\documentclass{beamer}
%\usetheme{AnnArbor}
\usepackage{hyperref}
\title{Modeling Falling Column of Water}
\subtitle{Remote Teaching  Module}
\author{Brian Winkel, Director\\Director@simiode.org}
\institute{SIMIODE, Cornwall NY USA}
\date{}
\begin{document}

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\titlepage
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\centerline{Class simulation followed by discussion.}

\begin{figure}
\centering
\includegraphics[width=10cm]{simiode-wordheader-Color}
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\vspace*{-.3 true cm}
\centerline{\bf Slides and video  at \href{https://www.simiode.org/modules}{SIMIODE's Remote Teaching Resource site}.}
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\begin{frame}

\begin{figure}
\centering
Remote Teaching  Modules on Teaching \\
Differential Equations with Modeling.


\vspace*{.3 true cm}

\includegraphics[width=7cm]{SIMIODE Banner}
\end{figure}
\centering

\vspace*{.3 true cm}
{\bf \href{https://www.simiode.org/register}{https://www.simiode.org/register}}
\vspace*{.3 true cm}

Registration is FREE. 

Resources are Open Education Resources (OER).


All are downloadable and fully adaptable.
\vspace*{.3 true cm}

SIMIODE is funded by the National Science Foundation 

Division of Undergraduate Education.
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 \includegraphics[width=1.8cm]{NSF Icon}
\end{figure}

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\begin{frame}


\begin{center}
{\bf Send questions to Director@simiode.org.}

\end{center}

\vspace*{.6 true cm}

During this session we  simulate a class activity on modeling with a differential equation. At the end we will conduct a discussion  on issues, both pedagogical and technical. 

\vspace*{.3 true cm}
All the resources used will be available at {\bf \href{https://www.simiode.org/modules}{SIMIODE's  website}}. 

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\begin{frame}
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\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: 

1-015-S-Torricelli {\bf \href{https://www.simiode.org/resources/488}{https://www.simiode.org/resources/488}}

1-015-T-Torricelli {\bf \href{https://www.simiode.org/resources/463}{https://www.simiode.org/resources/463}}

\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}


\begin{frame}
\begin{center}
\Huge {\bf Discussion, Questions, Ideas, please \ldots }


\end{center}

\vspace*{2 true cm} \hfill \large Special news and opportunities up next!
\end{frame}

\begin{frame}
A word from one of our sponsors . . . .

\begin{figure}[!h]
\centering
\includegraphics[width=3.25cm]{scudem-color.png}
\end{figure}
\vspace*{.3 true cm}

\centerline{\bf \href{https://www.simiode.org/scudem}{https://www.simiode.org/scudem}}



\vspace*{.3 true cm}
\hspace{1 true cm} OR  just search ``SCUDEM'' in Google.

\vspace*{.3 true cm}
Fall 2020 with Challenge Saturday due date 14 November 2020. 

\end{frame}




\begin{frame}
Features of SCUDEM V 2020 include
\small
\begin{itemize}
\item  teams of three high school, home school, individuals, or undergraduate students - undergraduate level or lower 
\item teams from same institution or assembled by SIMIODE from students around the world
\item mentor/coach faculty engage with teams and fellow mentors/coaches
\item mentoring period 1 September 2020 through 23 October 2020
\item three problems from physics/engineering, chemistry/life sciences, and social sciences/humanities released on 23 October 2020
\item students prepare 10 minute video  presentation and upload to YouTube by Challenge Saturday, 14 November 2020
\item faculty from around the world judge and give  feedback
\item Outstanding, Meritorious, and Successful awards
\item SIMIODE posts ALL student team submissions and essay by problem poser on student submissions.
\end{itemize}





\end{frame}


\begin{frame}


\begin{figure}

\centering {\Large \bf Discussions and Questions}

\vspace*{.6 true cm}


\includegraphics[width=9cm]{deferential equations}
\end{figure}

\end{frame}


\begin{frame}
Join us at SIMIODE and 

\hspace*{2 true cm} take the lead for modeling differential equations.

\vspace*{.3 true cm}
\begin{figure}


\includegraphics[width=3cm]{LeadDog2}
\end{figure}

\vspace*{.3 true cm}

Visit us at {\bf  \href{https://www.simiode.org}{SIMIODE}}   and register. 

\vspace*{ 1 true cm} \hfill All is FREE at SIMIODE!

\end{frame}




\end{document}