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\usepackage{amsmath,amssymb}
\usepackage{color}
\usepackage{graphicx}
\usepackage{hyperref}

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\begin{document}

\newpage
% You only need to change the NAME OF SCENARIO in the next line and delete the footnote. See above for instructions on toggling from Teacher Version to Student Version.
\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\ DRUG BOLUS}




\markboth{Drug Bolus}{Drug Bolus}






\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44023  USA\\
BrianWinkel@SIMIODE.org
}

\makeStitlePDFLaTex




\begin{abstract}
Given data on the concentration of a drug in the plasma of a human in mg/L at certain time intervals in hours can we determine the rate at which the drug leaves the plasma as well as the initial amount administered in a intravenous bolus of the drug?
\end{abstract}


\section*{SCENARIO DESCRIPTION}

\begin{figure}[!h]
\vspace*{-.5 true cm}
\begin{center}
\includegraphics[width=9cm]{1-85-Diagram.eps}


\parbox{8 true cm}{\caption{\label{fig:1-85-Diagram} Diagram of drug flow into plasma of a human.}}
\end{center}
\vspace*{-.4 true cm}
\end{figure}

A dose of 500 mg of a drug was administered to a healthy volunteer as an intravenous bolus 
(IV bolus), i.e. single injection. Figure~\ref{fig:1-85-Diagram} offers  a diagram which shows the path of the drug. This is called a one-compartment model and it is presumed that the drug enters the compartment all at once in the case of an IV Bolus and leaves the plasma (say for the tissues of the body where it is destined to perform its mission) at a rate proportional to the concentration in the plasma. See Figure~\ref{fig:1-85-Diagram}.


Seven blood samples were collected at 1, 2, 3, 4, 6, 10, and 12 hours. Plasma was separated from each blood sample and analyzed for drug concentration.  The collected data are shown in Table \ref{table:Data}. The first coordinate is time from intake of drug in hours, while the second coordinate is concentration in mg/L of the drug in the plasma.

\begin{table}[!h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
 Time (hr)&1 & 2 & 3 & 4 & 6 & 10 & 12 \\ \hline
 mg/L Drug in Plasma&12.72 & 10.8 & 8.26 & 7.02 & 4.63 & 1.99 &1.44 \\ \hline
\end{tabular}

\parbox{8 true cm}{\caption{\label{table:Data} Data on concentration of a drug in a patient's plasma in hours from intravenous bolus injection.}}
\end{center}
\vspace*{-.5 true cm}
\end{table}

Some basic information might help. The average human being has about 5 liters of blood of which some 55\% is referred to as plasma or serum. It is the yellow liquid found in blood when the platelets and red and white blood cells are separated out through centrifuge.






Use the data in Table~\ref{table:Data} and the information in Figure~\ref{fig:1-85-Diagram} to address these issues.

\begin{itemize}
\item [a)] build a differential equation model for this drug elimination,
\item [b)]estimate the initial concentration of the drug in the plasma, 
\item [c)] estimate the elimination rate constant, i.e. the rate constant with units 1/h which is the constant of proportionality in your differential equation model, and
\item [d)] confirm your model in some manner.
\end{itemize}

\teacher{
\section*{NOTES FOR TEACHERS}


The source for this modeling scenario is \cite{Bourne2015}. 
We consider the lesson, ``Estimate One Compartment Parameters, IV Bolus - Linear One Compartment."
 This material is part of  a site maintained by David W. A. Bourne, Associate Professor of Pharmaceutics, Department of Pharmaceutical Sciences, Skaggs School of Pharmacy, University of Colorado, Denver CO USA. The stipulation on use is ``Material on this website should be used for Educational or Self-Study Purposes Only.''

The model which is appropriate is a one compartment model
\vspace*{-.2 true cm}
\begin{equation}\label{eq:one}
s'(t) = - k s(t)\, , \quad s(0) = s_0\, ,
\end{equation}
where $s(t)$ is the concentration (mg/L) of the drug in the plasma at time $t$ in hr, $k$ is the rate constant with units 1/hr, and $s_0$ is the initial concentration in mg/L as a result of the IV bolus injection.

When we solve (\ref{eq:one}) we obtain the solution:
\vspace*{-.2 true cm}
\begin{equation}\label{eq:sol}
s(t) = s_0 e^{-k t} \, .
\end{equation}

Once we have the model we  form the sum of square errors  by differencing the model (solution of the differential equation) at each pint in time with the corresponding concentration at the same point in time.
\vspace*{-.4 true cm}

\begin{eqnarray}\label{eq:SSE}\nonumber
SSE(k,s_0) &=& \left(e^{-12 k}
   \text{s0}-1.44\right)^2+\left(e^{-10 k}
   \text{s0}-1.99\right)^2+\left(e^{-6 k}
   \text{s0}-4.63\right)^2+\left(e^{-4 k}
   \text{s0}-7.02\right)^2\\ 
&=&+\left(e^{-3 k}
   \text{s0}-8.26\right)^2+\left(e^{-2 k}
   \text{s0}-10.8\right)^2+\left(e^{-k}
   \text{s0}-12.72\right)^2
\end{eqnarray}

We can find the parameters $k$ and $s_0$ with reasonable software, e.g., Excel, Maple, Mathematica, and we find the parameters to be $k = 0.2039$ and $s_0 =15.739$. 

Substituting theses parameters into our model solution (\ref{eq:sol}) and plotting the resulting function over the data in Figure~\ref{fig:1-85-ModelOverData} from Table~~\ref{table:Data} renders a very satisfying and convincing plot that we have a good model.


\begin{figure}[!h]
\begin{center}
\includegraphics[width=9cm]{1-85-ModelOverData.eps}


\parbox{9 true cm}{\caption{\label{fig:1-85-ModelOverData} Plot of best fit model (\ref{eq:sol}) over the data from Table~\ref{table:Data}.}}
\end{center}
\vspace*{-.4 true cm}
\end{figure}

We offer a complete analysis in a Mathematica notebook 1-85-T-Mma-DrugBolis-TeacherVersion and a pdf version for those who do not red Mathematica in the Supplementary Documents for this Modeling Scenario.

\begin{thebibliography}{99}




\bibitem{Bourne2015}  Bourne,David. 2015  Estimate One Compartment Parameters, IV Bolus - Linear One Compartment.   \url{https://www.boomer.org/c/php/pk0504a.php}. Accessed  2 July 2016.




\end{thebibliography}
}





\end{document}
