I'd like help with only question 19. It involves looking at the
attached graphs below. The first graph is from question 7 and the
second graph is from question 9.
I D Like Help With Only Question 19 It Involves Looking At The Attached Graphs Below The First Graph Is From Question 1 (141.02 KiB) Viewed 21 times
The fraction k of infecteds recovering in a given day can be estimated from observation of infected individuals where k is the reciprocal of the number of days an individual is sick. enough to infect others. For many contagious diseases, the infectious time is approximately the same for most infecteds and is known by observations. However, there is no first way to observe b but there is an indirect way using the contact number, once the epidemic has run its course. 16. Recall that di ds = di dt ds dt and df = -bs(t)i(t) and di = bs(t)i(t) – ki(t). Use that to show that ds dt = = di ds 1 1+ CS where c is the contact number. 17. Show that the solution to the DE in (16) is i = -s +2 In(s) + q. 18. Note that in most epidemics i(0) = 0 and s(0) = 1. Use this to find the value of q. 19. After the epidemic has run its course, we have lim i(t) = iso = 0 = to Estimate lim s(t) = SOO too by examining the graphs in (7) and (9). It 0.03 a reasonable estimate?
09 0.8 0.7 0.01 0.5 04 0.3 02 0.1 0 0 20 30 8. If b = 1.2 and k = }, find a formula for ' in terms of S. (Notice that = . What rule from Cale I tells us this is = true?) 9. Below is the slopefield for in the s-i phase plane. Sketch the solution curve in the s-i phase plane for the case where s(0) = 0.9 and i(0) = 0.1. 0.81 0.61 1 0.4 021 02 0.4 06 0.8
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