# Calculus for Biological Scientists

## Section1.9Applications: The Lung Model and Competing Species

In our final section of the chapter we will survey two discrete-time dynamical systems, using the tools we’ve developed so far to analyze the behavior of each system over time.

### Warm-Up1.9.1.

Consider the updating function $$p_{t+1} = 0.7p_t + 0.6\text{.}$$
1. Determine the equilibrium value(s) of this system algebraically.
2. Create tables similar to that of Table 1.7.3 using the updating function and appropriate initial values to guess whether the equilibrium value(s) from the previous part are stable or unstable.
3. Graph the updating function and use cob-webbing to verify your conclusion regarding the stability of the equilibrium value(s).

### Subsection1.9.1The Lung Model

We will develop a basic DTDS measuring the concentration $$c_t$$ of a gas in the lung after $$t$$ breaths (exhale and inhale). As a starting place for developing our model, we will make several assumptions:
• There is a fixed maximum volume $$V$$ of air in the lung.
• The volume of air breathed out during exhale $$W$$ is the same as the volume of air breathed in during inhale.
• The concentration $$\beta$$ of the gas being measured is constant in the outside air.
This is a simplified model, but it is important to note that models are never perfect. It is important to analyze a model for its limitations, and adjust your model accordingly if there is a factor important to your study that is not being considered. An example of how one might adjust the lung model presented here is given in Exercise 1.9.4.1.
We will measure volume in liters (L) and the amount of gas particles in millimoles (mmol). Then the concentration of the gas will be measured in millimoles per liter (mmol/L). In order to develop our model, we will look at the concentration of gas at different stages of the breathing process, using unit conversions (Section 1.3) to guide our computations. The current concentration in the lung is represented by $$c_t\text{,}$$ and we are looking for the concentration after one more breath, $$c_{t+1}\text{.}$$
The final stage of a single breath is to analyze the air in the lung after inhale. During this stage, however, we are looking to compute the concentration instead of being given it as in previous stages. Concentration is computed as “amount” divided “volume”. The volume of air after inhale is the full volume of the lung, $$V$$ L. The amount of gas particles in the lung after inhale is the sum of of the amount left in the lung after exhale, plus the amount breathed in during inhale:
\begin{equation*} (V-W)c_t \text{ mmol} + W\beta \text{ mmol} \end{equation*}
Thus, the new concentration of gas in the lung after one breath is
\begin{align*} c_{t+1} \amp= \dfrac{(V-W)c_t + W\beta}{V} \\ \amp=\dfrac{(V-W)}{V}c_t + \dfrac{W}{V}\beta\\ \amp=\left( 1 - \dfrac{W}{V}\right)c_t + \dfrac{W}{V}\beta\text{.} \end{align*}
We summarize the development of the lung model below:

#### The Lung Model.

A DTDS which measures the concentration $$c_t$$ (in mmol/L) of a gas in the lungs after $$t$$ breaths is given by
\begin{equation*} c_{t+1} = (1-p)c_t + p\beta\text{,} \end{equation*}
where $$\beta$$ is the outside concentration of the gas being measured and $$p= \dfrac{W}{V}$$ is the percentage of the lung’s volume that is being filled with outside air during each inhale.

#### Activity1.9.2.

Consider a lung model with $$V = 6$$ L, $$W = 2$$ L, and $$\beta = 3$$ mmol/L.
1. Compute the equilibrium value(s) of this system algebraically.
2. Graph the updating function and use cob-webbing to classify the equilibrium value(s) as stable or unstable.
3. Interpret your previous answers in the context of the lung. Do your answers make sense?
The updating function for the lung model is an example of a weighted average, which we define now.

#### Definition1.9.2.Weighted Average.

A weighted average of two numbers $$a$$ and $$b$$ is a sum of the form
\begin{equation*} (1-p)a + pb\text{,} \end{equation*}
where $$p$$ is any number such that $$0 \leq p \leq 1\text{.}$$ The numbers $$p$$ and $$1-p$$ are called weights of the weighted average.

#### Remark1.9.3.

• When $$p=0.5\text{,}$$ this is just the average of two numbers:
\begin{equation*} (1-0.5)a+0.5b = 0.5a + 0.5b = \dfrac{a+b}{2} \end{equation*}
• The weights $$p$$ and $$1-p$$ represent a percentage, allowing us to place more importance (or weight) on one number over another.
• The weights $$p$$ and $$1-p$$ always sum to $$1$$ ($$100 \%$$).

#### Example1.9.4.Weighted Average.

1. The sum
\begin{equation*} 0.8(2) + 0.2(10) \end{equation*}
is a weighted average of the numbers $$2$$ and $$10\text{.}$$ The number $$2$$ is weighted as $$80\%$$ while the number $$10$$ is weighted as $$20\%\text{.}$$ The sum is equal to $$3.6\text{,}$$ which is much closer to $$2$$ than $$10$$ since $$2$$ is being given more importance.
2. The sum
\begin{equation*} 0.2(2) + 0.8(10) \end{equation*}
is a weighted average of the numbers $$2$$ and $$10\text{.}$$ The number $$2$$ is weighted as $$20\%$$ while the number $$10$$ is weighted as $$80\%\text{.}$$ The sum is equal to $$8.4\text{,}$$ which is much closer to $$10$$ than $$2$$ since $$10$$ is being given more importance.

#### Activity1.9.3.

The lung model $$c_{t+1} = (1-p)c_t + p\beta$$ is a weighted average.
1. What are the two numbers being averaged?
2. What are the weights of the weighted average?
3. Interpret the weighted average in the context of what is being modeled. What is being given more importance and why?

### Subsection1.9.2A Model for Competing Species

Consider a situation in which there are two different bacterial strains growing in the same environment. We’d like to describe the make-up of these two populations in a single discrete-time dynamical system. We’ll start by modeling each strain separately assuming an exponential growth model. Let the first strain’s population after $$t$$ minutes be represented by $$x_t$$ and the second strain’s population be represented by $$y_t\text{.}$$ Then each population can be modeled separately as
\begin{gather*} x_{t+1} = sx_t\\ y_{t+1}=ry_t\text{,} \end{gather*}
where $$s$$ and $$r$$ are the per capita growth rates of the first and second strain, respectively.
In order to create a single model that describes both of these populations, we will measure the population proportion of one of the strains. A population proportion is the fraction that a sub-population represents with respect to the total population. Let $$p_t$$ be the population proportion of the first strain after $$t$$ minutes. Then
\begin{equation*} p_t =\dfrac{\text{The population of strain 1 after t minutes}}{\text{The total population after t minutes}} = \dfrac{x_t}{x_t +y_t}\text{.} \end{equation*}
Note that even though $$p_t$$ is being measured with respect to the first strain’s population, we could also compute the population proportion of the second strain, $$\dfrac{y_t}{x_t+y_t}\text{,}$$ from this expression by computing $$1 - p_t\text{,}$$ since the sum of the two population proportions should equal $$1$$ ($$100\%$$).
We would like to have an updating function that describes the relationship $$p_t\text{.}$$ That is, we want an equation of the form $$p_{t+1} = f(p_t)\text{.}$$ Towards that end, we compute
\begin{align*} p_{t+1} \amp= \dfrac{x_{t+1}}{x_{t+1} + y_{t+1}}\\ \amp= \dfrac{sx_t}{sx_t + ry_t}\\ \amp= \dfrac{sx_t}{sx_t + ry_t} \cdot \dfrac{\frac{1}{x_t + y_t}}{\frac{1}{x_t + y_t}}\\ \amp= \dfrac{s\frac{x_t}{x_t + y_t}}{s\frac{x_t}{x_t + y_t} + r\frac{y_t}{x_t + y_t}} \\ \amp= \dfrac{sp_t}{sp_t +r(1-p_t)}\\ \amp= \dfrac{sp_t}{r +(s-r)p_t} \end{align*}
We summarize the development of this competing species model below:

#### Competing Species Model.

If species $$1$$ grows with per capita growth rate $$s$$ and species $$2$$ grows with per capita growth rate $$r\text{,}$$ the population proportion of species $$1$$ is modeled by the updating function
\begin{equation*} p_{t+1} = \dfrac{sp_t}{r+ (s-r)p_t}\text{.} \end{equation*}
We note here that the competing species model is an example of a non-linear updating function. That is, the updating function rule $$f(x) = \dfrac{sx}{r+(s-r)x}$$ is not a linear function. This can make finding equilibrium values algebraically more complex, and also allow us to have multiple different equilibrium values. The type of equations that are most relevant for us to be able to solve in this section are quadratic equations. You can find a nice review with solutions of solving quadratic equations at Paul’s Online Notes, Part 1 1  and Part 2 2 .

#### Activity1.9.4.

Use the interactive below to answer questions regarding the equilibrium values of the competing species model.
1. Do the equilibrium values of this model depend on the values of $$s$$ and $$r\text{?}$$ Explain.
2. Compute the equilibrium values of $$p_{t+1} = \dfrac{3p_t}{2+ (3-2)p_t}$$ algebraically, then use the interactive to verify your answers.
3. Compute the equilibrium values of $$p_{t+1} = \dfrac{sp_t}{r+ (s-r)p_t}$$ algebraically.
4. Does the stability of the equilibrium values depend on the values of $$s$$ and $$r\text{?}$$ Use cob-webbing to determine when each equilibrium value is stable/unstable. Try to do the cob-webbing on your own before using the button provided that illustrates the cob-webbing.
5. What do the equilibrium values represent in the context of the competing species model? How can you interpret the stability of each equilibrium value in the context of the competing species model?

### Subsection1.9.3Summary

• #### Question1.9.5.

How can we model gas exchange in the lung using a discrete-time dynamical system?
We can model gas exchange in the lung using the lung model.
• #### Question1.9.6.

What is a weighted average?
A weighted average between two numbers $$a$$ and $$b$$ is a sum of the form $$(1-p)a + pb\text{,}$$ where $$0\leq p \leq 1\text{.}$$ Weighted averages give us a way to place more importance on one number over another when computing the average value.
• #### Question1.9.7.

How can we model a system which involves two distinct populations that change over time?
We can form a DTDS that measures a population proportion. One example of this is when two distinct population exhibit exponential growth, we get the competing species model. One more example of this in the context of migration models is outlined in Exercise 1.9.4.5 and Exercise 1.9.4.4.

### Exercises1.9.4Exercises

#### 1.

Our current model for gas exchange in the lung does not account for absorption of the chemical in the body, which can occur for chemicals such as oxygen. If a fraction $$\alpha$$ of the chemical is absorbed before breathing out, the new DTDS becomes
\begin{equation*} c_{t+1} = (1 - p)(1-\alpha)c_t + p\beta\text{,} \end{equation*}
where $$p = \frac{W}{V}$$ and $$\beta$$ is the outside concentration of the chemical.
1. Explain in words why multiplying $$c_t$$ by $$(1-\alpha)$$ makes sense.
2. Find the equilibrium value(s) $$c^*$$ algebraically. Your answer should be in terms of $$p\text{,}$$ $$\alpha\text{,}$$ and $$\beta\text{.}$$
3. Let $$p = 0.5$$ and $$\beta= 0.2\text{.}$$ Use the table to test several values of $$\alpha$$ to see how absorption effects equilibria. As absorption increases, what happens to the equilibrium concentration?

#### 2.

Consider the lung model given by $$c_{t+1} = 0.4c_t + 1.2\text{,}$$ $$c_0 = 1\text{.}$$
1. Find the solution function for this DTDS.
2. Use the solution function to determine the approximate number of breaths it will take for the concentration to reach $$1.99$$ mmol/L.

#### 3.

Explain why the sum $$0.2(10) + 0.6(3)$$ is not a weighted average.

#### 4.

Suppose two nearby islands have populations of butterflies, with $$x_t$$ on the first island and $$y_t$$ on the second. Each year, $$20\%$$ of the butterflies from the first island fly to the second and $$30\%$$ of the butterflies from the second fly to the first.
Also suppose each butterfly that begins the year on the first island produces one offspring after migration (whether they find themselves on the first or the second island). Those that begin the year on the second island do not reproduce. Assume that no butterflies die.
1. As an initial example, suppose we start with 100 butterflies on each island. Find the number after migration and after reproduction on both islands.
Hint: The population of island 1 is $$190$$ and island 2 is $$110\text{,}$$ but it is how you calculate these numbers that you want to pay attention to.
2. Find equations for $$x_{t+1}$$ and $$y_{t+1}\text{,}$$ paying attention to how you calculated the previous part to identify patterns.
3. Let $$p_t$$ represent the proportion of butterflies on the first island after $$t$$ years. Find the updating function for $$p_t\text{.}$$
4. Find the equilibrium value of the system $$p_t\text{.}$$ There should be only one that makes biological sense. Then classify the equilibrium value as stable or unstable. You may use any method of your choice.
5. Interpret the meaning of the equilibrium value and its stability in the context of the model.

#### 5.

A population of salmon live in one of two territories in the Atlantic ocean. There are $$k_t$$ fish in territory 1 and $$h_t$$ fish in territory 2. Every year, $$80\%$$ of the fish from territory 1 migrate to territory 2, and $$75\%$$ of the fish from territory 2 migrate to territory 1.
1. Find an equation for $$k_{t+1}$$ in terms of $$k_t$$ and $$h_t\text{.}$$
2. Find an equation for $$h_{t+1}$$ in terms of $$k_t$$ and $$h_t\text{.}$$
3. Let $$p_t$$ represent the proportion of fish living in territory 1 after $$t$$ years. Find the updating function for $$p_t\text{.}$$
4. Find all equilibrium values of the system $$p_t$$ and classify each as stable or unstable. You may use any method of your choice.
5. Interpret the meaning of each equilibrium value and its stability in the context of the model.
tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsI.aspx
tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsII.aspx