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uring time in days, we should pick a timescale that is defined by the problem itself: the recovery time γ - 1 , or the reciprocal of the recovery rate. This al lows us to define a new variable τ = γ t , which plays the role of time but is a pure number with no units because γ is measured in day - 1 and the dimensional time t is measured in days. We then treat of our new unknown proportions s ( τ), i ( τ), and r ( τ) as functions of this new nondimensional time. Now, we focus on dR / dt in Eq. (1). It has two vari ables that we want to replace, R and t . From Eq. (4), it is simple to replace R with Nr . The compli cating factor, however, is replacing differentiation with respect to t with differentiation with respect to τ . The answer is straight out of introductory calcu lus—a careful application of the Chain Rule. If we view our definition of the nondimensional time as τ = yt being a function of t , we may differentiate this function to get d τ/ dt = γ . Also, if we think of r as a function of τ with τ being a function of t , then dif ferentiation of R with respect to t is actually now the derivative of a composition of functions using the Chain Rule:

While β and γ both still appear in these equations, they only appear in their ratio β/γ , which is dimen sionless because it is a measure of the rate of con tacts per day divided by the number of recoveries per day. That is, it is exactly how we described the reproduction number R 0 above—the average new infections per resolution of a single infection. Therefore, we define the nondimensional parameter R 0 as β/γ , and it becomes the only parameter in the system (5) - (7). We can now see how the earlier intuitive effects of R 0 being less than, equal to, or greater than 1 can be predicted from this model and with greater in sight. From Eq. (7), dr / d τ = R 0 is - i = ( R 0 s - 1) i . Since s is the proportion of the population that is susceptible, 0 ≤ s ( τ ) ≤ 1. The importance of R 0 < 1 is now clear: R 0 s ( τ) - 1 < 0 for all τ so that dr / d τ is always negative. Therefore, if R 0 < 1, the infectious proportion i ( τ ) is decreasing, and the outbreak dies out without growing. The endemic case with R 0 = 1 also has this property, but its decay rate will be slow enough that it may not be observed. Now let R 0 > 1. The initial susceptible proportion s (0) is close to 1 if the disease is new to the commu nity, and so we would expect s (0) > (1/ R 0 ) if the disease is new to the community. This means that di/d τ = ( R 0 s - 1) i is initially positive, indicating a rapid initial growth phase. However, from Eq. (6), di/d τ is negative, and so s decreases, which in turn slows the growth rate of i because ( R 0 s - 1) be comes smaller. Eventually, s will pass through a special value we will denote as s * = (1/ R 0 ), which makes di/d τ = 0 . As s continues to get smaller after that, di/d τ becomes negative and the infectious pro portion begins to decrease. Therefore, the infec tious proportion changes from increasing to de creasing, and the First Derivative Test says that i passes through its maximum value. The outbreak then begins to recede because it becomes increas- Herd Immunity

When we plug this into Eq. (1) and replace I with Ni , we see that both sides are multiplied by N γ . Di viding both sides by this eliminates all parameters from the new equation for the proportions.

Similarly, we must now replace the derivatives in Eqs. (2) and (3), but here after we divide by γ , the parameters do not all go away. We are left with:

Virginia Mathematics Teacher vol. 47, no. 1

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