Chapter 3.2, Problem 45E

Solution

a.

To find: The initial number of infected students.

The initial number of infected students is 16.

Given:

The function $P\left(t\right)=\frac{800}{1+49e^{-0.2t}}$ gives the number of infected students after $t$ days.

Calculation:

To find the initial number of infected students, put $t=0$ in the given function and simplify. That is,

  $\begin{array}{c}P\left(0\right)=\frac{800}{1+49e^{-0.2\left(0\right)}}\\ =\frac{800}{1+49e^0}\\ =\frac{800}{1+49\left(1\right)}\\ =\frac{800}{50}\\ =16\end{array}$

Hence, the initial number of infected students is 16.

b.

To find: The time taken to reach 200 infected students.

About 14 days.

Given:

The function $P\left(t\right)=\frac{800}{1+49e^{-0.2t}}$ gives the number of infected students after $t$ days.

Calculation:

Put $P\left(t\right)=200$ in the given function and solve it for $t$ . That is,

  $\begin{array}{c}200=\frac{800}{1+49e^{-0.2t}}\\ 1+49e^{-0.2t}=4\\ 49e^{-0.2t}=3\\ e^{-0.2t}=\frac{3}{49}\\ -0.2t=\ln \left(\frac{3}{49}\right)\\ t=-\frac{\ln \left(\frac{3}{49}\right)}{0.2}\\ t=\text{13}\text{.966}\\ t\approx 14\end{array}$

Hence, the number of infected students will reach 200 in about 14 days.

c.

To find: The time taken to close the school.

About 17 days.

Given:

The function $P\left(t\right)=\frac{800}{1+49e^{-0.2t}}$ gives the number of infected students after $t$ days.

The school will close when 300 of the 800-student body are infected.

Calculation:

Put $P\left(t\right)=300$ in the given function and solve it for $t$ . That is,

  $\begin{array}{c}300=\frac{800}{1+49e^{-0.2t}}\\ 1+49e^{-0.2t}=\frac{8}{3}\\ 49e^{-0.2t}=\frac{5}{3}\\ e^{-0.2t}=\frac{5}{147}\\ -0.2t=\ln \left(\frac{5}{147}\right)\\ t=-\frac{\ln \left(\frac{5}{147}\right)}{0.2}\\ t=\text{16}\text{.90}\\ t\approx 17\end{array}$

Hence, the school will close in about 17 days.