Chapter 3, Problem 93RE

Solution

a.

To find: The initial number of infected students.

The initial number of infected students is 16.

Given information:

The function determines the number of infected students after t days:

  $P(t)=\frac{1600}{1+99e^{-0.4t}}$

Calculation:

Substitute $t=0$ in $P\left(t\right)$ .

  $\begin{array}{c}P(0)=\frac{1600}{1+99e^0}\\ =\frac{1600}{1+99\left(1\right)}\\ =\frac{1600}{100}\\ =16\end{array}$

b.

To find: The time when 800 students get infected.

After $11\frac{1}{2}$ days 800 students will be infected.

Given information:

The function determines the number of infected students after t days:

  $P(t)=\frac{1600}{1+99e^{-0.4t}}$

Explanation:

Substitute $P\left(t\right)=800$ in the given function and solve for t .

  $\begin{array}{c}800=\frac{1600}{1+99e^{-0.4t}}\\ 1+99e^{-0.4t}=2\\ 99e^{-0.4t}=1\\ e^{-0.4t}=\frac{1}{99}\\ -0.4t=\ln \left(\frac{1}{99}\right)\text{ [Taking log on both sides]}\\ -0.4t=-4.595119\\ t\approx 11.4878\end{array}$

Hence, after $11\frac{1}{2}$ days 800 students will be infected.

c.

To find: The time when the school will close.

The school will be closed after 8 or 9 days.

Given information:

The function determines the number of infected students after t days:

  $P(t)=\frac{1600}{1+99e^{-0.4t}}$

The school will close when 400 students get infected.

Explanation:

Substitute $P\left(t\right)=400$ in the given function and solve for t .

  $\begin{array}{c}400=\frac{1600}{1+99e^{-0.4t}}\\ 1+99e^{-0.4t}=4\\ 99e^{-0.4t}=3\\ e^{-0.4t}=\frac{3}{99}\\ -0.4t=\ln \left(\frac{3}{99}\right)\text{ [Taking log on both sides]}\\ -0.4t=-3.49650756147\\ t\approx 8.7413\end{array}$

Hence, the school will be closed after 8 or 9 days.