Teacher Cheat Sheet

Teacher Cheat Sheet: What Does Math Have to Do with Getting Sick?

Day 1 Step #4: Sketch three graphs as an epidemic spreads through a totally susceptible population.

(a) As time increases from the outbreak of the epidemic (at time t = 0) susceptible individuals are falling ill. Thus we would expect the number S to start high on the graph (at time t = 0) and to show a steady decline as time increases. Unless we know more about the illness, we cannot decide whether everyone eventually catches it. In graphical terms, this means we don't know whether the graph of S levels off at zero or at a value above zero.

(b) When an epidemic begins, we say there is a “outbreak,'' which means that there is a (relatively) sudden increase in the number of cases. As time increases from the onset, the number of infected people climbs to a peak and then goes into a gradual decline. After some weeks or months, the illness virtually disappears.

(a) There may be some people who are immune to the disease, having already at it earlier, or some people who are removed, never coming into contact with infected individuals. In which case, the value of R would be slightly above the horizontal axis at time t = 0.

(b) Everybody is susceptible to the disease. Nobody has had it before and everybody has an equal chance to get infected. In which case, the value of R would be zero at time t = 0.

However, regardless of the value of R at time t = 0, we would expect more and more people in the recovered group as time passes. The graph of R should therefore climb from left to right, but it should flatten out because there is a limited number of people who can recover.

These sketches should look something like graphs found at the following site: http://www.biomedcentral.com/1471-2334/3/19/figure/F1 If all three graphs are combined on the same axes, the graph should look like the graph found at the following site: http://www.bondy.ird.fr/~bacaer/oldsars/node20.html Notice that the number of infected individuals is far less than the number of susceptible individuals or the number of recovered individuals.

Day 1 #7 Ask them what they know about the equation for Sў from the graph they made of S versus time.

When the value of Sў is positive, then the number of people in the susceptible group, S , increases. When the value of Sў is negative, then the number of people in the susceptible group, S , decreases. From their graph of S versus time, the students can see that the number of people in the susceptible group is decreasing as more of them become infected and move out of the susceptible group as time goes by. This means that the equation for Sў must include a negative sign.

Day 1 #8 Ask them to think about how to write an expression for the rate of transmission of the disease to a susceptible person from an infected person.

To understand how to get an expression for the rate of transmission of the disease, let’s consider a single susceptible person on a single day. a) On average, each day this one person will contact only a small fraction, p , of the infected population I . Finding the product of these variables, we get the following expression,

pI = Number of daily contacts per susceptible person. b) However, we don’t want an expression for the number of contacts for just one susceptible person. We want an expression that represents the average number of daily contacts made between the whole susceptible group, S , with the infected group, I . To find this expression, we can just multiply the average number of contacts per susceptible person by the total number of susceptible people. In other words,

(pI )S = pSI = Number of daily contacts the whole susceptible population will have with the infected population . c) But, with most diseases, not all contacts lead to new infections. Let’s assume that only a certain fraction q of the contacts between susceptible people and infected people do lead to new infections. We can now write an expression for the average number of new infections per day. To find this expression, we can just multiply the average number of daily contacts between the whole susceptible population and the infected population by q , the fraction of contacts that do lead to new infections. In other words,

(pSI )(q)= (pq)(SI )= aSI = Number of new infections per day = rate of transmission of the disease. (Note: since both p and q are constants, the product pq is also a constant and can be replaced by the constant a , where a > 0 ) d) Since each new infection decreases the number of susceptible people, we have the rate equation for S :

Sў= -aSI (The minus sign here tells us that S is decreasing since S , I , and a are positive).

Day 1 #9 Ask the students to think about which group of the population Rў is related to and why.

Rў is proportional to I, the number of infected people because the number of recovered/removed people depends upon the number of infected. In other words,

Rў= bI where b is a constant and b > 0

There is a way to approximate the value of b by using the length of the period of the infection, k 1 where b  . See http://www.math.smith.edu/Local/cicchap1/node5.html k

Day 1 #10 Ask the students to think about how the number of infected people changes. Tell them that these changes will lead them directly to an equation for I  The number of infected people is changing in two ways: (1) newly sick people from the susceptible population are added to the infected group and (2) infected people who have recovered or who have been removed are leaving the infected group. Thus,

I  = (Rate susceptible people get sick) – (Rate infected people recover/are removed)

But the newly sick people are exactly those people leaving the susceptible group, and we know that the rate of change for S is S  aSI . This means that the rate at which newly sick people join the infected group is aSI (with a positive sign this time because the number of infected people is increasing as newly sick people join the group).

Similarly, the rate at which infected people move out of the infected group is the same as the rate at which people move into the recovered/removed group. We know that rate of change for R is R  bI This means that the rate at which infected people join the recovered/removed group is  bI (with a negative sign this time because the number of infected people is decreasing as newly recovered/removed people leave the group).

Putting this all together, we get the following equation for I  : I   aSI  bI

Day 1 #11 Have the students write all three equations one under another and add them together to get the overall rate of change of the population. Ask the students to discuss what this means about our model for the epidemic.

S  aSI I   aSI  bI R  bI

The sum is zero. This means that the total population does not change, that nobody is dying from the epidemic and nobody is entering the population either.

Day 1 #12 Now we can see the predictive power of the SIR model. Have the students go back to the measles epidemic mentioned at the beginning of the lesson. Tell the students that the value of a  0.00003 and b  0.07 . Have the students enter these values into the rate equations.

S  0.00003SI I   0.00003SI  0.07I R  0.07I

Day 1 # 13 Calculating various values

Teacher Cheat Sheet:

a) Since S  I  R  50,000 , the total population, then S  50,000  2,100  2,500  45,400

b) The value of S' depends upon the values of S and I right now. This means

S  0.0000345,4002100  2860.2  2860 people c) The value of I  depends upon the values of S, I, and R right now. This means

I   0.00003(45,400)(2,100)  (0.07)(2,100)  2860 147  2713

d) The value of R depends upon the value of I right now. This means

R  (0.07)(2,100)  147

e) The value of S tomorrow is S1  S  S  45,400  2860  42,539

f) The value of I tomorrow is I1  I  I   2100  2713  4,813

g) The value of R tomorrow is R1  R  R  2500 147  2647

(If you add these numbers, the sum is 49,999 due to rounding.)

Day 1 #14

Have the students use tomorrow’s values of S1 , I1 , and R1 to calculate S2 , I 2 , and R2 , the values of S, I, and R for two days from now.  S2  S1  S1   S1   aS1I1   42,539   0.0000342,5394,813  42,539  6,142  36,397

 I  I  I   I  aS I  bI  2 1 1 1 1 1 1  4,813  0.0000342,5394,813 0.074,813  4,813  6,142  337  10,618

 R2  R1  R1   R1  bI1   2,647  0.074,813 2,647  337  2984

Day 1 Homework

Problem (1): We know from the work in class that a  pq , where p is the fraction of the infected population that one susceptible person will contact daily and q is the fraction of contacts 1 that do lead to new infections. In the given situation, p  0.3%  0.003 and q  , which 6  1  means the value of a is a  pq  0.003   0.0005 .  6  The value of b is b  25%  0.25

Problem (2): We know the equations for S, I , and R are the following: S  aSI , I   aSI  bI , R  bI

Plugging in the values of a and b , the equations become S  0.0005SI I   0.0005SI  .25I R  .25I

Problem (3): Since S  I  R  1,800 , the total population of the college, S  1800  I  R  1,800  80  31  1800 111  1689

Problem (4) For today’s values, we know S  1689, I  80, R  31, a  0.0005, b  0.25 . This means:

S  aSI   0.0005168980  68 I   aSI  bI  0.0005168980 .2580  68  20  48 R  bI  .2580  20

Problem (5) i) The value of S tomorrow is S1  S  S  1689  68  1621

ii) The value of I tomorrow is I1  I  I   80  48  128

iii) The value of R tomorrow is R1  R  R  31 20  51

Problem (6) i) The value of S two days from today is:

 S2  S1  S1   S1   aS1I1   1621  0.00051621128  1621104  1517

ii) The value of I two days from today is

 I  I  I   I  aS I  bI  2 1 1 1 1 1 1  128  0.00051621128 0.25128  128  104  32  200

iii) The value of R two days from today is

 R2  R1  R1   R1  bI1   51  0.25128 51  32  83

Problem (7) The new equations for S, I , and R would be: S  aSI  cR I   aSI  bI R  bI  cR The equation for S would change because the number of susceptible people would drop as people get newly sick, but it would grow at the same time as recovered people lose immunity and become susceptible again. Thus,

S = -(Rate susceptible people get newly sick) + (Rate recovered people lose immunity)

The equation for I  would not change because Immunity Lose only affects the flow of people from the recovered population to the susceptible population. Immunity lose does not affect the flow from the susceptible population to the infected population nor the flow from the infected population to the recovered population

The equation for R would change because the number of recovered people would grow as people get recover from being infected, but it would drop at the same time as recovered people lose immunity and become susceptible. Thus,

R = (Rate infected people newly recover) - (Rate recovered people lose immunity)