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\begin{document}


% You only need to change the NAME OF SCENARIO in the next line and delete the footnote. See above for instructions on toggling from Teacher Version to Student Version.
\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\Barging Ahead: Optimizing a Trip Upriver}
%\footnote{DELETE THIS FOOTNOTE BEFORE SUBMISSION.  In the \TeX \, source code there is a toggle/option to produce a Teacher Version or a Student Version by ``carving'' out with  $\backslash$teacher\{\} that material which should only appear in the Teacher version. See the source code.}




\markboth{Barging Ahead}{Barging Ahead}






\author{Timothy Pennings\\
Davenport University\\
Grand Rapids MI  49512 USA\\
tpennings@davenport.edu}


\makeStitlePDFLaTex



\begin{abstract} As captain of a barge, you need to determine how fast to transport
your barge up river against the current in order to minimize the expended energy.
Since expended energy is proportional to the force, and since the force is proportional to the speed, traveling too fast is inefficient. However, if traveling too slow, it will almost  take an infinite amount of time to arrive, which is also inefficient. How fast then should you go? We find an answer to this question, and to the related question of minimizing cost of travel, where the cost is a linear combination of the energy and the time. 

%Offer an abstract which describes what the students will do and learn, including technical descriptions of mathematics, area of application, and outcomes expected for the scenario. This should help the teacher decide if the material is suitable for a given class.   This will appear in Teacher Version only.
\end{abstract}


\section*{SCENARIO DESCRIPTION}
%
%This is the STATEMENT of the scenario for the student, with pointers to data sets, supplementary files,  videos, etc. for student consumption. It should have clear statements, instructions, hints, necessary materials and possible reference to original source material, but not to material which ``gives away'' a solution. E.g., if there is historical material cited then that should be given in a set of references \cite{Hutchinson1978}.

Barges are an integral component of the freight transportation system. 
The capacity of a  barge is sixty times that of a semi truck, and in tonnage per mile, barges are nine times as fuel efficient, saving shippers and consumers
over seven billion dollars annually. Here's your opportunity, as the CEO of a freight company,  to
help optimize the efficiency of barges.
\vspace*{.2 true cm}

\textbf{I)} \ How much energy is needed for a barge to cross  Lake Macatawa (with no current)?

The energy needed is equal to the work accomplished which is equal to the force (in the direction of the movement) times the distance. The primary (we assume the $only$) force to counteract in moving the barge is due to friction
with the water, denoted $F_{\rm friction}$.  As is noticed by anyone with their hand out the window of a moving car, the force required  to counteract friction is in the direction of travel and  increases with speed; in fact, for slow speeds the force is proportional to the speed.   Thus letting $v$ represent the speed of the barge through the water,  we have
\vspace*{-.1 true cm}  
$$ F_{\rm friction} = kv\, $$

\noindent where $k$ is a proportionality constant.
 
Then, the energy, $E$,  needed to move a distance from point \emph{0} to
point $p$ (distance of $p$) is:

\vspace*{-.4 true cm}
\begin{eqnarray*}
E &=& ({\rm Force}) ({\rm Distance})\\
&=& (F_{\rm friction})(p)\\
&=& k v p. 
\end{eqnarray*}


We see that as $v$ decreases to $0$, so does $E$. This
implies that there is no minimum amount of energy needed to cross
the lake. The required energy will become arbitrarily small as $v$
approaches $0$. \\

\noindent 1) Why is this not a realistic model for optimizing the efficiency of barges?  
\vspace{25mm}

\teacher{\vspace{-25mm} 
\begin{center}\parbox{12 true cm}{
[Customers cannot wait an infinite time (or even half that long!) to get their products. So the company would go bankrupt.]
}
\end{center}}

\vspace{.2 cm}
\noindent \textbf{II)} \ Suppose, more realistically,  that your company wants a barge to cross Lake Macatawa (distance $p$) with a minimum cost, where $T$ is the time of travel.

\noindent 1) Why might the following be a good model for the cost, $C$, of the trip?
\vspace{-.2 cm}
$$C = \alpha E + \beta T$$ 
\vspace{25mm}

\teacher{\vspace{-28mm} 
\begin{center}\parbox{12 true cm}{
[Two significant costs of transporting cargo are  cost of fuel and  time needed (cost of time).]
}
\end{center}}

\vspace{.2 cm}
\noindent 2) What factors are important in determining the
values of $\alpha$ and $\beta$?

\vspace{50mm}
\teacher{\vspace{-50mm} 
\begin{center}\parbox{12 true cm}{
[$\alpha$:  Cost of fuel - efficiency of engines;

$\beta$:  Cost of time - hourly wage for employees, number of employees on the barge, supplies, etc.]}
\end{center}}


\noindent 3) Find the value of $v$ that minimizes $C$.
(Hint: It will be in terms of $\alpha$, $\beta$, and $k$.) Explain why your answer is physically reasonable. 

\vspace{50mm} 
\teacher{\vspace{-50mm} 
\begin{center}\parbox{12 true cm}{[Since $T = \frac{p}{v}$ we know

\vspace*{-.8 true cm}
\begin{eqnarray*}
C &=& \alpha E + \beta T \\
&=& \alpha (k v p) + \beta \left(\frac{p}{v}\right) \, .
\end{eqnarray*}

So in order to minimize $C = C(v)$ we need to take the first derivative with respect to $v$ and set it equal to 0.
\vspace*{-.2 true cm}
$$C'(v) = \alpha k p - \frac{\beta p}{v^2} = 0 $$
then yields $v^2 = \frac{\beta}{\alpha K}$ which gives $v = \sqrt{\frac{\beta}{\alpha k}}$.

\vspace*{.2 true cm}
We see reasonable notions from this mathematical conclusion, for as the cost of time, $\beta$, increases, so does the speed; as the cost of fuel, $\alpha$, increases, the speed decreases; and also, the greater the proportionality constant, $k$, for the energy, the slower the speed should be. 
]}
\end{center}}

\newpage
\noindent \textbf{III)} \ Now let's suppose that your company  wants to minimize the energy needed to 
transport coal a distance $p$ upstream along a straight portion of the Grand River from Lake Michigan to  Grand Rapids, where the current of the river has a
constant speed, $v_{c}$.

\begin{figure}[h]
\begin{center}
\includegraphics[width=13 true cm]{1-50-RiverImage.eps}\\
%\begin{center} \includegraphics[width=250pt]{awesomeness.eps} \end{center}
\vspace*{-14.2 true cm}
\parbox{8.5 true cm}{\caption{Depiction of river-barge configuration.}}

\end{center}

\vspace*{-.9 true cm}
\end{figure}

Since the energy is required  to push the barge through the water, the  ``distance" we need for $E = ({\rm Force}) ({\rm Distance})$ is not how far the barge moves relative to the land, but  how far it moves relative to the water, i.e.,  the linear  distance of water that
passes the front of the barge.  We will call this the {\em Effective Distance} and denote it, $D_E$. 
Then,   $$E = (F_{\rm friction})( D_E )\, .$$


\noindent 1) Express  $D_E$ in terms of $v, v_c,$ and $p$. (Hint: Find $T$, the travel time from $0$ to $p$.)

\vspace{12mm}
\teacher{\vspace{-10mm} \vspace*{-.2 true cm}
\begin{center}\parbox{12 true cm}{[
Since $T = \frac{P}{v - v_c}$ we can say $D_E = vT = \frac{P v}{v-v_c}\, .$
]}\vspace*{-.5 true cm}
\end{center}}

\vspace{1 true cm}
\noindent 2) Express the energy, $E$, needed
to travel from $0$ to $p$ as a function of $v$.

\vspace{30mm}
\teacher{\vspace{-30mm} 
\begin{center}\parbox{12 true cm}{[
We see $E = F_{\rm friction} D_E = k v \frac{p v}{v-v_c} = \frac{k p v^2}{v-v_c}\, .$
]}
\end{center}}

\noindent 3) Find the value of $v$ that minimizes the total
energy. If the water in the Grand  River flows at 3 mi/hr, at what rate should the barge move through the water to minimize the energy of transport? 

\vspace{55mm} 
\teacher{\vspace{-55mm} 
\begin{center}\parbox{12 true cm}{[To minimize the energy, $E(v)$, we need to take its first derivative with respect to $v$ and set it equal to 0:
\vspace*{-.5 true cm}

\begin{eqnarray*}
E'(v) &=& \frac{(2 k p v)(v-v_c) - k p v^2(1)}{(v-v_c)^2}\\
 &=& \frac{2 k p v^2 - 2 k p v_c v - k p v^2}{(v-v_c)^2}\\
&=& \frac{k p v (v-2 v_c)}{(v-v_c)^2} = 0\, .
\end{eqnarray*} 

We obtain two values of $v$ for which $D(v) = 0$, namely $v = 0$, which yields the trivial case in which the barge goes backward, and $v = 2 v_c$. Thus we have $v = 2 v_c$ which means on the Grand River with $v_c = 3$ that our barge should move at 6 mi/hr through the water to minimize expended energy, $E(v)$.]}
\end{center}}

\noindent 4) Since we do not want the barge going backwards, we are concerned only with $v>v_c$.  

Graph  $E$ vs $v$ for $v>v_c$. (Choose, e.g.,  $k=1$, $p=2$ and $v_c=3$.)

\vspace*{5 true cm} 
\teacher{\vspace*{-5 true cm} \begin{figure}[h]
\begin{center}
\includegraphics[width=13 true cm]{1-50-EvPlot.eps}


\parbox{8.5 true cm}{\caption{\label{fig:EvPlot}Plot of Energy, $E(v)$, as a function of velocity, $v$.}}

\end{center}
\vspace*{-.5 true cm}
\end{figure}}

 

\noindent 5) Explain why the shape of the graph is physically reasonable. In particular,  explain  why the graph is shaped as it is  for  values of $v$ slightly larger than $v_c$. 



\vspace{12mm}
\teacher{\vspace{-10mm} 
\begin{center}\parbox{12 true cm}{
[It takes a great amount of energy - approaching an infinite amount - to transport items when moving through the water just above $v_c = 3$ mi/hr. This makes sense because it would take a long, long time to travel $p$ distance since $v-v_c$ is a very small positive number.]
}
\end{center}}


\noindent \textbf{IV)} \ Suppose now that your company is in danger of going bankrupt due to
your sub-par management skills. Thus we need to minimize the
 cost of transporting freight upriver rather than just the energy. Remember that the cost function is of the form

\vspace*{-.5 true cm}
$$C = \alpha E + \beta T.$$


\noindent 1) Express $C$ as a function  of $v$.

\vspace{30mm}


\teacher{\vspace{-30mm} 
\begin{center}\parbox{12 true cm}{\vspace*{-.6 true cm}
$$[C = \alpha \left(\frac{k p v^2}{v-v_c}\right) + \beta\left(\frac{p}{v-v_c}\right) = \frac{\alpha k p v^2 + \beta p}{v-v_c}\, .]$$}
\end{center}}

\noindent 2) Find the velocity that minimizes the cost. Simplify.

\vspace{105mm}

\teacher{\vspace{-105mm} 
\begin{center}
\parbox{12 true cm}{
[To minimize the cost, $C(v)$, we need to take the derivative with respect to $v$ and set this equal  to 0. Thus
$$C'(v) = \frac{(2 \alpha k p v)(v-v_c)- (\alpha k p v^2 + \beta p)}{(v-v_c)^2} = 0\, ,$$
yields this quadratic form
\vspace*{-.2 true cm}
$$\alpha k p v^2 - 2 \alpha k p v_c v - \beta P = 0 \, .$$
Solving for $v$ we find
\begin{eqnarray*}
v &=& \frac{2 \alpha k p v_c \pm \sqrt{4 \alpha^2 k^2 p^2 v_c^2 + 4 \alpha \beta k p^2}}{2 \alpha k p}\\
&= &v_c \pm \sqrt{v_c^2 + \frac{\beta}{\alpha k}}\, .\\
\end{eqnarray*}

\vspace*{-.3 true cm}
We select the positive speed so
\vspace*{-.2 true cm}
$$v =v_c + \sqrt{v_c^2 + \frac{\beta}{\alpha k}}\, .]$$
\vspace*{-.2 true cm}
}
\end{center}}



\noindent 3) How does the optimal velocity compare with the optimal
velocity from Part III? Explain why your answer is physically reasonable.

\vspace{55mm}


\teacher{\vspace{-50mm} 
\begin{center}
\parbox{12 true cm}{
[We notice the following:

\begin{enumerate}
\item There is an $\frac{\beta}{\alpha k}$ term as in Part II
\item If $\beta = 0$, then $v= v_c + v_c = 2 v_c$ as in Part III.
\item As in Part II, as cost of time increases, so does $v$ and as cost of energy increases, $v$ decreases, as expected.]
\end{enumerate}
}
\end{center}}

\noindent{\textbf{V)}} \ The previous model explored the case of
traveling upstream against a current of constant speed, $v_c$. More realistically, the speed of the current may change along the river due to widening or narrowing of the river or the addition of water from tributaries.  We now model traveling  upstream against a variable current,  expressing  the speed of the current as $v_c(x)$,
where $x$ represents the position along the bank of the river. (We  assume $v> v_c(x)$ for all values of $x$.) As before, the
required energy, $E$, is the product of the force and the Effective Distance, $D_E$:

\vspace*{-.3 true cm}
$$\hspace{6mm}E = (F_{\rm friction})( D_E )$$
%$$\ \hspace{25mm}= Force * (Rate * Time) $$
$$\ = (kv)(vT).$$


The time, $T$,  it takes to go distance $p$ can no longer be found
through a simple algebraic method because $v_c(x)$ is no longer
constant. Instead, the variable rate
must now be expressed using a differential equation.

\vspace*{.2 true cm}\noindent  1) Express $\frac{dx}{dt}$ in terms of $v$ and $v_c(x)$,
then use the equation to find, $T$,  the time of   travel. Use this to determine an expression for $E$
(which will include an integral).

\vspace{65mm}
\teacher{\vspace{-65mm} 
\begin{center}
\parbox{12 true cm}{
[$\frac{dx}{dt} = v - v_c$ implies $dt = \frac{dx}{v-v_c(x)}$ and thus $T = \int_0^T dt= \int_0^p \frac{dx}{v - v_c(x)}$. Then we have $E = k v^2  \int_0^p \frac{dx}{v - v_c(x)}$.
]
}
\end{center}}

\noindent 2) Consider a river whose current is increasing as you
travel upstream so that  $v_c(x)=\alpha x$ for $\alpha >0$. Simplify the expression for $E$ by
evaluating the integral.


\vspace{65mm}
\teacher{\vspace{-65mm} 
\begin{center}
\parbox{12 true cm}{

\begin{eqnarray*}
[E &=& k v^2 \int_0^p \frac{dx}{v - \alpha x} = k v^2\left( -\frac{1}{\alpha} \ln\vert v - \alpha x\vert \left.\right|_0^p\right)\\
&=& k v^2\left(\frac{1}{\alpha}(\ln\vert v\vert - \ln \vert v - \alpha P\vert)\right)\\
&=& \frac{k v^2}{\alpha}(\ln(v) - \ln (v - \alpha p)) = \frac{k v^2}{\alpha}\ln\left(\frac{v}{v -\alpha p}\right)\, .]
\end{eqnarray*}

}
\end{center}}


\noindent 3) Find the value of $v$ (in terms of $\alpha$ and $p$) that minimizes the energy
$E(v)$. (Hint: Show $E'(v) =(kv/\alpha)(1-\beta +2 \ln(\beta))$, where
$\beta = \frac{v}{v-\alpha p}$. Solve numerically for a value of $\beta$
between $3$ and $4$ and then use that value to solve for $v$. $\beta=1$ is the other solution, but notice that it requires $p=0$.)

\vspace{175mm}

\teacher{\vspace{-170mm} 
\begin{center}\vspace*{-.2 true cm}
\parbox{12 true cm}{
[\vspace*{-1 true cm}
\begin{eqnarray*}
0 &=& E'(v) =  \frac{2 v k}{\alpha} \ln\left(\frac{v}{v - \alpha p}\right) + \frac{k v^2}{\alpha} \left(\frac{1}{v} - \frac{1}{v- \alpha p}\right)\\
&=& \frac{v k}{\alpha}\left(2\ln\left(\frac{v}{v-\alpha p
}\right) + 1 - \frac{v}{v-\alpha p}\right)\\
&=& \frac{k v}{\alpha} (1 - \beta + 2 \ln(\beta)) \quad {\rm where} \quad \beta =\frac{v}{v - \alpha p}\, .
\end{eqnarray*}

Thus $\beta = 3.513$ and so $\frac{v}{v - \alpha p} = 3.513$ which implies that $v = 3.513 v - 3.513 \alpha p$. Hence, $2.513 v = 3.513 \alpha p$. And this means $v = \frac{3.513 \alpha p}{2.513} = 1.398 \alpha p$.
]
}
\end{center}}
\vspace*{.1true cm}

\noindent 4) For $\alpha = 1/5$ and a $10$ mile trip up the river,
what value of $v$ will minimize the expended energy of the trip?

\vspace{35mm}
\teacher{\vspace{-35mm} 
\begin{center}\vspace*{-.1true cm}
\parbox{12 true cm}{
[$v = (1.398)\left(\frac{1}{5}\right)(10) = 2.796$ mi/hr.]
}
\end{center}}


During this $10$ mile trip, the speed of the
river  increases uniformly from $0$ to $2$ mi/hr, so (as a function of position) the current has an average speed of $1$ mi/hr.  Recall from  ({\bf III}) that for constant $v_c$, the energy is
minimized when $v$ is twice the speed of
the current.  Thus one might  naively expect the optimal speed for this variable current model to be  $v=2$ mi/hr which is 
twice the average speed of the current.
  
\vspace*{.2 true cm}
\noindent 5) Explain why the optimal speed is, in fact, 
significantly greater than $2$ mi/hr.  (Hint: Refer to the graph
from {\bf III} and consider the rate of energy  increase on both sides of the minimum at $v=2v_c$.)

\vspace{55mm}
\teacher{\vspace{-55mm} 
\begin{center}
\parbox{12 true cm}{
[The required energy increases much more quickly as $v$ decreases from $2v_c$ than when it increases from $2v_c$. We can see this from the plot of $E(v)$ in Figure~\ref{fig:EvPlot}. Thus it is better to err on the side of too large a value of $v$. Notice, also that if $v = 2$, then the barge would not be able to reach  point $p$ where $v_c=2$.]
}
\end{center}}

\noindent\textbf{VI)} Recap:  What simplifying assumptions were used in these models?  As the CEO of the freight company, how might you make a better - more realistic - model?  



\vspace{85mm}
\teacher{\vspace{-85mm} 
\begin{center}
\parbox{12 true cm}{
[Simplifying assumptions include:

\begin{enumerate}
\item The mass of the barge stays constant.
\item Engine efficiency is independent of the speed of the barge, $v$.
\end{enumerate}

A better model would consider the following:

\begin{enumerate}
\item Take into account that as fuel is used the barge becomes lighter, so the value of $k$ will decrease, since the barge will then sit higher in the water.
\item Take into account that the efficiency of the engine depends on its revolution per minute rate, so as $v$ changes so will $\alpha$ in the formula, $C = \alpha E + \beta T$.]
\end{enumerate} 
}
\end{center}}
\section*{Acknowlegements}

The author was assisted with this project by Ryan TerLouw and Ryan Weaver, two students in an NSF-REU project during the summer of 2004.





\end{document}
