Chapter 3, Problem 78RE

(a).

To determine

The initial number of students infected with measles.

Answer to Problem 78RE

The initial number of students infected with the measles is $1$ .

Explanation of Solution

Given:

  $P\left(t\right)=\frac{200}{1+e^{5-t}}$

Where $t$ is the number of days since the measles first appeared and $P\left(t\right)$ is the total number of students who have caught the measles to date.

Calculation:

To estimate the initial number of students infected with the measles., put $t=0$ :

  $\begin{array}{c}P\left(0\right)=\frac{200}{1+e^{5-0}}\\ =\frac{200}{1+e^5}\\ =\frac{200}{1+148.4132}\\ =\frac{200}{149.4132}\\ =1.3385\\ \approx 1\end{array}$

Conclusion:

The initial number of students infected with the measles is $1$ .

(b).

To determine

Number of students in all will get the measles.

Answer to Problem 78RE

Therefore, $200$ students will be infected by measles over time.

Explanation of Solution

Given:

  $P\left(t\right)=\frac{200}{1+e^{5-t}}$

Where $t$ is the number of days since the measles first appeared and $P\left(t\right)$ is the total number of students who have caught the measles to date.

Calculation:

The given function is a logistic model, so the numerator is the upper limit.

The total number that the function will reach approaches this number.

Conclusion:

Therefore, $200$ students will be infected by measles over time.

(c).

To determine

The time when the rate of spread of measles be greatest. Also, the rate.

Answer to Problem 78RE

The rate of speed of measles will be greatest when it is at $50$ students per day.

Explanation of Solution

Given:

  $P\left(t\right)=\frac{200}{1+e^{5-t}}$

Where $t$ is the number of days since the measles first appeared and $P\left(t\right)$ is the total number of students who have caught the measles to date.

Concept Used:

• The differentiation of $\frac{d}{dx}\left[e^{f\left(x\right)}\right]=e^{f\left(x\right)}\cdot \frac{d}{dx}\left[f\left(x\right)\right]$

Calculation:

The rate of spread of the measles will be the greatest at the maximum of $P\text{'}\left(t\right)$ is as follow:

  $\begin{array}{c}P\left(t\right)=\frac{200}{1+e^{5-t}}\\ P\text{'}\left(t\right)=200\cdot \frac{d}{dt}\left[\frac{1}{1+e^{5-t}}\right]\\ =-200\cdot \frac{\frac{d}{dt}\left[1+e^{5-t}\right]}{{\left(1+e^{5-t}\right)}^2}\\ =-200\cdot \frac{0+e^{5-t}\cdot \frac{d}{dt}\left(5-t\right)}{{\left(1+e^{5-t}\right)}^2}\\ =-200\frac{e^{5-t}\left[0-1\right]}{{\left(1+e^{5-t}\right)}^2}\\ =\frac{200{\text{e}}^{5-t}}{{\left({\text{e}}^{5-t}+1\right)}^2}\end{array}$

Also,

  $\begin{array}{c}P\text{'}\text{'}\left(t\right)=0\\ \frac{d}{dt}\left[\frac{200{\text{e}}^{5-t}}{{\left({\text{e}}^{5-t}+1\right)}^2}\right]=0\\ 200\cdot \frac{d}{dt}\left[\frac{{\text{e}}^{5-t}}{{\left({\text{e}}^{5-t}+1\right)}^2}\right]=0\\ \frac{200e^{5-t}\left(e^{5-t}-1\right)}{{\left({\text{e}}^{5-t}+1\right)}^3}=0\\ 200e^{5-t}\left(e^{5-t}-1\right)=0\\ 200e^{5-t}\ne 0\text{for}\text{all}\text{values}\text{of}t.\\ \left(e^{5-t}-1\right)=0\\ e^{5-t}=1\\ \ln \left(e^{5-t}\right)=\ln \left(1\right)\\ 5-t=0\\ t=5\end{array}$

To find the maximum rate, find $P\text{'}\left(t\right)$ at $t=5$ :

  $\begin{array}{c}P\text{'}\left(5\right)=\frac{200{\text{e}}^{5-5}}{{\left({\text{e}}^{5-5}+1\right)}^2}\\ =\frac{200}{{(1+1)}^2}\\ =\frac{200}{8}\\ =50\end{array}$

Conclusion:

The rate of speed of measles will be greatest when it is at $50$ students per day.