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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\Heart Death Rate}



\markboth{Heart Death Rate}{Heart Death Rate }

\author{Arati Nanda Pati\\

University of St. Thomas\\
Houston TX 77006 USA\\
patia@stthom.edu
}

\makeStitlePDFLaTex


\begin{abstract}
In this modeling scenario, we offer students simulation experience from a given data set which reprents the heart death rate during the period 2000 - 2010 using several approaches to include exponential decay, difference equation, differential equation, and parameter estimation using EXCEL. We offer students to understand the role of a parameter and esimate it using EXCEL spreadsheet in the differential equation setting.  
This is an introductory modeling activity to experiment with a parameter to get a good sense of it on the solution of the first order differential equation.
\end{abstract}


\section*{SCENARIO DESCRIPTION}


Heart disease is the leading cause of death in the United States. It is not a single disease. Rather, a collection of diseases of the heart and blood vessels collectively  known as cardiovascular disease. Since 1950, death rates from heart disease and stroke have declined sharply representing one of the most important health achievements of the 20$^{\rm th}$ century. Intensive investigations led to finding major risks factors of the disease are cigarette smoking, high blood pressure, elevated total cholesterol, obesity, diabetes, and physical inactivity.  Subsequent development 
in early detection, prevention, and treatment of these risks  factors due to medical care have been very effective in decreasing the heart death rate. In the 21st century, still heart disease is the leading cause of death though it is following the decreasing trend.   We received the following data set to explore the mathematical modeling opportunity.



{\bf{Data set:}}
This example data is taken from  \cite[ Ch 4, Sec 4.2, Exercise 11, pp 198]{Lial}, Health: The table shows the age-adjusted death rates per 100,000 American for heart disease. (Data from: U.S. Center for Health Statistics.)



\begin{table}[h!]\label{table:data1}
\begin{center}
\begin{tabular}{|c|c|c||} \hline
\multicolumn{1}{|c|}{Year} & \multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c||}{Death Rate}\\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
2000& 0 & 257.6\\ \hline
2002& 2 & 240.8 \\ \hline
2004& 4 & 217 \\ \hline
2006& 6 & 210.2 \\ \hline
2008& 8 & 186.5 \\ \hline
2010& 10 & 178.5 \\ \hline
\end{tabular}

\parbox{10.7 true cm}{\caption{\label{table:data1} Heart death rate between the years 2000 and 2010.}}
\end{center}
\end{table}




\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRate.jpg}
\parbox{10.3 true cm}{\caption{\label{fig:dataplot}  Plot of the data from Table~\ref{table:data1} in which x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}





\subsection*{In-Class Activity}

We found the data as an exercise in \cite{Lial}. The data displays a downward trend. Although death from heart diseases is the leading cause of death in the United States, the decreasing pattern is encouraging to model it mathematically. We propose to develop  three modeling activities for the same data set as an individual class activity. Therefore, our priority is to model it initially as an algebraic exponential decay and  difference equation method, and finally as a  differential equation to predict the downturn disposition. 



\subsection*{(a) Exponential Function Model}
Consider the data set given in Table~\ref{table:data1}. Assume that the year 2000 represents the initial time 0 for data collection. Further, it shows that data has been collected in every two years and reported under the column death rate.  Insert the data on a spreadsheet and plot a scatter plot.  Analyze the graph and discuss your observations.

\vspace{0.5cm}

\noindent
Build a model for $H(t)$, the heart death rate after $t$ units of time, where $H(0)=257.6$. State your assumptions and variables clearly and show how they are used in the modeling process to obtain an algebraic equation. Compare your model predictions with the data and graph your results on the same plot. 



\subsection*{(b) Difference Equation Model}
From Figure~{\ref{fig:dataplot}, observe that the death rate is decreasing. So we want to model it mathematically by comparing the consecutive terms and recognizing a pattern between them. In order to do that first state your assumptions and variables clearly to write heart death rate in discrete notation.  If $H_{n}$ represents the death rate  at time step $n$, then $H_{n+1}$ represents the future value. The goal is to determine $H_{n+1}$ from the previous discrete value $H_n$ as $H_{n+1} = f(H_{n})$.  To find the relationship between two consecutive terms, one can write either the difference or ratio between the terms as $H_{n+1} - H_{n}$ or $H_{n+1}/H_{n}$, respectively.

\vspace{0.5cm}

\noindent
Investigate the difference and ratio of the heart death rate of consecutive terms.  Display your findings on a spreadsheet in column form and analize your results to draw your conclusions. Use the ratio pattern to write a mathematical model. Compare your model predictions with the data and display the graphs on the same plot.


\subsection*{(c) Differential Equation Model} 

From the scatter plot Figure~{\ref{fig:dataplot}, observe that the death rate is decreasing. Build a model with differential equation as a continuous death rate for the heart disease after $t$ units of time. Assume that the rate of change of the heart death rate is  directly proportional to the size of the heart death rate. Investigate the choice of the proportionality constant as a decay model with negative value. Hence,  build a model for a parameter estimation for the differential equation.  Using EXCEL spreadsheet, guess the value of the constant  by hit and trial method to find an admissible value of the parameter. Compare your model predictions with the data and show your results with graphs.
 
 
 \subsection*{(d) Parameter Estimation} 
 Guessing a value of the constant or parameter  is an {\it ad hoc\/} process. The procedure is time consuming and inaccurate. In order to get the best fit, one can use a parameter estimation procedure using EXCEL spreadsheet. Use the EXCEL built-in command `solve' to find the best parameter automatically for the differential equations modeling. Similarly,  investigate the parameter estimation procedure for the algebraic exponential decay and difference method. At the end, compare all three methods with the original  data and visualize the graphs on the same plot.
  
  
 
 \section*{NOTES FOR TEACHERS}



 We found the data as an exercise in \cite{Lial}. The data displays a downward trend. Although death from heart diseases is the leading cause of death in the United States, the decreasing pattern is encouraging to model it mathematically. We propose to develop  three modeling activities for the same data set as an individual class activity. Therefore, our priority is to model it initially as an algebraic exponential decay and  difference equation method, and finally as a  differential equation to predict the downturn disposition. In order to get the best fit, we use a parameter estimation procedure using EXCEL spreadsheet.  Hence, to motivate the importance of the parameter estimation technique, we suggest exploring the  differential equation modeling with parameter estimation first, where steps are discussed in detail. This experience  leads one to investigate the parameter estimation procedure for algebraic exponential decay and difference method with a few EXCEL commands. At the end, we compare all three methods with the original  data for best fitting results.
 
 
 
 
 \section*{Model Development}
\subsection*{(a) Exponential Function Model}
We consider the data set given in Table~\ref{table:data1}. Assume that the year 2000 represents the initial time 0 for data collection. Further, it shows that data has been collected every two years and reported under the column death rate. For mathematical modeling, we introduce our variables as follows. We assume the independent variable time as $t$ and heart death rate as dependent variable $H(t)$. Using EXCEL spreadsheet, we visualized the graph of the data as a scatter plot.

From Figure~{\ref{fig:dataplot}, we observe that the death rate is decreasing. So, algebraically we can model it as an exponential decay function as 




\begin{equation} \label{eq:exp}
H(t) =H_0*b^t\, , 
\end{equation}


\noindent 
where $H_0=257.6$, heart death rate at time $t=0$ and  $b$ is the unknown base and $t=$ time. Our aim is to find $b$. We take an elementary approach using one data point. Since $H(t)$ is known at $t=0$ and $t=10$, we can compare the values to find the following:
\noindent
at $t=10$

\[
H(t)=H_0*b^t=178.5
\]
yields

\[
H(10)=257.6*b^{10}=178.5
\]

\noindent
or


\[
b^{10}= \frac{178.5}{257.6}.
\]
\noindent
Taking the 10th root of both sides, we obtain

\[
b= \left(\frac{178.5}{257.6}\right)^{\frac{1}{10}}=0.963982.
\]

\noindent
Therefore, we can write our exponential model~ (\ref{eq:exp}) for heart death rate as 

\begin{equation} \label{eq:exp1}
\begin{split}
H(t) & =257.6 \ (0.963982)^t.
\end{split}
\end{equation}



We have displayed  the data and values obtained using the exponential decay model~(\ref{eq:exp1}) in Table~\ref{table:expdata1}. Figure~\ref{fig:expdecaydataplot} shows the data with the predictions of the model~(\ref{eq:exp1}). This model gives a very good  comparison with the data from the graph. However, we want to find the best value of $b$, so the error between the data and the model predictions will be minimum. This process is called {\it parameter estimation\/}. Although the actual mathematics is rigorous, we can use an EXCEL spreadsheet to find it easily. We have described the detailed  procedure in the differential equation modeling section.  We encourage you to follow the steps described there. Below we give the results using EXCEL. 


\begin{table}[h!]  \label{table:expdata1}
\begin{center}
\begin{tabular}{|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Year} & \multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate} & \multicolumn{1}{|c||}{Exponential Model} \\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
2000& 0 & 257.6 & 257.6  \\ \hline
2002& 2 & 240.8 & 239.3777 \\ \hline
2004& 4 & 217  &  222.4444  \\ \hline
2006& 6 & 210.2 & 206.709  \\ \hline
2008& 8 & 186.5 & 192.0867 \\ \hline
2010& 10 & 178.5 & 178.4987  \\ \hline
\end{tabular}
\parbox{10.7 true cm}{\caption{\label{table:expdata1}  
Comparison of heart death rate and exponential decay model $H(t)=257.6 (0.963982)^t$ between the years 2000 and 2010.}}
\end{center}
\end{table}



\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRateExpDecay.jpg}
\parbox{12 true cm}{\caption{\label{fig:expdecaydataplot}  Plot of the data from Table~\ref{table:expdata1} comparing the actual data with exponential decay 
method~(\ref{eq:exp1}) in which the x-axis represents the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}



\subsection*{Parameter Estimation} 

We can write the exponential decay model as 


\[
H(t)=257.6 \ b^t\, .
\]

\noindent
EXCEL has a built in command called `solver'  which yields the best value of $b=0.962977$.
With this choice, we can rewrite  the exponential model as 
\begin{equation} \label{eq:exp2}
\begin{split}
H(t) & =257.6 \ (0.962977)^t.
\end{split}
\end{equation}
This model will predict the heart death rate and we can then compare this result with the data as shown in Table~\ref{table:exppardata1} and Figure~\ref{fig:expdecaypardataplot}, respectively. EXCEL also yields  
 the Sum Square Error, $SSE=66.31917$ and $R^2=0.985744$ signifying a better fit using all the data in an exponential decay model. 
 



\begin{table}[h!]\label{table:exppardata1}
\begin{center}
\begin{tabular}{|c|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Year} & \multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate} & \multicolumn{1}{|c|}{Exponential Model} & \multicolumn{1}{|c||}{Error Square} \\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
2000& 0 & 257.6 & 257.6  & 0 \\ \hline
2002& 2 & 240.8 & 238.8788 & 3.690932 \\ \hline
2004& 4 & 217  &  221.5182 & 20.41421 \\ \hline
2006& 6 & 210.2 &  205.4193 & 22.85521 \\ \hline
2008& 8 & 186.5 & 190.4904 & 15.92298 \\ \hline
2010& 10 & 178.5 & 176.6464 & 3.435833 \\ \hline
\end{tabular}
\parbox{13.2 true cm}{\caption{\label{table:exppardata1} Comparison of the heart death rate data with exponential decay model~(\ref{eq:exp2}) with parameter estimation $H(t)=257.6 (0.962977)^t$  between the years 2000 and 2010. The last column displays the square of the error by finding the difference between the data and model values and then squaring it to find the best choice of $b$ using EXCEL spreadsheet.}}
\end{center}
\end{table}



\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRateExpDecayPara.jpg}
\parbox{12.2 true cm}{\caption{\label{fig:expdecaypardataplot}  Plot of the data from Table~\ref{table:exppardata1} that compares the actual data with the exponential model ~(\ref{eq:exp2})  $H(t)=257.6 (0.962977)^t$  with best choice of $b$.  Here, the x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}




\subsection*{(b) Difference Equation Model}
From Figure~{\ref{fig:dataplot}, we observe that the death rate is decreasing. So we want to model it mathematically. Our goal is to determine $H_{n+1}$ from previous discrete value $H_n$ as $H_{n+1} = f(H_{n})$ where  $H_{n}$ represents the death rate  at time step $n$. To find the relationship between two consecutive terms, we can write either the difference or ratio between the terms as $H_{n+1} - H_{n}$ or $H_{n+1}/H_{n}$, respectively. With two years of time step, $n=0$ denotes the initial time and  $n=1$ denotes time after two years. So we can denote the heart death rate as 
$H_{0}= 257.6$, $H_{1}= 240.8$, $H_{2}= 217$, $H_{3}= 210.2$, $H_{4}= 186.5$, and $H_{5}= 178.5$.


\begin{table}[h!]\label{table:ddata2}
\begin{center}
\begin{tabular}{|c|c|c||} \hline
\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{$H_{n+1} - H_{n}$} & \multicolumn{1}{|c||}{$H_{n+1}/H_{n}$}\\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
0 & -16.8 & 0.934782609 \\ \hline
1 & -23.8 & 0.901162791\\ \hline
2 & -6.8  & 0.968663594\\ \hline
3 & -23.7 & 0.887250238\\ \hline
4 & -8    & 0.957104558\\ \hline
\end{tabular}

\parbox{12.3 true cm}{\caption{\label{table:ddata2} Difference and ratio of heart death rate in between the years 2000 and 2010.}}
\end{center}
\end{table}

From Table~\ref{table:ddata2}, we notice that the difference between consecutive heart death rates does not show any particular pattern. However, the ratio remains relatively constant as $n$ increases. So for any constant $k$, we can write

\[
\frac{H_{n+1}}{H_n} = k  \Rightarrow
H_{n+1} = k {H_n}. 
\]

\noindent
The constant $k$ is unknown. We will assign the value to $k$ as follows:
From Table~\ref{table:ddata2}, column 3, considering only the first digit after the decimal point, we choose $k=0.9$. Therefore, we can write our difference equation model  (1) due to this common ratio as 


\begin{equation} \label{eq:difr1}
\begin{split}
H_{n+1} = 0.9 {H_n}\, . 
\end{split}
\end{equation}



\noindent
We  can also write the difference equation model (2)  as 


\begin{equation} \label{eq:difr2}
\begin{split}
H_{n+1} = 0.93 {H_n} 
\end{split}
\end{equation}


\noindent
where $k=0.93$ is the average value of the 3rd column of Table~\ref{table:ddata2}. With these two models, we can compare the data using $H_0=257.6$ as shown in Table~\ref{table:ddata3} and Figure~\ref{fig:differencedataplot}.




\begin{table}[h!]\label{table:ddata3}
\begin{center}
\begin{tabular}{|c|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Year} & \multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate} & \multicolumn{1}{|c|}{Model 1} & \multicolumn{1}{|c||}{Model 2}\\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
2000& 0 & 257.6 & 257.6 & 257.6 \\ \hline
2002& 2 & 240.8 & 231.84 & 239.568\\ \hline
2004& 4 & 217  &  208.656 & 222.7982 \\ \hline
2006& 6 & 210.2 & 187.7904 & 207.2024 \\ \hline
2008& 8 & 186.5 & 169.0114 & 192.6982 \\ \hline
2010& 10 & 178.5 &  152.1102 & 179.2093 \\ \hline
\end{tabular}

\parbox{10.7 true cm}{\caption{\label{table:ddata3} Comparing data with two difference equation models given by equations~(\ref{eq:difr1}) and~(\ref{eq:difr2}) due to ratio of two consecutive terms for the heart death rate in between the years 2000 and 2010.}}
\end{center}
\end{table}






\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRateDifferenceM1M2A1.jpg}
\parbox{11.8 true cm}{\caption{\label{fig:differencedataplot}  Plot of the data from Table~\ref{table:ddata3} comparing actual data with two models of difference method given by equations~(\ref{eq:difr1}) and~(\ref{eq:difr2}) in which x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}

\subsection*{Parameter Estimation}

To find the best value of $k$, we  do parameter estimation. We rewrite our difference model (3)  as 


\begin{equation} \label{eq:difr3}
\begin{split}
H_{n+1} = k {H_n}. 
\end{split}
\end{equation}

\noindent
Using EXCEL and the procedure described in the differential equation modeling section (c), we  find the best choice of $k=0.928082$ with Sum Square Error  $SSE =67.65204 $ and $ R^2 = 0.985725$. Model prediction values are displayed in Table~\ref{table:ddata4}
and visualized in Figure~\ref{fig:differencedataplot1}.


\begin{table}[h!]\label{table:ddata4}
\begin{center}
\begin{tabular}{|c|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Year} & \multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate} & \multicolumn{1}{|c|}{Model 3} & \multicolumn{1}{|c||}{Error Square}\\ \hline
%\multicolumn{1}{|c|}{n} & \multicolumn{1}{|c|}{a(n)} \\ \hline
2000& 0 & 257.6 & 257.6 & 0 \\ \hline
2002& 2 & 240.8 & 239.074 &  2.979195265\\ \hline
2004& 4 & 217  &  221.8803 & 23.81716426 \\ \hline
2006& 6 & 210.2 & 205.9231 &  18.29158792\\ \hline
2008& 8 & 186.5 & 191.1136 &  21.28518746\\ \hline
2010& 10 & 178.5 & 177.3691& 1.278908476 \\ \hline
\end{tabular}

\parbox{11.8 true cm}{\caption{\label{table:ddata4} Comparing data with difference model given by equation~(\ref{eq:difr3}) with $k=0.928082$ obtained by parameter estimation for the heart death rate in between the years 2000 and 2010. The last column displays the square of the errors.}}
\end{center}
\end{table}



\begin{figure}[h!]
\begin{center}
\includegraphics[width=10.5cm]{HeartRateDifferenceParaA1.jpg}
\parbox{12 true cm}{\caption{\label{fig:differencedataplot1}  Plot of the data from Table~\ref{table:data4} comparing actual data with difference method equation~(\ref{eq:difr3})  with parameter $k=0.928082$, in which the $x$-axis represents   time in years and the $y$-axis represents the death rate for heart disease.}}
\end{center}
\end{figure}






\subsection*{(c) Differential Equation Model} \label{sec:secc}



From the scatter plot in Figure~{\ref{fig:dataplot}, we observe that the death rate is decreasing. So, we can model it as a differential equation model with negative coefficient. Additionally, the heart death rate is known when $t=0$. Therefore, we give the initial value differential equation as



\begin{equation} \label{eq:diff}
\begin{split}
\frac{dH}{dt} & = -kH \\
H(0) & = 257.6
\end{split}
\end{equation}


 
 \noindent
where $k>0$ is unknown and we need a value for it so the solution will match the original data.  The negative sign indicates the decrease in heart death rate. We can use the separation of variables method to find the exact solution of the above differential equation. By bringing all $H$ terms to one side, and $t$ terms to the other and then integrating, we get



 
\begin{eqnarray*}\label{eq:sep}
\frac{dH}{dt} & =& -kH \quad {\rm or \,\,\ equivalently} \quad \frac{dH}{H} = -k dt \\
\int \frac{dH}{H} &=& -k \int dt \quad {\rm and \,\,\, integrating \,\,\,  both \,\,\,  sides}\\
\ln H & =& -kt +C,  \quad     C= \text{constant} \\
e^{\ln H} & =& e^{-kt +C} \\
H(t) & =& Ce^{-kt},  \quad   {{\rm using \,\,\,  replacement} \,\,\, C=e^C} \\
H(t) & =& 257.6\ e^{-kt}.
\end{eqnarray*}





\noindent
Since $H(0) = 257.6$ is given, the general constant can be evaluated to obtain $C=257.6$. However, we can get the exact solution quickly, using the {\it{WolframAlpha}} command, {\tt solve h'(t)=-k*h, h(0)=257.6}, which displays the solution as {$h(t) = 257.6 e^{-k t}$}. Here the symbol, ' , denotes derivative. We observe that the parameter $k$ is unknown. We can find the value of $k$ using parameter estimation with solver in EXCEL.







%\subsection*{(d) Parameter Estimation}
\vspace{0.5cm}
\noindent
{\bf{Parameter Estimation}}

\noindent
We found the exact solution to  the differential equation~(\ref{eq:diff}). We do not know the value of $k$. Let's start with an  arbitrary choice of $k$ and try to observe the behavior of the exact solution. We take $k=0.3$ as an initial guess and compare the data with the exact solution given in Table~\ref{table:data2} and Figure~\ref{fig:dataplot1}.



\begin{table}[h]\label{table:data2}
\begin{center}
\begin{tabular}{|c|c|c||} \hline
\multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate:Data} & \multicolumn{1}{|c||}{Differential Equation Model} \\ 
\hline
\multicolumn{1}{|c|}{t} & \multicolumn{1}{|c|}{H(t)}  & \multicolumn{1}{|c||}{H(t)=$257.6e^{-k*t}, k=0.3$} \\ \hline
0 & 257.6 & 257.6\\ \hline
2 & 240.8 & 141.3739\\ \hline
4 & 217 & 77.58763\\ \hline
6 & 210.2 & 42.58099\\ \hline
8 & 186.5 & 23.36894\\ \hline
10 & 178.5 & 12.82515\\ \hline
\end{tabular}
\parbox{8.7 true cm}{\caption{\label{table:data2} Comparing the data with the exact solution of the differential equation~(\ref{eq:diff}) for $k=0.3$.}}
\end{center}
\end{table}







\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRatePara1.jpg}
\parbox{12.4 true cm}{\caption{\label{fig:dataplot1}  Initial guess of the parameter, $k=0.3$. Plot of the data from Table~\ref{table:data2} and the exact solution of the differential equation~(\ref{eq:diff})  given by H(t)=$257.6e^{-k*t}$. Here x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}





\noindent
Let's take $k=0.03$ as a second guess and observe the change in the exact solution.  
In Table~\ref{table:data3}, we have see the data with the exact solution and this is visualized in Figure~\ref{fig:dataplot2}.
Now, comparing Figures~\ref{fig:dataplot1} and  \ref{fig:dataplot2}, we see there is a  great improvement of the exact solution for corresponding values of $k=0.3$ and $k=0.03$, respectively.  The value of $k=0.03$ is a better choice  than $k=0.3$ as it compares well with the given data. This suggests that the choice of $k$ is very important to get a better result. However, here we are trying to find a value of $k$ that would perform better than a trial and error method. In order to get the best choice of $k$, we will use EXCEL spreadsheet that will best fit with the actual data automatically.



\begin{table}[h]\label{table:data3}
\begin{center}
\begin{tabular}{|c|c|c||} \hline
\multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate:Data} & \multicolumn{1}{|c||}{Differential Equation Model} \\ 
\hline
\multicolumn{1}{|c|}{t} & \multicolumn{1}{|c|}{H(t)}  & \multicolumn{1}{|c||}{H(t)=$257.6e^{-k*t}, k=0.03$} \\ \hline
0 & 257.6 & 257.6\\ \hline
2 & 240.8 & 242.5985\\ \hline
4 & 217 & 228.4707\\ \hline
6 & 210.2 & 215.1656\\ \hline
8 & 186.5 & 202.6353\\ \hline
10 & 178.5 & 190.8348\\ \hline
\end{tabular}
\parbox{10.7 true cm}{\caption{\label{table:data3} Comparing the data with the exact solution of the differential equation~(\ref{eq:diff}) for $k=0.03$.}}
\end{center}
\end{table}




\begin{figure}[h!]
\centering
\includegraphics[width=10.5cm]{HeartRatePara2.jpg}
\parbox{12.7 true cm}{\caption{\label{fig:dataplot2}  Another guess of the parameter, $k=0.03$. Plot of the data from Table~\ref{table:data3} and the exact solution of the differential equation~(\ref{eq:diff})  given by H(t)=$257.6e^{-k*t}$. Here x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}


\subsection*{Parameter Estimation: Finding Best Value of $k$}

\noindent
We solve the differential equation (\ref{eq:diff}) using {\it WolframAlpha} command  
\vspace*{.2 true cm}

\centerline{\tt solve h'(t)=-k*h(t), h(0)=257.6\, {\rm ,}}
\vspace*{.2 true cm}

\noindent  and it displays the solution as {$h(t) = 257.6 e^{-kt}$} where a value of $k$ needs to be determined. Initially we guess a value for $k$, to determine the solution of the modified differential equation for each time unit $t$. We can find out the differences of the errors by subtracting the model data from the original given data and get the error squared. Next we find the sum of the squared error. Now our goal is to minimize this total squared error. EXCEL has a built in command, 'solver', which minimizes this error for the value of $k$.  We recommend to watch the video \cite{APMonitor}  for step by step guidance to find the best value of the unknown $k$.  For our experiment we obtained $k=0.037726$. This yields a Sum of Squared Error, SSE = $66.31917$, which suggests a great match between the data and the model.The value of $R^2$ measures how well data and model agree with each other. A value close to $1$ is considered to be good. In our example, we get $R^2=0.985744$. The data table displays a great agreement which is displayed in Figure~\ref{fig:dataplot3}.








%Parameter: DE

\begin{table}[h]\label{table:data4}
\begin{center}
\begin{tabular}{|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate:Data} & \multicolumn{1}{|c|}{Differential Equation Model} & \multicolumn{1}{|c||}{Error Square} \\ 
\hline
\multicolumn{1}{|c|}{t} & \multicolumn{1}{|c|}{H(t)}  & \multicolumn{1}{|c|}{H(t)=$257.6e^{-k*t}$} &  \multicolumn{1}{|c||}{Error Square}\\ \hline
0 & 257.6 & 257.6 & 0\\ \hline
2 & 240.8 & 238.8788 & 3.690929\\ \hline
4 & 217 & 221.5182 & 20.41423\\ \hline
6 & 210.2 & 205.4193 & 22.85519\\ \hline
8 & 186.5 & 190.4904 & 15.923\\ \hline
10 & 178.5 & 176.6464 & 3.435821\\ \hline
\end{tabular}
\parbox{10.7 true cm}{\caption{\label{table:data4} Comparing the data with the exact solution of the differential equation~(\ref{eq:diff}) for the best value of $k=0.037726$.}}
\end{center}
\end{table}





\begin{figure}[h!]
\centering
\includegraphics[width=9.5cm]{HeartRatePara3.jpg}
\parbox{12.3 true cm}{\caption{\label{fig:dataplot3}  Best value of the parameter, $k=0.037726$. Plot of the data from Table~\ref{table:data4}  and the exact solution of the differential equation~(\ref{eq:diff})  given by H(t)=$257.6e^{-k*t}$. Here x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}
 


\newpage
\noindent
{\bf{ Conclusion}}

\noindent
To summarize our results, let's compare all three methods, exponential decay, difference method, and differential equation with best choices of their parameters.  Table~\ref{table:dataAll} displays the actual and model data. Figure~\ref{fig:AllMethod} visualizes the corresponding plots. We see a very good agreement of all three methods with the actual data. In this paper our goal is to model with differential equation modeling scenario for the heart death rate with a parameter estimation and therefore, we conclude that the results are quite encouraging. 

\begin{table}[h]\label{table:dataAll}
\begin{center}
\begin{tabular}{|c|c|c|c|c||} \hline
\multicolumn{1}{|c|}{Time} & \multicolumn{1}{|c|}{Death Rate:Data} & \multicolumn{1}{|c|}{Model DE} & \multicolumn{1}{|c|}{Difference Model}  & \multicolumn{1}{|c||}{Exponential Decay}  \\ 
\hline
\multicolumn{1}{|c|}{t} & \multicolumn{1}{|c|}{H(t)}  & \multicolumn{1}{|c|}{$H(t)=257.6e^{-k*t}$} &  \multicolumn{1}{|c|}{$H_{n+1}=k*H_{n}, H_0=257.6$}    &  \multicolumn{1}{|c||}{$H(t)=257.6*b^t$}\\ 
\multicolumn{1}{|c|}{} & \multicolumn{1}{|c|}{}  & \multicolumn{1}{|c|}{$k= 0.037725798 $} &  \multicolumn{1}{|c|}{$k=0.928082164$}    &  \multicolumn{1}{|c||}{$b=0.962976953$}\\ \hline
0 & 257.6 & 257.6 & 257.6 & 257.6\\ \hline
2 & 240.8 & 238.8788 & 239.073965 & 238.8788\\ \hline
4 & 217 & 221.5182 & 221.880283 & 221.5182\\ \hline
6 & 210.2 & 205.4193 & 205.923133 & 205.4193 \\ \hline
8 & 186.5 & 190.4904 & 191.113587 & 190.4904 \\ \hline
10 & 178.5 & 176.6464 &  177.369112 & 176.6464 \\ \hline
\end{tabular}
\parbox{12.7 true cm}{\caption{\label{table:dataAll} Comparing the data with all methods with best parameter values obtained for Differential Equations, Difference Methods, and Exponential Decay methods. The differential equation modeling and algebraic exponential modeling yiled the results.}}
\end{center}
\end{table}



\begin{figure}[h!]
\centering
\includegraphics[width=9.5cm]{HeartRateAllMethods.jpg}
\parbox{12.3 true cm}{\caption{\label{fig:AllMethod}  Visualizing all the methods with parameter estimation with the original data for heart death rate.  The data is shown in  Table~\ref{table:dataAll}. Here x-axis represents  the time in years and y-axis represents the death rate for heart disease.}}
\end{figure}




\newpage

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 \bibitem{APMonitor}  APMonitor.com. 2017. ODE Parameter Estimation in Excel. 
\href{https://www.youtube.com/watch?v=eXco8\_3MmjI}\href{https://www.youtube.com/watch?v=eXco8\_3MmjI}. Accessed 16 September 2021.
 
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