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\newpage
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\WATER FLOWING FROM TANK}




\markboth{Water Flowing from Tank}{Water Flowing from Tank}




\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
}
\makeStitlePDFLaTex



\begin{abstract}
We help students develop a model (Torricelli's Law) for the height of a falling column of water with a small hole in the container at the bottom of the column of water through which water exits the column. We offer several sources of simulations on YouTube from which one can collect data and ask students to verify their model through parameter estimation.
\end{abstract}

\section*{STATEMENT}



In Figure~1 we see a screen capture from a video of an experiment \cite{Alicante2013} to display and  measure the height (cm) of a column of water in a graduated cylinder at time (s) when there is a tiny hole at the bottom of the cylinder through which the water exits. The time elapsed is 29 s while the water level is at 17.5 cm. Table~\ref{table:Alicante1} shows sample data from the same run of the experiment in which the column of water was 35 cm high originally, i.e. at $t = 0$.


\begin{figure}[t]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=6cm]{1-15-TorricelliAlicante.eps}


\vspace*{-3.5 true cm}
\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorriccelliAlicante} Screen capture \cite{Alicante2013} from a video of an experiment to display and  measure the height (cm) of a column of water in a graduated cylinder at time (s) when there is a tiny hole at the bottom of the cylinder through which the water exits. }}
\vspace*{-.3 true cm}
\end{figure}


\begin{table}[h]

\begin{center}
\begin{tabular}{|c|c|}\hline
Time (s) & Height (cm) \\ \hline
0 & 35 \\ \hline
8 & 30 \\ \hline
16 & 25 \\ \hline
25 & 20 \\ \hline
35.5 & 15 \\ \hline
48.7 &10 \\ \hline
66.8 & 5\\ \hline
\end{tabular}

\caption{\label{table:Alicante1} Time vs. height data for column of water experiment (FourthTrial) from \cite{Alicante2013}.}
\end{center}
\vspace*{-.5 true cm}
\end{table}

Finally, we offer a plot (Figure~\ref{fig:1-15-TorricelliAlicantePlot}) of the data in Table~\ref{table:Alicante1}.

\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliAlicantePlot.eps}


\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorricelliAlicantePlot} Plot of the data in Table~\ref{table:Alicante1}. }}
\vspace*{-.5 true cm}
\end{figure}

\begin{itemize}
\item [(1)] 
Form a mathematical model for the height of water in the column  in Table~\ref{table:Alicante1} and compare it to your data. How good is your model? Your physics? What measure will you use?
\end{itemize}


\section*{Derivation of Model from First Principles}


We examine a model based on first principles, not just on empirical fit.  Given a column of water  in a cylinder with a tiny hole at the bottom of the cylinder through which the water exits, how long does it take to empty the cylinder down to the level of the hole? More specifically, if we are given the cross sectional area of the cylinder as a function of height and the area of the tiny hole at the bottom of the cylinder can we model the outflow of the water from the cylinder.

Now for some basic physics background.  We first consider a law of physics which can help us.  The Law of Conservation of Energy says that the sum of the potential energy and the kinetic energy of a particle of mass $m$ is constant, i.e. total energy is conserved. So if we consider a particle of water of mass $m$ initially atop  a cylinder of water, some $h$  meters above a small, sharp-edged opening in the side wall of the cylinder through which the water can exit the cylinder, this mass of  water has initial potential energy $PE_i=m\cdot  g\cdot  h$, where $g$ is the acceleration due to gravity, and initial kinetic energy  $\frac{1}{2}m v_i^2$ ($KE_i$), where $v_i$ is the initial velocity of the mass.  Thus, we have an 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*} 

When this mass of water reaches the opening it has height 0 meters above the opening and a final velocity of $v_f$. Hence, the total energy at the final time ($TE_f$) the mass reaches the opening is 
\begin{equation*}
 TE_f = KE_f + PE_f = \frac{1}{2} m v_f^2 + m g\cdot 0 = \frac{1}{2} m v_f^2  \, . 
\end{equation*}
Now, by the Law Conservation of Energy we know $TE_i = TE_f$ and thus we have  
\begin{equation*}
\frac{1}{2} m v_i^2 + m g h = \frac{1}{2} m v_f^2 \, ,
\end{equation*}
which when one solves for $v_f$ yields $v_f = \sqrt{2 g h + v_i^2}$. In particular, if our element of mass of water was at rest initially, i.e. $v_i$ = 0 we have the classical Torricelli's Law

\begin{equation*}
v_f = \sqrt{2 g h} \, ,
\end{equation*}
where $v_f$ is the speed of the element of mass of water as it exits the opening in the cylinder. 

In particular, we can use this velocity of an element of water at the bottom of the cylinder given the height of the colulmn of water in the cylinder, $h$, to do some accounting in the change of the volume of the column of water in the tank at time $t$ and hence the height of the column of water, $h(t)$, in the tank at time $t$.

We equate two ways of computing the change in the volume of water in the tank at time $t$.

\begin{equation}\label{eq:way1}
A(t) h'(t)
\end{equation}
where $A(h(t))$ is the cross sectional area of the column at height $h(t)$, often constant, so $A = A(h(t))$ in our case, and
\begin{equation}\label{eq:way2}
v_f\cdot a  \cdot \alpha= \sqrt{2 g h} \cdot  a\cdot \alpha
\end{equation}
where $v_f$ is the velocity of the water exiting the column of water when the water is at height $h(t)$, $a$ is the cross section of the small bore hole through which the water exits, and $\alpha$ is an empirical number which indicates the perentage of the  maximum flow rate which actually gets through the small bore hole, the reduction due to friction and constriction in the small hole.  $\alpha$, is called the {\it discharge or contraction coefficient\/}.

Now equating (\ref{eq:way1}) and  (\ref{eq:way2}) 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 $t$:
\begin{equation}\label{eq:Torricelli}
A(h(t))\cdot h'(t) =- a \alpha  \sqrt{2 g h(t)} \, ,
\end{equation} 
where $h(t)$ is the height of the column of water at time $t$. Note the negative sign on the right hand side of (\ref{eq:Torricelli}) as water is leaving the column of water.

If we solve for $h(t)$ and note that $A(h(t)) = A$, while gathering all the constants except $g$ into one big constant, $b$, we have

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

\begin{itemize}
\item [(2)] 
Solve (\ref{eq:Torr}) for $h(t)$ if we know the initial height of the column of water,  i.e.  $h(0) = h_0$ and we presume the cross sectional area of the column of water at height $h$ is constant.
\end{itemize}




\subsection*{Modeling the outflow from a cylinder of water}

 Suppose we have a cylinder of height 0.20 m and radius  0.05 m and it is filled with water. There is an  opening in the side of the cylinder at the bottom of the cylinder in the form of a sharp edged circle of radius $r = 0.01$ m. 
 
Empirically, it is known that while the area of the opening is $\pi r^2 = \pi (.01)^2$ m$^2$ the effective area for fluid flow through the opening is only a fraction of the area. That fraction, $\alpha$, is called the {\it discharge or contraction coefficient\/}. Let us assume $\alpha = 0.7$, which is a reasonable value from the literature for various opening shapes and sizes. 

We now build a differential equation to determine the height of the water in the cylinder as a function of time, $h(t)$ in meters -- in this case. To do this we need to use accounting!! This will prove to be a nice way to build models, namely compute some quantity in two different ways and set the two different computations equal to each other. If the computations involve derivatives we have built a differential equation!

So let us compute, in two different ways,  the rate at which the volume of water changes at time $t$. We shall do this for a general cross sectional area of our cylinder as a function of height, $A(h(t))$.  

\begin{itemize}
\item [(L)] At time $t$ a small element of height changed, $h'(t)$, over a cross sectional area $A(h(t))$. Hence we will have lost 
\begin{equation}\label{eq:slab}
A(h(t))\cdot h'(t)
\end{equation}
 volume of water. In our case, for our circular cylinder of constant radius 0.05 m, we have the following computation:
\begin{equation}
A(h(t))\cdot h'(t) = \pi (.05)^2 h'(t)
\end{equation}
\item [(R)]  Since the water is flowing out of the opening with a velocity (according to Torricelli's Law) of $\sqrt{2 g h(t)}$ m/s and our opening's cross sectional area is $a = \pi r^2 = \pi (.01)^2$ m$^2$ then we are losing water at a rate of 
\begin{equation}\label{eq:opening}
\alpha a \sqrt{2 g h(t)} = \alpha \pi r^2 \sqrt{2 g h(t)} = .7 \pi (.01)^2 \sqrt{2 g h(t)} \quad {\rm m}^3{\rm /s}\, ,
\end{equation} 
where we recall the effective area for fluid flow through the opening is only a fraction ($\alpha = .7$) of the area.  
\end{itemize}
Now if we equate the computations for the loss of water found in (\ref{eq:slab}) with that computed in (\ref{eq:opening}) we obtain the following differential equation (adding our initial height $h(0) = h_0 = 0.20$ m as an initial condition):
\begin{equation}\label{eq:changevol}
A(h(t))\cdot h'(t)= \alpha\, a \sqrt{2 g h(t)}\, ,\quad h(0) = h_0
\end{equation}
in general, and in our case, using $g = 9.8$ m/s$^2$,
\begin{equation}\label{eq:drain}
\pi (.05)^2 h'(t)= .7 \pi (.01)^2 \sqrt{2 (9.8) h(t)}\, , \quad h(0) = 0.20\, .
\end{equation}

We note that we could collect all the constants in (\ref{eq:drain}) into on constant $b$ inside the square root sign and write:
\begin{equation}\label{eq:bcon}
h'(t) = -b \sqrt{ g \cdot h(t)} \, .
\end{equation}
Here $ b =\displaystyle{\frac{ .7 \pi (.01)^2}{\pi (.05)^2} \sqrt{2}} = 0.039598$.

\begin{itemize}
\item [(3)] Solve the differential equation~(\ref{eq:bcon}) (be sure you pick the correct branch of the solution) and determine how long it takes to empty the cylinder of water. Plot the height $h(t)$ as a function of time, $t$, for the duration of the draining. 
\item [(4)] Suppose $\alpha = 0.6$ instead of 0.7, then how long would it  take to empty the cylinder of water?  Plot the height $h(t)$ as a function of time, $t$, for the duration of the draining. Compare this answer and the plot with those from (a). Explain why this is reasonable?
\item [(5)] Continuing with (3), determine the velocity of the surface of the water and plot it over the  duration of the draining. Explain why what you see is reasonable?
\end{itemize}

Now let us play some ``what if" games with our model.


\begin{itemize}
\item [(6)] Vary the radius of the {\it opening\/}, $r$,   and determine the time $T(r)$ at which $h(T(r)) = 0$ for each r, i.e.  the time $T(r)$ to empty the tank as a function of $r$. Plot $T(r)$ and comment on its reasonableness. Use $\alpha = 0.7$ and the same parameters as in the discussion above. 
\item [(7)] Vary the radius of the {\it cylinder\/}, $r$,  in (3)  and determine the time $TT(r)$ at which $h(TT(r)) = 0$ for each r, i.e.  time $TT(r)$ to empty the tank as a function of $r$. Plot $TT(r)$ and comment on its reasonableness. Use $\alpha = 0.7$ and the same parameters as in (3) above. 
\item [(8)] Suppose the tank of water were in the shape of a frustum of a right cone, i.e. we change the radius of the bottom circular base, call it $s$, out (or in!) from the original radius. Again we have a cylinder of water of height $h_0 = 0.2$ m, a radius of 0.05 m at the top of the cylinder, an opening of radius 0.01 m at the bottom of the cylinder, but the base of the cylinder is now a circle with radius $s$ m. Use $\alpha = 0.7$ here, also. Collect some data on how long it takes to empty the cylinder for $s$ values in the region $s \in [0, 0.2]$, say, and try to come up with a function that models the time it takes to drain the cylinder as a function of the bottom radius, $s$.  NB:  If $s = 0.2$ then we have already done this analysis, as this is the constant cross sectional area case. Explain why your obtained function offers a reasonable relationship? Incidentally, if $s=0$ we have an inverted cone.

\end{itemize}


  
\section*{Two Sources of Data For Model Confirmation}

One could collect data with a modest apparatus of a graduated cylinder with a tiny hole near the bottom and a clock, all in the video frame at once to see height and time at the same time. However, we go to several sources on the web for videos and we collect data.


One source for data is from the University of Alicante physics experiments \cite{Alicante2013}. At the 3:41 mark of the 10:58 video there is a fourth run of data collection of water running out of a graduated cylinder of circular constant cross section   with radius 1.68 cm.  The initial height of the liquid in the tube is $h = 35$ cm.

\begin{itemize}
\item [(9)]
Here is the data we collected from our first video source \cite{Alicante2013}. How does it compare to yours?  How well can you model this data set? Offer up analysis to determine parameters from your model. Compare data with model.

\begin{table}[h]\label{table:Alicante}

\begin{center}
\begin{tabular}{|c|c|}\hline
Time (s) & Height (cm) \\ \hline
0 & 35 \\ \hline
8 & 30 \\ \hline
16 & 25 \\ \hline
25 & 20 \\ \hline
35.5 & 15 \\ \hline
48.7 &10 \\ \hline
66.8 & 5\\ \hline
\end{tabular}

\caption{Time vs. height data for column of water experiment (FourthTrial) from \cite{Alicante2013}.}
\end{center}

\vspace*{-.5 true cm}
\end{table}
\end{itemize}

A second source for data is the video \cite{Masooraahs2011}. Here, we see four trials of data collection of water running out of a graduated cylinder of constant circular cross section. The data comes from the FourthTrial which begins at 5:34 of the 7:23 video \cite{Masooraahs2011}.  We can also obtain the radius at the 9:34 time in to a second video \cite{Masooraahs2011a}. There is a picture which enables us to estimate the diameter of the cylinder as 19 cm or a radius of 9.5 cm. 

\begin{itemize}
\item [(10)]
In Table~\ref{table:FourthTrial} there is the data we collected from our second source \cite{Masooraahs2011}. How does it compare to yours? What data did you get from the First, Second, and Third Trial?  How well can you model each data set? Offer up analysis to determine parameters from one of the models above. Compare data with model.

\begin{table}[h]

\begin{center}
\begin{tabular}{|c|c|}\hline
Time (s) & Height (cm) \\ \hline
0 & 13 \\ \hline
.58 &  12 \\ \hline
1.35 &  11 \\ \hline
1.95 &  10 \\ \hline
2.85 &  9 \\ \hline
3.65 &  8 \\ \hline
4.55 & 7 \\ \hline
5.55 & 6 \\ \hline
6.55 & 5 \\ \hline
7.55 & 4 \\ \hline
8.55 & 3 \\ \hline
10.45 & 2 \\ \hline
13.45 & 1 \\ \hline

\end{tabular}

\caption{\label{table:FourthTrial} Time vs. height data for column of water experiment (FourthTrial) from \cite{Masooraahs2011}.}
\end{center}
\vspace*{-.8 true cm}

\end{table}
\end{itemize}


\section*{Additional Questions to Consider}

Here are some additional questions from \cite{KriederLahrDiesel2005}.


\begin{itemize}
\item [(11)]
Suppose we have a  cylindrical tank with the area of the tiny hole at the bottom being $a$ and cross sectional area of the cylinder being $A$. 

\begin{itemize}
\item [a)] 
How long does it take to empty the tank?
\end{itemize}

For each of the following physical changes in the tank, describe the effect on the time it takes to empty the tank. 

\begin{itemize}
\item [b)] Double the area, $a$, of the tiny hole at the bottom of the cylinder.

\item [c)] Change $g$ by considering the tank to be on the Moon. There  gravity is one sixth that of the Earth.
\end{itemize}
\end{itemize}




\section*{Building a Water Clock}

\begin{itemize}
\item [(12)] Design a cylinder of cross-sectional area $A(h)$ at height $h$ so that the water falls at a constant rate? In this way, we could  display equal intervals, say quarter hour, in equal vertical distance,  marking the time as it passes. To model a constant rate of fall for the height,  $h(t)$, of water we need to set $h'(t) = c$ for some constant we desire, e.g., 1 centimeter per 15 minutes. This clock cylinder  could then be marked at vertical heights of 1 cm with the knowledge that for ever centimeter the water falls 15 minutes has elapsed.

So, what shape must we elect; namely what is the cross sectional area at height $h$, i.e. $A(h)$,  which actually  gives us a constantly decreasing height in our water level?

When done you can run right down to your 3D printer shop and make your own water clock.  
\end{itemize}




\section*{Video sources for data collection}



We summarize our data sources. 

We used two videos of columns of water falling, without knowledge given as to the radius of the opening at the bottom of the cylinder. These were from YouTube and we cite them properly below. These are referred to in Activity 13.

  We also built experimental apparatus to collect data using a free digital  clock on a computer ({\tt www.on-line-stopwatch.com}) which measures and display nicely and in large fonts real time to the thousandth of a second, a clear sparkling water plastic container, a collection bin for the exiting water, and an iPad to produce video of the experiments. We did this for small holes  at the bottom of the cylinder of diameters 13/16'', 9/64'', 11/64'', and 7/64''. In the latter case we used two data sets - sampling from the entire run and then sampling from a smaller section of the run. We placed the following videos  for your  use in this section for this scenario 1-15-Torricelli on YouTube.  These are referred to in Activity 14.



\vspace*{.3 true cm}
\noindent Source 1:


Name: Ley de Torricelli Vaciado de un dep\'{o}sito 

URL:  {\tt http://www.youtube.com/watch?v=6LOgdOfFAbI} 

 

\vspace*{.3 true cm}

\noindent Source 2:



Name: Numec project Sem1 2011 THE Massooraahs! 

URL: {\tt http://www.youtube.com/watch?v=44fZdi7QC8E}   




\vspace*{.3 true cm}
Sources 3 - 6 are available at YouTube and at this Modeling Scenario site. 

{\tt https://www.simiode.org/resources/488}.

\noindent Source 3:



Name: 13Over64Inch Small Hole 

URL: {\tt https://www.youtube.com/watch?v=gsNdsuQ1ZCo}  

and {\tt https://www.youtube.com/watch?v=SCNqxFe3eIs}

Also available at {\tt https://www.simiode.org/resources/488/video?resid=2085}.

\vspace*{.3 true cm}
\noindent Source 4:



Name: 9Over64 Inch Small Hole  

URL: {\tt https://www.youtube.com/watch?v=gsNdsuQ1ZCo}

and {\tt https://www.youtube.com/watch?v=6cUuNUjlaPI}


Also available at {\tt https://www.simiode.org/resources/488/video?resid=2083}.


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=13cm]{1-15-TorricelliSetUp.eps}

\vspace*{-10.7 true cm}
\parbox{13.7 true cm}{\caption{\label{fig:1-15-TorricelliSetUp} Experimental apparatus for collecting data  on water flowing from cylindrical column of water. Left to right we see laptop computer with {\tt www.online-stopwatch.com} running to get time to the nearest thousandth of a second, marked circular cylinder of water  
(in cm of height) in the form of a sparkling water plastic container, water exiting to the right from small hole of radius 9/64 '', and receiving container for water leaving the cylinder.}}
\vspace*{-.3 true cm}
\end{figure}

\vspace*{.3 true cm}
\noindent Source 5:



Name: 11Over64Inch Small Hole  

URL: {\tt https://www.youtube.com/watch?v=76diQ1eW8i4}

and {\tt https://www.youtube.com/watch?v=NBr4DOj4OTE}

Also available at {\tt https://www.simiode.org/resources/488/video?resid=2087}.

\vspace*{.3 true cm}
\noindent Source 6:

Name: 7Over64 Inch Small Hole 

URL: {\tt https://www.youtube.com/watch?v=k7TSAqRfHgg}

and {\tt https://www.youtube.com/watch?v=xDyDSPydN\_E}\, .

Also available at {\tt https://www.simiode.org/resources/488/video?resid=2084}.

\begin{itemize}
\item [(13)] and (14) 
Using one of the videos from Source 1 or Source 2 and then one of the videos from Sources 3-6 collect data on time and height of the column of water and model it, first with a Power Rule and then if possible with Torricelli's Law differential equation. Estimate all parameters possible, and confirm your model against the data you have collected.
\end{itemize}

\section*{Beginning and End Observations - You Mess with Mr.~In Between}

\begin{itemize}
\item [(15)]
\begin{quote}
A small hole of radius $s = 0.138906$ cm is drilled in the side of a right circular cylindrical container of radius $r = 4.16986$ cm and the height of the water level (above the hole) goes from 12.5 cm at a stop watch time of 1:40.419 min to 3 cm at a stop watch time of 2:34.116  min. Use two different models to estimate the height at intermediate times. These models are (1) linear model:  $h'(t) = - k h(t)$, $h(0) =  12.5$ with $h(143.286) = 2.0$ and (2) Torricelli's Law: $A h'(t) = - \alpha \cdot  a\cdot  \sqrt{2\cdot  g\cdot  h(t)}$, $h(0) =  12.5$ with $h(143.286) = 2.0$. Here $A$ is the cross sectional area of the cylinder and $a$ is the area of the small hole for drainage, while $\alpha$ is the discharge or contraction coefficient.

After you have completed your analyses, go to the video, entitled 11Over64 Inch Small Hole from the from this Scenario - 1-15-Torricelli, and compare your observations from the video with those from each of your model. Which model  is the better one?  Why? 

\end{quote}

\vspace*{.2 true cm}

\noindent We offer a table (Table~\ref{table:TorricelliData}) of clock times, relative times, and known observations. You fill in the rest from your modeling activity results.


\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|} \hline
Clock Time & Relative Time & Linear & Torricelli & Observed\\
$t$ (min) & $t$ (s)  & $h$ (cm)& $h$ (cm) & $h$ (cm)\\  \hline 
1:40.419 = 100.419 & 0&12.5 & 12.5 & 12.5 \\ \hline
1:45.598 = 105.598 &5.179 & \hspace*{2 true cm} & \hspace*{2 true cm}&  \hspace*{2 true cm} \\ \hline
1:47.568 = 107.788 & 7.369& & & \\  \hline
1:49.623 = 109.523 & 9.104& & &  \\ \hline 
1:50.985 = 110.985 & 10.566 & & & \\  \hline 
1:56.388 = 116,388 & 15.969& & & \\  \hline 
2:02.057 = 122.057 & 21.638& & & \\  \hline 
2:08.032 = 128.032 & 27.613& & &  \\ \hline
2:16.386 = 136.386 & 35.967& & &  \\ \hline
2:25.568 = 145.568 & 45.149& & &  \\ \hline
2:34.116 = 154.286 & 53.867& & &2.0  \\ \hline
\end{tabular}

\parbox{12.1 true cm}{\caption{\label{table:TorricelliData} Clock Time and Relative Time for sample Torricelli Law run.}}
\end{center}
\end{table}
\end{itemize}


\begin{thebibliography}{99}
\bibitem{Alicante2013}  University of Alicante. 2013. {\it Experienceas de F\'{i}sica.  Demostraciones y Pr\'{a}cticas de Laboratorio\/}. \href{http://www.dfists.ua.es/experiencias\_de\_fisica/index1.html}{http://www.dfists.ua.es/experiencias\_de\_fisica/index1.html}.  Also available as  {\it Ley de Torricelli Vaciado de un dep\'{o}sito\/} at  {\tt http://www.youtube.com/watch?v=6LOgdOfFAbI}. Accessed 30 March 2013.

\bibitem{FarmerGass1992} Farmer, T. and F. Gass. 1992. Physical Demonstrations in the Calculus Classroom. {\it The College Mathematics Journal\/}.  23(2): 146-148.

\bibitem{KriederLahrDiesel2005} Krieder, D., D. Lahr, and S. Diesel. 2005. {\it Principles of Calculus Modeling:  An Interactive approach\/}. From the Quiz in Unit on Torricelli's Law. \href{http://www.math.dartmouth.edu/~klbooksite/2.18/218quiz/218quiz\_p02.htm}{http://www.math.dartmouth.edu/~klbooksite/2.18/218quiz/218quiz\_p02.htm}.  Accessed 2 April 2013.

\bibitem{Masooraahs2011}  Masooraahs. 2011. Numec project Sem1 2L011 THE Masooraahs!  \href{http://www.youtube.com/watch?v=44fZdi7QC8E}{http://www.youtube.com/watch?v=44fZdi7QC8E}. Accessed 31 March 2013. Video length 7:23.

\bibitem{Masooraahs2011a}  Masooraahs. 2011a. Numerical methods extended video.   \href{http://www.youtube.com/watch?v=-HsVq3KJIGA}{http://www.youtube.com/watch?v=-HsVq3KJIGA}. Accessed 31 March 2013. Video length 9:58.


\end{thebibliography}

\teacher{
\section*{COMMENTS}


We offer comments with  possible responses for Activities from this scenario.  Activities 13-15 are discussed  in the Mathematica notebook 1-15-T-Mma-Torricelli-TeacherVersion.nb and in pdf format. In addition there are other data sets  in this notebook which are of use.

We turn now to Activities 1-12.

\subsection*{Activity 1}

\begin{itemize}
\item [(1)] 
Form a mathematical model for the height of water in the column offered in the data in Table~\ref{table:Alicante1} and compare it to your data. How good is your model? Your physics? What measure will you use?
\end{itemize}

We offer a more complete opportunity to respond to this data set and video in Activity 13 after we have developed more modeling efforts, in fact, after we have constructed a differential equation from Torricelli's Law based on the Law of Conservation of Energy. However, for now we would expect to respond with an exponential decay fitting effort in which students DO NOT go after a differential equation, but rather try to fit something like $h(t) = A e^{-k t}$ to the data, with perhaps $A = h(0) = 35$.

Using any reasonable approach ({\tt FindFit} command in Mathematica, direct minimization of sum of square errors for function model $h(t) = a e^{-b t}$, or exponential function in TrendLine in EXCEL students will obtain something close to this function: $h(t) = 36.1103 e^{-0.0252994 t}$ and when they plot the exponential model over their data they will get something like that seen in Figure~\ref{fig:1-15-Torricelli-Activity1-ExpFit}.


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9cm]{1-15-Torricelli-Activity1-ExpFit.eps}


\parbox{12.1 true cm}{\caption{\label{fig:1-15-Torricelli-Activity1-ExpFit} For the modeling request in Activity 1 we offer a typical student response to fitting exponential decay function,   $h(t) = a e^{-b t}$, to the falling water level data.}}
\vspace*{-.3 true cm}
\end{figure}

This not too perfect model leads students in class to discussions about what the exponential function really comes from, with respect to a rate of change, namely, $h'(t) = - k\cdot  h(t)$ - the more we have the more we lose per unit time. That certainly seems to be what is going on in the water falling model, for when the column is bigger we would expect more weight to apply more pressure to the water at the small hole at the bottom of the cylinder through which the water exits, so the higher the column of water the faster the flow through the small hole and hence (assuming the cross sectional area of the cylinder) is a constant) the faster the water level decreases in the column. Aha, but ``Wait a minute." students say, ``There could be a more general model at work, say $h'(t) = - k\cdot  h(t)^r$ where $r$ is some power and not necessarily $r = 1$ as in the exponential decay case.''

This begins the real modeling journey, even before the introduction of supporting physics, and students can do a two parameter estimation  ($k$ and $r$) after solving the differential equation $h'(t) = - k\cdot  h(t)^r$. We are usually in Mathematica and so there is little fear of solving  and obtaining a solution for the differential equation,

\begin{equation}
\label{eq:genmodel}
h(t) = \left((r-1) \left(k
   t+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}   \, ,
\end{equation}

\noindent and forming sum of square errors 

\small
\begin{eqnarray*}
SSE(k,r) &=& \left(30-\left((r-1) \left(8
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2+\left(25-\left((
   r-1) \left(16
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2\\
&& +\left(20-\left((
   r-1) \left(25
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2+\left(15-\left((
   r-1) \left(35.5
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2\\
&& +\left(10-\left((
   r-1) \left(48.7
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2+\left(5-\left((r
   -1) \left(66.8
   k+\frac{35^{1-r}}{r-1}\right)\right)^{\
   \frac{1}{1-r}}\right)^2\\
&&  +\left(35-\left(3
   5^{1-r}\right)^{\frac{1}{1-r}}\right)^2
\end{eqnarray*} \normalsize
between the observed data and the model predictions with both $k$ and $r$ unknown parameters in this complex sum to be minimized in order to find the best parameters ($k$ and $r$). We felt it important to show here how complex the sum of squares expression can be. Of course, with more data the expression's length increases.  Using Mathematica's FindMinimu command on  $SSE(k,r)$ we obtain values of $k = 0.0931652,$ and $r = 0.563884$ and a plot of this general model, over the data appears in Figure~\ref{fig:1-15-Torricelli-Activity1-GenFit.eps} . Now this looks better to students. 



\begin{figure}[t]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9cm]{1-15-Torricelli-Activity1-GenFit.eps}


\parbox{11.7 true cm}{\caption{\label{fig:1-15-Torricelli-Activity1-GenFit.eps} In Activity 1 we see a plot of the data and a plot of the general solution (\ref{eq:genmodel}) to the differential equation $h'(t) = - k \cdot  h(t)^r$ with best fitting parameters, $k = 0.0931652,$ and $r = 0.563884$, to the falling water level data.}}
\vspace*{-.8 true cm}
\end{figure}

Incidentally, if this model, sum of squares, and minimize for best parameter estimate approach is new to the students then this is the time to spend effort to build confidence, for this will be the main approach to parameter estimation in differential equation modeling  in our approach.

 
\subsection*{Activity 2}

To solve the differential equation obtained from the Law of Conservation of Energy:  $h'(t) = \sqrt{2g h(t)}\, , h(0) = h_0$ one can separate variables 

$$\frac{1}{h(t)^{1/2}} \frac{dh}{dt} = -\sqrt{2 g} \, , \qquad {\rm etc.} \,$$ or use something like {\tt DSolve} in Mathematica to obtain two possible solution:

$$h(t) = \frac{1}{2} \left(\pm 2 \sqrt{2} \sqrt{g}
   \sqrt{h_0} t+g t^2+2
   h_0\right)\, .$$

\vspace*{-.2 true cm}
The solution,

\vspace*{-.2 true cm}
$$h(t) = \frac{1}{2} \left(- 2 \sqrt{2} \sqrt{g}
   \sqrt{h_0} t+g t^2+2
   h_0\right)\, ,$$
with the minus sign is the correct one, otherwise $h(t)$ just keeps increasing, and we know the water level in our situation decreases!


\subsection*{Activity 3}

This is really the same as Activity 2, only with numbers in it and meaning from the derivation of the differential equation for a column of water of height $h(t)$ at time $t$ with cross sectional area $A(h(t))$. Notice the negative in front of the right had side of (\ref{eq:Activity3DE}) due to the fact that the tank does drain and the constants cannot reflect the negative nature of the situation; hence, we have to do our own accounting and make the right hand side negative.

So a solution to

\vspace*{-.3 true cm}
\begin{equation}\label{eq:Activity3DE}
\pi (.05)^2 h'(t) = -0.7 \pi (0.01)^2 \sqrt{2\cdot  9.8 h(t)}\, 
\end{equation}
\vspace*{-.7 true cm}

\noindent is 

\vspace*{-.4 true cm}
$$h(t) = 0.0038416  t^2 - 0.0554372 t + 0.2 \, .$$    

To determine when all the water has run out  set $h(t) = 0$ and we see that the time at which the water has run out is $t_f = 7.21538$ s. A plot of the height of water in the cylinder over the time interval $[0, 7.21538]$ is shown in Figure~\ref{fig:1-15-TorricelliActivity3Plot}.

\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9cm]{1-15-TorricelliActivity3Plot.eps}


\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorricelliActivity3Plot} We plot the solution for  (\ref{eq:Activity3DE}).}}
\vspace*{-.3 true cm}
\end{figure}

\subsection*{Activity 4}

This  the same as Activity 3, only changing $\alpha$ from 0.7 to 0.6, i.e. the effective percent of the tiny hole's area which actually permits water to flow out has gone from 70\% to 60\%.

So a solution to

\vspace*{-.3 true cm}
\begin{equation}\label{eq:Activity4DE}
\pi (.05)^2 h'(t) = -0.6 \pi (0.01)^2 \sqrt{2\cdot  9.8 h(t)}\, 
\end{equation}
\vspace*{-.3 true cm}
is 

$$h(t) = 0.0028224 t^2 - 0.0475176  t + 0.2 \, .$$    

To determine when all the water has run out  set $h(t) = 0$ and we see that the time at which the water has run out is $t_f = 8.41794$ s. A plot of the height of water in the cylinder over the time interval $[0, 8.41794]$ is show in Figure~\ref{fig:1-15-TorricelliA3A4Plots}. We actually plot the cases where $\alpha = 0.7$ (thin line) and $\alpha = 0.6$ (thick line) on the same axes and there is no surprise; it takes longer to empty when the $\alpha$ value is lessened.


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliA3A4Plots.eps}

\vspace*{-6.4 true cm}
\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorricelliA3A4Plots} We plot the solution for  (\ref{eq:Activity3DE}) [thin] and (\ref{eq:Activity4DE}) [thick].}}

\end{figure}


\subsection*{Activity 5}

Here we are asked to, ``\ldots with (3) determine the velocity of the surface of the water and plot it over the  duration of the draining. Explain why what you see is reasonable?"

By differentiation the formula for the height (P$h(t)$) of the column of water from Activity 3 we see the velocity:

$$v(t) = h'(t) = 0.0076832 t-0.0554372 \, ,$$ 
whose plot is rendered in Figure~\ref{fig:1-15-TorricelliActivity5Plot}.
\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliActivity5Plot.eps}

\vspace*{-6.8 true cm}
\parbox{13.4 true cm}{\caption{\label{fig:1-15-TorricelliActivity5Plot} We plot the solution for  velocity of the height of the water described in (\ref{eq:Activity3DE}). We note that the velocity of the height is negative (the height is decreasing) and that the rate is getting smaller in size as time goes on, for the cross section stays the same, but initially the height is higher - hence the pressure at the small hole at the bottom of the cylinder where the water exits is greater, hence the larger sized rate of decrease of the height of he column of water.}}

\end{figure}


\subsection*{Activity 6}

Here we are asked to `` Vary the radius of the {\it opening\/}, $r$,   and determine the time $T(r)$ at which $h(T(r)) = 0$ for each r, i.e.  time $T(r)$ to empty the tank as a function of $r$. Plot $T(r)$ and comment on its reasonableness. Use $\alpha = 0.7$ and the same parameters as in the discussion above.''

When we solve (\ref{eq:Activity3DE}) with $r$ as a variable in place of 0.01 as the radius of the small hole at the bottom of the cylinder through which the water exits, we obtain

$$h(t,r) = 0.2 - 554.372 r^2 t + 384160.0 r^4 t^2\, ,$$
from which we find that 
$h(t,r) = 0$ when the time to empty the cylinder, in this case, is 
$$T(r) = 0.000721538/r^2\, .$$

We plot the time to empty the cylinder as a function of the small hole radius in Figure~\ref{fig:1-15-TorricelliTimeToEmptySmallHoleRadius}. We note that if $r = 0$, i.e. there is no hole at the bottom then the time to empty the tank is infinite, which makes sense.

\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliTimeToEmptySmallHoleRadius.eps}


\parbox{13.7 true cm}{\caption{\label{fig:1-15-TorricelliTimeToEmptySmallHoleRadius} We plot the time to empty the cylinder as a function of $r$ the radius in cm of the small hole at the bottom of the cylinder through which the water exits  given the height of the water described in (\ref{eq:Activity3DE}). We are not surprised that the larger the radius the quicker the tank empties.}}

\end{figure}


\subsection*{Activity 7}

Here we are asked, ``Vary the radius of the {\it cylinder\/}, $r$,  in (3)  and determine the time $TT(r)$ at which $h(TT(r)) = 0$ for each r, i.e.  time $TT(r)$ to empty the tank as a function of $r$. Plot $TT(r)$ and comment on its reasonableness. Use $\alpha = 0.7$ and the same parameters as in (3) above."

When we solve (\ref{eq:Activity3DE}) with $r$ as a variable in place of 0.05 as the radius of the cylinder, we obtain

$$h(t,r) = \frac{2.401 \times 10^{-8}}{r^4}
   t^2-\frac{0.000138593
   }{r^2}t +0.2\, ,$$
from which we find that 
$h(t,r) = 0$ when the time to empty the cylinder, in this case, is $$TT(r) = 2886.15 r^2-2.91909r^2 = 2883.23 r^2\, .$$


We plot the time to empty the cylinder as a function of the radius of the cylinder   in Figure~\ref{fig:1-15-TorricelliTimeToEmptyCylinderRadius}.  We note that if $r = 0$, i.e. there is no column of water then the time to empty the tank is 0, which makes sense.
\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliTimeToEmptyCylinderRadius.eps}


\parbox{13.7 true cm}{\caption{\label{fig:1-15-TorricelliTimeToEmptyCylinderRadius} We plot the time to empty the cylinder as a function of $r$ the radius in cm of the cylinder   given the height of the water described in (\ref{eq:Activity3DE}). We are not surprised that the larger the radius of the cylinder the longer it takes to empty. }}

\end{figure}


\vspace*{-.7 true cm}
\subsection*{Activity 8}

Here we are asked, ``Suppose the tank of water were in the shape of a frustum of a right cone, i.e. we change the radius of the bottom circular base, call it $s$, out (or in!) from the original radius. Again we have a cylinder of water of height $h_0 = 0.2$ m, a radius of 0.05 m at the top of the cylinder, an opening of radius 0.01 m at the bottom of the cylinder, but the base of the cylinder is now a circle with radius $s$ m. Use $\alpha = 0.7$ here, also. Collect some data on how long it takes to empty the cylinder for $s$ values in the region $s \in [0, 0.2]$, say, and try to come up with a function that models the time it takes to drain the cylinder as a function of the bottom radius, $s$.  NB:  If $s = 0.2$ then we have already done this analysis, as this is the constant cross sectional area case. Explain why your obtained function offers a reasonable relationship? Incidentally, if $s=0$ we have an inverted cone.''


We have decided to address issues for this Activity in a separate notebook:  1-15-T-Mma-Activity8-Comments-Torricelli-TeacherVersion.nb and its accompanying pdf form.

However, we outline the approach here and leave the execution and details to the notebook.


Using similar figures we can determine a relationship between $r$, the radius of the cross sectional area, at a given height $h$ by

\begin{equation}
\label{eq:hs}
h=\frac{4 r-4 s}{1\, -20 s}\, ,
\end{equation}
where 0.05 is the original radius of the circular cross section of the surface of the water at the top, $h(0) = 0.2$ is the height of the water  at the start, $r$ and $h$ are typical radius and height of a circular cross section of the frustum at any given time.

Now solving (\ref{eq:hs}) we have that at height $h$  of the surface of the water the radius is $r = h (0.25 -5  s)+  s$ and the cross sectional area of the surface of the water at height $h$ is $$A(h) = \pi (h (0.25 -5  s))^2\, . $$

This means that the differential equation for Torricelli's Law

\begin{equation}\label{eq:TorrLaw}
A(h(t)) h'(t) = \alpha a \sqrt{2 \cdot  g \cdot  h(t)} \, ,
\end{equation}
in our situation is
\begin{equation}\label{eq:frustum}
 \pi (h(t) (0.25 -5  s))^2 h'(t) = 0.7 (\pi (0.01)^2)  \sqrt{2 \cdot  9.8 \cdot  h(t)} \, ,
\end{equation}

When attempting to solve, in closed form, (using {\tt DSolve} in Mathematica) for $h(t)$, the solution to the differential equation (\ref{eq:frustum}) has 10 branches. After some investigation the correct branch for our model is determined - it is a large and we only produce it in the Mathematica notebook! We called it $hs6(t,s)$ as it was the sixth branch that meet the model criteria of positive real numbers doing the correct things in our modeling situation.

When we tried to solve $hs6(t,s) = 0$ for $t$ to see if we could ascertain the time it takes to empty the tank of base radius $s$ we were met with Mathematica's $\{\}$. However,  if we sought out specific solutions for $t$   to $hs6(t,s) = 0$ where $s$ ran over the interval $[0,.2]$ in step sizes of 0.0005 we obtained the plot in Figure~\ref{fig:1-15-TorricelliTimeToDrainActivity8}. This plot is very reassuring for it tells us that as the base radius, $s$, of our frustum shaped tank increases  the time to empty the tank increases. This make sense, for as $s$ increases the amount of water in the tank increases, and hence the time it takes to empty increases.

\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliTimeToDrainActivity8.eps}


\parbox{13.7 true cm}{\caption{\label{fig:1-15-TorricelliTimeToDrainActivity8} A summary plot of the data on how long it takes the frustum shaped tank to drain if the initial height is $h_0 = 0.2.$ m with surface area radius of 0.05 m at the top of the water as a function of $s$, the radius of the base of the frustum in m.}}

\end{figure}

\subsection*{Activity 9}

A complete analysis of this data is offered in the Mathematica notebook 1-15-T-Mma-Torricelli-TeacherVersion.nb and in pdf format. It is listed as First Source:  Data for Torricelli's Law.

\subsection*{Activity 10}

A complete analysis of this data is offered in the Mathematica notebook 1-15-T-Mma-Torricelli-TeacherVersion.nb and in pdf format. It is listed as Second Source:  Data for Torricelli's Law.

\subsection*{Activity 11}

We are asked the following questions:



Suppose we have a  cylindrical tank with the area of the tiny hole at the bottom being $a$ and cross sectional area of the cylinder being $A$. 

\begin{itemize}
\item [a)] 
How long does it take to empty the tank?

\end{itemize}

For each of the following physical changes in the tank, describe the effect on the time it takes to empty the tank. 

\begin{itemize}
\item [b)] Double the area, $a$, of the tiny hole at the bottom of the cylinder.
\item [c)] Change $g$ by considering the tank to be on the Moon. There  gravity is one sixth that of the Earth.
\end{itemize}

For (a) we need to solve the general Torricelli's Law differential equation (\ref{eq:GenTorr})

\begin{equation}\label{eq:GenTorr}
A h'(t) = \alpha a \sqrt{2 g h[t]}
\,. 
\end{equation}

The solution to (\ref{eq:GenTorr}) is (\ref{eq:GenTorrSol}).

\begin{equation}\label{eq:GenTorrSol}
h(t) = \frac{a^2 \alpha ^2 g t^2-2
  a \alpha  A \sqrt{2 g h_0}
   \, t+2 A^2
   h_0}{2 A^2}\, .
\end{equation}
 
Thus setting $h(t)$ in (\ref{eq:GenTorrSol}) equal to 0 and solving for $t$ we have 

$$ t_e = \frac{A
   \sqrt{2 h_0}}{a \alpha 
   \sqrt{g}}\, ,$$
as the time it takes the tank to empty. Before we leave this formula examine it to see how the variables work in determining the time at which the column of water empties. If we were to increase $A$, the cross sectional area of the cylinder, we would have more water and hence it would take longer to empty the tank. We see that $A$ is in the numerator of $t_e$ so that makes sense. While if we were to increase $a$, the area of the small hole at the bottom of the tank through which the water exits, the value of $t_e$ would go down and this makes sense because of the exit hole is bigger more water flows out and the $t_e$ would get smaller. 

\vspace*{.2 true cm}

For (b)  we first find the time it takes to empty the cylinder with cross section value $a$ for the area of the hole at the bottom of the cylinder 

\begin{equation}\label{eq:a}
 A h'(t) = \alpha a \sqrt{2 g h(t)}\\
\end{equation}
\begin{equation}\label{eq:aDouble}
 A h'(t) = \alpha (2a) \sqrt{2 g h(t)}
\end{equation}

The time it takes to empty the cylinder in the case (\ref{eq:a}) is 

$$t_{f(a)} = \frac{\sqrt{2} A
   \sqrt{h_0}}{a \alpha 
   \sqrt{g}}\, ,$$
while the time it takes to empty the cylinder in the case (\ref{eq:aDouble}) is
$$t_{f(2a)} = \frac{ A
   \sqrt{h_0}}{\sqrt{2}a \alpha 
   \sqrt{g}}\, ,$$
or exactly half that of the case where the  hole at the bottom of the cylinder of area $a$.  So double the area of the small exit hole and we halve the time to empty the tank.

\vspace*{.2 true cm}

For (c), where we are on the Moon and $g$ is one sixth of its value on Earth, the time it takes for the cylinder to empty  is $\sqrt{\frac{3}{2}} = 1.22474$ times as great:

$$t_f = \frac{\sqrt{3} A
   \sqrt{h_0}}{a \alpha 
   \sqrt{g}} \, .$$

This makes sense conceptually, for in the Moon's environment there is less pull on the column of water and hee it will take longer to empty.

\subsection*{
Activity 12 - Building a Water Clock}
\begin{itemize}
\item [(12)] Design a cylinder of cross-sectional area $A(h)$ at height $h$ so that the water falls at a constant rate? In this way, we could  display equal intervals, say quarter hour, in equal vertical distance  markings the time as it passes. To model a constant rate of fall for the height,  $h(t)$, of water we need to set $h'(t) = c$ for some constant we desire, e.g., 1 centimeter per 15 minutes. This clock cylinder  could then be marked at vertical heights of 1 cm with the knowledge that for ever centimeter the water falls 15 minutes has elapsed.

So, what shape, namely what is the cross sectional area at height $h$ ($A(h)$)  which actually will give us a constantly decreasing height in our water level?

When done you can run right down to your 3D printer shop and make your own water clock.  
\end{itemize}

So let us begin with (\ref{eq:changevol}) and  solve for $h'(t)$. Setting $h'(t) = c$ for some constant in  (\ref{eq:changevol}) gives (\ref{eq:constant}) 

\begin{equation}\label{eq:constant}
c= h'(t)= \alpha\, a \frac{\sqrt{2 g h(t)}}{A(h(t))}\, ,\quad h(0) = h_0\, .
\end{equation}

\noindent Now in (\ref{eq:constant}) we can solve for $A(h(t)$:

\begin{equation}\label{eq:solveA}
A(h(t))= \alpha\, a \frac{\sqrt{2 g h(t)}}{c}= {\alpha\, a \sqrt{2g}}{c} \sqrt{h(t)} = K \sqrt{h(t)}\, ,\quad h(0) = h_0\, .
\end{equation}

So we see that in order to create a cylinder of cross-sectional area $A(h)$ at height $h$ so that the water falls at a constant rate we must make the cross sectional area proportional to the square root of $h(t)$. Indeed, should we desire to make the cross-sectional area circular, i.e. $A(h(t)) = \pi r^2$ then we can specify our radius $r$ in terms of our height:

\begin{equation}\label{eq:radius}
r = \sqrt{\frac{\alpha\, a \sqrt{2 g h}}{\pi c}} \, ,
\end{equation}\noindent where $r$ is the radius of a circular cross sectional clock at height $h$; $a$ is the cross sectional area of the tiny hole near the bottom of our clock; $\alpha$ is the discharge or contraction coefficient, i.e. the effective percent of the tiny hole's area which actually permits water to flow out; $g$ is the acceleration due to gravity - different  clocks for different altitudes(!); and $c$ is the actual rate in cm/(unit time) we want our water level to fall.

Thus we see a simple power relationship between $h$ and $A(h)$ is all it takes to design a clock whose water falls at a constant rate. So run right down to your 3D printer shop and make your own water clock.


\section*{Video sources for data collection}

Here we discuss Activities (13-14), use of video sources for modeling with Power Rule and then with Torricelli's Law.

We summarize our data analyses. We used two videos of columns of water falling, without knowledge given as to the radius of the opening at the bottom of the cylinder. These were from YouTube and we cite them properly below.  We also built experimental apparatus to collect data using a free digital clock on a computer ({\tt www.on-line-stopwatch.com}) which measures and displays nicely and in large fonts real time to the thousandth of a second; a clear sparkling water plastic container; a collection bin for the exiting water; and an iPad to produce exceptionally clear HD video of the experiments. We did this for small holes  at the bottom of the cylinder of diameters 13/16'', 9/64'', 11/64'', and 7/64''. In the latter case we used two data sets - sampling from the entire run and then sampling from a smaller section of the run. We placed the following videos  for student use in the Student section in SIMIODE for this scenario 1-15-Torricelli on YouTube.  



\vspace*{.3 true cm}
\noindent Source 1:



Name: Ley de Torricelli Vaciado de un dep\'{o}sito 

URL:  {\tt http://www.youtube.com/watch?v=6LOgdOfFAbI} 



\vspace*{.3 true cm}

\noindent Source 2:



Name: Numec project Sem1 2011 THE Massooraahs! 

URL: {\tt http://www.youtube.com/watch?v=44fZdi7QC8E}   




\vspace*{.3 true cm}
Sources 3 - 6 are available at YouTube.


\vspace*{.3 true cm}
\noindent Source 3:



Name: 13Over64Inch Small Hole 

URL: {\tt https://www.youtube.com/watch?v=gsNdsuQ1ZCo}  

and {\tt https://www.youtube.com/watch?v=SCNqxFe3eIs}


Also available at {\tt https://www.simiode.org/resources/488/video?resid=2085}.

\vspace*{.3 true cm}
\noindent Source 4:



Name: 9Over64 Inch Small Hole  

URL: {\tt https://www.youtube.com/watch?v=gsNdsuQ1ZCo}

and {\tt https://www.youtube.com/watch?v=6cUuNUjlaPI}


Also available at {\tt https://www.simiode.org/resources/488/video?resid=2083}.

\vspace*{.3 true cm}
\noindent Source 5:



Name: 11Over64Inch Small Hole  

URL: {\tt https://www.youtube.com/watch?v=76diQ1eW8i4}

and {\tt https://www.youtube.com/watch?v=NBr4DOj4OTE}

Also available at {\tt https://www.simiode.org/resources/488/video?resid=2087}.

\vspace*{.3 true cm}
\noindent Source 6:



Name: 7Over64 Inch Small Hole 

URL: {\tt https://www.youtube.com/watch?v=k7TSAqRfHgg}

and {\tt https://www.youtube.com/watch?v=xDyDSPydN\_E}

Also available at {\tt https://www.simiode.org/resources/488/video?resid=2084 }.
  
\subsection*{Report on results of data analyses}

We report here summary results from our analyses. We ran complete analyses on all data sets including a shortened version of the sixth video, i.e. we sampled less from each. These videos are all available in the Mathematica notebook 1-15-T-Mma-Torricelli-TeacherVersion.nb and in pdf format.

In each case we considered two approaches for building a differential equation model for  $h(t)$, the height of the water in cm at time $t$ in seconds.

\begin{enumerate}
\item General Power Law
$$ h'(t) = - k h(t)^r\, , \quad h(0) = y_0\, , $$
where $k$ is a rate constant with appropriate units
\item Torricelli's Law
$$A\, h'(t) = a\, \alpha\, \sqrt{2 g h(t)} \, , \quad h(0) = y_0\, ,$$
where $A$ is the cross sectional area of the column of water  - $A$ was constant and the shape was circular in all Sources; $g$ was the gravitation constant, presumed to be $9.80 m/s^2$. In Sources 3-6 $a$ is the cross sectional area of the small hole at the bottom of the column of water through which the water exits and $a = \pi r^2$ where $r$ is the radius of the hole in bottom of the cylinder of water given in the title of the video, e.g.,   13/16'', 9/64'', 11/64'', and 7/64''; $\alpha$ is   the discharge or contraction coefficient, i.e. the effective percent of the tiny hole's are which actually permits water to flow out. In 
Sources 1 and 2 where the shape of the small hole at the bottom of the cylinder was now known we presumed that $a\, \alpha = k$, a constant.
\end{enumerate}

We offer the sum of square errors ($SSE$) as a measure of how good our model is.   in each case we give the best parameters and the resulting $SSE$.

\vspace*{.3 true cm}
\noindent Source 1:

Power Rule: $SSE = 0.130256$, $k = 0.0931652$, $r = 0.56388$.

Torricelli's Law (presumes $r = 0.5$)  $SSE = 0.365508$, $k = 7.25116$.


\vspace*{.3 true cm}
\noindent Source 2:

Power Rule: $SSE = 0.206507$, $k = 0.362918$, $r = 0.573579$

Torricelli's Law (presumes $r = .5$)  $SSE = 0.302832$, $k = 27.033991$.

\vspace*{.2 true cm}

We note in Sources 1 and 2 that the $SSE$ is higher for the  Torricelli's model than for the Power Rule model. In source 1 it is almost 300\% higher, while in Source 2 it is almost 50\% higher. The $k$ values are not comparable so we are not concerned about their disparity. In Figure~\ref{fig:1-15-TorricelliSource2} we note that both models are quite good; however, the dashed Power Rule model seems to hold better, especially near the end of the run.



\vspace*{.3 true cm}
\noindent Source 3: (small hole of radius 13/64'')

Power Rule: $SSE = 0.00327907$, $k = 0.109294$, $r = 0.563275$. 

Torricelli's Law (presumes $r = 0.5$)  $SSE = 0.00456251$, $\alpha =k  = 0.716054$. 

\vspace*{.3 true cm}
\noindent Source 4: (small hole of radius 9/64'')

Power Rule: $SSE = 0.020492$, $k = 0.053235$, $r = 0.501133$. 

Torricelli's Law (presumes $r = 0.5$)  $SSE = 0.0205103$,  $\alpha = k = 0.656871$.
 
\vspace*{.3 true cm}
\noindent Source 5: (small hole of radius 11/64'')

Power Rule: $SSE = 0.0508712$, $k = 0.0781168$, $r = 0.503808$.

Torricelli's Law (presumes $r = 0.5$)  $SSE = 0.0513088$,  $\alpha = k = 0.648622$.

\vspace*{.3 true cm}
\noindent Source 6: (small hole of radius 7/64'')

Power Rule: $SSE = 0.0151031$, $k  = 0.0237784$, $r = 0.593436$.
 
Torricelli's Law (presumes $r = 0.5$)  $SSE = 0.180729$, $\alpha = k = 0.583483$.


\vspace*{.3 true cm}
\noindent Source 6: (small hole of radius 7/64'' - using only 15 data points)

Power Rule: $SSE = 0.00304714$, $k  =  0.026167$, $r = 0.549328$.  

Torricelli's Law (presumes $r = 0.5$)  $SSE =  0.00696854$, $\alpha = k = 0.593613.$.

\vspace*{.3 true cm}


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliSource2.eps}


\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorricelliSource2} Plot of the results of Source 2 data analyses. }}
\vspace*{-.3 true cm}
\end{figure}





We note that in all Sources 3 - 7 from our own experimentation that the Power Rule has a lower (slightly) $SSE$. We also note that the $r$ values in the broader approach of the Power Rule are close to that which we presume in Torricelli's Law, namely $r = 0.5$.  The fact that our Power Rule does not quite converge to $r = 0.5$ (the value developed and presumed in Torricelli's Law), but stays close to $r = 0.5$, we attribute to the extra degree of freedom we introduce in the Power Rule model wherein $r$ is permitted to vary in order to optimize, i.e. minimize the $SSE$. Thus, we would not expect the optimum to occur exactly at $r = 0.5$ since the readings are not perfect, i.e. there is error in readings! Moreover, the plots for Sources 3-6 (we show final plots for Source 5 in Figure~\ref{fig:1-15-TorricelliSource5} as an example), show both models predict quite well.


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=8cm]{1-15-TorricelliSource5.eps}


\parbox{11.7 true cm}{\caption{\label{fig:1-15-TorricelliSource5} Plot of the results of Source 5 data analyses. }}
\vspace*{-.3 true cm}
\end{figure}



Also, in the literature the value of $\alpha$,    the discharge or contraction coefficient, is quite often measured to be and presumed in situations to be about 0.7 or 70\% of the maximum flow rate is obtained. Thus with our values of $\alpha = 0.716054,\, 0.656871,\, 0.648622,\, 0.583483$ and $0.593613$ we are in the right neighborhood with our analyses in these cases. 

\section*{An Interesting Take on Torricelli's Law - Linear vs. Torricelli}

This is a discussion and write-up on Activity 15 - Beginning and End Observation - You Mess with Mr. In Between.  In a very fine article \cite{FarmerGass1992} the authors pose the following problem:

\begin{quote}
A small hole is drilled in the side of a cylindrical container and the height of the water level (above the hole) goes from 10 cm to 3 cm in 68 seconds. Estimate the height at intermediate times.
\cite[p.~147]{FarmerGass1992}
\end{quote}

They offer two models to their students (1) a linear model $h'(t) = k$, $h(0) = 10$, $h(68) = 3$, and (2) a model based on Torricelli's Law, $h'(t) = k \sqrt{h}$, $h(0) = 10$, $h(68) = 3$ and ask them to use each in predicting the height of the water at intermediate times.  The next class they bring the very apparatus they used to collect their original data, i.e. start and finish data, and they have the class (with assigned experimenter rolls,  i.e. digital clock watcher, recorder, falling water level observer) conduct an experiment with call outs as to height of water at times, in their case, $ t = 0,10, 20,30, 40,50,60,70$ seconds. They then compare the observations with the two models. Voil{\`a}!

Here we provide a comparable activity using one of the videos, say with water height at 12.5 cm above the small exit hole at clock time of 1 minute, 40.419 seconds, or 1:40.419 minutes,  i.e. $h(100.419) = 12.5$ and then  with a final observation of water height  at 2 cm above the small exit hole at clock time 2 minute, 34.286 seconds or 2:34.216 minutes, i.e. $h(154.286) = 2.0$.

So a version for students of SIMIODE for the problem as  posed in \cite[p.~147]{FarmerGass1992} is Activity (1) entitled Beginning and End Observation - You Mess with Mr. In Between (to paraphrase a line from Johnny Mercer's tune {\it Accentuate the Positive\/}.
\begin{quote}
A small hole of radius $s = 0.138906$ cm is drilled in the side of a right circular cylindrical container of radius $r = 4.16986$ cm and the height of the water level (above the hole) goes from 12.5 cm at a stop watch time of 1:40.419 min to 3 cm at a stop watch time of 2:34.116  min. Use two different models to estimate the height at intermediate times. These models are (1) linear model:  $h'(t) = - k h(t)$, $h(0) =  12.5$ with $h(143.286) = 2.0$ and (2) Torricelli's Law: $A h'(t) = -\alpha \cdot  a\cdot  \sqrt{2\cdot  g\cdot  h(t)}$, $h(0) =  12.5$ with $h(143.286) = 2.0$. Here $A$ is the cross sectional area of the cylinder and $a$ is the area of the  hole for drainage, while $\alpha$ is the discharge or contraction coefficient.

After you have completed your analyses, go to the video, entitled 11Over64 Inch Small Hole  from this Scenario - 1-15-Torricelli on YouTube, and compare your observations from the video with those from each of your models. Which model  is the better one?  Why? 

\end{quote}
 

 We ask students to fill in the table below with $t = 0$ being at time 154.286 seconds where the water is at level 2 cm above the exit hole and   $ t = 154.286 - 100.419 = 53.867$ being our stopping point or length of our run  where the water is at level 2 cm above the exit hole.



In the Mathematica notebook 1-15-T-Mma-Torricelli-TeacherVersion.nb we offer a complete analysis. Here are the results:

\vspace*{.2 true cm} \noindent 
Linear Model (best fit in terms of minimum sum of square errors $SSEL$):

$yL(t) = 12.5 - 0.213651 t$, $SSEL = 3.01273$.

\vspace*{.2 true cm} \noindent
Torricelli Model (best fit in terms of minimum sum of square errors $SSET$):

$yT(t) = 0.00154308 t^2-0.277766 t+12.5$, $SSET = 0.00754184$



We offer the unfilled table of data information first.
\begin{center}
\small
\begin{tabular}{|c|c|c|c|c|} \hline
Clock Time & Relative Time & Linear & Torricelli & Observed\\
$t$ (min) & $t$ (s)  & $h$ (cm)& $h$ (cm) & $h$ (cm)\\  \hline 
1:40.419 = 100.419 & 0&12.5 & 12.5 & 12.5 \\ \hline
1:45.598 = 105.598 &5.179 & \hspace*{2 true cm} & \hspace*{2 true cm}&  \hspace*{2 true cm} \\ \hline
1:47.568 = 107.788 & 7.369& & & \\  \hline
1:49.623 = 109.523 & 9.104& & &  \\ \hline 
1:50.985 = 110.985 & 10.566 & & & \\  \hline 
1:56.388 = 116,388 & 15.969& & & \\  \hline 
2:02.057 = 122.057 & 21.638& & & \\  \hline 
2:08.032 = 128.032 & 27.613& & &  \\ \hline
2:16.386 = 136.386 & 35.967& & &  \\ \hline
2:25.568 = 145.568 & 45.149& & &  \\ \hline
2:34.116 = 154.286 & 53.867& & &2.0  \\ \hline
\end{tabular}
\end{center}




\vspace*{.2 true cm} \noindent Table~\ref{table:LinearVsTorricelliData} is the filled in data set using the results of the linear and Torricelli models - compare that with the actual observed data.



Finally,  in Figure~\ref{fig:1-15-TorricelliVsLinearPlot} we compare both linear and Torricelli models to the actual data, the data the students would obtain from the video. This is very convincing evidence that Torricelli's Law actually does model the level of the falling water in a cylinder!


\section*{CLOSING}

We have offered a good many activities here, more than students can or should do in routine assignments. However, this material can be sampled and assigned in part and also used as for course projects.  Torricell's Law and modeling, using data, is a rich area for a differential equations course and we believe the instructor will find ready material for students to study here.
 
\vspace*{.2 true cm}
\begin{table}[h]
\small
\begin{center}
\begin{tabular}{|c|c|c|c|c|} \hline
Clock Time & Relative Time & Linear & Torricelli & Observed\\
$t$ (min) & $t$ (s)  & $h$ (cm)& $h$ (cm) & $h$ (cm)\\  \hline 
1:40.419 = 100.419 & 0&12.5 & 12.5 & 12.5 \\ \hline
1:45.598 = 105.598 &5.179 & 11.3935& 11.1028&  11.1 \\ \hline
1:47.568 = 107.788 & 7.369& 10.9256& 10.5369& 10.5\\  \hline
1:49.623 = 109.523 & 9.104& 10.5549& 10.0991& 10.1 \\ \hline 
1:50.985 = 110.985 & 10.566 & 10.2426& 9.7374& 9.8\\  \hline 
1:56.388 = 116,388 & 15.969& 9.08821& 8.45786& 8.5\\  \hline 
2:02.057 = 122.057 & 21.638& 7.87702& 7.21218& 7.2\\  \hline 
2:08.032 = 128.032 & 27.613&6.60046 & 6.00662& 6.0 \\ \hline
2:16.386 = 136.386 & 35.967& 4.81562& 4.50576& 4.5 \\ \hline
2:25.568 = 145.568 & 45.149& 2.85388& 3.10461& 3.1 \\ \hline
2:34.116 = 154.286 &53.867& 0.991273& 2.01507& 2.0 \\ \hline
\end{tabular}


\parbox{13.4 true cm}{\caption{\label{table:LinearVsTorricelliData} Computations with the linear model and Torricelli  model  compared to   observed data. }}
\end{center}

\vspace*{-.6 true cm}
\end{table}


\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9cm]{1-15-TorricelliVsLinearPlot.eps}


\parbox{12.9 true cm}{\caption{\label{fig:1-15-TorricelliVsLinearPlot} Plot of the observed data, the linear model (solid line) and the Torricelli  model (dashed). We see the linear plat as very inadequate and the Torricelli plot as right on!}}

\end{figure}


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\bibitem{FarmerGass1992} Farmer, T. and F. Gass. 1992. Physical Demonstrations in the Calculus Classroom. {\it The College Mathematics Journal\/}.  23(2): 146-148.

\bibitem{KriederLahrDiesel2005} Krieder, D., D. Lahr, and S. Diesel. 2005. {\it Principles of Calculus Modeling:  An Interactive approach\/}. From the Quiz in Unit on Torricelli's Law. \href{http://www.math.dartmouth.edu/~klbooksite/2.18/218quiz/218quiz\_p02.htm}{http://www.math.dartmouth.edu/~klbooksite/2.18/218quiz/218quiz\_p02.htm}.  Accessed 2 April 2013.

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\end{thebibliography}
}
\end{document}
