Math 311 Spring 2016
Author:
Jason Adkins
Last Updated:
há 9 anos
License:
Creative Commons CC BY 4.0
Abstract:
Two Snow Plows Problem
\begin
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Two Snow Plows Problem
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[9pt]{beamer}
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\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{quest}[thm]{Question}
\newtheorem{prob}[thm]{Problem}
\usetheme{CambridgeUS}
\usecolortheme{wolverine}
\title[Math 311 Spring 2016]{The Two Snow Plows\\ Group Problem 2E}
\date{May 2, 2016}
\begin{document}
%--------------------New Frame---------------------------
\begin{frame}
\titlepage
\end{frame}
%--------------------New Frame---------------------------
\AtBeginSubsection[]
{
\begin{frame}<beamer>
\frametitle{Layout}
\tableofcontents[currentsection,currentsubsection]
\end{frame}
}
%\begin{frame}{Agenda}
%Today we will talk about:
%\begin{itemize}
%\pause \item mathematical models
%\pause \item examples
%\end{itemize}
%\end{frame}
\section{Direction Fields}
%--------------------New Frame---------------------------
\begin{frame}{The Problem}
\begin{itemize}
\item[] Suppose it starts snowing at a constant rate at 12:00 pm and two snowplows are to be dispatched to clear a long road from the same garage. If Plow X leaves at 1:00 pm and Plow Y leaves at 2:00 pm, will they collide? If so, when? Assume the snowplows can clear snow at a rate inversely proportional to the depth of snow.\vskip 3mm
\pause \item[] Using information from the problem, we can determine that: \vskip 1mm
\begin{itemize}
\pause \item[a)] The snow is falling at a constant rate $r$, \\
\pause \item[b)] At $t = 1$, Plow X departs, making $x(1) = 0$, \\
\pause \item[c)] At $t = 2$, Plow Y departs, making $y(2) = 0$.
\end{itemize}
\vskip 3mm
\pause \item[] Since the rate of snowfall is constant, we symbolize this with the constant $r$.
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{Setting Up the DEs}
\begin{itemize}
\item[] Since the snowplows can clear at a rate inversely proportional to the depth of snow, we can set up their differential equations:
\pause \item[]\begin{eqnarray*}
\dfrac{dx}{dt}&=&\dfrac{k}{h_x(t)}, \\
\dfrac{dy}{dt}&=&\dfrac{k}{h_y(t)}.
\end{eqnarray*}
\pause \item[] Where
\begin{eqnarray*}
h_x(t) &=& rt, \\
h_y(t) &=& r(t-T), \\
A &=& \dfrac{k}{r}.
\end{eqnarray*}
\pause \item[] Substituting:
\begin{eqnarray*}
\dfrac{dx}{dt}&=&\dfrac{A}{t}, \\
\dfrac{dy}{dt}&=&\dfrac{A}{(t-T)}.
\end{eqnarray*}
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{Solving for $x(t)$}
\begin{itemize}
\pause \item[]
\begin{eqnarray*}
\int \dfrac{dx}{dt} &=& \dfrac{A}{t}, \\
\pause x(t) &=& A\ln|t| + C.
\end{eqnarray*}
\pause \item[] Solving for C using the initial condition $x(1) = 0$:
\begin{eqnarray*}
x(1) = 0 &=& A\ln|1| + C, \\
\pause 0 &=& C.
\end{eqnarray*}
\pause \item[] This yields our position function for Plow X:
\begin{eqnarray*}
x(t) &=& A\ln|t|.
\end{eqnarray*}
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{Solving for $y(t)$}
\begin{itemize}
\pause \item[] Inverting the DE for Plow Y:
\pause \item[] \begin{eqnarray*}
\pause \dfrac{dt}{dy}&=&\dfrac{(t-T)}{A}, \\
\pause B &=& \dfrac{1}{A}.
\end{eqnarray*}
\pause \item[]Solving the linear DE:
\begin{eqnarray*}
\pause \int \dfrac{dt}{dy} &=& \int B(t-T), \\
\pause t(y) &=& e^{By}(D-By).
\end{eqnarray*}
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{Solving for $y(t)$}
\begin{itemize}
\item[] Solving for D using the initial condition $y(2) = 0$:
\begin{eqnarray*}
\pause t(0) = 2 &=& e^{0}(D-0), \\
\pause 2 &=& D.
\end{eqnarray*}
\pause \item[] This yields Plow Y's position function:
\begin{eqnarray*}
t(y) &=& e^{By}(2-By).
\end{eqnarray*}
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{The Collision}
\begin{itemize}
\item[] The plows will collide at time $T$ if:
\begin{eqnarray*}
x(T) &=& y(T), \\
\pause A\ln|T| &=& y(T), \\
\pause T &=& e^{By(T)}.
\end{eqnarray*}
\pause \item[] Solving for $t(y)$ where $t = T$:
\begin{eqnarray*}
T &=& e^{By}(2-By), \\
T &=& T(2-\ln|T|), \\
T &=& e^{1} = e.
\end{eqnarray*}
\pause \item[] Plow Y will collide with Plow X $e$ hours after it begins snowing, or approximately 2:43 PM.
\end{itemize}
\end{frame}
%--------------------New Frame---------------------------
\begin{frame}{The Collision, Part B}
\begin{itemize}
\item[] If Plow Y leaves 2 hours after Plow X instead of 1 hour, will they still collide? \vskip 3mm
\pause \item[] The initial condition changes so that $y(3) = 0$. \vskip 5mm
\pause \item[] Solving for $t(y)$ where $t = T$:
\begin{eqnarray*}
T &=& e^{By}(3-By), \\
T &=& T(3-\ln|T|), \\
T &=& e^{2} = e^2.
\end{eqnarray*}
\pause \item[] Plow Y will collide with Plow X $e^2$ hours after it begins snowing, or approximately 7:23 PM.
\end{itemize}
\end{frame}
\end{document}