\documentclass[12pt,letterpaper]{amsart}
\usepackage{times}
\usepackage{geometry}
\usepackage{fullpage}
\usepackage{booktabs}
\usepackage{url}
\usepackage{hyperref}
\usepackage{overpic}
\newcommand{\VX}{\operatorname{VX}}
\newcommand{\VY}{\operatorname{VY}}
\newcommand{\PX}{\operatorname{PX}}
\newcommand{\PY}{\operatorname{PY}}
\newcounter{lscount}
\newenvironment{problems}
{\begin{list}{\arabic{lscount}.}{\usecounter{lscount}\setlength{\leftmargin}{15pt}\setlength{\rightmargin}{0pt}\setlength{\itemsep}{5pt}}}
{\end{list}}
\begin{document}
\setlength{\parskip}{12pt}
\begin{center}
\large{\textbf{Math 2250 Lab \#4}}
\end{center}
\bigskip
This writeup clarifies the fourth lab assignment. The setup for the lab is that we have built a robot to throw a ball by attaching the ball to a rotating arm using an electromagnet. The arm has length 27 VEX holes (at 0.5 cm each) and rotates around a center at $(0,h)$ where h is 33 VEX holes. The rotational speed of the arm varies between 0 and 150 rpm.
At some time $t$, we release the ball. The ball then follows a parabolic path as shown in the pictures
\begin{center}
\begin{tabular}{cccc}
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_1} &
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_4} &
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_3} \\
ball starts at $t=0$ \dots &
\dots released at $t$ \dots &
\dots will land at $x(t)$. \\
\end{tabular}
\end{center}
\section{Problems}
\begin{problems}
\item We have solved previously for the landing position (as a function of release time):
\begin{equation*}
x(t) = \VX \left( \frac{-\VY - \sqrt{\VY^2 + 2 \times 9.8 \times \PY}}{-9.8} \right) + \PX.
\end{equation*}
where the functions are given by
\begin{align*}
\PX &= r \cos \left( \frac{\pi k}{30} t \right) \\
\PY &= r \sin \left( \frac{\pi k }{30} t \right) + h \\
\VX &= -r \sin \left( \frac{\pi k}{30} t \right) \frac{\pi k}{30} \\
\VY &= r \cos \left( \frac{\pi k}{30} t \right) \frac{\pi k}{30}
\end{align*}
\begin{enumerate}
\item Differentiate $\PX, \PY, \VX$, and $\VY$ with respect to $t$ to arrive at equations for $\PX', \PY', \VX'$ and $\VY'$. Now express your answers in terms of $\PX$, $\PY$, $\VX$, and $\VY$. For example,
$\PX' = \VX$.
\item Differentiate $x(t)$ with respect to $t$ \textbf{without} substituting the equations above for $\PX$, $\PY$, $\VX$, and $\VY$. This is like an implicit differentiation problem: when you need to differentiate $\PX$, for example, just write $\PX'$.
\item Now substitute the answers you arrived at in part (1) for $\PX'$, $\VX'$, and so forth in order to obtain an expression for $x'(t)$ \emph{in terms of $\PX$, $\PY$, $\VX$, and $\VY$}.
\end{enumerate}
\item We have decided that the best way to hit a target at distance $D$ is to adjust $k$ so that $D$ is the \emph{maximum} range of the robot for that rotation speed $k$. Describe how to use Newton's method to solve the equation $x'(t) = 0$ and find the maximum value of $x(t)$ for a given $k$. (This will require a formula for $x''(t)$. You can refer to this as ``$x''(t)$''. You don't have to actually differentiate.)
\item (Bonus Credit) Actually find a formula for $x''(t)$ in terms of $\PX$, $\PY$, $\VX$, and $\VY$ either by hand or using \emph{Mathematica}. (I don't think Wolfram Alpha is powerful enough to do this.)
\item (Bonus Credit) Use \emph{Mathematica} to actually find the maximum range for $k = 150$ rpm.
\item The solution we arrived at earlier by Newton's method defines a function $M(k)$ which tells us the maximum range for the robot for a given rotation speed $k$. Given a target at distance $x_0$ from the robot, we need to solve an equation in the form
\begin{equation*}
M(k) = x_0
\end{equation*}
to figure out the right speed for the rotating arm. Our usual approach would be to solve this equation by Newton's method. However, to do that, we'd need to know $M'(k)$. Since the values of $M(k)$ are (basically) arrived at by Newton's method already, we can compute $M(k)$ for any given $k$, but we don't really have any hope of figuring out $M'(k)$.
Read the Wikipedia page on the ``bisection method'' for solving equations \url{http://en.wikipedia.org/wiki/Bisection_method} and explain in your own words how to use the bisection method to solve $M(k) - x_0 = 0$ for $k$.
\item (Bonus Credit) Use \emph{Mathematica} to implement the bisection method and solve for the correct rotation speed $k$ for a target $1$ meter from the robot.
\item Once we have arrived at the correct rotation speed $k$ so that the maximum value of $x(t)$ is equal to our target distance $x_0$, we must still decide when to release the ball. Given the information you've already come up with, describe explicitly how to find the correct $t$ value to release the ball.
\item (Bonus Credit) The error formula for a function $f(x)$ at a point where $f'(x)$ is not $0$ is
\begin{equation*}
\Delta f = f'(x) \Delta x.
\end{equation*}
Derive the corresponding formula at a point where $f'(x) = 0$ but $f''(x)$ is not $0$. Assume that $t_0$ is the $t$ value so that $x(t_0) = 1$ and $x'(t_0) = 0$. Write a formula for the range of acceptable $t$ values ($\Delta t$) around $t_0$ so that the ball lands within $5$ cm and $1$ cm of the target. The (unknown) value $x''(t_0)$ should appear in your answer.
\item (Bonus Bonus Credit) Assuming that you've done all the previous extra credit problems, plug in everything and find $k$, $t_0$ and $\Delta t$ for a target at $x_0 = 1$ meter with $5$ cm and $1$ cm accuracy.
\end{problems}
\end{document}