\documentclass{beamer}
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\usepackage{hyperref}

\begin{document}



\begin{frame}
\vspace*{-.5 true cm}
\begin{figure}
\centering
\includegraphics[width=10cm]{simiode-wordheader-Color}
\end{figure}
\vspace*{.5 true cm}

\begin{center}
{\Huge \bf Modeling Falling \\ Column of Water}

\vspace*{.9 true cm}

Brian Winkel, Director@simiode.org

\vspace*{.4 true cm}
SIMIODE, Cornwall NY USA
\end{center}
\end{frame}


\begin{frame}
\frametitle{Overview}
%\tableofcontents
\vspace*{-1 true cm}
\centerline{\Large \bf Modeling Falling Column of Water}
\vspace*{.4 true cm}
We collect data on a falling column of water and model the height using first principles from physics with a differential equation.

\vspace*{.4 true cm}

\begin{enumerate}
\item Video, data, qualitative behavior, empirical model
\vspace*{.1 true cm}\item First principles  analytic model, Torricelli's Law 
\vspace*{.1 true cm}\item Differential equation, estimate parameters,   validate model
\vspace*{.1 true cm}\item Discussions
\end{enumerate}

\vspace*{.4 true cm}
Source: 

 Modeling Scenario in SIMIODE - {\bf \href{https://qubeshub.org/community/groups/simiode/publications?id=3305\&v=5}{1-015-Torricelli}}


\end{frame}


\begin{frame}
\noindent  We use data  taken from  video at SIMIODE YouTube Channel
\begin{figure}
\centering
\includegraphics[width=5cm]{TorrDataYouTubeVideo}
\end{figure}

\vspace*{.4 true cm}
\centerline{{\bf \href{https://www.youtube.com/watch?v=NBr4DOj4OTE}{https://www.youtube.com/watch?v=NBr4DOj4OTE}}\, . }
\vspace*{.4 true cm}

\noindent Cylindrical column (radius = 4.17 cm) of water empties through a  hole (diameter = 11/16''= 0.218281 cm) in  bottom of column. Exit hole  at bottom of column - height is 0 cm. 
\vspace*{.3 true cm}

We seek to model $h(t)$, the height of the column of water.

\vspace*{.2 true cm}
\centerline{Just visit. No run yet.}
\end{frame}










\begin{frame}
\noindent Here is data we collected. What do you see or notice?
\vspace*{.2 true cm}

Make some observations now.

\vspace*{.4 true cm}
\begin{minipage}[t]{5 true cm}
\begin{table}[h]\tiny

\begin{center}
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}

\end{center}

\end{table}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{TorrData}
\end{figure}
\end{minipage}

\end{frame}

\begin{frame}
\noindent Linear Fit?


\vspace*{.4 true cm}

\begin{minipage}[t]{5 true cm}
\begin{table}[h]\tiny

\begin{center}
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}

\end{center}

\end{table}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{TorrDataLine}
\end{figure}
\end{minipage}

\end{frame}

\begin{frame}
\noindent Exponential Decay  Fit?


\vspace*{.4 true cm}

\begin{minipage}[t]{5 true cm}
\begin{table}[h]\tiny

\begin{center}
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}

\end{center}

\end{table}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{TorrDataExp}
\end{figure}
\end{minipage}


\end{frame}

\begin{frame}
\Large
\centerline{All  are empirical fits with no understanding.}


\vspace*{.4 true cm}

\centerline{They just fit a function to data.}

\vspace*{.4 true cm}

\centerline{And neither line nor exponential are good.}

\vspace*{.4 true cm}
\end{frame}

\begin{frame}
Run  video of  falling column and observe.



\vspace*{.3 true cm}
\centerline{{\bf \href{https://www.youtube.com/watch?v=NBr4DOj4OTE}{https://www.youtube.com/watch?v=NBr4DOj4OTE}}}

\vspace*{.3 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=6cm]{TorrDataYouTubeVideo}
\end{figure}

\vspace*{.3 true cm}

\centerline{What happens to height $h(t)$?}
 
 \vspace*{.3 true cm}
\centerline{How fast is column of water falling?   Early and later?}


 
\end{frame}

\begin{frame}
\noindent From  the video of the falling column what can we see?
\vspace*{.3 true cm}

 For large $h(t)$  the column of water falls faster or slower or same \ldots
\vspace*{.3 true cm}

For small $h(t)$  falls faster or slower or same \ldots


\begin{table}[h]\small

\begin{center}
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}
\end{center}

\end{table}

\vspace*{.3 true cm} What is your conclusion about $\displaystyle{\frac{d h(t)}{dt}}$? 

\end{frame}

\begin{frame}
\noindent So might something like this be true?


\begin{equation*}
\frac{dh(t)}{dt} = f(h(t))\, , \quad h(0) = h_0\, .
\end{equation*}
\vspace*{.2 true cm}

\noindent Where for large $h(t)$ we have large $f(h(t))$
\vspace*{.2 true cm}

\noindent Where for small $h(t)$ we have small $f(h(t))$



\vspace*{.6 true cm} And, of course, $f(h(t))$ is negative. Why?


\vspace*{.6 true cm}

\noindent Let's check out  change in height over various intervals of time. 
\end{frame}


\begin{frame}

\begin{center}\tiny
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}
\end{center}




\noindent  Check out the average rate of falling of the height of the column of water in several intervals, say, $[0, 2.187]$,


$$\frac{10.6-11.1}{2.187-0} = -0.2286\, ,$$


\noindent or in the interval $[39.503, 47.663]$,


$$\frac{2.0-3.0}{47.663 - 39.503} = -0.122549\, .$$

What do you see?  What can you say about $h'(t)$?
\end{frame}



\begin{frame}


\centerline{Let's find a model from some first principles.}
\vspace*{.1 true cm}

\centerline{This would be an analytic model.}
\vspace*{.4 true cm}

\centerline{{\bf NOT} just fit a function to data.}

\vspace*{.1 true cm}
\centerline{{\bf NOT} just ``it looks like it falls faster or slower.''}
\vspace*{.4 true cm}

\begin{minipage}[t]{5 true cm}
\begin{table}[h]\tiny

\begin{center}
\begin{tabular}{|c|c|} \hline 
Time (s)	&Height (cm)\\ \hline \hline
0.0&11.1\\ \hline
2.187&10.6\\ \hline
6.933&9.3\\ \hline
9.717&8.6\\ \hline
17.102&7.0\\ \hline
22.968&	5.75\\ \hline
30.603&	4.4\\ \hline
39.503&	3.0\\ \hline
47.663&	2.0\\ \hline
\end{tabular}

\end{center}

\end{table}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{TorrData}
\end{figure}
\end{minipage}

\end{frame}



\begin{frame}
\begin{center}
\parbox{10 true cm}{Enter Evangelista Torricelli  1608–1647,  an Italian physicist and mathematician, and a student of Galileo. Best known for his invention of the barometer.}
\end{center}

\vspace*{.3 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=4.5cm]{Torricelli}
\end{figure}

\end{frame}

\begin{frame}
\centerline{Torricelli's Law to the rescue!}


\begin{equation*}
\frac{dh(t)}{dt} = - b \sqrt{g \cdot  h(t)}\, , \quad h(0) = h_0\,  \quad b >0.
\end{equation*}

\vspace*{.2 true cm}

\noindent Say it out loud in  sentence form. 

\hspace*{4 true cm} Explain to yourself what it means.
\vspace*{.6 true cm}

\noindent Does Torricelli's Law agree with  observations?
\vspace*{.7 true cm}

\noindent For large $h(t)$  the column of water falls  faster or slower \ldots .

\vspace*{.2 true cm}

 \noindent For small $h(t)$  the column of water falls faster or slower \ldots .
 
 
\end{frame}

\begin{frame}
\centerline{Torricelli's Law}


\begin{equation*}
\frac{dh(t)}{dt} = - b \sqrt{g \cdot  h(t)}\, , \quad h(0) = h_0\,  \quad b >0.
\end{equation*}
\vspace*{.6 true cm}

\noindent Does Torricelli's Law agree with  observations?
\vspace*{.7 true cm}

\noindent For large $h(t)$  falls {\bf faster} or slower \ldots  YES!

\vspace*{.2 true cm}

 \noindent For small $h(t)$  falls faster or {\bf slower} \ldots YES!

 \vspace*{.6 true cm} We build the model that IS Torricelli's Law from First Principles. 
 \vspace*{.2 true cm}
 
\hspace*{4 true cm}   This will be an analytic model.
\end{frame}

\begin{frame}
\noindent Given the cross sectional area of the cylinder of water as a function of height and the area of the tiny exit hole at the bottom of the cylinder can we model the outflow of the water from the cylinder?

\vspace*{.5 true cm}
We will apply this First Principle

\vspace*{.5 true cm}

\centerline{\bf The Law of Conservation of Energy.}

\vspace*{.5 true cm}

We could ask you for a short statement of  The Law of Conservation of Energy.


%Poll

\end{frame}

\begin{frame} 

Thank you!  
\vspace*{.3 true cm}

Basically, The Law of Conservation of Energy says  that total energy is conserved.

\vspace*{.3 true cm} We will apply it to a slab of water, first at the surface of the column of water and then at the bottom of the column ($h = 0$)

\vspace*{.4 true cm}Total Energy is the the sum of the {\bf potential energy} and the {\bf kinetic energy} of a particle of mass $m$ and this sum is constant at each instance in time, $t$. 

\vspace*{1 true cm} Care to share  formula for {\bf potential energy} and for {\bf kinetic energy}? 

\hfill Go ahead and jump the gun!
\end{frame}

\begin{frame}
\vspace*{.6 true cm}

Consider  mass of water  $m$ initially atop  a cylinder of water, some $h$  meters above exit hole, this mass has 
\vspace*{.5 true cm}

\noindent {\bf initial potential energy} $PE_i=m\cdot  g\cdot  h$, where $g$ is the acceleration due to gravity  and

\vspace*{.5 true cm}
\noindent {\bf initial kinetic energy}  $ KE_i = \frac{1}{2}m v_i^2$, where $v_i$ is the initial velocity of the mass.   

\vspace*{.5 true cm}
Thus we have initial total energy of $TE_i$  when the mass of water is on the top of the cylinder of water:


\begin{equation*}
 TE_i = KE_i + PE_i = \frac{1}{2} m v_i^2 + m g h\, . 
\end{equation*} 

\end{frame}


\begin{frame}
When this mass of water reaches the exit hole  it has height $0$ meters and a final velocity of $v_f$. 


\vspace*{.3 true cm} 
Hence, the total energy at the final time, $TE_f$, the mass reaches the exit hole where $h=0$ is 

\vspace*{.3 true cm}


\begin{equation*}
 TE_f = KE_f + PE_f =  \frac{1}{2} m v_f^2 + m g\cdot h =  \frac{1}{2} m v_f^2 + m g\cdot 0 = \frac{1}{2} m v_f^2  \, . 
\end{equation*}
\vspace*{.3 true cm}

Now applying The  Law Conservation of Energy,  $TE_i = TE_f$, we build an equation - see the equal sign!! 

\end{frame}



\begin{frame}
Now by The  Law Conservation of Energy,  

\begin{equation*}
TE_i = \frac{1}{2} m v_i^2 + m g h = \frac{1}{2} m v_f^2 = TE_f \, ,
\end{equation*}

\vspace*{.2 true cm}
\noindent Divide both sides by $m$ and multiply by $2$ - to solve for $v_f$:

\vspace*{.4 true cm}

 \centerline{$v_f = \sqrt{2 g h + v_i^2}\, .$} 

\vspace*{.4 true cm}
\noindent Since $v_i$ = 0 we have one classical form of  Torricelli's Law
\begin{equation*}
v_f = \sqrt{2 g h} \, ,
\end{equation*}

\noindent where $v_f$ is the speed of the water as it leaves the exit hole. 


\end{frame}

\begin{frame}

We employ an effective tool in modeling: {\bf equate two ways of computing}  loss in  volume of water in  tank at time $t$. {\bf First},


\begin{equation}\label{eq:way1}
A(h(t)) h'(t)
\end{equation}


\large \noindent where $A(h(t))$ is  cross sectional area of  column at height $h(t)$, often constant, so $A(h(t)) = A$ in our case, and {\bf second}


\begin{equation}\label{eq:way2}
-v_f\cdot a  \cdot \alpha= -\sqrt{2\, g\, h} \cdot  a\cdot \alpha = -a\cdot \alpha\cdot   \sqrt{2\, g\, h} \, .
\end{equation}

\vspace*{-.3 true cm}

\noindent 
\begin{itemize}
\item $v_f$ is  velocity of  water exiting the column of water when the water is at height $h(t)$, 
\item $a$ is the cross sectional area of the small bore hole through which the water exits, and 
\item $\alpha$ is an empirical number, percent of   maximum flow rate  through  small  hole,   due to friction and constriction.  $\alpha$, is called  {\it discharge or contraction coefficient\/}. 
\end{itemize}

\end{frame}

\begin{frame}
 Now {\bf equating} (\ref{eq:way1}) and  (\ref{eq:way2})  - two different ways to calculate the rate at which the volume of water in   the cylinder decreases - gives rise to a working version of Torricelli's Law for the height of a  constant cross section column of water, $h(t)$, at time $t$:
\vspace*{.3 true cm}

\begin{equation}\label{eq:TorricelliLaw}
A \cdot h'(t) = A(h(t))\cdot h'(t) =- a \alpha  \sqrt{2 g \cdot h(t)} \, ,
\end{equation} 
\vspace*{.3 true cm}

\noindent where $h(t)$ is the height of the column of water at time $t$. 

\vspace*{.5 true cm}In our case $A(h(t)) = A$ as  cross section area of  water  is constant.  


\end{frame}


\begin{frame}

\noindent General form of Torricelli's Law in our case

\vspace*{.3 true cm}
\begin{equation*}\label{eq:Torricelli}
A \cdot h'(t) = A(h(t))\cdot h'(t) =- a \alpha  \sqrt{2 g \cdot h(t)} \, ,
\end{equation*} 

\vspace*{.3 true cm}
\noindent Solve  (\ref{eq:TorricelliLaw}) for $h'(t)$ and note that $A(h(t)) = A$.  Gather all the constants except $g$ into one big constant, $b$, we have

\begin{equation}\label{eq:Torr}
h'(t) = - b \sqrt{g \cdot h(t)}\, .
\end{equation}

\vspace*{.3 true cm}
\begin{center}
How  could we use our water videos  to validate this model.

 Take your time, think before you share your ideas.
 \end{center}

\end{frame}

\begin{frame}
So we have an analytic model (differential equation!) for $h(t)$.


\begin{equation*}
\frac{dh(t)}{dt} = - b \sqrt{g \cdot  h(t)}\, , \quad h(0) = h_0\, .
\end{equation*}

\vspace*{.5 true cm}
\noindent We solve this differential equation for $h(t)$ to realize a model.
\vspace*{.3 true cm}

\noindent What strategy/technique can we employ? What technology?

\vspace*{.3 true cm} We use this solution and our data to estimate  parameter $b$ and validate our model by comparing model predictions to data.
\end{frame}

\begin{frame}
\begin{equation*}
\frac{dh(t)}{dt} = - b \sqrt{g \cdot  h(t)} = -b\sqrt{g} \cdot   (h(t))^{1/2}.
\end{equation*}

\vspace*{.2 true cm}
{\bf Separate the variables}

\vspace*{.1 true cm}
\begin{equation*}
 (h(t))^{-1/2}\cdot \frac{dh(t)}{dt} =   -b\sqrt{g}.
\end{equation*}

\noindent OR 
\vspace*{.1 true cm}
\begin{equation*}
 (h(t))^{-1/2}\cdot dh =   -b\sqrt{g}\cdot dt.
\end{equation*}
\vspace*{.2 true cm}

\noindent Integrate both sides. (What is C?)

\vspace*{.1 true cm}
\begin{equation*}
 \int (h(t))^{-1/2}\cdot \frac{dh(t)}{dt}\, dt =   \int -b\sqrt{g}\, dt + C\, ,
\end{equation*}

\vspace*{.1 true cm}

\begin{equation*}
2 (h(t))^{1/2} =   -b\sqrt{g}\cdot  t + C.
\end{equation*}
\end{frame}

\begin{frame}
Now to find $C$ using Initial Conditions:
\vspace*{.1 true cm}

\begin{equation*}
2 (h(t))^{1/2} =   -b\sqrt{g}\cdot  t + C.
\end{equation*}

\begin{equation*}
2 (h(0))^{1/2} =   -b\sqrt{g}\cdot 0 + C = C.
\end{equation*}



Thus we have 


\begin{equation*}
2 (h(t))^{1/2} =   -b\sqrt{g}\cdot  t +  2 (h(0))^{1/2}.
\end{equation*}

\vspace*{.1 true cm}

\noindent Divide both sides by 2 and then square both sides yields:
\begin{equation}\label{eq:Know}
h(t) =   \left( -\frac{b\sqrt{g}}{2}\cdot  t + (h(0))^{1/2}\right)^2.
\end{equation}


\noindent This is model for  height of the column of water, $h(t)$,  at time $t$.

\vspace*{.3 true cm}
\noindent What do we know and what do we need to estimate $b$ in (\ref{eq:Know})?

\vspace*{.3 true cm}
\noindent  $h(0) = 11.1$ cm and $g = 980$ cm/s$^2$

\end{frame}

\begin{frame}
 Thus from  $h(0) = 11.1$ cm and $g = 980$ cm/s$^2$ 
\begin{equation*}
h(t) =   \left( -\frac{b\sqrt{g}}{2}\cdot  t +  (h(0))^{1/2}\right)^2 
\end{equation*}

\noindent becomes
\begin{equation*}
h(t) =   \left( -\frac{b\sqrt{980}}{2}\cdot  t +  (11.1)^{1/2}\right)^2\, ,
\end{equation*}

\noindent and expanded in decimals we have
\vspace*{-.3 true cm}

\begin{equation}
h(t) =   \left( - 15.6525\cdot b \cdot  t + 3.33166\right)^2.
\end{equation}

We have arrived at our model and now we seek to determine $b$ and validate our model.
\end{frame}

\begin{frame}
We turn to our Excel spreadsheet and seek to determine the parameter $b$ which  minimizes the sum of the squared errors between our data ($h_i$) and our model ($h(t_i)$) over  our data points. 



\begin{equation*}
SSE(b) = \sum_{i=1}^9 \left(h_i - h(t_i)\right)^2 \ .
\end{equation*}

\vspace*{.2 true cm}

\noindent Minimize as a function of the parameter $b$:

\begin{equation*}
SSE(b) = \sum_{i=1}^9 \left(h_i - h(t_i)\right)^2 \ .
\end{equation*}


\noindent where 
\begin{itemize}
\item $t_i$ is the $i^{\rm th}$ time observation,  
\item $h_i$ is the observed height at time $t_i$,
\item $h(t_i)$ is our model's prediction of the height at time $t_i$, and 
\item $n = 9$ is the number of data points we have.
\end{itemize}

\end{frame}


\begin{frame}

\centerline{Model Analysis in Excel Using Solver}

\vspace*{.1 true cm}
\begin{minipage}[t]{5 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelData}
\end{figure}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelPlot}
\end{figure}
\end{minipage}
\vspace*{.1 true cm}

\noindent We can use Excel's Solver to minimize the TOTAL SEE or SSE which is currently  7.679799546 with parameter $b = 0.002$ by asking Solver to minimize SEE or SSE as a function of $b = 0.002$ cell. 
\end{frame}



\begin{frame}
\centerline{Parameter Estimation with Excel Solver - Results}


\vspace*{.4 true cm}
\begin{minipage}[t]{5 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelDataOpt}
\end{figure}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelPlotOpt}
\end{figure}
\end{minipage}

\vspace*{.6 true cm}
Note: TOTAL SEE was at 7.679799546.

\vspace*{1 true cm}


\hspace*{ 5 true cm}\Large{\bf Go \href{https://www.simiode.org/resources/7769/download/CompleteModel11Over64InchHole.xlsx}{live to Excel}}. 

\end{frame}


\begin{frame}
\centerline{\bf Assignment}

\vspace*{.3 true cm}

\begin{enumerate}

\item Write an overview of the modeling process to obtain $h'(t)$ using first principles - not all details, just highlights. Arrive at the model

\begin{equation*}
h(t) =   \left( -\frac{b\sqrt{g}}{2}\cdot  t +  (h(0))^{1/2}\right)^2\, . 
\end{equation*}

\item Collect data for your team's  cylinder from 

\hspace{1.5 true cm} {\bf  \href{https://www.simiode.org/resources/488}{https://www.simiode.org/resources/488}}\, .

\vspace*{.3 true cm}\noindent What if we took many data points? Few data points?  Try both by taking a subset of ALL the data points you took for ``few data'' points and see how your parameter $b$ fares.
\end{enumerate}
\end{frame}

\begin{frame}


\noindent In Excel we  do these steps to determine  best fit parameter $b$:
\vspace*{.2 true cm}

\begin{enumerate}\setcounter{enumi}{2}
\item Compute our model value of the height $h(t_i)$ at time $t_i$.
\item Take the differences between actual data and model prediction, i.e. $h_i - h(t_i)$, and square these differences, $(h_i - h(t_i))^2$.
\item Sum these square errors to obtain $SSE(b)$.
\item Use Excel's Solver  to minimize $SSE(b)$.
\item Read  the value of $b$ and put it in our model as  best parameter estimate of  $b$.
\item Plot our best model values on the same axes as our data and compare.
\item Collect the parameters $b$ for the various outflow hole sizes from different videos selected by teams and see if there is any relationship between hole size and $b$. 
\end{enumerate}
\end{frame}

\begin{frame}
\centerline{Parameter Estimation with Excel Solver - Results}


\vspace*{.4 true cm}
\begin{minipage}[t]{5 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelDataOpt}
\end{figure}
\end{minipage}
\  \,  \ 
\begin{minipage}[t]{5.25 true cm}
\begin{figure}[!h]
\centering
\includegraphics[width=5.25cm]{ExcelModelPlotOpt}
\end{figure}
\end{minipage}

\end{frame}


\begin{frame}
\noindent {\bf CONGRATULATIONS!!!}
\vspace*{.3 true cm}

\noindent We went from
 
\begin{itemize}
\item seeing and collecting data, 
\item to conjecturing  empirical models, 
\item to building an analytical model from first principles,  
\item to realizing a differential equation model, 
\item to solving of the differential equation, 
\item to estimating our parameter, 
\item to comparing our model with the actual data.
\end{itemize}
\end{frame}






\end{document}