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Differential Calculus: From Practice to Theory

Eugene Boman, Robert Rogers

Section 11.2 Selected Modeling Problems

Life is an incurable disease.
―Abraham Cowley
 4 
https://www.britannica.com/biography/Abraham-Cowley
(1618—1667)

Subsection 11.2.1 Epidemics, The SIR Model

The SIR Model is a fairly simple model that is frequently used to understand the spread of an infectious disease through a population. As with any mathematical model, it is over–simplified but, just like IVP (11.1), it provides a foundation which can be tweaked as needed for better predictions.
We will assume that every member the population falls into one of three categories: (S)usceptible, (I)nfected, or (R)ecovered. We assume that any population member who has recovered from the disease is immune to it, and those that have not are susceptible to infection.
We let
\begin{align*} S\amp =S(t)=\text{ The fraction of the population susceptible to infection.}\\ I\amp =I(t)=\text{ The fraction of the population currently infected.}\\ R\amp =R(t)=\text{ The fraction of the population no longer susceptible to infection. } \end{align*}
Note that \(R(t)\) includes those victims who have died.

Drill 11.2.1.

Assuming that no members are entering or leaving the population (by births, deaths from other diseases, or migration), explain why
\begin{equation*} S+I+R=1. \end{equation*}
Since the disease spreads by contact between a susceptible and an infected individual, we will assume that the number of susceptible population members is decreasing (you cannot get the disease twice) and the rate of decrease is proportional to the number of susceptible and the number of infected currently present. This says that
\begin{equation*} \dfdx{S}{t}=-aSI \end{equation*}
for some positive constant, \(a\text{.}\) The constant \(a\) is called the transmission rate. (Why?)

Drill 11.2.2.

Explain how we know that \(\dfdx{S}{t}\) must be negative.
Since the only way to become immune is to recover from the disease, we also assume that the rate of change of \(R\) is proportional to the number of infected individuals present. This means that
\begin{equation*} \dfdx{R}{t}=bI \end{equation*}
for some positive constant \(b\text{.}\) The constant \(b\) is called the recovery rate. (Why?)

Problem 11.2.3.

(a)
Explain how we know that \(\dfdx{R}{t}\) must be positive.
(b)
Show that
\begin{equation*} \dfdx{I}{t}=(aS-b)I. \end{equation*}
(c)
Use the information in part (b) to show that the number of infected is increasing when \(S\gt \frac{b}{a}\) and decreasing when \(S\lt \frac{b}{a}.\)
(d)
Show that
\begin{equation*} \dfdxn{I}{t}{2}=a^2I\left[S^2-\left(\frac{2b}{a}+I\right)S+\left(\frac{b}{a}\right)^2\right] \end{equation*}
and use this to show that \(\dfdxn{I}{t}{2}\lt 0\) when
\begin{equation*} \frac{b}{a}+\left(\frac{I}{2}-\sqrt{\left(\frac{I}{2}\right)^2+\frac{bI}{a}}\right)\lt S\lt \frac{b}{a}+\left(\frac{I}{2}+\sqrt{\left(\frac{I}{2}\right)^2+\frac{bI}{a}}\right) \end{equation*}
and is positive otherwise.
Putting this together we see that the number of infected is increasing and accelerating when
\begin{equation*} \frac{b}{a}+\left(\frac{I}{2}+\sqrt{\left( \frac{I}{2}\right)^2+\frac{bI}{a}}\right)\lt S\lt 1 \end{equation*}
and increasing, but slowing down, when
\begin{equation*} \frac{b}{a}\lt S\lt \frac{b}{a}+\left(\frac{I}{2}+\sqrt{\left(\frac{I}{2}\right)^2+\frac{bI}{a}}\right). \end{equation*}
When \(S\lt \frac{b}{a}\) it is decreasing.

Subsection 11.2.2 The Tractrix 

               Truckin', got my chips cashed in
               Keep truckin', like the do-dah man
               Together, more or less in line
               Just keep truckin' on.
―The Grateful Dead
 5 
https://en.wikipedia.org/wiki/Grateful_Dead
(1967—1995)
We first looked at the curve called the tractrix Problem 7.3.8 of Section 7.3. Consider reviewing that problem before you proceed.
Recall that initially the center of the rear axle of the tractor is at the origin and the center of the rear axle of the trailer is at the point \((1,0)\text{.}\) Also we assumed that the tractor pulls the front wheels vertically up the \(y\)-axis and that the rear wheels don’t slip.

Problem 11.2.4.

In Problem 7.3.8 you showed that the path followed by the center of the rear axle of the trailer follows, \(y=y(t)\text{,}\) must satisfy the IVP:
\begin{align} \dfdx{y}{x}=-\frac{\sqrt{1-x^2}}{x},\amp{}\amp{} y(1)=0.\tag{11.12} \end{align}
(a)
Show that \(y(x)=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}\) satisfies IVP (11.12).
  1. Graph \(y(x)\text{.}\)
  2. Does your graph in part (a) match your intuition about the shape of the curve?
  3. Does it match the graph you found using Euler’s Method in Problem 7.3.8?
(b)
How far will the tractor have gone (in trailer lengths) before the trailer is within one degree of vertical?
You will learn how to derive this solution from IVP (11.12) when you take Integral Calculus.

Subsection 11.2.3 The Pursuit Problem

Our scientific power has outrun our spiritual power. We have guided missiles and misguided men.
―Martin Luther King, Jr.
 6 
https://en.wikipedia.org/wiki/Martin_Luther_King_Jr.
(1929—1968)
Like the the tractrix problem above, we first looked at this pursuit curve in Problem 7.3.9 of Section 7.3. Consider reviewing that problem before you proceed.

Problem 11.2.5.

As in Problem 7.3.9 we assume that a rocket \(R\text{,}\) is traveling vertically up the line \(x=1\) at a constant speed \(v\text{.}\) When the rocket reaches the point \((1,0)\text{,}\) a missile \(M\) is fired from the origin directly at the rocket. Assuming that the missile is always aimed directly at the rocket, and that it travels at a speed which is \(k\) times the speed of the rocket (\(k\gt1\)) we showed in Problem 7.3.9 that the curve the missile follows will satisfy the IVP:
\begin{align} \dfdx{y}{x}=\frac{\frac{s}{k}-y}{1-x},\amp{}\amp{} y(0)=0.\tag{11.13} \end{align}
where \(s\) denotes the length of the the missile’s path at time \(t\text{.}\)
(a)
Use (11.13) to show that the missile’s path must satisfy the (second order) differential equation
\begin{equation*} (1-x)\dfdxn{y}{x}{2}=\frac{1}{k}\dfdx{s}{x}=\frac{1}{k}\sqrt{1+\left(\dfdx{y}{x}\right)^2} \end{equation*}
with the initial conditions: \(y(0)=0\) and \(\dfdxat{y}{x}{0}=0\text{.}\)
(b)
Show that
\begin{equation*} y=\frac12\left(\frac{k}{k+1}\left(1-x\right)^{\frac{k+1}{k}}-\frac{k}{k-1}\left(1-x\right)^{\frac{k-1}{k}}\right)+\frac{k}{k^2-1} \end{equation*}
satisfies the differential equation and initial conditions in part (a).
(c)
Find how long it takes for the missile to catch the rocket for \(k=3\text{,}\) \(k=2\text{,}\) and \(k=1.1\text{.}\) Does this make sense to you physically? Explain
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