\documentclass{article} \usepackage{Mathematics} \makeatletter \@fleqnfalse \makeatother \pdftitle{SW5-Math1A-tutoring} % === TITLE === \title{\textbf{Mathematics 1A Tutoring session\\ Semester Week 5}} \author{Matteo Frongillo} \date{} \newcommand{\exercise}[2][]{% \par\noindent\textbf{Exercise \thesection.#1:\\} #2\par \vspace*{.25cm} } \newcommand{\solution}[2][]{% \par\noindent\textbf{Solution #1:\\} #2\par \vspace*{.25cm} } % === TEXT === \begin{document} \maketitle \part*{Exercises} \section{Trigonometry} \exercise[1]{ Find the hypotenuse of a right triangle with adjacend side $b = 36.4$ cm and opposite side $c = 41.8$ cm. } \exercise[2]{ Solve a right triangle where only the hypothenuse $a=3$ cm and the angle $\beta = 34,7^\circ$ are known. } \exercise[3]{ Determine domain and range of the following trigonometric functions: \begin{center} \[ \sin\left(x\right)\quad ; \quad \cos^{-1}\left(x\right)\quad ; \quad \tan\left(x\right)\quad ; \quad \arctan\left(x\right) \] \end{center} } \exercise[4]{ A block of mass $m = 2\,\mathrm{kg}$ rests on an inclined plane that forms an angle $\theta$ with the horizontal. The gravitational acceleration is $g = 9.81\,\mathrm{m\,s^{-2}}$. \begin{enumerate}[label=\alph*.] \item Express the components of the force parallel and perpendicular to the plane, $F_\parallel$ and $F_\perp$, as functions of $\theta$, knowing that $F = m g$. \item Compute $F_\parallel$ and $F_\perp$ for $\theta = 25^\circ$. \item Find the value of $\theta$ for which $F_\parallel = \dfrac{1}{2} F_\perp$. \item Determine the length $L$ of the inclined plane knowing that the height is $h = 1.5\,\mathrm{m}$ and the horizontal projection is $x = 3.3\,\mathrm{m}$. \end{enumerate} } \exercise[5 {\color{red}BONUS}]{\\ \begin{minipage}{0.3\textwidth} \begin{center} \includegraphics[width=.8\textwidth]{media/trig_sw5.png} \end{center} \end{minipage} \hfill \begin{minipage}{0.65\textwidth} A cube $ABCDEFGH$ with edge length 8 cm is given.\\ Point $M$ is defined so that $\overrightarrow{EM} = \dfrac{2}{5}\,\overrightarrow{EG}$.\\ Calculate the measure of the sides and angles of triangle $ACM$ \end{minipage} } \newpage \section{Exponential functions} \exercise[1]{ Determine the half-life of a radioactive substance $X$, knowing that after one year its mass has decreased to one third. } \exercise[2]{ A sample of St-89 (Strontium-89, a radioactive element with $T=50.5$ days) has lost 2 g of mass after one month. Determine the mass of the same initial sample after one year. Hint: for decays, $k = -\dfrac{\ln(2)}{T}$ } \exercise[3]{ Solve in $\mathbb{R}$ \textbf{without the calculator}: \[ \log\left(x^2\right) = 3\quad ; \quad \ln\left(x+1\right)=\frac{1}{2}\quad ; \quad \ln\left(x+1\right)=0.001^2\quad ; \quad \log_2\left(x^2-4x+4\right)=2\quad ; \quad \left(x^2+x-2\right)\ln(2x)=0 \] } \exercise[4]{ A hot object with an initial temperature $T_0$, placed at time $t=0$ in an environment with lower temperature $T_1$, cools according to Newton's law of cooling: \figbox{$T=T_1 + \left(T_0 - T_1\right) e^{-kt}$}\\ where $T$ is the temperature of the object at time $t$, and $k$ is a constant depending on the material of the object. Consider a steel sphere heated to $133^\circ$C and then placed to cool in a room where the air temperature is $22^\circ$C.\\ Calculate: \begin{enumerate}[label=\alph*.] \item the temperature of the sphere after 20 minutes, knowing that after 10 minutes it had cooled to $108^\circ$C; \item the time required for the temperature to drop to $66.5^\circ$C. \end{enumerate} } \section{Composite functions} \exercise[1]{ Let $f: \mathbb{R} \to \mathbb{R}$ \[x \mapsto y = \begin{cases} \dfrac{1}{2}x+1\ , &\text{for } x \geq 4\\ x-1\ , &\text{for } -1 < x < 4\\ -2x-4\ , &\text{for } x\leq -1 \end{cases} \] Solve: \noindent \begin{minipage}[t]{0.3\textwidth} \begin{enumerate}[label=\alph*.,itemsep=1em] \item $f(0)$ \item $f(-11)$ \end{enumerate} \end{minipage}% \hfill \begin{minipage}[t]{0.3\textwidth} \begin{enumerate}[label=\alph*.,resume,itemsep=1em] \item $f\!\left(\tfrac{9}{2}\right)$ \item $f\!\left(f\!\left(-\tfrac{1}{2}\right)\right)$ \end{enumerate} \end{minipage}% \hfill \begin{minipage}[t]{0.3\textwidth} \begin{enumerate}[label=\alph*.,resume,itemsep=1em] \item $f(6)-f(-6)$ \item $f\!\left(3f(-1)-2f(2)\right)$ \end{enumerate} \end{minipage} } \exercise[2]{ Let $f: \mathbb{R} \to \mathbb{R}$ \[x \mapsto y = \begin{cases} \dfrac{1}{2}x\ , &\text{for } x < -2\\[2.5ex] \dfrac{2}{3}x + 1\ , &\text{for } -2 \leq x \leq 2\\[2.5ex] \dfrac{x+2}{3}\ , &\text{for } x > 2 \end{cases} \] Solve: \[ a.\ f\left(\frac{6}{5}\right)\quad ; \quad b.\ f\big(f\left(-3\right)\big)\quad ; \quad c.\ 2\big(f\left(7\right)-1\big)^2 \] } \section{Limits} \exercise[1]{ Solve the following limits using the dominant term method: \begin{gather*} \dm \lim_{x\to\infty} \frac{\left(x+1\right)^2}{x^2+1}\quad ; \quad \lim_{x\to\infty} \frac{\sin(x)}{x}\quad ; \quad \lim_{x\to\infty} \frac{1000x}{x^2+1}\quad ; \quad \lim_{x\to\infty} \frac{x^2-5x+1}{3x+7}\\[2.5ex] \lim_{x\to\infty} \frac{2x^2-x+3}{x^3-8x+5}\quad ; \quad \lim_{x\to\infty} \frac{\left(2x+3\right)^3\left(3x-2\right)^2}{x^5+5}\quad ; \quad \lim_{x\to\infty} \frac{2x^2-3x-4}{\sqrt[3]{x^4+1}} \end{gather*} } \exercise[2]{ For each of the following, determine whether the statement is true or false: \begin{center} \begin{tikzpicture} \begin{axis}[ axis lines=middle, width=.8\textwidth, height=10cm, xmin=-8, xmax=8, ymin=-12, ymax=12, xtick={-10,-8,...,10}, ytick={-12,-10,...,12}, grid=both, major grid style={gray!30}, minor grid style={gray!15}, minor tick num=1, tick style={black}, xlabel={$x$}, ylabel={$y$}, restrict y to domain=-15:15, unbounded coords=jump, clip=true ] \addplot[dashed, blue] coordinates {(-10,2) (10,2)}; \addplot[dashed, blue] coordinates {(-1,-15) (-1,15)}; \addplot[dashed, blue] coordinates {(4,-15) (4,15)}; \addplot[thick, red, samples=1500, domain=-10:10] {2+(8*x - 4)/((x + 1)*(x - 4))}; \node[red] at (.8,3) {$f(x)$}; \end{axis} \end{tikzpicture} \end{center} \begin{gather*} \text{a}) \lim_{x\to\infty} f(x)=0\quad , \quad \text{b}) \lim_{x\to 0} f(x)=2\quad , \quad \text{c}) \lim_{x\to 4^-} f(x) = +\infty\quad , \quad \text{d}) \lim_{x\to-1} f(x)=f(-1)\\[2ex] \text{e}) \lim_{x\to0} f(x) = 3\quad , \quad \text{f}) \lim_{x\to\infty} f(x)=-1\quad , \quad \text{g})\ f(x)=0 \iff x \in \left\{-3,4\right\} \end{gather*} } \exercise[3 (From Question 4.1, Homeworks Week 4)]{ Are the following claims true or false? Explain why: \begin{enumerate}[label=\alph*.] \item If a function $y$ of $x$ is not defined at the position $x=x_0$, then $\dm \lim_{x\to x_0}$ does not exist either \item If $\dm \lim_{x\to x_0}$ does not exist, then $y$ is also not defined at the point $x=x_0$ \item If a function $y$ of $x$ is defined at the position $x=x_0$, then $\dm \lim_{x\to x_0} y = y \big|_{x=x_0}$ \item If $x$ approaches $1.000.000$ from the left, $1/x$ gets closer to the value 0. Therefore, $\dm \lim_{x \to 1.000.000^-} \frac{1}{x}=0$ \item If $y$ has limit value 5 when $x$ approaches 3 from the left, then $y$ must already assume the value 5 in the range $x<3$ \end{enumerate} } \newpage \part*{Quick solutions} \solution[1.1]{ a = 55.43 cm } \solution[1.2]{ b = 1.71 cm, c = 2.47 cm, $\alpha$ = 55.3$^\circ$ } \solution[1.3]{\vspace*{-.5cm} \begin{enumerate} \item $\mathcal{D}_f = \forall x \in \mathbb{R}, \quad Im_f = \left[-1,1\right]$ \item $\mathcal{D}_f = \left[-1,1\right], \quad Im_f = \left[\pi, 0\right]$ \item $\mathcal{D}_f = \forall x \in \mathbb{R} \ \left\{\dfrac{\pi}{2}+k\pi \sht k \in \mathbb{Z}\right\}, \quad Im_f = \left[-1,1\right]$ \item $\mathcal{D}_f = \forall x \in \mathbb{R}, \quad Im_f = \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ \end{enumerate} } \solution[1.4]{ \vspace*{-0.5cm} \begin{enumerate}[label=\alph*.] \item $F_\parallel = mg\cos(x)\ , \quad F_\perp = mg\sin(x)$ \item $F_\parallel = 17.78$ N, $\quad F_\perp = 8.29$ N \item $\theta \approx 1.11 \text{ rad} = 63.4^\circ$ \item $L = 3.6$ m \end{enumerate} } \solution[1.5]{ $\overrightarrow{AC} = 8\sqrt(2)$ cm $\approx 11.3$ cm\\[2ex] $\overrightarrow{AM} = \dfrac{8\sqrt(33)}{5}$ cm $\approx 9.19$ cm\\[2ex] $\overrightarrow{MC} = \dfrac{8\sqrt(43)}{5}$ cm $\approx 10.49$ cm\\[2ex] $\widehat{MAC}: \alpha \approx 60.5^\circ $\\[2ex] $\widehat{ACM}: \beta \approx 49.7^\circ$\\[2ex] $\widehat{AMC}: \gamma \approx 69.8^\circ$ } \solution[2.1]{ Half-time: 0.63 years $\approx$ 230 days } \solution[2.2]{ $m_0 = 5.93$ g } \solution[2.3]{ \begin{enumerate}[label=\alph*.] \item $\dm x = \pm \sqrt{10^3}$ \item $\dm x = e^{\tfrac{1}{2}}-1$ \item $\dm x = e^{10^{-6}}-1 \approx 0$ \item $\dm x_1 = 0,\ x_2 = 4$ \item $\dm x \in \left\{-2, \dfrac{1}{2}, 1\right\}$ \end{enumerate} } \newpage \solution[2.4]{ \begin{enumerate}[label=\alph*.] \item $T = 88.63^\circ$C \item $t \approx 35.2$ min \end{enumerate} } \solution[3.1]{ \begin{enumerate}[label=\alph*.] \item $f(0) = 1$ \item $f(-11)=18$ \item $f\left(\dfrac{9}{2}\right) = \dfrac{13}{4}$ \item $f\left(f\left(-\dfrac{1}{2}\right)\right) = -1$ \item $f(6) - f(-6) = -4$ \item $f\big(3f(-1)-2f(2)\big) = 12$ \end{enumerate} } \solution[3.2]{ \begin{enumerate}[label=\alph*.] \item $f\left(\dfrac{6}{5}\right) = \dfrac{9}{5}$ \item $f\big(f(-3)\big) = 0$ \item $2\big(f(7)-1\big)^2=8$ \end{enumerate} } \solution[4.1]{ \begin{enumerate}[label=\alph*.] \item = 1 \item = 0 \item = 0 \item = 0 \item = 72 \item = $\infty$ \end{enumerate} } \solution[4.2]{ \begin{enumerate}[label=\alph*.] \item False \item False \item False \item False \item True \item False \item False \end{enumerate} } \solution[4.3]{ All the statements are False } \end{document}