi}{2})) \\ 268 | &\approx 0.977503 269 | \end{align*} 270 | 271 | \end{document} 272 | -------------------------------------------------------------------------------- /ps7.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps7.pdf -------------------------------------------------------------------------------- /ps7.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{hyperref} 12 | \graphicspath{{images/ps7/}} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 7 Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \textbf{Q1)} (from PS6) The voltage $V$ of house current is given by\\ 30 | \begin{center} 31 | $V(t) = Csin(120\pi t)$ 32 | \end{center} 33 | \begin{center} 34 | \end{center} 35 | where $t$ is time, in seconds and $C$ is a constant amplitude. The square root of the average value of $V^2$ over one period of $V(t)$ (or cycle) is called the \emph{root-mean-square} voltage, abbreviated RMS. This is what the voltage meter on a house records. For house current, find the RMS in terms of the constant $C$. (The peak voltage delivered to the house is $\pm C$. The units of $V^2$ are square volts; when we take the square root again after averaging, the units become volts again.) 36 | \end{tcolorbox} 37 | 38 | Average value of $V^2$ over $1$ period of $V(t)$ is 39 | 40 | \begin{align*} 41 | \frac{1}{60} \int_0^{\frac{1}{60}} C^2 sin^2(120\pi t) dt &= \frac{C^2}{60} \int_0^{\frac{1}{60}} sin^2(120\pi t) dt \\ 42 | &= \frac{C^2}{60} \int_0^{\frac{1}{60}} \frac{1 - cos(240\pi t)}{2} dt \\ 43 | &= \frac{C^2}{120} \int_0^{\frac{1}{60}} 1 - cos(240\pi t) dt \\ 44 | &= \frac{C^2}{120} (t - \frac{sin(240\pi t)}{240\pi}) \bigg]_0^{\frac{1}{60}} \\ 45 | &= \frac{C^2}{120} (\frac{1}{60} - \frac{sin(240\pi \cdot \frac{1}{60})}{240\pi}) \\ 46 | &= \frac{C^2}{120} (\frac{1}{60} - \frac{sin(4\pi)}{240\pi}) \\ 47 | &= \frac{C^2}{120} (\frac{1}{60}) \\ 48 | &= \frac{C^2}{7200} 49 | \end{align*} 50 | 51 | Square root of average value of $V^2$ over 1 period of $V(t) = \sqrt{\frac{C^2}{7200}} = \frac{C}{\sqrt{3600 * 2}} = \frac{C}{60\sqrt{2}}$ 52 | 53 | 54 | \begin{tcolorbox} 55 | \textbf{Q2)} The solid torus is the figure obtained by rotating the disk $(x - b)^2 + y^2 \leq a^2$ around the $y$-axis. Find its volume by the method of shells. (Hint: Substitute for $x - b$. As noted p. 229/11, the answer happens to be the area of the disk multiplied by the distance travelled by the center as it revolves.) 56 | \end{tcolorbox} 57 | 58 | \begin{center} 59 | \includegraphics[scale=0.5]{q2.jpg} 60 | \end{center} 61 | 62 | For a circle centered at $x = b, y = 0$ with radius $a$, the volume of the torus is: 63 | 64 | \begin{align*} 65 | 2 \int_{b-a}^{b+a} 2 \pi x (\sqrt{a^2 - (x - b)^2}) dx &= 4 \pi \int_{b-a}^{b+a} x \sqrt{a^2 - (x - b)^2} dx 66 | \end{align*} 67 | 68 | Let $u = x - b$. Then $du = dx$. Also, $x = u + b$. Substitute those into the above: 69 | 70 | \begin{align*} 71 | 4 \pi \int_{b-a}^{b+a} x \sqrt{a^2 - (x - b)^2} dx &= 4 \pi \int_{-a}^{a} (u + b) \sqrt{a^2 - u^2} du \\ 72 | &= 4 \pi (\int_{-a}^{a} u \sqrt{a^2 - u^2} du + b \int_{-a}^{a} \sqrt{a^2 - u^2} du) \\ 73 | &= 4\pi ( -\frac{1}{2} \cdot \frac{({a^2 - u^2})^{3/2}}{\frac{3}{2}} \bigg]_{-a}^{a} + b \int_{-a}^{a} \sqrt{a^2 - u^2} du ) \\ 74 | &= 4 \pi b \int_{-a}^{a} \sqrt{a^2 - u^2} du \ \ \ \ \text{(area of semicircle of radius $a$ centered at origin)} \\ 75 | &= 4 \pi b (\frac{1}{2} \pi a^2) \\ 76 | &= 2 \pi^2 a^2 b 77 | \end{align*} 78 | 79 | 80 | \begin{tcolorbox} 81 | \textbf{Q3a)} For any integer $n \geq 0$, use the substitution $tan^2 x = sec^2 x - 1$ to show that \\ 82 | \begin{center} 83 | $\int tan^{n+2} x \ dx = \frac{1}{n + 1} tan^{n+1} x - \int tan^n x \ dx$ 84 | \end{center} 85 | \end{tcolorbox} 86 | 87 | \begin{align*} 88 | \int tan^{n+2} x \ dx &= \int tan^2 x \ tan^n x \ dx \\ 89 | &= \int (sec^2 x - 1) tan^n x \ dx \\ 90 | &= \int sec^2 x \ tan^n x - tan^n x \ dx \\ 91 | &= \frac{tan^{n+1} x}{n + 1} - \int tan^n x \ dx 92 | \end{align*} 93 | 94 | 95 | \begin{tcolorbox} 96 | \textbf{Q3b)} Deduce a formula for $\int tan^4 x \ dx$ 97 | \end{tcolorbox} 98 | 99 | \begin{align*} 100 | \int tan^4 x \ dx &= \int tan^{2+2} x \ dx \\ 101 | &= \frac{1}{2 + 1} tan^{2+1} x - \int tan^2 x \ dx \\ 102 | &= \frac{1}{3} tan^3 x - \int tan^{0+2} x \ dx \\ 103 | &= \frac{1}{3} tan^3 x - (\frac{1}{0+1} tan^{0+1} x - \int tan^0 x \ dx \\ 104 | &= \frac{1}{3} tan^3 x - tan \ x + \int 1 dx \\ 105 | &= \frac{1}{3} tan^3 x - tan \ x + x + C 106 | \end{align*} 107 | 108 | Verify: 109 | 110 | \begin{align*} 111 | \frac{d}{dx}(\frac{1}{3} tan^3 x - tan \ x + x + C) &= tan^2 x \ sec^2 x - sec^2 x + 1 \\ 112 | &= tan^2 x (1 + tan^2 x) - (1 + tan^2 x ) + 1 \\ 113 | &= tan^2 x + tan^4 x - 1 - tan^2 x + 1 \\ 114 | &= tan^4 x 115 | \end{align*} 116 | 117 | 118 | \begin{tcolorbox} 119 | \textbf{Q4a)} Derive a formula for $\int sec \ x \ dx$ by writing $sec \ x = \frac{cos \ x}{1 - sin^2 x}$ (verify this), and then making a substitution for $sin x$ and using partial fractions. (Your final answer must be expressed in terms of $x$.) 120 | \end{tcolorbox} 121 | 122 | 123 | \begin{align*} 124 | \frac{cos \ x}{1 - sin^2 x} &= \frac{cos \ x}{cos^2 x} \\ 125 | &= \frac{1}{cos \ x} \\ 126 | &= sec \ x 127 | \end{align*} 128 | 129 | For $\int sec x \ dx = \int \frac{cos \ x}{1 - sin^2 x} dx$, let $u = sin x$. Then $du = cos \ x \ dx$ 130 | 131 | \begin{align*} 132 | \frac{cos \ x}{1 - sin^2 x} dx &= \int \frac{du}{1 - u^2} \\ 133 | &= \int \frac{1}{(1 + u)(1 - u)} du \\ 134 | &= \int \frac{1 / 2}{1 + u} + \frac{1 / 2}{1 - u} du \\ 135 | &= \frac{1}{2} ln(1 + u) - \frac{1}{2} ln(1 - u) + C \\ 136 | &= \frac{1}{2} ln(1 + sin \ x) - \frac{1}{2} ln(1 - sin \ x) + C \\ 137 | \end{align*} 138 | 139 | Verify: 140 | 141 | \begin{align*} 142 | \frac{1}{2} ln(1 + sin \ x) - \frac{1}{2} ln(1 - sin \ x) + C &= \frac{1}{2} (\frac{cos \ x}{1 + sin \ x}) - \frac{1}{2} (\frac{- cos \ x}{1 - sin x}) \\ 143 | &= \frac{1}{2} (\frac{cos \ x (1 - sin \ x) + cos \ x (1 + sin \ x)}{(1 + sin \ x)(1 - sin \ x)}) \\ 144 | &= \frac{1}{2} (\frac{cos \ x - sin \ x \ cos \ x + cos \ x + sin \ x \ cos \ x}{1 - sin^2 x}) \\ 145 | &= \frac{1}{2} (\frac{2 cos \ x}{cos^2 x}) \\ 146 | &= \frac{1}{cos \ x} \\ 147 | &= sec \ x 148 | \end{align*} 149 | 150 | 151 | \begin{tcolorbox} 152 | \textbf{Q4b)} Convert the formula into the more familiar one by multiplying the fraction in the answer on both top and bottom by $1 + sin \ x$. (Note that $(1 / 2) ln \ u = ln \sqrt{u}$ 153 | \end{tcolorbox} 154 | 155 | \begin{align*} 156 | \frac{1}{2} ln(1 + sin \ x) - \frac{1}{2} ln(1 - sin \ x) + C &= \frac{1}{2} (ln(1 + sin \ x) - ln(1 - sin \ x)) \\ 157 | &= \frac{1}{2} ln(\frac{1 + sin \ x}{1 - sin \ x}) \\ 158 | &= \frac{1}{2} ln(\frac{1 + sin \ x}{1 - sin \ x} \cdot \frac{1 + sin \ x}{1 + sin \ x}) \\ 159 | &= \frac{1}{2} ln(\frac{sin^2 x + 2 sin \ x + 1}{1 - sin^2 x}) \\ 160 | &= \frac{1}{2} ln(\frac{sin^2 x + 2 sin \ x + 1}{cos^2 x}) \\ 161 | &= \frac{1}{2} ln(\frac{sin^2 x}{cos^2 x} + \frac{2 sin \ x}{cos^2 x} + \frac{1}{cos^2 x}) \\ 162 | &= \frac{1}{2} ln(tan^2 x + 2 sec \ x \ tan \ x + sec^2 x) \\ 163 | &= \frac{1}{2} ln((sec \ x + tan \ x)^2) \\ 164 | &= ln(\sqrt{(sec \ x + tan \ x)^2}) \\ 165 | &= ln(sec \ x + tan \ x) 166 | \end{align*} 167 | 168 | 169 | \begin{tcolorbox} 170 | \textbf{Q5)} Find the volume under the first hump of the function $y = cos \ x$ rotated around the $y$-axis by the method of shells. 171 | \end{tcolorbox} 172 | 173 | 174 | \begin{align*} 175 | \text{Volume} &= \int_0^{\pi/2} 2 \pi \ x \ cos \ x \ dx \\ 176 | &= 2 \pi \int_0^{\pi / 2} x \ cos \ x \ dx \\ 177 | &= 2 \pi (x \ sin \ x \bigg]_0^{\pi/2} - \int_0^{\pi/2} sin \ x \ dx) \\ 178 | &= 2 \pi (\frac{\pi}{2} sin(\frac{\pi}{2}) - (- cos \ x) \bigg]_0^{\pi / 2}) \\ 179 | &= 2 \pi (\frac{\pi}{2} - (- cos(\frac{\pi}{2}) - (- cos 0))) \\ 180 | &= 2 \pi (\frac{\pi}{2} - (- 0 - (- 1))) \\ 181 | &= 2 \pi (\frac{\pi}{2} - (1)) \\ 182 | &= 2 \pi (\frac{\pi}{2} - 1) 183 | \end{align*} 184 | 185 | \end{document} 186 | -------------------------------------------------------------------------------- /ps8a.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/yanhan/mit-1801-ps-unofficial-sol/e285de85221593c349fc7f9147edd14c5687a3e0/ps8a.pdf -------------------------------------------------------------------------------- /ps8a.tex: -------------------------------------------------------------------------------- 1 | \documentclass[9pt]{article} 2 | 3 | \usepackage{amsmath} 4 | \usepackage{tcolorbox} 5 | % `parskip` removes indentation for all paragraphs: http://tex.stackexchange.com/a/55016 6 | \usepackage{parskip} 7 | % Allows us to color rows / cols of a table. 8 | % See https://texblog.org/2011/04/19/highlight-table-rowscolumns-with-color/ 9 | \usepackage{color, colortbl} 10 | 11 | \usepackage{hyperref} 12 | \graphicspath{{images/ps8a/}} 13 | 14 | \leftmargin=0.25in 15 | \oddsidemargin=0.25in 16 | \textwidth=6.0in 17 | \topmargin=-0.25in 18 | \textheight=9.25in 19 | 20 | \definecolor{Gray}{gray}{0.9} 21 | 22 | \begin{document} 23 | 24 | \begin{center} 25 | \large\textbf{MIT 18.01 Problem Set 8A Unofficial Solutions} 26 | \end{center} 27 | 28 | \begin{tcolorbox} 29 | \textbf{4E-2} Find the rectangular equation for $x = t + 1/t$ and $y = t - 1/t$ (compute $x^2$ and $y^2$). 30 | \end{tcolorbox} 31 | 32 | \begin{align*} 33 | x^2 &= t^2 + 2 + 1/t^2 \\ 34 | y^2 &= t^2 - 2 + 1/t^2 \\ 35 | y^2 &= x^2 -4 \\ 36 | x^2 &- y^2 = 4 37 | \end{align*} 38 | 39 | This is an equation of a hyperbola centred at the origin with width 2. 40 | 41 | 42 | \begin{tcolorbox} 43 | \textbf{4E-3} Find the rectangular equation for $x = 1 + sin\ t, y = 4 + cos\ t$ 44 | \end{tcolorbox} 45 | 46 | \begin{align*} 47 | x^2 &= 1 + 2sin\ t + sin^2\ t \\ 48 | y^2 &= 16 + 8sin\ t + cos^2\ t \\ 49 | x^2 + y^2 &= 17 + 2sin\ t + 8cos\ t + sin^2\ t + cos^2\ t \\ 50 | &= 18 + 2(sin\ t + 4cos\ t) \\ 51 | &= 2(1 + sin\ t) + 8(4 + cos\ t) - 16 \\ 52 | &= 2x + 8y - 16 53 | \end{align*} 54 | 55 | Then 56 | 57 | \begin{align*} 58 | x^2 - 2x + y^2 - 8y + 16 &= 0 \\ 59 | (x-1)^2 + (y-4)^2 - 1 &= 0 \\ 60 | (x-1)^2 + (y-4)^2 &= 1 61 | \end{align*} 62 | 63 | This is an equation of a circle with radius 1 centred at $(1, 2)$. 64 | 65 | 66 | \begin{tcolorbox} 67 | \textbf{4E-8} At noon, a snail starts at the center of an open clock face. It creeps at a steady rate along the hour hand, reaching the end of the hand at 1:00 PM. The hour hand is 1 meter long. Write parametric equations for the position of the snail at time t, in some reasonable xy-coordinate system. 68 | \end{tcolorbox} 69 | 70 | The average velocity of the snail is 1 metre / h. Let the centre of the clock be the origin. 71 | 72 | Treat the positive x-axis as 0 radians. When the hour hand is 12pm, the angle with respect to the positive x-axis is $\pi / 2$ radians. When the hour hand is at 1pm, the angle with respect to the positive x-axis is $\pi / 3$ radians. 73 | 74 | Hence, the angle traversed by the hour hand from 12pm to 1pm is $\pi / 6$ radians. 75 | 76 | \begin{align*} 77 | x = t cos\ \theta = t\ cos(\frac{\pi}{2} - \frac{\pi}{6}t) \\ 78 | y = t sin\ \theta = t\ sin(\frac{\pi}{2} - \frac{\pi}{6}t) 79 | \end{align*} 80 | 81 | \begin{center} 82 | \includegraphics[scale=0.8]{p1_4e8.jpg} 83 | \end{center} 84 | 85 | Verification: 86 | 87 | At $t = 0$ (12pm), the snail should be at $(0, 0)$. 88 | 89 | \begin{align*} 90 | x &= 0\ cos(\frac{\pi}{2} - \frac{\pi}{6} \cdot 0) = 0 \\ 91 | y &= 0\ sin(\frac{\pi}{2} - \frac{\pi}{6} \cdot 0) = 0 92 | \end{align*} 93 | 94 | At $t = 1$ (1pm), the snail should be at $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ (draw out a right angle triangle with hypothenus 1 and the angle being $\pi / 3$ radians and you will see why these numbers): 95 | 96 | \begin{align*} 97 | x &= 1\ cos(\frac{\pi}{2} - \frac{\pi}{6} \cdot 1) = cos(\frac{\pi}{3}) = \frac{1}{2} \\ 98 | y &= 1\ sin(\frac{\pi}{2} - \frac{\pi}{6} \cdot 1) = sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} 99 | \end{align*} 100 | 101 | 102 | \begin{tcolorbox} 103 | \textbf{4F-1d} Find the arclength of $y = (1/3)(2 + x^2)^{3/2},\ 1 \leq x \leq 2$. 104 | \end{tcolorbox} 105 | 106 | \begin{align*} 107 | \frac{dy}{dx} &= \frac{1}{2}(2x)(2 + x^2)^{1/2} = x(2 + x^2)^{1/2} \\ 108 | \sqrt{1 + (\frac{dy}{dx})^2} &= \sqrt{1 + (x(2 + x^2)^{1/2})^2} \\ 109 | &= \sqrt{1 + x^2(2 + x^2)} \\ 110 | &= \sqrt{1 + 2x^2 + x^4} \\ 111 | &= \sqrt{(x^2 + 1)^2} \\ 112 | &= x^2 + 1 \\ 113 | \\ 114 | ds &= x^2 + 1\ dx 115 | \end{align*} 116 | 117 | Arc length: 118 | 119 | \begin{align*} 120 | \int_{1}^{2} x^2 + 1\ dx &= \frac{x^3}{3} + x \bigg]_{1}^{2} \\ 121 | &= \frac{2^3}{3} + 2 - (\frac{1}{3} + 1) \\ 122 | &= \frac{8}{3} + 2 - \frac{1}{3} - 1 \\ 123 | &= \frac{10}{3} 124 | \end{align*} 125 | 126 | 127 | \begin{tcolorbox} 128 | \textbf{4F-4} Find the length of the curve $x = t^2$, $y = t^3$ for $0 \leq t \leq 2$. 129 | \end{tcolorbox} 130 | 131 | \begin{align*} 132 | \frac{dx}{dt} &= 2t \\ 133 | \frac{dy}{dt} &= 3t^2 \\ 134 | ds &= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \\ 135 | &= \sqrt{(2t)^2 + (3t^2)^2} \\ 136 | &= \sqrt{4t^2 + 9t^4} \\ 137 | &= t\sqrt{4 + 9t^2} \\ 138 | \int_{s_0}^{s_1} ds &= \int_{0}^{2} t\sqrt{4 + 9t^2} dt \\ 139 | &= \frac{2/3 \cdot (4 + 9t^2)^{3/2}}{18} \bigg]_{0}^{2} \\ 140 | &= \frac{1}{27} (4 + 9t^2)^{3/2} \bigg]_{0}^{2} \\ 141 | &= \frac{1}{27} ((4 + 9(2)^2)^{3/2} - (4 + 9(0)^2)^{3/2}) \\ 142 | &= \frac{1}{27} ((4 + 36)^{3/2} - 4^{3/2}) \\ 143 | &= \frac{1}{27} (40^{3/2} - 8) \\ 144 | \end{align*} 145 | 146 | 147 | \begin{tcolorbox} 148 | \textbf{4F-5} Find an integral for the length of the curve given parametrically in Exercise 4E-2 for $1 \leq t \leq 2$. Simplify the integrand as much as possible but do not evaluate. 149 | \end{tcolorbox} 150 | 151 | What is given in 4E-2: $x = t + 1 / t$, $y = t - 1 / t$. 152 | 153 | \begin{align*} 154 | \frac{dx}{dt} &= 1 - \frac{1}{t^2} \\ 155 | \frac{dy}{dt} &= 1 + \frac{1}{t^2} \\ 156 | ds &= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt \\ 157 | &= \sqrt{(1 - \frac{1}{t^2})^2 + (1 + \frac{1}{t^2})^2} dt \\ 158 | &= \sqrt{1 - \frac{2}{t^2} + \frac{1}{t^4} + 1 + \frac{2}{t^2} + \frac{1}{t^4}} dt \\ 159 | &= \sqrt{2 + \frac{2}{t^4}} dt \\ 160 | &= \sqrt{\frac{2t^4 + 2}{t^4}} dt \\ 161 | \int_{s_0}^{s_1} ds &= \int_{1}^{2} \sqrt{\frac{2t^4 + 2}{t^4}} dt 162 | \end{align*} 163 | 164 | 165 | \begin{tcolorbox} 166 | \textbf{4F-8)} Find the length of the curve $x = e^t cos\ t$, $y = e^t sin\ t$ for $0 \leq t \leq 10$. 167 | \end{tcolorbox} 168 | 169 | \begin{align*} 170 | x^2 + y^2 &= e^{2t} cos^2\ t + e^{2t} sin^2\ t = e^{2t} \\ 171 | \frac{dx}{dt} &= e^t cos\ t - e^t sin\ t \\ 172 | \frac{dy}{dt} &= e^t sin\ t + e^t cos\ t \\ 173 | \end{align*} 174 | 175 | \begin{align*} 176 | (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 &= e^{2t} cos^2\ t - 2e^{2t} cos\ t\ sin\ t + e^{2t}sin^2\ t + e^{2t}sin^2\ t + 2e^{2t}sin\ t\ cos\ t + e^{2t}cos^2\ t \\ 177 | &= e^{2t} cos^2\ t + e^{2t}sin^2\ t + e^{2t}sin^2\ t + e^{2t}cos^2\ t \\ 178 | &= 2 e^{2t} 179 | \end{align*} 180 | 181 | \begin{align*} 182 | ds &= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}\ dt \\ 183 | &= \sqrt{2e^{2t}}\ dt \\ 184 | &= \sqrt{2}\ e^{t}\ dt \\ 185 | \int_{s_0}^{s_1} ds &= \int_{0}^{10} \sqrt{2}\ e^t\ dt \\ 186 | &= \sqrt{2}\ e^t\ \bigg]_{0}^{10} \\ 187 | &= \sqrt{2}\ (e^{10} - e^{0}) \\ 188 | &= \sqrt{2}\ (e^{10} - 1) 189 | \end{align*} 190 | 191 | 192 | \begin{tcolorbox} 193 | \textbf{4G-2} Find the area of the segment of $y = 1 - 2x$ in the first quadrant revolved around the x-axis. 194 | \end{tcolorbox} 195 | 196 | \begin{align*} 197 | ds &= \sqrt{1 + (\frac{dy}{dx})^2} \ dx \\ 198 | &= \sqrt{1 + (-2)^2} \ dx \\ 199 | &= \sqrt{5} \ dx 200 | \end{align*} 201 | 202 | Surface area: 203 | 204 | \begin{align*} 205 | \int_{s_0}^{s_1} 2 \pi y \ ds &= \int_{0}^{1/2} 2 \pi (1 - 2x) \sqrt{5} dx \\ 206 | &= 2 \sqrt{5} \pi \int_{0}^{1/2} 1 - 2x \ dx \\ 207 | &= 2 \sqrt{5} \pi (x - x^2) \bigg]_{0}^{1/2} \\ 208 | &= 2 \sqrt{5} \pi (\frac{1}{2} - \frac{1}{4})) \\ 209 | &= \frac{\sqrt{5}}{2} \pi 210 | \end{align*} 211 | 212 | 213 | \begin{tcolorbox} 214 | \textbf{4G-5} Find the area of $y = x^2$, $0 \leq x \leq 4$ revolved around the y-axis. 215 | \end{tcolorbox} 216 | 217 | \begin{align*} 218 | ds &= \sqrt{1 + (\frac{dy}{dx})^2} \ dx \\ 219 | &= \sqrt{1 + (\frac{1}{2}y^{-\frac{1}{2})^2}} \ dy \\ 220 | &= \sqrt{1 + \frac{1}{4y}} \ dy 221 | \end{align*} 222 | 223 | Surface area: 224 | 225 | \begin{align*} 226 | \int_{s_0}^{s_1} 2 \pi x \ ds &= \int_{0}^{16} 2 \pi x \sqrt{1 + \frac{1}{4y}} \ dy \\ 227 | &= 2 \pi \int_{0}^{16} \sqrt{y} \sqrt{1 + \frac{1}{4y}} \ dy \\ 228 | &= 2 \pi \int_{0}^{16} \sqrt{y + \frac{1}{4}} \ dy \\ 229 | &= 2 \pi \frac{2}{3} (y + \frac{1}{4})^{3/2} \bigg]_{0}^{16} \\ 230 | &= \frac{4 \pi}{3} (y + \frac{1}{4})^{3/2} \bigg]_{0}^{16} \\ 231 | &= \frac{4 \pi}{3} ((\frac{65}{4})^{3/2} - \frac{1}{8}) 232 | \end{align*} 233 | 234 | 235 | \begin{tcolorbox} 236 | \textbf{4H-1b} Give the polar coordinates for the rectangular coordinate $(-2, 0)$ 237 | \end{tcolorbox} 238 | 239 | $r = 2$, $\theta = \pi$ 240 | 241 | Verify: 242 | 243 | \begin{align*} 244 | x &= r\ cos \theta = 2 cos\ \pi = 2(-1) = 2 \\ 245 | y &= r\ sin \theta = 2 sin\ \pi = 0 246 | \end{align*} 247 | 248 | 249 | \begin{tcolorbox} 250 | \textbf{4H-1f} Give the polar coordinates for the rectangular coordinate $(0, -2)$ 251 | \end{tcolorbox} 252 | 253 | $r = 2$, $\theta = \frac{3\pi}{2}$ 254 | 255 | Verify: 256 | 257 | \begin{align*} 258 | x &= r\ cos \theta = 2 cos\ \frac{3\pi}{2} = 0 \\ 259 | y &= r\ sin \theta = 2 sin\ \frac{3\pi}{2} = 2(-1) = -2 \\ 260 | \end{align*} 261 | 262 | Alternatively, $r = 2$, $\theta = -\frac{\pi}{2}$ 263 | 264 | 265 | \begin{tcolorbox} 266 | \textbf{4H-1g} Give the polar coordinates for the rectangular coordinate $(\sqrt{3}, -1)$ 267 | \end{tcolorbox} 268 | 269 | $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$ 270 | 271 | We know that $cos\ \theta = \frac{\sqrt{3}}{2}$ and $sin\ \theta = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$ 272 | 273 | But in this case, we are in the 4th quadrant. So $\theta = 2 \pi - \frac{\pi}{6} = \frac{11 \pi}{6}$ or equivalently, $\theta = -\frac{\pi}{6}$. 274 | 275 | Verify: 276 | 277 | \begin{align*} 278 | x &= r\ cos \theta = 2 cos\ \frac{11\pi}{6} = 2 cos\ \frac{\pi}{6} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \\ 279 | y &= r\ sin \theta = 2 sin\ \frac{11\pi}{6} = -2 sin\ \frac{\pi}{6} = -2 \cdot \frac{1}{2} = -1 280 | \end{align*} 281 | 282 | 283 | \begin{tcolorbox} 284 | \textbf{4H-2a} Find using two different methods the equation in polar coordinates for the circle of radius $a$ with center at $(a, 0)$ on the x-axis, as follows: \\ 285 | \\ 286 | (i) write its equation in rectangular coordinates, and then change it to polar coordinates (substitute $x = r\ cos\ \theta$ and $y = r\ sin\ \theta$, and then simplify). \\ 287 | \\ 288 | (ii) treat it as a locus problem: let $OQ$ be the diameter lying along the x-axis, and $P: (r, \theta)$ a point on the circle; use $\Delta OPQ$ and trigonometry to find the relation connecting $r$ and $\theta$. 289 | \end{tcolorbox} 290 | 291 | For part (i) 292 | 293 | Equation of the circle is $(x - a)^2 + y^2 = a^2$. 294 | 295 | Let $x = r\ cos\ \theta$, $y = r\ sin\ \theta$. Then 296 | 297 | \begin{align*} 298 | (r\ cos\ \theta - a)^2 + (r\ sin\ \theta)^2 &= a^2 \\ 299 | r^2\ cos^2\ \theta - 2ar\ cos\ \theta + a^2 + r^2\ sin^2\ \theta &= a^2 \\ 300 | r^2 - 2ar\ cos\ \theta &= 0 \\ 301 | r &= 2a\ cos\ \theta 302 | \end{align*} 303 | 304 | For part (ii) 305 | 306 | \begin{center} 307 | \includegraphics[scale=0.8]{p1_4h2.jpg} 308 | \end{center} 309 | 310 | \begin{align*} 311 | OQ &= 2a \\ 312 | \angle OPQ &= \frac{\pi}{2} \\ 313 | \angle POQ &= \theta \\ 314 | \angle PQO &= \frac{\pi}{2} - \theta 315 | \end{align*} 316 | 317 | \begin{align*} 318 | cos\ \theta &= \frac{r}{2a} \\ 319 | r &= 2a\ cos\ \theta 320 | \end{align*} 321 | 322 | 323 | \begin{tcolorbox} 324 | \textbf{4H-3f} For $r = a\ cos(2\theta)$ (4-leaf rose) \\ 325 | (i) give the corresponding equation in rectangular coordinates; \\ 326 | (ii) draw the graph; indicate the direction of increasing $\theta$ 327 | \end{tcolorbox} 328 | 329 | For part (i) 330 | 331 | \begin{align*} 332 | x = r\ cos\ \theta &= a\ cos(2 \theta)\ cos(\theta) \\ 333 | y = r\ sin\ \theta &= a\ cos(2 \theta)\ sin(\theta) \\ 334 | \\ 335 | x &= a\ cos(2 \theta)\ cos(\theta) \\ 336 | &= a(cos^2 \theta - sin^2 \theta)\ cos\ \theta \\ 337 | &= a(cos^3 \theta - cos\ \theta\ sin^2 \theta) \\ 338 | &= a(cos^3 \theta - cos\ \theta\ (1 - cos^2 \theta)) \\ 339 | &= a(cos^3 \theta - cos\ \theta\ + cos^3 \theta) \\ 340 | &= a(2\ cos^3 \theta - cos\ \theta) \\ 341 | \\ 342 | y &= a\ cos(2 \theta)\ sin(\theta) \\ 343 | &= a\ (cos^2 \theta - sin^2 \theta)\ sin(\theta) \\ 344 | &= a\ (1 - sin^2 \theta - sin^2 \theta)\ sin(\theta) \\ 345 | &= a\ (1 - 2\ sin^2 \theta)\ sin(\theta) \\ 346 | &= a\ (sin\ \theta - 2\ sin^3 \theta) \\ 347 | \\ 348 | x^2 &= a^2 (2\ cos^3 \theta - cos\ \theta)^2 \\ 349 | &= a^2(4\ cos^6 \theta - 4\ cos^4 \theta + cos^2 \theta) \\ 350 | \\ 351 | y^2 &= a^2\ (sin\ \theta - 2\ sin^3 \theta)^2 \\ 352 | &= a^2\ (4\ sin^6\ \theta - 4\ sin^4 \theta + sin^2 \theta) \\ 353 | \\ 354 | r &= a\ cos(2\theta) = a(cos^2 \theta - sin^2 \theta) \\ 355 | r^3 &= ar^2\ cos(2\theta) \\ 356 | &= a(r^2 cos^2 \theta - r^2 sin^2 \theta) \\ 357 | &= a(x^2 - y^2) 358 | \end{align*} 359 | 360 | Using polar coordinates, we know that $r = \sqrt{x^2 + y^2}$. Therefore $r^3 = (\sqrt{x^2 + y^2})^3$. 361 | 362 | Equating $r^3 = a(x^2 - y^2)$ and $r^3 = (\sqrt{x^2 + y^2})^3$: 363 | 364 | \begin{align*} 365 | a(x^2 - y^2) &= \sqrt{(x^2 + y^2)}^{3} \\ 366 | (x^2 + y^2)^{3/2} &= a(x^2 - y^2) 367 | \end{align*} 368 | 369 | For part (ii) 370 | 371 | \begin{center} 372 | \begin{tabular}{|c|c|} 373 | \hline 374 | \rowcolor{Gray} 375 | $\theta$ & $r = a\ cos(2 \theta)$ \\ \hline 376 | $0$ & $a\ cos(0) = a$ \\ \hline 377 | $\pi / 6$ & $a\ cos(\pi / 3) = a / 2$ \\ \hline 378 | $\pi / 4$ & $a\ cos(\pi / 2) = 0$ \\ \hline 379 | $\pi / 3$ & $a\ cos(2 \pi / 3) = -a / 3$ \\ \hline 380 | $\pi / 2$ & $a\ cos(\pi) = -a$ \\ \hline 381 | $2 \pi / 3$ & $a\ cos(4 \pi / 3) = -a / 2$ \\ \hline 382 | $3 \pi / 4$ & $a\ cos(3 \pi / 2) = 0$ \\ \hline 383 | $5 \pi / 6$ & $a\ cos(5 \pi / 3) = a / 2$ \\ \hline 384 | $\pi$ & $a\ cos(2 \pi) = a$ \\ \hline 385 | $7 \pi / 6$ & $a\ cos(7 \pi / 3) = a / 2$ \\ \hline 386 | $5 \pi / 4$ & $a\ cos(5 \pi / 2) = 0$ \\ \hline 387 | $3 \pi / 2$ & $a\ cos(3 \pi) = -a$ \\ \hline 388 | $5 \pi / 3$ & $a\ cos(10 \pi / 3) = -a / 2$ \\ \hline 389 | $7 \pi / 4$ & $a\ cos(7 \pi / 2) = 0$ \\ \hline 390 | $11 \pi / 6$ & $a\ cos(11 \pi / 3) = a / 2$ \\ \hline 391 | $2 \pi$ & $a\ cos(4 \pi) = a$ \\ \hline 392 | \end{tabular} 393 | \end{center} 394 | 395 | \begin{center} 396 | \includegraphics[scale=0.8]{p1_4h3.jpg} 397 | \end{center} 398 | 399 | 400 | \begin{tcolorbox} 401 | \textbf{4I-2} Find the area of one leaf of a three-leaf rose $r = a\ cos(3 \theta)$. 402 | \end{tcolorbox} 403 | 404 | \begin{center} 405 | \begin{tabular}{|c|c|} 406 | \hline 407 | \rowcolor{Gray} 408 | $\theta$ & $r = a\ cos(3 \theta)$ \\ \hline 409 | $0$ & $a\ cos(0) = a$ \\ \hline 410 | $\pi / 12$ & $a\ cos(\pi / 4) = \sqrt{2} \ a / 2$ \\ \hline 411 | $\pi / 6$ & $a\ cos(\pi / 2) = 0$ \\ \hline 412 | $\pi / 3$ & $a\ cos(\pi) = -a$ \\ \hline 413 | $\pi / 2$ & $a\ cos(3 \pi / 2) = 0$ \\ \hline 414 | $2 \pi / 3$ & $a\ cos(2 \pi) = a$ \\ \hline 415 | $3 \pi / 4$ & $a\ cos(9 \pi / 4) = \sqrt{2} \ a / 2$ \\ \hline 416 | $5 \pi / 6$ & $a\ cos(5 \pi / 2) = 0$ \\ \hline 417 | $11 \pi / 12$ & $a\ cos(11 \pi / 4) = -\sqrt{2} \ a / 2$ \\ \hline 418 | $\pi$ & $a\ cos(3 \pi) = -a$ \\ \hline 419 | \end{tabular} 420 | \end{center} 421 | 422 | \begin{center} 423 | \includegraphics[scale=0.8]{p1_4i2.jpg} 424 | \end{center} 425 | 426 | Assume the petals are of equal area. We will find the area of the curve from $\theta = 0$ to $\theta = \pi / 6$ and multiply by 2. 427 | 428 | \begin{center} 429 | \includegraphics[scale=0.8]{p1_4i2-integration.jpg} 430 | \end{center} 431 | 432 | \begin{align*} 433 | 2 \int_0^{\pi / 6} \frac{1}{2} r^2 d \theta &= \int_0^{\pi / 6} r^2 d \theta \\ 434 | &= \int_0^{\pi / 6} a^2\ cos^2(3 \theta) d \theta \\ 435 | &= a^2 \int_0^{\pi / 6} cos^2(3 \theta) d \theta \\ 436 | &= a^2 \int_0^{\pi / 6} \frac{1 + cos(6 \theta)}{2} d \theta \\ 437 | &= \frac{a^2}{2} \int_0^{\pi / 6} 1 + cos(6 \theta) d \theta \\ 438 | &= \frac{a^2}{2} (\theta + \frac{sin(6 \theta)}{6}) \bigg]_0^{\pi / 6} \\ 439 | &= \frac{a^2}{2} (\frac{\pi}{6} + \frac{sin(6 \cdot \pi / 6)}{6}) \\ 440 | &= \frac{a^2}{2} (\frac{\pi}{6} + \frac{sin(\pi)}{6}) \\ 441 | &= \frac{a^2 \pi}{12} 442 | \end{align*} 443 | 444 | 445 | \begin{tcolorbox} 446 | \textbf{4I-3} Find the area of the region $0 \leq r \leq e^{3 \theta}$ for $0 \leq \theta \leq \pi$ 447 | \end{tcolorbox} 448 | 449 | Skipped. 450 | 451 | 452 | \begin{tcolorbox} 453 | \textbf{Q1a)} Find the algebraic equation in $x$ and $y$ for the curve \\ 454 | \\ 455 | $x = a\ cos^k t, y = a \ sin^k t$ \\ 456 | \\ 457 | Draw the portion of the curve $0 \leq t \leq \pi / 2$ in the three cases $k = 1, k = 2, k = 3$. 458 | \end{tcolorbox} 459 | 460 | Raising both $x$ and $y$ to the power $2/k$, we get 461 | 462 | $x^{2/k} = a^{2/k} \ cos^2 t$ 463 | 464 | $y^{2/k} = a^{2/k} \ sin^2 t$ 465 | 466 | Summing them, we get $x^{2/k} + y^{2/k} = a^{2/k} cos^2 t + a^{2/k} sin^2 t = a^{2/k}$ 467 | 468 | For $k = 1$, this is $x^2 + y^2 = a^2$ 469 | 470 | \begin{center} 471 | \includegraphics[scale=0.8]{1_keq1.jpg} 472 | \end{center} 473 | 474 | For $k = 2$, this is $x + y = a$ or equivalently, $y = a -x$. 475 | 476 | Assuming $a > 0$, we get: 477 | 478 | \begin{center} 479 | \includegraphics[scale=0.8]{1_keq2.jpg} 480 | \end{center} 481 | 482 | For $k = 3$, this is $x^{2/3} + y^{2/3} = a^{2/3}$ 483 | 484 | \begin{center} 485 | \includegraphics[scale=0.8]{1_keq3.jpg} 486 | \end{center} 487 | 488 | 489 | \begin{tcolorbox} 490 | Q1b) Without calculation, find the arclength in the cases $k = 1$ and $k = 2$. 491 | \end{tcolorbox} 492 | 493 | Frankly speaking, we need calculations for these. 494 | 495 | Arc length for $k = 1$: $\frac{\pi a}{2}$ 496 | 497 | Arc length for $k = 2$: $\sqrt{2a^2}$ 498 | 499 | 500 | \begin{tcolorbox} 501 | Q1c) Find a definite integral formula for the length of the curve for general $k$. Then evaluate the integral in the three cases $k = 1, k = 2$ and $k = 3$. (Your answer in the first two cases should match what you found in part (b), but the calculation takes more time.) 502 | \end{tcolorbox} 503 | 504 | \begin{align*} 505 | x^{2/k} + y^{2/k} &= a^{2/k} \\ 506 | y^{2/k} &= a^{2/k} - x^{2/k} \\ 507 | y &= (a^{2/k} - x^{2/k})^{k/2} \\ 508 | \frac{dy}{dx} &= \frac{k}{2}(a^{2/k} - x^{2/k})^{k/2 - 1}(-\frac{2}{k}x^{(2/k) - 1}) \\ 509 | \\ 510 | ds &= \sqrt{1 + (dy/dx)^2} \ dx \\ 511 | &= \sqrt{1 + (\frac{k}{2}(a^{2/k} - x^{2/k})^{k/2 - 1}(-\frac{2}{k}x^{(2/k) - 1}))^2} \ dx \\ 512 | &= \sqrt{1 + (\frac{k^2}{4}(a^{2/k} - x^{2/k})^{k - 2} \ (\frac{4}{k^2}x^{2(2 - k)/k})} \ dx \\ 513 | &= \sqrt{1 + (a^{2/k} - x^{2/k})^{k - 2} \ (x^{2(2 - k)/k})} \ dx \\ 514 | \\ 515 | \int_{s_0}^{s_1} ds &= \int_{0}^{a} \sqrt{1 + (a^{2/k} - x^{2/k})^{k - 2} \ (x^{2(2 - k)/k})} \ dx \\ 516 | \end{align*} 517 | 518 | For $k = 1$ 519 | 520 | \begin{align*} 521 | \int_{0}^{a} \sqrt{1 + (a^2 - x^2)^{-1} \ x^2} \ dx &= \int_{0}^{a} \sqrt{1 + \frac{x^2}{a^2 - x^2}} \ dx \\ 522 | &= \int_{0}^{a} \sqrt{\frac{a^2}{a^2 - x^2}} \ dx \\ 523 | &= a \int_{0}^{a} \sqrt{\frac{1}{a^2 - x^2}} \ dx \\ 524 | \end{align*} 525 | 526 | Let $x = a\ sin(u)$. Then $dx = a\ cos(u)\ du$. Substitute into above. 527 | 528 | \begin{align*} 529 | a \int_{0}^{\pi / 2} \sqrt{\frac{1}{a^2 - a^2 sin^2(u)}} \ a \ cos(u) \ du &= a^2 \int_{0}^{\pi / 2} \sqrt{\frac{1}{a^2 cos^2(u)}} \ cos(u) \ du \\ 530 | &= a^2 \int_{0}^{\pi / 2} \frac{cos(u)}{a\ cos(u)} \ du \\ 531 | &= a \int_{0}^{\pi / 2} du \\ 532 | &= a\ u \bigg]_0^{\pi / 2} \\ 533 | &= a\pi / 2 534 | \end{align*} 535 | 536 | For $k = 2$ 537 | 538 | \begin{align*} 539 | \int_{0}^{a} \sqrt{1 + (a^{2/k} - x^{2/k})^{k - 2} \ (x^{2(2 - k)/k})} \ dx &= \int_0^a \sqrt{1 + (a - x)^0 \ x^{2(2 - 2) / 2}} \ dx \\ 540 | &= \int_0^a \sqrt{1 + 1 \cdot \ x^0} \ dx \\ 541 | &= \int_0^a \sqrt{2} \ dx \\ 542 | &= \sqrt{2}\ x \bigg]_0^a \\ 543 | &= \sqrt{2} \ a 544 | \end{align*} 545 | 546 | For $k = 3$ 547 | 548 | \begin{align*} 549 | \int_{0}^{a} \sqrt{1 + (a^{2/k} - x^{2/k})^{k - 2} \ (x^{2(2 - k)/k})} \ dx &= \int_{0}^{a} \sqrt{1 + (a^{2/3} - x^{2/3})^{3 - 2} \ (x^{2(2 - 3)/3})} \ dx \\ 550 | &= \int_{0}^{a} \sqrt{1 + (a^{2/3} - x^{2/3}) \ (x^{-2/3})} \ dx \\ 551 | &= \int_{0}^{a} \sqrt{1 + a^{2/3}x^{-2/3} - 1} \ dx \\ 552 | &= \int_{0}^{a} \sqrt{a^{2/3}x^{-2/3}} \ dx \\ 553 | &= \int_{0}^{a} a^{1/3}x^{-1/3} \ dx \\ 554 | &= a^{1/3} \int_{0}^{a} x^{-1/3} \ dx \\ 555 | &= a^{1/3} \cdot \frac{3}{2} x^{2/3} \bigg]_{0}^{a} \\ 556 | &= a^{1/3} \cdot \frac{3}{2} a^{2/3} \\ 557 | &= \frac{3}{2} a 558 | \end{align*} 559 | 560 | 561 | \begin{tcolorbox} 562 | Q2) The hyperbolic sine and cosine are defined by \\ 563 | \\ 564 | $cosh \ x = \frac{e^x + e^{-x}}{2}$, $sinh \ x = \frac{e^x - e^{-x}}{2}$ \\ 565 | \\ 566 | a) Show that \\ 567 | \\ 568 | i) $\frac{d}{dx}sinh \ x = cosh \ x$ and $\frac{d}{dx}cosh \ x = sinh \ x$. \\ 569 | \\ 570 | ii) $\frac{d}{dx}cosh^2 x = 1 + sinh^2 x$ \\ 571 | \\ 572 | iii) $cosh^2 x = \frac{1 + cosh \ 2x}{2}$. 573 | \end{tcolorbox} 574 | 575 | \begin{align*} 576 | \frac{d}{dx}sinh \ x &= \frac{d}{dx} \frac{e^x - e^{-x}}{2} \\ 577 | &= \frac{e^x}{2} - (-1) \frac{e^{-x}}{2} \\ 578 | &= \frac{e^x + e^{-x}}{2} \\ 579 | &= cosh \ x 580 | \end{align*} 581 | 582 | \begin{align*} 583 | cosh^2 x &= (\frac{e^x + e^{-x}}{2})^2 \\ 584 | &= \frac{e^2x + 2 + e^{-2x}}{4} \\ 585 | sinh^2 x &= (\frac{e^x - e^{-x}}{2})^2 \\ 586 | &= \frac{e^2x - 2 + e^{-2x}}{4} \\ 587 | sinh^2 x + 1 &= \frac{e^2x - 2 + e^{-2x}}{4} + \frac{4}{4} \\ 588 | &= \frac{e^2x + e^{-2x} + 2}{4} \\ 589 | &= cosh^2 x 590 | \end{align*} 591 | 592 | \begin{align*} 593 | cosh\ 2x &= \frac{e^{2x} + e^{-2x}}{2} \\ 594 | &= \frac{2e^{2x} + 2e^{-2x}}{4} \\ 595 | &= \frac{2e^{2x} + 2e^{-2x} + 4 - 4}{4} \\ 596 | &= \frac{2(e^{2x} + e^{-2x} + 2 - 2)}{4} \\ 597 | &= \frac{2(e^{2x} + e^{-2x} + 2)}{4} - \frac{4}{4} \\ 598 | &= 2cosh^2 x - 1 \\ 599 | cosh^2 x &= \frac{cosh\ 2x + 1}{2} 600 | \end{align*} 601 | 602 | 603 | \begin{tcolorbox} 604 | Q2) The hyperbolic sine and cosine are defined by \\ 605 | \\ 606 | $cosh \ x = \frac{e^x + e^{-x}}{2}$, $sinh \ x = \frac{e^x - e^{-x}}{2}$ \\ 607 | \\ 608 | b) What curve is described parametrically by $x = cosh\ t$, $y = sinh\ t$? (Give the equation and its name.) 609 | \end{tcolorbox} 610 | 611 | Using $cosh^2 x = 1 + sinh^2 x$, $x = cosh\ x$ and $y = sinh \ t$, we have 612 | 613 | $x^2 = 1 + y^2$ 614 | 615 | $x^2 - y^2 = 1$ 616 | 617 | This is a hyperbola. 618 | 619 | 620 | \begin{tcolorbox} 621 | Q2) The hyperbolic sine and cosine are defined by \\ 622 | \\ 623 | $cosh \ x = \frac{e^x + e^{-x}}{2}$, $sinh \ x = \frac{e^x - e^{-x}}{2}$ \\ 624 | \\ 625 | c) The curve $y = cosh \ x$ is known as a \textit{catenary}. It is the curve formed by a chain whose two ends are held at the same height.\\ 626 | i) Sketch the curve \\ 627 | ii) Find its arclength from the lowest point to the point $(x_1, cosh\ x_1)$ for a fixed $x_1 > 0$ 628 | \end{tcolorbox} 629 | 630 | $\frac{dy}{dx} cosh\ x = sinh\ x = \frac{e^x - e^{-x}}{2}$. This is $0$ when $e^x - e^{-x} = 0$, or $e^x = e^{-x}$, or $x = -x$, when $x = 0$. 631 | 632 | When $x = 0$, $y = \frac{e^0 + e^{-0}}{2} = 1$. Hence $(0, 1)$ is a critical point. 633 | 634 | $\frac{d^2 y}{dx^2} = cosh\ x = y$ . This is $0$ when $e^x + e^{-x} = 0$ which is impossible. There are no points of inflection. 635 | 636 | Now we calculate the intercepts. 637 | 638 | We know from above that when $x = 0$, $y = 1$. Because $e^x > 0$ and $e^{-x} > 0$, $y > 0$. 639 | 640 | $\lim_{x \to -\infty} \frac{e^x + e^{-x}}{2} = \lim_{x \to -\infty} \frac{e^x}{2} + \lim_{x \to -\infty} \frac{e^{-x}}{2} = \infty$ 641 | 642 | Similarly, the limit as $x \to \infty$ is also $\infty$. 643 | 644 | When $x = 0$, $\frac{d^2 y}{dx^2} = cosh \ x = \frac{e^x + e^{-x}}{2} = \frac{e^0 + e^{-0}}{2} = 1 > 0$ . Hence $(0, 1)$ is the minimum point. 645 | 646 | \begin{center} 647 | \includegraphics[scale=0.8]{2c.jpg} 648 | \end{center} 649 | 650 | \begin{align*} 651 | ds &= \sqrt{1 + (\frac{dy}{dx})^2} \ dx \\ 652 | &= \sqrt{1 + sinh^2 \ x} \ dx \\ 653 | &= \sqrt{cosh^2 x} \ dx \\ 654 | &= cosh \ x \ dx 655 | \\ 656 | \int_{s_0}^{s_1} ds &= \int_0^{x_1} cosh \ x \ dx \\ 657 | &= sinh \ x \bigg]_0^{x_1} \\ 658 | &= \frac{e^{x_1} + e^{-x_1}}{2} - \frac{e^0 - e^{-0}}{2} \\ 659 | &= \frac{e^{x_1} + e^{-x_1}}{2} \\ 660 | &= sinh \ x_1 661 | \end{align*} 662 | 663 | 664 | \begin{tcolorbox} 665 | Q2) The hyperbolic sine and cosine are defined by \\ 666 | \\ 667 | $cosh \ x = \frac{e^x + e^{-x}}{2}$, $sinh \ x = \frac{e^x - e^{-x}}{2}$ \\ 668 | \\ 669 | d) FInd the area of the surface of revolution formed by revolving the portion of the curve from part (c) around the $x$-axis This surface is known as a catenoid. It is interesting because it is the surface of least area connecting the two circles that form its edges. If you dip two circles of wire in a soap solution, then (with some caoxing) a soap film will form in this shape. In general, the soap films try to span a frame of wires with a surface with the least area possible. 670 | \end{tcolorbox} 671 | 672 | Surface area: 673 | 674 | \begin{align*} 675 | \int_0^{x_1} 2 \pi y \ dx &= 2 \pi \int_0^{x_1} y \ dx \\ 676 | &= 2 \pi \int_0^{x_1} cosh \ x \ dx \\ 677 | &= 2 \pi sinh \ x \bigg]_0^{x_1} \\ 678 | &= 2 \pi (\frac{e^{x_1} - e^{-x_1}}{2} - \frac{e^0 - e^{-0}}{2}) \\ 679 | &= 2 \pi sinh \ x_1 680 | \end{align*} 681 | 682 | which is $2\pi$ times the arc length. 683 | 684 | 685 | \begin{tcolorbox} 686 | Q3a) Find the area of the right triangle with vertices at $(x, y) = (0, 0)$, $(x, y) = (a, 0)$ and at $(x, y) = (a, h)$, using polar coordinates. (One of the less convenient ways to find the area of a triangle.) 687 | \end{tcolorbox} 688 | 689 | \begin{center} 690 | \includegraphics[scale=0.8]{3a.jpg} 691 | \end{center} 692 | 693 | Suppose $a > 0$, $h > 0$. 694 | 695 | $r = \sqrt{a^2 + h^2}$ 696 | 697 | $\theta = tan^{-1} \frac{h}{a}$. 698 | 699 | $tan \theta = \frac{h}{a}$. 700 | 701 | \begin{align*} 702 | \int_{0}^{tan^{-1}(h/a)} \frac{1}{2} r^2 d \theta &= \frac{1}{2} \int_{0}^{tan^{-1}(h/a)} r^2 \ d \theta \\ 703 | &= \frac{1}{2} \int_{0}^{tan^{-1}(h/a)} a^2 + h^2 \ d \theta \\ 704 | &= \frac{a^2}{2} \int_{0}^{tan^{-1}(h/a)} 1 + \frac{h^2}{a^2} \ d \theta \\ 705 | &= \frac{a^2}{2} \int_{0}^{tan^{-1}(h/a)} 1 + (tan \ \theta)^2 \ d \theta \\ 706 | &= \frac{a^2}{2} \int_{0}^{tan^{-1}(h/a)} sec^2 \ \theta \ d \theta \\ 707 | &= \frac{a^2}{2} tan \ \theta \bigg]_{0}^{tan^{-1}(h/a)} \\ 708 | &= \frac{a^2}{2} [tan(tan^{-1}(h/a)) - tan(0)] \\ 709 | &= \frac{a^2}{2} [\frac{h}{a} - 0] \\ 710 | &= \frac{ah}{2} 711 | \end{align*} 712 | 713 | 714 | \begin{tcolorbox} 715 | Q3b) Find the equation in $(x, y)$ coordinates for the curve $r = 1 / (1 + sin\ \theta)$ and sketch it. 716 | \end{tcolorbox} 717 | 718 | \begin{align*} 719 | r &= \frac{1}{1 + sin\ \theta} \\ 720 | r &+ r\ sin\ \theta = 1 \\ 721 | r &+ y = 1 \\ 722 | x &= r\ cos\ \theta = \frac{cos\ \theta}{1 + sin\ \theta} \\ 723 | y &= r\ sin\ \theta = \frac{sin\ \theta}{1 + sin\ \theta} \\ 724 | x^2 + y^2 &= \frac{cos^2 \theta + sin^2 \theta}{(1 + sin\ \theta)^2} = \frac{1}{(1 + sin\ \theta)^2} = r^2 \\ 725 | r &= (x^2 + y^2)^{1/2} \\ 726 | r + y &= (x^2 + y^2)^{1/2} + y = 1 727 | \end{align*} 728 | 729 | Hence the equation is $(x^2 + y^2)^{1/2} + y = 1$. 730 | 731 | Skipping the sketch. 732 | 733 | 734 | \begin{tcolorbox} 735 | Q3c) Find the area of the region $0 \leq r \leq 1 / (1 + sin\ \theta)$, $0 \leq \theta \leq \pi$, using the \textit{rectangular} coordinate formula you found in part (b). 736 | \end{tcolorbox} 737 | 738 | \begin{align*} 739 | (x^2 + y^2)^{1/2} + y &= 1 \\ 740 | (x^2 + y^2)^{1/2} &= 1 - y \\ 741 | x^2 + y^2 &= (1 - y)^2 \\ 742 | x^2 = (1 - y)^2 - y^2 &= 1 - 2y \\ 743 | y &= \frac{1 - x^2}{2} 744 | \end{align*} 745 | 746 | Area: 747 | 748 | \begin{align*} 749 | \int_{-1}^1 \frac{1 - x^2}{2} dx &= (\frac{x}{2} - \frac{x^3}{6}) \bigg]_{-1}^1 \\ 750 | &= \frac{1}{2} - \frac{1}{6} - (\frac{-1}{2} - \frac{(-1)^3}{6}) \\ 751 | &= \frac{1}{2} - \frac{1}{6} + \frac{1}{2} - \frac{1}{6} \\ 752 | &= \frac{2}{3} 753 | \end{align*} 754 | 755 | 756 | \begin{tcolorbox} 757 | Q3d) Find the area of the region in part (c) using polar coordinates. One way to evaluate the integral is to change variables to $u = \theta - \pi / 2$, and then use the half angle formula \\ 758 | \\ 759 | $1 + cos\ u = 2 cos^2(u/2)$ \\ 760 | \\ 761 | This area was already computed in part (c), but the polar coordinate formula is still valuable because it gives the area of any sector, not just the one that is bounded by a horizontal line. The area swept out from the viewpoint of the focus of an ellipse, parabola, or hyperbola is related by Kepler's law to the speed of planets and comets. 762 | \end{tcolorbox} 763 | 764 | Area: 765 | 766 | \begin{align*} 767 | \int_0^{\pi} \frac{1}{2} r^2 d\theta = \frac{1}{2} \int_0^{\pi} (\frac{1}{1 + sin\ \theta})^2 d\theta 768 | \end{align*} 769 | 770 | Let $u = \theta - \pi / 2$. Then $du = d \theta$ 771 | 772 | When $\theta = 0$, $u = -\pi / 2$. When $\theta = \pi$, $u = \pi / 2$. 773 | 774 | Then $sin\ \theta = sin(u + \pi / 2) = cos\ u$ 775 | 776 | \begin{align*} 777 | \frac{1}{2} \int_0^{\pi} (\frac{1}{1 + sin\ \theta})^2 d\ \theta &= \frac{1}{2} \int_{-\pi / 2}^{\pi / 2} (\frac{1}{2cos^2(u / 2)})^2\ du \\ 778 | &= \frac{1}{2} \int_{-\pi / 2}^{\pi / 2} \frac{1}{4} sec^4(u / 2)\ du \\ 779 | \end{align*} 780 | 781 | Let $w = tan(u / 2)$. Then $dw = \frac{1}{2} sec^2(u / 2) \ du$ 782 | 783 | \begin{align*} 784 | \int_{-\pi / 2}^{\pi / 2} sec^4(u / 2)\ du &= \int_{tan(-\pi / 2 / 2)}^{tan(\pi / 2 / 2)} sec^2(u / 2) \cdot sec^2(u / 2) \ du \\ 785 | &= \int_{-1}^{1} (1 + tan^2(u / 2)) \cdot (2 \cdot \frac{1}{2} sec^2(u / 2)) \ du \\ 786 | &= \int_{-1}^{1} (1 + w^2) \cdot 2 \ dw \\ 787 | &= 2 \int_{-1}^{1} 1 + w^2 dw \\ 788 | &= 2 (w + \frac{w^3}{3}) \bigg]_{-1}^1 \\ 789 | &= 2 (1 + \frac{1}{3} - (-1 + \frac{(-1)^3}{3})) \\ 790 | &= 2 (1 + \frac{1}{3} + 1 + \frac{1}{3}) \\ 791 | &= 2 (2 + \frac{2}{3}) \\ 792 | &= \frac{16}{3} \\ 793 | \\ 794 | \frac{1}{2} \int_{-\pi / 2}^{\pi / 2} \frac{1}{4} sec^4(u / 2)\ du &= \frac{1}{8} \int_{-\pi / 2}^{\pi / 2} sec^4(u / 2)\ du \\ 795 | &= \frac{1}{8} \cdot \frac{16}{3} \\ 796 | &= \frac{2}{3} 797 | \end{align*} 798 | 799 | \end{document} 800 | --------------------------------------------------------------------------------