Question

The rate, r, at which people get sick during an epidemic of the flu can be approximated by r = 1600te^(−0.5t), where r is measured in people/day and t is measured in days since the start of the epidemic.

(a) When are people getting sick the fastest?

(b) How many people get sick altogether?

Answer 1

Let S(t) be the number of sick people on day t. Thus\[\frac{dS}{dt}=r=1600te^{-0.5t}\]this rises linearly ( like 1600t ) when t is small ( as the exponential is about 1 ) but then falls later due to the fact that the exponential has a negative term. So when 'people are getting sick the fastest' is when the rate (r) is a maximum. Take another derivative :\[\frac{dr}{dt}=1600[t(-0.5)e^{-0.5t}+e^{-0.5t}]\]and set that to zero :\[\frac{dr}{dt}=0=1600[t(-0.5)e^{-0.5t}+e^{-0.5t}]\]\[t(-0.5)e^{-0.5t}+e^{-0.5t}=0\]\[(-0.5t + 1)e^{-0.5}=0\]\[-0.5t + 1=0\]\[0.5t=1\]\[t=2\]so that's day 2. Now for the total number that get sick is an integral of r over the whole time, so the limits are going to be between day zero and infinity. Although in practice, since you only get whole people - not fractions of people - getting sick, then there will actually be a day of finite date when the last person became ill. But hey it's a model ! \[\int_{0}^{\infty}r dt = \int_{0}^{\infty}1600te^{-0.5t}dt\]using integration by parts \[=1600[te^{-0.5t}/(-0.5)]_{0}^{\infty}-\frac{1600}{(-0.5)}\int_{0}^{\infty}e^{-0.5t}dt\]\[=\frac{1600}{-0.5}[0-0]+3200[e^{-0.5t}/(-0.5)]_{0}^{\infty}=6400\]

Answer 2

Actually maybe you'd call it the third day if the first day is 'day zero' ....