From 22cbe0dbb566345370be9babcf7b9bf531c1c296 Mon Sep 17 00:00:00 2001 From: Mark Date: Sun, 13 Nov 2022 13:17:43 -0800 Subject: [PATCH] Added vector handouts --- Intermediate/Newton's Laws/main.tex | 534 ++++++++++++++++++++++++ Intermediate/Vectors 1/main.tex | 494 +++++++++++++++++++++++ Intermediate/Vectors 2/main.tex | 601 ++++++++++++++++++++++++++++ 3 files changed, 1629 insertions(+) create mode 100755 Intermediate/Newton's Laws/main.tex create mode 100755 Intermediate/Vectors 1/main.tex create mode 100755 Intermediate/Vectors 2/main.tex diff --git a/Intermediate/Newton's Laws/main.tex b/Intermediate/Newton's Laws/main.tex new file mode 100755 index 0000000..ca74990 --- /dev/null +++ b/Intermediate/Newton's Laws/main.tex @@ -0,0 +1,534 @@ +% https://git.betalupi.com/Mark/latex-packages +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +% Last built with version 1.0.6 +\documentclass[solutions, simplefooter]{ormc_handout} + + +\begin{document} + + \maketitle + + + {Newton's Laws of Motion} + { + Prepared by Mark on \today \\ + Based on a handout by Oleg Gleizer + } + + \section{Newton's First Law} + If the net force acting on an object is zero, the velocity of that object does not change. \\ + Conversely, if the velocity of an object doesn't change, the net force acting on it is zero. + + \medskip + + In the context of vectors, the ``net force'' is the sum of all the force vectors acting on the object. ``Speed'' is the length (or \textit{magnitude}) of the velocity vector. + + + \problem{} + There are no forces acting on the object + $A$ below. The current velocity of the object, + in meters per second, is represented by the vector $\overrightarrow{v}$. + Draw the position of the object two seconds later. + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-11,-7) grid (1,1); + \draw [line width = 1.5pt, ->] (-9,0) -- (-5,-3); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-6.9,-1.5); + \filldraw (-9,0) circle (3pt); + \coordinate [label = left:{$A$}] (a) at (-9,0); + \end{tikzpicture} + \end{normalsize} + \end{center} + + The sides of the grid squares + on the picture above are one metre long. + What is the speed of the object? + \vfill + + What distance would the object cover + in two seconds? \\ + \vfill + + \pagebreak + + \problem{} + There are no forces acting on the object + $A$ below. + The current velocity of this object, + in miles per hour, + is represented by the vector $\overrightarrow{v}$. + Draw the position of the object half an hour later. + \vspace{20pt} + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-11,-7) grid (1,1); + \draw [line width = 1.5pt, ->] (-9,0) -- (-1,-6); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-4.8,-3); + \filldraw (-9,0) circle (3pt); + \coordinate [label = left:{$A$}] (a) at (-9,0); + \end{tikzpicture} + \end{normalsize} + \end{center} + + The sides of the grid squares + on the picture above are ten miles long. + What is the speed of this object? + + \vfill + + What distance would the object cover + in half an hour? \\ + + \vfill + + What distance would the object cover + in three hours? \\ + + \vfill + + \pagebreak + + \definition{Acceleration} + \textit{Acceleration} is the rate at which velocity changes. + Let's represent acceleration by the vector $\overrightarrow{a}$. + + If $\overrightarrow{a}$ does not change over time, then the speed + of an object at time $t$ is given by the following equation: + + \begin{equation} \label{eq:ac} + \overrightarrow{v_t} = \overrightarrow{v_0} + + t \overrightarrow{a}. + \end{equation} + + \problem{} + It takes a minivan seven seconds to accelerate from 0 to 60 miles per hour. Find its acceleration in meters per second squared. \\ + \hint{1 mile $\approx$ 1600 meters} + + \vfill + + Note that in the previous problem, motion is one-dimensional (it happens on a straight line). In this case, both velocity and acceleration are one-dimensional vectors---in other words, (real) numbers! \\ + + In general, velocity and acceleration are \textit{not} numbers, but vectors. You'll see this in the next few problems. + + \pagebreak + + + Now, let's try a few examples with vectors: \\ + \medskip + Consider an object, currently at $P_0$, moving along the vector $\overrightarrow{v_0}$. + + As before, let $\overrightarrow{a}$ represent the acceleration of the object. This could be caused by gravity, current, or any other constant force. + + \medskip + + One second later, the object will be at $P_1$, and has the velocity vector $\overrightarrow{v_1} = \overrightarrow{v_0} + \overrightarrow{a}$. + + \medskip + + Two seconds later, the object will be positioned at $P_2$ and will have the velocity vector $\overrightarrow{v_2} = \overrightarrow{v_0} + 2\overrightarrow{a}$. + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-19,-15) grid (1,1); + + \draw [line width = 1.5pt, ->] (-18,-1) -- (-16,-4); + \coordinate [label = left:{$\overrightarrow{v_0}$}] + (v0) at (-17.1,-2.6); + \draw [line width = 1.5pt, ->] (-18,-1) -- (-15,-2); + \coordinate [label = above:{$\overrightarrow{a}$}] + (a) at (-16.4,-1.5); + \filldraw (-18,-1) circle (3pt); + \coordinate [label = above:{$P_0$}] (p0) at (-18,-1); + + \filldraw (-14.5,-4.5) circle (3pt); + \coordinate [label = right:{$P_1$}] (p1) at (-14.45,-4.4); + \draw [line width = 1.5pt, ->] (-14.5,-4.5) -- (-9.5,-8.5); + \coordinate [label = right:{$\overrightarrow{v_1}$}] + (v1) at (-11.75,-6.5); + \draw [->] (-14.5,-4.5) -- (-12.5,-7.5); + \coordinate [label = left:{$\overrightarrow{v_0}$}] + (v01) at (-13.6,-6); + \draw (-12.5,-7.5) -- (-9.5,-8.5); + \coordinate [label = left:{$\overrightarrow{a}$}] + (a1) at (-11,-8.3); + + \filldraw (-8,-9) circle (3pt); + \coordinate [label = above:{$P_2$}] (b) at (-8,-9); + \draw [->] (-8,-9) -- (-6,-12); + \coordinate [label = left:{$\overrightarrow{v_0}$}] + (v02) at (-7,-10.5); + \draw (-6,-12) -- (0,-14); + \coordinate [label = left:{$2\overrightarrow{a}$}] + (a1) at (-3.5,-13); + \draw [line width = 1.5pt, ->] (-8,-9) -- (0,-14); + \end{tikzpicture} + \end{normalsize} + \end{center} + + $t$ seconds later, the object + will have the velocity + \[ + \overrightarrow{v_t} = \overrightarrow{v_0} + t \overrightarrow{a} + \] + + Note that the word {\it acceleration} has two different meanings. + One is a vector, as in the example above. + The other meaning is the length + of the vector that represents the magnitude + of the velocity change. In this case, + we do not put an arrow above the letter + $a$ representing acceleration. + In other words, $a = |\overrightarrow{a}|$. + Similarly, speed is the length + of the velocity vector, + $v = |\overrightarrow{v}|$. + + \vfill + \pagebreak + + + \example{} + Moving with a constant acceleration, + an object moves from point $A$ to point $B$ + in one second. The velocities of the motion, + in metres per second, are represented + by the vectors $\overrightarrow{v_0}$ and + $\overrightarrow{v_1}$. + Find the acceleration vector. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-18,-8) grid (1,1); + \draw [line width = 1.5pt, ->] (-17,-1) -- (-13,-4); + \coordinate [label = above:{$\overrightarrow{v_0}$}] + (v0) at (-14.9,-2.5); + \filldraw (-17,-1) circle (3pt); + \coordinate [label = left:{$A$}] (a) at (-17,-1); + \filldraw (-10,-3) circle (3pt); + \coordinate [label = above right:{$B$}] (b) at (-10,-3.1); + \draw [line width = 1.5pt, ->] (-10,-3) -- (0,-4); + \coordinate [label = above:{$\overrightarrow{v_1}$}] + (v1) at (-5,-3.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + + According to formula on the previous page, + $\overrightarrow{v_1} = \overrightarrow{v_0} + (1\text{ sec}) \times \overrightarrow{a}$. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-18,-8) grid (1,1); + \draw [line width = 1.5pt, ->] (-17,-1) -- (-13,-4); + \coordinate [label = above:{$\overrightarrow{v_0}$}] + (v0) at (-14.9,-2.5); + \filldraw (-17,-1) circle (3pt); + \coordinate [label = left:{$A$}] (a) at (-17,-1); + \filldraw (-10,-3) circle (3pt); + \coordinate [label = above right:{$B$}] (b) at (-10,-3.1); + \draw [line width = 1.5pt, ->] (-10,-3) -- (0,-4); + \coordinate [label = above:{$\overrightarrow{v_1}$}] + (v1) at (-5,-3.5); + \draw [line width = 1.5pt, ->] (-10,-3) -- (-6,-6); + \coordinate [label = below:{$\overrightarrow{v_0}$}] + (v01) at (-8.2,-4.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \problem{} + Complete \ref{accel_example}: \\ + Draw the acceleration vector on the above diagram. + + + \vfill + \pagebreak + + \problem{} + At an initial moment in time, + the object at point $A$ + is moving with the velocity vector + $\overrightarrow{v_0}$, + measured in metres per second. + The constant acceleration + acting on the object + is represented by the vector + $\overrightarrow{a}$, + measured in meters per second squared. + Two seconds later, the object is located + at point $B$. Draw its velocity vector + at that moment. + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-19,-15) grid (1,1); + \draw [line width = 1.5pt, ->] (-18,-1) -- (-16,-4); + \coordinate [label = left:{$\overrightarrow{v_0}$}] + (v0) at (-17.1,-2.6); + \draw [line width = 1.5pt, ->] (-18,-1) -- (-15,-2); + \coordinate [label = above:{$\overrightarrow{a}$}] + (a) at (-16.4,-1.5); + \filldraw (-18,-1) circle (3pt); + \coordinate [label = left:{$A$}] (a) at (-18,-1); + \filldraw (-8,-9) circle (3pt); + \coordinate [label = left:{$B$}] (b) at (-8,-9); + %\draw [line width = 1.5pt, ->] (-8,-9) -- (0,-14); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vspace{12pt} + + + The side length of the grid squares + on the picture above is one metre. + Find the speed of the object when it is + at point $B$. + + \vfill + \pagebreak + + \section{Newton's Second Law} + + The second law is simple: + \[ + \overrightarrow{F} = m \times \overrightarrow{a} + \] + + Here, $\overrightarrow{F}$ is the net force acting on an object with mass $m$, and $\overrightarrow{a}$ is the acceleration the object experiences as a result of this action. Mass is a measure of an object's \textit{interia}: the heavier an object is, the more effort it takes to change its velocity. \\ + + In civilized countries, mass is measured in grams and force is measured in \textit{newtons}. One newton is the force it takes to accelerate 1 kg of mass to 1 meter per second. In other words, + + \[ + 1\ N = (1\ kg) (1\ \frac{m}{s}) + \] + + \problem{} + The {\it Millenium Falcon}, at point $A$ at the moment, + is trying to escape from the Death Star, which is trying + to arrest the ship using its attracting beam. + The thrust of the Falcon's engines, + 200,000 kN in total, is represented + by the vector $\overrightarrow{T}$. + The force of the beam is represented + by the vector $\overrightarrow{B}$. + In addition, a nearby star + exerts a gravitational force of + $\overrightarrow{G}$ on the ship. + Draw the vector of the net force + acting on the vessel. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-16,-9) grid (1,1); + \draw [line width = 1.5pt, ->] (-8,-5) -- (-12,-5); + \coordinate [label = below:{$\overrightarrow{T}$}] (v) at (-10,-5.1); + \filldraw (-8,-5) circle (3pt); + \coordinate [label = below:{$A$}] (a) at (-8,-5.1); + \draw [line width = 1.5pt, ->] (-8,-5) -- (-3,-6); + \coordinate [label = above:{$\overrightarrow{B}$}] (b) at (-5.4,-5.5); + \draw [line width = 1.5pt, ->] (-8,-5) -- (-9,-2); + \coordinate [label = left:{$\overrightarrow{G}$}] (g) at (-8.6,-3.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + + The mass of the ship + is 400,000 kg. What acceleration, + in metres per second squared, + does the ship experience? + \vfill + + + + The purpose of this daring mission + was to find out the force + that the the attracting beam exerts. + However, Han Solo is not particularly + good with vectors. Please help him complete the mission. + \vfill + \pagebreak + + + \section{Newton's Third Law} + + Newton's third law also concerns forces. It states that \textit{every action has an equal and opposite reacton}. + + In other words, this means that when one object exerts force on another, the second simultaneously exerts a force equal in magnitude and opposite in direction to the force exerted on it by the first. + + \medskip + + In fact, we saw this in our first handout! + + \begin{tcolorbox}[ + colback=white, + colframe=gray!75!black, + title={ Handout 1, Page 7 } + ] + + + Here is an important example of an inverse vector. When you stand still, the floor pushes you up with the force opposite to the force of the gravitational pull, a.k.a. \textit{weight}. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw (0,0) -- (6,0); + \draw (2.5,0) -- (3,1); + \draw (3.5,0) -- (3,1); + \draw (3,1) -- (3,2); + \draw (3,2.3) circle (.3); + \draw (2.6,1.7) -- (3.4,1.7); + \draw [->, line width = 2pt] (2.5,0) -- (2.5,-1); + \coordinate [label = below: {weight}] + (g) at (2.5,-1.1); + \draw [->, line width = 2pt] (3.5,0) -- (3.5,1); + \coordinate [label = right: {floor reaction}] + (w) at (3.7,1); + \end{normalsize} + \end{tikzpicture} + \end{center} + The two opposing vectors add up to the zero vector, and therefore you don't move. + + \end{tcolorbox} + + \problem{} + In a second, the truck and car on the picture below will collide in a crash test. + The weight of the truck is 20 metric tons. The weight of the car is 2 metric tons. + Find the ratio of the accelerations, + $a_t$ (acceleration of the truck) and + $a_c$ (that of the car), + the vehicles will undergo. \\ + + \begin{center} + \begin{tikzpicture} + \draw [gray!50!black] (-7,0) -- (7,0); + + \filldraw [gray!80!blue] (-3.6,1.5) -- (-4.4,1.5) -- + (-4.4,2) -- (-6.8,2) -- (-6.8,.3) -- (-3.6,.3) -- + (-3.6,1.5); + \filldraw (-6,.3) circle (.3); + \filldraw [gray] (-6,.3) circle (.15); + \filldraw (-4,.3) circle (.3); + \filldraw [gray] (-4,.3) circle (.15); + + \filldraw [red!80!white] (4,.2) -- (6.6,.2) -- (6.55,.6) -- + (6.2,.6) -- (6.1,.8) -- (4.9,.8) -- (4.7,.6)-- + (4.2,.6) -- (4,.45) -- (4,.2); + \filldraw (6,.2) circle (.2); + \filldraw [gray] (6,.2) circle (.1); + \filldraw (4.6,.2) circle (.2); + \filldraw [gray] (4.6,.2) circle (.1); + \end{tikzpicture} + \end{center} + + \medskip + + \begin{huge} + $a_t \div a_c =$ + \end{huge} + + \vfill + \newpage + + \section{Bonus Problems} + + \problem{} + Use vector algebra to prove + that diagonals of a parallelogram + in the Euclidean plane + split each other in halves. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw (0,0) -- (4,-2); + \coordinate [label = left: {$A$}] (a) at (0,0); + \coordinate [label = right: {$B$}] (b) at (4,-2); + \coordinate [label = above: {$\overrightarrow{v}$}] (v1) at (2,-.9); + \draw (-1,-3) -- (3,-5); + \coordinate [label = left: {$C$}] (c) at (-1,-3); + \coordinate [label = right: {$D$}] (d) at (3,-5); + \coordinate [label = below: {$\overrightarrow{v}$}] (v2) at (1,-4.1); + \draw (-1,-3) -- (0,0); + \draw (3,-5) -- (4,-2); + \coordinate [label = left: {$\overrightarrow{w}$}] (w1) at (-.55,-1.5); + \coordinate [label = right: {$\overrightarrow{w}$}] (w2) at (3.55,-3.5); + \draw (0,0) -- (3,-5); + \draw (-1,-3) -- (4,-2); + \coordinate [label = above: {$M$}] (m) at (1.7,-2.5); + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + \newpage + + \problem{} + Use vector algebra to prove that + all the three medians of a triangle + in the Euclidean plane intersect + at one point that splits each of the medians + in the ratio 2:1, counting from the vertices. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) -- (6,-1); + \draw (0,0) -- (3,-5); + \draw (3,-5) -- (6,-1); + \draw [color = white] (3,-5) -- (9,-6); + \draw [color = white] (6,-1) -- (9,-6); + \coordinate [label = left: {$A$}] (a) at (0,0); + \coordinate [label = right: {$B$}] (b) at (6,-1); + \coordinate [label = below: {$C$}] (c) at (3,-5); + \coordinate [label = above: {$\overrightarrow{v}$}] + (v) at (3,-.45); + \coordinate [label = left: {$\overrightarrow{w}$}] + (w) at (1.45,-2.5); + \filldraw (4.5,-3) circle (1.5pt); + \coordinate [label = right: {$M_A$}] (ma) at (4.65,-2.85); + \draw (0,0) -- (4.5,-3); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \vfill + \newpage + + \problem{} + Find the area of an equilateral triangle + with the side length $a$. + \vfill + + \problem{} + Does there exist an equilateral triangle + in the Euclidean plane + such that all of its vertices + have integral coordinates? + Why or why not? \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-15,-9) grid (1,1); + \draw [line width = 1.5pt, ->] (-7,-9) -- (-7,1); + \coordinate [label = above:{$y$}] (y) at (-7,1); + \draw [line width = 1.5pt, ->] (-15,-4) -- (1,-4); + \coordinate [label = right:{$x$}] (x) at (1,-4); + \coordinate [label = below:{$1$}] (ox) at (-6,-4); + \coordinate [label = left:{$1$}] (oy) at (-7,-3); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + +\end{document} \ No newline at end of file diff --git a/Intermediate/Vectors 1/main.tex b/Intermediate/Vectors 1/main.tex new file mode 100755 index 0000000..badc084 --- /dev/null +++ b/Intermediate/Vectors 1/main.tex @@ -0,0 +1,494 @@ +% https://git.betalupi.com/Mark/latex-packages +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +% Last built with version 1.0.4 +\documentclass[solutions]{ormc_handout} +\usepackage{adjustbox} + +\begin{document} + + \begin{adjustbox}{minipage=0.7\textwidth, margin=0pt \smallskipamount,center} + \begin{center} + \textsc{Intermediate 2 \hfill ORMC Summer Sessions} \\ + \rule{\linewidth}{0.2mm}\\ + + \huge + Vectors 1\\ + \normalsize + \vspace{1ex} + Prepared by Mark on \today. \\ + Based on a handout by Oleg Gleizer. + \rule{\linewidth}{0.2mm}\\ + \end{center} + \end{adjustbox} + + + \section{Warm-Up} + + \problem{} + Simplify the following fraction: \\ + + $\displaystyle{ + \frac{ + \displaystyle{\frac12} + }{ + \phantom{..} + \displaystyle{\frac13} - + \displaystyle{\frac14} + \phantom{..} + } + } =$ + \vfill + + \problem{} + Simplify the following fraction: \\ + $\displaystyle{ + \frac{ + \displaystyle{\frac{a}{b}} - + \displaystyle{\frac{c}{d}} + }{ + \phantom{..} + \displaystyle{\frac{a}{d}} + + \displaystyle{\frac{c}{b}} + \phantom{..} + } + } =$ + \vfill + + + \problem{} + The point $A$ is placed inside a circle. + + \begin{center} \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) circle (2cm); + \filldraw (40:1.2cm) circle (2pt); + \coordinate [label = left: {$A$}] (a) at (40:1.2cm); + \filldraw (0,0) circle (2pt); + \end{tikzpicture} + \end{normalsize} \end{center} + + Cut the circle into two parts so that you can move one to make a circle centered at $A$. + + \vfill + \pagebreak + + \section{Vectors} + + \definition{} + A vector in the Euclidean plane is a directed line segment. + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-2); + \coordinate [label = left: {$A$}] (a) at (0,0); + \coordinate [label = right: {$B$}] (b) at (4,-2); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + \end{normalsize} + \coordinate [label = left: + {$v = \overrightarrow{AB}$}] (c) at (10,-1); + \end{tikzpicture} + \end{center} + \vspace{30pt} + + For the vector $v = \overrightarrow{AB}$, point $A$ is called {\it initial} and point $B$ is called {\it terminal}. \\ + Two vectors, $v = \overrightarrow{AB}$ and $w = \overrightarrow{CD}$ are considered equivalent if the quadrilateral $ABDC$ is a parallelogram. + \vspace{20pt} + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-2); + \coordinate [label = left: {$A$}] (a) at (0,0); + \coordinate [label = right: {$B$}] (b) at (4,-2); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + \draw [->, line width = 2pt] (-1,-3) -- (3,-5); + \coordinate [label = left: {$C$}] (c) at (-1,-3); + \coordinate [label = right: {$D$}] (d) at (3,-5); + \coordinate [label = below: {$w$}] (w) at (1,-4.1); + \draw (-1,-3) -- (0,0); + \draw (3,-5) -- (4,-2); + \end{normalsize} + \end{tikzpicture} + \end{center} + \vspace{20pt} + + In other words, two vectors are equivalent if they have the same length and direction. If this is the case, we write $v = w$. + + \note{ + Convince yourself that this is true. Why are these two definitions of vector equivalence interchangable? + } + + \note{ + A vector is characterized by its direction and length. One cannot make a formal definition out of this observation, because a ``direction'' is formally defined in terms of a vector. + } + + \vfill + \pagebreak + + \theorem{} + If two distinct straight lines in the Euclidean plane form the angles of equal size with a third straight line in the plane, then they are parallel. + + In other words, to check that the lines $a$ and $b$ on the picture below have no common point, you don't need to travel to infinity. + All you need to do is to measure the angles $\alpha$ and $\gamma$. If $\alpha = \gamma$, then $a$ is parallel to $b$. \\ + + \begin{center} + \begin{footnotesize} + \begin{tikzpicture} + \draw [line width=1pt] (-5,0) -- (5,0); + \draw [line width=1pt] (-4,-2) -- (2,4); + \draw[blue] (-1,0) arc (0:45:1); + \coordinate [label=right:{$\gamma$}] (g) at (-1.1,0.5); + \draw [line width=1pt] (-5,2) -- (5,2); + \draw[blue] (1,2) arc (0:45:1); + \coordinate [label=right:{$\alpha$}] (a) at (0.9,2.5); + \coordinate [label=right:{$b$}] (b) at (5,2); + \coordinate [label=right:{$a$}] (a) at (5,0); + \coordinate [label=right:{$c$}] (c) at (2,4); + \end{tikzpicture} + \end{footnotesize} + \end{center} + + \vfill + + \theorem*{Euclid's 5th postulate:} + For any straight line in the (Euclidean) plane and for any point away from it, there exists a unique straight line that passes through the point and is parallel to the original line. \\ + + \begin{center} + \begin{footnotesize} + \begin{tikzpicture} + \draw [line width=1pt] (-6,0) -- (6,0); + \filldraw [black] (3,2) circle (2pt); + \draw [line width=1pt] (-6,2) -- (6,2); + \end{tikzpicture} + \end{footnotesize} + \end{center} + + \vfill + + \problem{} + Use a compass and a ruler to draw a straight line parallel to the one given below and passing through the given point not lying on the original straight line. + + \begin{center} + \begin{footnotesize} + \begin{tikzpicture} + \draw [line width=1pt] (-6,0) -- (6,0); + \filldraw [black] (3,4) circle (2pt); + \end{tikzpicture} + \end{footnotesize} + \end{center} + + \vfill + \newpage + + + \problem{} + Use a compass and a ruler to construct a vector $w$ with initial point $C$ equal to the vector $v$ below. + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (6,-2); + \coordinate [label = above: {$v$}] (v) at (3,-.9); + \filldraw (-4,-5) circle (1.5pt); + \coordinate [label = below left: {$C$}] (c) at (-4,-5.1); + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + + Physical forces, such as the force of gravity or the force that pulls together two magnets are vectors in three dimensions. The direction of a vector shows the direction in which the corresponding force is acting. The length of the vector shows the strength of the force. \\ + + \problem{} + On the picture below, draw the vectors of the gravitational pull the Earth exerts on you and on your Math Circle leader. + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) circle (3); + \coordinate [label = above: {Earth}] (a) at (0,2); + \filldraw (0,0) circle (1pt); + \coordinate [label = below: {centre}] (c) at (0,-.1); + \draw [line width = 1pt] (30:3) -- (25:3.5); + \draw [line width = 1pt] (20:3) -- (25:3.5); + \draw [line width = 1pt] (25:4) -- (25:3.5); + \draw (25:4.15) circle (.15); + \draw [line width = 1pt] (3.3,1.9) -- (3.6,1.3); + \coordinate [label = above right: {Oleg}] (o) at (25:4.3); + \draw [line width = 1pt] (190:3) -- (188:3.2); + \draw [line width = 1pt] (186:3) -- (188:3.2); + \draw [line width = 1pt] (188:3.5) -- (188:3.2); + \draw (188:3.65) circle (.15); + \draw [line width = 1pt] (-3.35,-.23) -- (-3.28,-.68); + \coordinate [label = above left: {Me}] (m) at (188:3.85); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \begin{itemize} + \item Where do the gravitational force vectors point? Why? + \item If Oleg is twice as heavy as you are, how would you draw the gravitational pull vectors on the above picture? + \end{itemize} + + \vfill + \pagebreak + + \vspace{60pt} + + Motion can be represented by a vector, too. The direction of the velocity vector shows the direction in which an object is moving at the moment. The length of the velocity vector represents the object's \textit{speed}. It shows how fast an object is moving at the moment. \\ + + \problem{} + The truck on the picture below + is going 30 mph. + The car on the same picture + is speeding at 90 mph the opposite way. + Draw the corresponding velocity vectors. \\ + + \begin{center} + \begin{tikzpicture} + \draw [gray!50!black] (-7,0) -- (7,0); + + \filldraw [gray!80!blue] (-3.6,1.5) -- (-4.4,1.5) -- + (-4.4,2) -- (-6.8,2) -- (-6.8,.3) -- (-3.6,.3) -- + (-3.6,1.5); + \filldraw (-6,.3) circle (.3); + \filldraw [gray] (-6,.3) circle (.15); + \filldraw (-4,.3) circle (.3); + \filldraw [gray] (-4,.3) circle (.15); + + \filldraw [red!80!white] (4,.2) -- (6.6,.2) -- (6.55,.6) -- + (6.2,.6) -- (6.1,.8) -- (4.9,.8) -- (4.7,.6)-- + (4.2,.6) -- (4,.45) -- (4,.2); + \filldraw (6,.2) circle (.2); + \filldraw [gray] (6,.2) circle (.1); + \filldraw (4.6,.2) circle (.2); + \filldraw [gray] (4.6,.2) circle (.1); + \end{tikzpicture} + \end{center} + \vspace{40pt} + + Velocities and forces of the real world are vectors in three dimensions. To keep things simple, we'll begin our study of vectors in two dimensions. Everything we are going to learn about vectors in two-dimensional space is be valid in a Euclidean space of three---or more---dimensions. + + \vfill + + \section{Adding Vectors} + + To find the sum of two vectors $v$ and $w$, one needs to take $w$ so that the initial point of $w$ coincides with the terminal point of $v$. The vector originating at the initial point of $v$ and terminating at the terminal point of $w$ is the sum $v + w$. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-2); + %\coordinate [label = left: {$A$}] (a) at (0,0); + %\coordinate [label = right: {$B$}] (b) at (4,-2); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + \draw [->, line width = 2pt] (4,-2) -- (2,-4); + %\coordinate [label = left: {$D$}] (d) at (-2,-2); + %\coordinate [label = right: {$C$}] (c) at (2.15,-4); + \coordinate [label = right: {$w$}] (w) at (3.1,-3.1); + \draw [->, line width = 2pt] (0,0) -- (2,-4); + \coordinate [label = left: + {$v + w$}] (p) at (.9,-2.15); + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + \pagebreak + + \problem{} + Use a compass and a ruler to construct the sum $v + w$ of the vectors $v$ and $w$ given below. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (5,-2); + \coordinate [label = above: {$v$}] (v) at (2.5,-.95); + \draw [->, line width = 2pt] (-3,-3) -- (-2,-6); + \coordinate [label = left: {$w$}] (w) at (-2.55,-4.5); + \end{normalsize} + \end{tikzpicture} + \end{center} + + \definition{The Zero Vector} + A vector that has coinciding initial and terminal points is called the {\it zero vector} and is denoted as $\overrightarrow{0}$. According to the above definition of the vector addition, + \begin{equation} + v + \overrightarrow{0} = + \overrightarrow{0} + v = + v + \end{equation} + for any vector $v$. + + \vfill + + \definition{Inverses of Vectors} + A vector $w$ such that $w + v = \overrightarrow{0}$ is called the \textit{inverse} of $v$ and is denoted as $-v$. The vector $-v$ lies either on the same straight line as $v$ or on a parallel one, has the same length as $v$, but points in the opposite direction: \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-2); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + \draw [<-, line width = 2pt] (-1,-3) -- (3,-5); + \coordinate [label = below: {$-v$}] (n) at (.85,-4.1); + \end{normalsize} + \end{tikzpicture} + \end{center} + \vspace{20pt} + + Note that $-v + v = \overrightarrow{0}$ by definition, but the validity of the equation $v + (-v) = \overrightarrow{0}$ follows from the definitions of vector addition and the zero vector. We can combine both into the following: + \begin{equation} + -v + v = + v + (-v) = + \overrightarrow{0} + \end{equation} + + \vfill + \pagebreak + + Here is an important example of an inverse vector. When you stand still, the floor pushes you up with the force opposite to the force of the gravitational pull, a.k.a. \textit{weight}. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw (0,0) -- (6,0); + \draw (2.5,0) -- (3,1); + \draw (3.5,0) -- (3,1); + \draw (3,1) -- (3,2); + \draw (3,2.3) circle (.3); + \draw (2.6,1.7) -- (3.4,1.7); + \draw [->, line width = 2pt] (2.5,0) -- (2.5,-1); + \coordinate [label = below: {weight}] + (g) at (2.5,-1.1); + \draw [->, line width = 2pt] (3.5,0) -- (3.5,1); + \coordinate [label = right: {floor reaction}] + (w) at (3.7,1); + \end{normalsize} + \end{tikzpicture} + \end{center} + The two opposing vectors add up to the zero vector, and therefore you don't move. + + \problem{} + Give an example of a different pair of opposite forces. + + \vfill + + + The following is the last thing we'll mention about the opposite vectors. The formula $w - v$ is defined as $w + (-v)$ for any vectors $v$ and $w$: + \begin{equation} + w - v = + w + (-v) + \end{equation} + + \newpage + + \problem{Dividing a segment into parts} + Using your compass and ruler, divide a segment into three equal parts. How would you split it into four? five? Have an instructor check your answer before moving on. + + \hint{Don't skip this problem, you'll need it later. Make sure you check your answer!} + + \vfill + + \begin{center} + \begin{tikzpicture} \begin{normalsize} + \filldraw (0,0) circle (1.5pt); + \coordinate [label = left: {$A$}] (a) at (0,0); + \filldraw (12,0) circle (1.5pt); + \coordinate [label = right: {$B$}] (b) at (12,0); + \draw [line width = 1.5 pt] (0,0) -- (12,0); + \end{normalsize} \end{tikzpicture} + \end{center} + \vfill + + \begin{center} + \begin{tikzpicture} \begin{normalsize} + \filldraw (0,0) circle (1.5pt); + \coordinate [label = left: {$A$}] (a) at (0,0); + \filldraw (12,0) circle (1.5pt); + \coordinate [label = right: {$B$}] (b) at (12,0); + \draw [line width = 1.5 pt] (0,0) -- (12,0); + \end{normalsize} \end{tikzpicture} + \end{center} + + \vfill + \newpage + + \problem{} + Is it possible to check whether $v = w$ on the picture below using only a compass? Why or why not? \\ + + \begin{center} + + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-2); + %\coordinate [label = left: {$A$}] (a) at (0,0); + %\coordinate [label = right: {$B$}] (b) at (4,-2); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + \draw [->, line width = 2pt] (-1,-3) -- (3,-5); + %\coordinate [label = left: {$C$}] (c) at (-1,-3); + %\coordinate [label = right: {$D$}] (d) at (3,-5); + \coordinate [label = below: {$w$}] (w) at (1,-4.1); + + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + + \problem{} + Use a compass and a ruler to construct the vector $w = -.75 v$ for the vector $v$ given below such that point $C$ is its terminal point. + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (6,-2); + \coordinate [label = above: {$v$}] (v) at (3,-.9); + \filldraw (-4,-5) circle (1.5pt); + \coordinate [label = below left: {$C$}] (c) at (-4,-5.1); + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + + \note{ + With the tools we have thus far, we can multiply vectors by any rational number using only a compass and a ruler. Multiplying a vector by an irrational number is a bit more tricky, but it is doable... + } + + \newpage + + \problem{} + Use a compass and a ruler to construct the vector $w = \sqrt{3} v$ for the vector $v$ given below so that $C$ is its initial point. \\ + \hint{Pythagoras.} + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (5,-2); + \coordinate [label = above: {$v$}] (v) at (2.5,-.9); + \filldraw (-5,-6) circle (1.5pt); + \coordinate [label = below left: {$C$}] (c) at (-5,-6.1); + \end{normalsize} + \end{tikzpicture} + \end{center} + + + + \vfill + \newpage + + + \section{Bonus} + + \problem{Oldaque de Freitas' Puzzle} + Two ladies are sitting in a street caf\'e, talking about their children. One lady says that she has three daughters. The product of the girls' ages equals 36 and the sum of their ages is the same as the number of the house across the street. The second lady replies that this information is not enough to figure out the age of each child. The first lady agrees and adds that her oldest daughter has beautiful blue eyes. The second lady then solves the puzzle. Please do the same. + + \vfill + + \problem{There must be a better way...} + Using pen and paper, sum up all the integers from $1$ to $1000$. + + \vfill + +\end{document} diff --git a/Intermediate/Vectors 2/main.tex b/Intermediate/Vectors 2/main.tex new file mode 100755 index 0000000..340b644 --- /dev/null +++ b/Intermediate/Vectors 2/main.tex @@ -0,0 +1,601 @@ +% https://git.betalupi.com/Mark/latex-packages +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +% Last built with version 1.0.6 +\documentclass[solutions]{ormc_handout} + + +\begin{document} + + \maketitle + + + {Vectors 2} + { + Prepared by Mark on \today \\ + Based on a handout by Oleg Gleizer + } + + + + \section{Review} + + \definition{} + A \textit{vector} is a directed line segment. In the vector below, $A$ is its initial point and $B$ is its terminal point. + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-1); + \coordinate [label = left: {$A$}] (a) at (0,0); + \coordinate [label = right: {$B$}] (b) at (4,-1); + \coordinate [label = above: {$v$}] (v) at (2,-1); + \end{normalsize} + \end{tikzpicture} + \end{center} + + + \definition{Equivalence} + We say two vectors are equal if the quadrilateral they form is a parallelogram. In other words, two vectors are equal if they have the same length and direction. + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4, -1); + \coordinate [label = left: {$A$}] (a) at (0, 0); + \coordinate [label = right: {$B$}] (b) at (4, -1); + \coordinate [label = above: {$v$}] (v) at (2, -1); + + \draw [->, line width = 2pt] (-1,-1) -- (3,-2); + \coordinate [label = left: {$C$}] (c) at (-1,-1); + \coordinate [label = right: {$D$}] (d) at (3,-2); + \coordinate [label = below: {$w$}] (w) at (1,-1.6); + \draw (-1,-1) -- (0,0); + \draw (3,-2) -- (4,-1); + \end{normalsize} + \end{tikzpicture} + \end{center} + + + + \definition{Addition} + To add two vectors $v$ and $w$, we move $w$ so that the initial point of $w$ coincides with the terminal point of $v$. The vector originating at the initial point of $v$ and terminating at the terminal point of $w$ is the sum $v + w$. \\ + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4,-1); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + + \draw [->, line width = 2pt] (4,-1) -- (2,-2); + \coordinate [label = right: {$w$}] (w) at (3, -1.7); + + \draw [->, line width = 2pt] (0,0) -- (2,-2); + \coordinate [label = left: {$v + w$}] (p) at (1,-1.2); + \end{normalsize} + \end{tikzpicture} + \end{center} + + + Note that $v+w = w+v$. If we create a parallelogram with sides $w$ and $v$ (\ref{vec_eq}), the sums create the same diagonal: + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 1pt] (0,0) -- (4,-1); + \coordinate [label = above: {$v$}] (v) at (2,-.9); + + \draw [->, line width = 1pt] (4,-1) -- (2,-2); + \coordinate [label = right: {$w$}] (w) at (3, -1.7); + + \draw [<-, line width = 1pt] (2,-2) -- (-2,-1); + \coordinate [label = above: {$v$}] (v) at (0.2,-2); + + \draw [<-, line width = 1pt] (-2,-1) -- (0, 0); + \coordinate [label = right: {$w$}] (w) at (-1.5, -0.3); + + \draw [->, line width = 2pt] (0,0) -- (2,-2); + \coordinate [label = left: {$v + w$}] (p) at (1,-1.2); + + + \end{normalsize} + \end{tikzpicture} + \end{center} + + \vfill + \pagebreak + + \definition{The Zero Vector} + A vector that has coinciding initial and terminal points is called the {\it zero vector} and is denoted as $\overrightarrow{0}$. According to the above definition of the vector addition, + \begin{equation*} + v + \overrightarrow{0} = + \overrightarrow{0} + v = + v + \end{equation*} + for any vector $v$. + + + \definition{Inverse Vectors} + A vector $w$ such that $w + v = \overrightarrow{0}$ is called the \textit{inverse} of $v$ and is denoted as $-v$. The vector $-v$ lies either on the same straight line as $v$ or on a parallel one, has the same length as $v$, but points in the opposite direction: \\ + + + \begin{center} + \begin{tikzpicture} + \begin{normalsize} + \draw [->, line width = 2pt] (0,0) -- (4, -1); + \coordinate [label = above: {$v$}] (v) at (2, -1); + \draw [<-, line width = 2pt] (-1,-1) -- (3,-2); + \coordinate [label = below: {$-v$}] (w) at (1,-1.6); + \end{normalsize} + \end{tikzpicture} + \end{center} + + + + \vfill + \pagebreak + + \section{Vectors} + + \problem{} + For the given vector $\overrightarrow{v}$ + and point $A$, construct the vector + $\overrightarrow{w} = \overrightarrow{v}$ + having $A$ as its initial point + on the graph paper below. + Use the grid instead of a compass and ruler. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-8,-8) grid (1,1); + \draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1); + \filldraw (-6,-4) circle (2pt); + \coordinate [label = left:{$A$}] (a) at (-6,-4); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + For the given vector $\overrightarrow{v}$ + and point $A$, construct the vector + $\overrightarrow{w} = -\overrightarrow{v}$ + having $A$ as its initial point + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-12,-7) grid (1,1); + \draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1); + \filldraw (-6,-4) circle (2pt); + \coordinate [label = below:{$A$}] (a) at (-6,-4.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + \problem{} + For the given vector $\overrightarrow{v}$ + and point $A$, construct the vector + $\overrightarrow{w} = 1.5 \overrightarrow{v}$ + having $A$ as its initial point + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-9,-9) grid (1,1); + \draw [line width = 1.5pt, ->] (-6,-1) -- (-2,-3); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.1); + \filldraw (-7,-4) circle (2pt); + \coordinate [label = left:{$A$}] (a) at (-7,-4.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + For the given vector $\overrightarrow{v}$ + and point $A$, construct the vector + $\overrightarrow{w} = -2 \overrightarrow{v}$ + having $A$ as its initial point + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-16,-8) grid (1,1); + \draw [line width = 1.5pt, ->] (-5,-2) -- (-1,-4); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-3.1); + \filldraw (-6,-5) circle (2pt); + \coordinate [label = below:{$A$}] (a) at (-6,-5.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + \problem{} + For the given vector $\overrightarrow{v}$ + and point $A$, construct the vector + $\overrightarrow{w} = -\frac13 \overrightarrow{v}$ + having $A$ as its initial point + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-11,-6) grid (1,1); + \draw [line width = 1.5pt, ->] (-7,-1) -- (-1,-4); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.4); + \filldraw (-6,-4) circle (2pt); + \coordinate [label = below:{$A$}] (a) at (-6,-4.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + For the given vectors $\overrightarrow{v}$ + and $\overrightarrow{w}$, + construct the vector + $\overrightarrow{w} + \overrightarrow{v}$ + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-12,-10) grid (1,1); + \draw [line width = 1.5pt, ->] (-5,0) -- (0,-4); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.4,-2.1); + \draw [line width = 1.5pt, ->] (-10,-1) -- (-9,-5); + \coordinate [label = left:{$\overrightarrow{w}$}] (w) at (-9.5,-3.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + \problem{} + For the given vectors $\overrightarrow{v}$ + and $\overrightarrow{w}$, + construct the vector + $\overrightarrow{w} - \overrightarrow{v}$ + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-12,-6) grid (1,1); + \draw [line width = 1.5pt, ->] (-11,0) -- (-6,-4); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-8.4,-2.1); + \draw [line width = 1.5pt, ->] (-1,-1) -- (0,-5); + \coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + For the given vectors $\overrightarrow{v}$ + and $\overrightarrow{w}$, + construct the vector + $2\overrightarrow{v} - 3\overrightarrow{w}$ + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-13,-10) grid (1,1); + \draw [line width = 1.5pt, ->] (-12,-1) -- (-7,-5); + \coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-9.6,-3.1); + \draw [line width = 1.5pt, ->] (-1,-2) -- (0,-5); + \coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.6); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + \problem{} + For the given vectors $\overrightarrow{v}$ + and $\overrightarrow{w}$, + construct the vector + $1.75\overrightarrow{v} - \frac23 \overrightarrow{w}$ + on the graph paper below. + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-19,-9) grid (1,1); + \draw [line width = 1.5pt, ->] (-18,-1) -- (-10,-5); + \coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-14.4,-3.1); + \draw [line width = 1.5pt, ->] (-3,0) -- (0,-6); + \coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-1.5,-3); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + For the given vectors $\overrightarrow{v}$ + and $\overrightarrow{w}$, + construct the vector + $\overrightarrow{v} + \overrightarrow{w}$ + originating at the same point + as the vector $\overrightarrow{v}$ + and the vector + $\overrightarrow{w} + \overrightarrow{v}$ + originating at the same point + as the vector $\overrightarrow{w}$. + Is $\overrightarrow{v} + \overrightarrow{w} = + \overrightarrow{w} + \overrightarrow{v}$? + Why or why not? + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-17,-7) grid (1,1); + \draw [line width = 1.5pt, ->] (-16,-1) -- (-11,-5); + \coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-13.6,-3.1); + \draw [line width = 1.5pt, ->] (-8,-6) -- (-5,-1); + \coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-6.5,-4.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + + \section{Pythagoras' Theorem} + + \problem{} + Formulate and prove the Pythagoras' theorem. + \vfill + \pagebreak + + \problem{} + Use the Pythagors' theorem to find $x$ + for the following right triangles. \\ + + \noindent a.~~ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) -- (3,0) -- (0,3) -- (0,0); + \coordinate [label = left:{$1$}] (a) at (0,1.5); + \coordinate [label = below:{$1$}] (b) at (1.5,-.1); + \coordinate [label = right:{$x$}] (x) at (1.6,1.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + \bigskip + + \noindent b.~~ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) -- (4,0) -- (0,3) -- (0,0); + \coordinate [label = left:{$3$}] (a) at (0,1.5); + \coordinate [label = below:{$x$}] (x) at (2,-.1); + \coordinate [label = right:{$5$}] (b) at (2.1,1.6); + \end{tikzpicture} + \end{normalsize} + \end{center} + \bigskip + + \noindent c.~~ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,0) -- (4.5,0) -- (0,3) -- (0,0); + \coordinate [label = left:{$x$}] (x) at (0,1.5); + \coordinate [label = below:{$\sqrt{2}$}] (a) at (2,-.1); + \coordinate [label = right:{$\sqrt3$}] (b) at (2.2,1.7); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \vfill + \pagebreak + + + \problem{} + Construct a segment of length $\sqrt{2} a$ + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-7,-7) grid (1,1); + \draw [line width = 1.5pt] (-5,-1) -- (-1,-1); + \coordinate [label = above:{$a$}] (a) at (-3,-.9); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + Construct a segment of length $\sqrt{5} a$ + on the graph paper below. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-12,-6) grid (1,1); + \draw [line width = 1.5pt] (-5,-1) -- (-1,-1); + \coordinate [label = above:{$a$}] (a) at (-3,-.9); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + + \problem{} + The side length of a grid square below + is one unit. Find the length + the vector $\overrightarrow{v}$. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-5,-3) grid (1,1); + \draw [line width = 1.5pt, ->] (-4,0) -- (0,-2); + \coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.7,-2.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + \pagebreak + + \problem{} + The side length of a grid square below + is three units. Find the length + the vector $\overrightarrow{w}$. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} [scale=.7] + \draw[step=1cm, gray, very thin] (-6,-3) grid (1,1); + \draw [line width = 1.5pt, <-] (-5,0) -- (0,-2); + \coordinate [label = above:{$\overrightarrow{w}$}] (w) at (-2.5,-2); + \end{tikzpicture} + \end{normalsize} + \end{center} + \medskip + + + \section{Rationals} + + A number is called {\it rational} + if it can be represented as a ratio $p/q$ + of an integer $p$ and a positive integer $q$ + such that $p$ and $q$ have no common + factors. Otherwise, that number is called {\it irrational} \\ + + \problem{} + Decide whether the following numbers + are rational or irrational. + In each case, give a reason. \\ + + \begin{enumerate}[itemsep=2mm] + \item $\displaystyle{\frac{375}{376}}$ + \item $10$ + \item $0.5$ + \item $-5$ + \item $1.2345$ + \item $0.111111111...$ + \item $\sqrt{2}$ + \item $\sqrt[3]{10}$ + \end{enumerate} + + \vfill + \pagebreak + + \problem{} + Find $\left\lfloor \sqrt[3]{10} \right\rfloor$ + and $\left\lceil \sqrt[3]{10} \right\rceil$. + \vfill + + \problem{} + Simplify $\sqrt{8}$. + \vfill + \pagebreak + + \problem{} + A man is crossing a river in a boat. + The speed of the boat is three meters per second. + The speed of the water in the river + is one meter per second. + In what direction should the man steer the boat, + if he wants the vessel to move + perpendicular to the banks? + Construct the velocity vector. \\ + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \draw (0,10) -- (10,10); + \draw (0,0) -- (10,0); + \draw (4.5,1) -- (5.5,1) --(5.5,3) -- (5,3.5) -- (4.5,3) --(4.5,1) ; + \filldraw (5,2.1) circle (2pt); + \draw [line width = 2pt, ->] (5,2.1) -- (7,2.1); + \coordinate [label = below: {$1\frac{m}{s}$}] (w) at (6.3,1.9); + \end{tikzpicture} + \end{normalsize} + \end{center} + \bigskip + + The width of the river + is $10\sqrt{2}$ meters. + How long would it take the man + to cross the river? + + \vfill + \pagebreak + + + \problem{} + You need to slide a heavy box over + the floor from point $A$ to point $B$. + The box is about twice as tall as you are. + Which way is easier, to push or to pull? + Why? \vspace{60pt} + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \filldraw [gray!90!blue] (1,0) -- (1,2) -- + (4,2) -- (4,0) -- (1,0); + \draw (0,0) -- (10,0); + \filldraw (2.5,0) circle (1.5pt); + \coordinate [label = below: {$A$}] (a) at (2.5,-.1); + \filldraw (7.5,0) circle (1.5pt); + \coordinate [label = below: {$B$}] (b) at (7.5,-.1); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \vfill + + \problem{} + The dot on the picture below + represents a spaceship. + There are three forces acting on the ship. + $\overrightarrow{T}$ is the thrust + of the ship's engine. + $\overrightarrow{P}$ is the gravitational pull + of the neighbouring planet. + $\overrightarrow{S}$ is the gravitational pull + of the planet's home star. + You are the captain. + Use a compass and a ruler to figure out + where the resulting force would steer the ship. + \vspace{100pt} + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \filldraw (0,0) circle (2pt); + \draw [line width = 1pt, ->] (0,0) -- (225:6); + \coordinate [label = below: {$\overrightarrow{T}$}] + (t) at (230:3); + \draw [line width = 1pt, ->] (0,0) -- (4,0); + \coordinate [label = above: {$\overrightarrow{P}$}] + (p) at (2,.1); + \draw [line width = 1pt, ->] (0,0) -- (150:3); + \coordinate [label = right: {$\overrightarrow{S}$}] + (s) at (140:1.5); + \end{tikzpicture} + \end{normalsize} + \end{center} + \vfill + +\end{document} \ No newline at end of file