3.4: Mathematical models using the derivative

The rate of change of a function provides a powerful new way of thinking about models of biological processes.

The changes in biological and physical properties that were measured in discrete packages \(\left(P_{t+1}-P_{t}, A_{t+1}-A_{t}, I_{t+1}-I_{t}\right)\) in Chapter 1 can be more accurately represented as instantaneous rates of change \(\left(P^{\prime}(t), A^{\prime}(t), I^{\prime}(t)\right)\) using the derivative. In Section 1.1, Table 1.1, the V. natriegens population was measured at 16 minute intervals but was growing continuously. The kidneys filter continuously, not in 5-minute spurts. Here we begin a process of using the derivative to interpret mathematical models that continues through out the book.

 

3.4.1 Mold growth.

We wrote in Section 1.5 that the daily increase in the area of a mold colony is proportional to the circumference of the colony at the beginning of the day. Alternatively, we might say:

Mathematical Model 3.4.1 Mold growth. The rate of increase in the area of the mold colony at time t is proportional to the circumference of the colony at time t.

Letting \(A(t)\) be area and \(C(t)\) be circumference of the mold colony at time \(t\), we would write

\[A^{\prime}(t)=k C(t)\]

Because \(C(t)=2 \sqrt{\pi} \sqrt{A(t)}\) (assuming the colony is circular)

\[A^{\prime}(t)=k 2 \sqrt{\pi} \sqrt{A(t)}=K \sqrt{A(t)} \label{3.24}\]

where \(K=k 2 \sqrt{\pi}\).

Equation \ref{3.24}, \(A^{\prime}(t)=K \sqrt{A(t)}\), is a statement about the function \(A\). We recall also that the area of the colony on day 0 was 4 \(mm^2\). Now we search for a function \(A\) such that

\[A(0)=4 \quad A^{\prime}(t)=K \sqrt{A(t)} \quad t \geq 0 \label{3.25}\]

Warning: Incoming Lightning Bolt. Methodical ways to search for functions satisfying conditions such as Equations \ref{3.25} are described in Chapter 17. At this stage we only write that the function

\[A(t)=\left(\frac{K}{2} t+2\right)^{2} \quad t \geq 0 \quad \text { Bolt Out of Chapter } 17 . \label{3.26}\]

satisfies Equations \ref{3.25} and is the only such function. In Exercise 3.6.7 you are asked to confirm that \(A(t)\) of Equation \ref{3.26} is a solution to Equation \ref{3.25}.

Finally, we use an additional data point, \(A(8) = 266\) to find an estimate of \(K\) in \(A(t)=\left(\frac{K}{2} t+2\right)^{2}\).

\[\begin{gathered}
A(8)=\left(\frac{K}{2} 8+2\right)^{2} \\
266=(4 K+2)^{2} \\
K \doteq 3.58
\end{gathered}\]

Therefore, \(A(t)=(1.79 t+2)^{2}\) describes the area of the mold colony for times \(0 \leq t \leq 9\). Furthermore, A has a quadratic expression as suggested in Section 1.5 based on a discrete model. A graph of the original data and A appears in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): Graph of \(A(t)=(1.79 t+2)^{2}\) and the area of a mold colony from Table 1.5

 

3.4.2 Difference Equations and Differential Equations.

We have just seen our first example of a differential equation, Equation \ref{3.24}. It writes the derivative, \(A^{\prime}(t)\), of a function in terms of the function, \(=K \sqrt{A(t)}\). We prefer the term derivative equation but defer to the almost universal use of differential equation. The comparison is

\[\begin{aligned}\\
A_{t+1}-A_{t}=K \sqrt{A_{t}} \quad A_{0}=4 &  & A^{\prime}(t)=K A(t) \quad A(0)=4\\
& \quad \quad vs &\\
\text{Difference Equation} \quad \quad& & \text{Differential Equation}\\
\end{aligned}\]

Differential equations often more accurately represent the biological, physical chemical, economic, or social system under study because the equation and solution are defined throughout a time interval or space interval, whereas difference equations and solutions are defined only for discrete sequences of numbers, and maybe a finite sequence. Deer populations that breed only annually, however, may be better represented by a difference equation

Difference equations are inherently simpler than differential equations. Solutions to difference equations require only arithmetic, whereas solutions to differential equations involve calculus. The solution to the difference equation above seems to only have an arithmetically derived solution. The corresponding differential equation has a simple algebraic solution, \(\left(\frac{K}{2} t+2\right)^{2}\). You have to wait 14 chapters before we describe its derivation, but we will write and solve differential equations in Chapters 5, 7 and 13.

Some computers will return solutions to differential equations, much as looking up addresses in a telephone book. Give ’White Pages’ on the internet a name and area code, and it will return the address, telephone number, age and close associates of a person, and perhaps another one or two people, with that name and area code. Computers are very adept, however, at computing approximate solutions to some very complex differential equations, such as those that describe the path of a manned space craft traveling to the moon. Ironically, what they actually do is compute solutions to difference equations closely associated with the a differential equation; computers are arithmetic whizzes.

 

3.4.3 Chemical kintetics.

Note

Chemists use the rate of change in the amount of product from a chemical reaction as a measure of the reaction rate. You will compute some rates of chemical reactions from discrete data of chemical concentration vs time.

Chemical reactions in which one combination of chemicals changes to another are fundamental to the study of chemistry. It is important to know how rapidly the reactions occur, and to know what factors affect the rate of reaction. A reaction may occur rapidly as in an explosive mixture of chemicals or slowly, as when iron oxidizes on cars in a junk yard. Temperature and concentration of reactants often affect the reaction rate; other chemicals called catalysts may increase reaction rates; in many biological processes, there are enzymes that regulate the rate of a reaction. Consider

\[\mathrm{A} \longrightarrow \mathrm{B}\]

to represent (part of) a reaction in which a reactant, A, changes to a product, B. The rate of the reaction may be measured as the rate of disappearance of A or the rate of appearance of B.

Figure \(\PageIndex{2}\): Conductivity of a water and butyl chloride solution at times after butyl chloride was added to water.

Butyl chloride.

When butyl chloride, C₄H₉Cl, is placed in water, the products are butyl alcohol and hydrochloric acid. The reaction is

\[\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}^{+}+\mathrm{Cl}^{-}\]

As it takes one molecule of C₄H₉Cl to produce one atom of Cl⁻, the rate at which butyl chloride disappears is the same as the rate at which hydrochloric acids appears. The presence of Cl⁻ may be measured by the conductivity of the solution. Two students measured the conductivity of a solution after butyl chloride was added to water, and obtained the results shown in Figure \(\PageIndex{2}\). The conductivity probe was calibrated with 8.56 mmol NaCl, and conductivity in the butyl chloride experiment was converted to mmol Cl⁻. The experiment began with butyl chloride being added to water to yield 9.6 mmol butyl chloride.

The average rate of change over the time interval [30,40]

\[m_{30,40}=\frac{1.089-0.762}{40-30}=0.0327 \text { , }\]

and the average rate of change over the time interval [40,50]

\[m_{40,50}=\frac{1.413-1.089}{50-40}=0.0324 .\]

both approximate the reaction rate. A better estimate is the average of these two numbers, \(\frac{0.0327+0.0324}{2}=0.03255 \). We only use the average when the backward and forward time increments, -10 seconds and +10 seconds, are of the same magnitude. The average can be computed without computing either of the backward or forward average rates, as

\[m_{30,50}=\frac{1.413-0.762}{50-30}=0.03255\]

In the case of t = 150, knowledge of the forward time increment is not available, and we use the backward time increment only.

\[m_{140,150}=\frac{3.355-3.255}{150-140}=0.01\]

Explore 3.4.1 Estimate the reaction rate at time t = 80 seconds.

Example 3.4.1 It is useful to plot the reaction rate vs the concentration of Cl⁻ as shown in Figure \(\PageIndex{3}\). The computed reaction rates for times \(t = 0, 10, \cdots 40\) are less than we expected. At these times, the butyl chloride concentrations are highest and we expect the reaction rates to also be highest. Indeed, we expect the rate of the reaction to be proportional to the butyl chloride concentration. If so then the relation between reaction rate and Cl⁻ concentration should be linear, as in the parts corresponding to times \(t = 60, 70, \cdots 160\) s. The line in Figure \(\PageIndex{3}\) has equation \(y = 0.0524 − 0.0127x\). We can not explain the low rate of appearance of Cl⁻ at this time.

Figure \(\PageIndex{3}\): Rate at which Cl⁻ accumulates as a function of Cl⁻ concentration after butyl chloride is added to water.

The reaction is not quite so simple as represented, for if butyl alcohol is placed in hydrochloric acid, butylchloride and water are produced. You may see from the data that the molarity of Cl⁻ is tapering off and indeed later measurements showed a maximum Cl⁻ concentration of 3.9 mmol. If all of the butyl chloride decomposed, the maximum Cl⁻ concentration would be 9.6, the same as the initial concentration of butyl chloride. There is a reverse reaction and the total reaction may be represented

\[\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftarrows}} \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}(\mathrm{aq})+\mathrm{H}^{+}+\mathrm{Cl}^{-}(\mathrm{aq})\]

The numbers, \(k_1\) and \(k_2\), are called rate constants of the reaction. The number \(k_1\) is the negative of the slope of the line computed in Example 3.4.1 (that is, 0.0127).

 

Exercises for Section 3.4 Mathematical models using the derivative.

Exercise 3.4.1 Write a derivative equation that describes the following model of mold growth.

Mathematical Model. A mold colony is growing in a circular pattern. The radius of the colony is increasing at a constant rate.

Exercise 3.4.2 Write a derivative equation that describes the following model of light depletion below the surface of a lake.

Mathematical Model. The rate at which light intensity decreases at any depth is proportional to the intensity at that depth.

Exercise 3.4.3 Write a derivative equation that describes the following model of penicillin clearance.

Mathematical Model. The rate at which the kidneys remove penicillin is proportional to the concentration of penicillin.

Exercise 3.4.4 Data from Michael Blaber of Florida State University College of Medicine⁶ for the butyl chloride experiment

\[\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq})\]

are shown in Table 3.4.4. These are more nearly what one would expect from this experiment.

1. Graph the data.
2. Estimate the rate of change of the concentration of C₄H₉Cl for each of the times shown.
3. Draw a graph of the rate of reaction versus concentration of C₄H₉Cl.

Table for Exercise 3.4.4 Data for Ex. 3.4.4.

Time (sec)	0	50	100	150	200	300	400	500	800
C[4H9Cl] (M)	0.1	0.0905	0.0820	0.0741	0.0671	0.0549	0.0448	0.0368	0.0200

 

Exercise 3.4.5 Data from Purdue University⁷ for the decrease of the titration marker phenolphthalein (Hln) in the presence of excess base are shown in Table 3.4.5. The data show the concentration of phenolphthalein that was initially at 0.005 M in a solution with 0.61 M OH⁻ ion.

1. Graph the data.
2. Estimate the rate of change of the concentration of phenolphthalein (Hln) for each of the times shown.
3. Draw a graph of the rate of reaction versus concentration of phenolphthalein.

Table for Exercise 3.4.5 Data for Ex. 3.4.5.

Time (sec)	0	10.5	22.3	35.7	51.1	69.3	91.6	120.4	160.9
Hln (M)	0.005	0.0045	0.0040	0.0035	0.0030	0.0025	0.0020	0.0015	0.0010

 

Exercise 3.4.6 Data from Michael J. Mombourquette, Queens University, Kingston, Ontario, Canada⁸ for the decrease of CO in the reaction 

\[\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})\]

are shown in Table 3.4.6A. and the decrease of N₂O₅ in the reaction

\[2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}\]

are shown in Table 3.4.6B. Initially, 0.1 g/l of CO was mixed with 0.1 g/l of NO₂.

For each table,

1. Graph the data.
2. Estimate the rate of change of the concentration of CO or of N₂O₅ for each of the times shown.
3. Draw a graph of the rate of reaction versus concentration of reactant.

Table for Exercise 3.4.6 Data for Ex. 3.4.6.

A. Time (sec) [CO] g/l 0 0.100 10 0.067 20 0.050 30 0.040

B. Time (sec) [N2O5] 0 0.0172 10 0.0113 20 0.0084 30 0.0062 40 0.0046 50 0.0035 60 0.0026

 

 

⁶ http://wine1.sb.fsu.edu/chem1046/not...s/Rxnrates.htm

⁷ http://chemed.chem.purdue.edu/genche...ch22/rate.html

⁸ http://www.chem.queensu.ca/people/fa...tics/index.htm