Chapter 4, Problem 78RE

a.

To determine

To estimate: the initial number of students infected with measles.

Answer to Problem 78RE

About 1 student was infected.

Explanation of Solution

Given information: The spread of measles in a certain school is given by,

  $P(t)=\frac{200}{1+e^{5-t}},$ Where t is the number of days since the measles first appeared, and P(t) is total number of students who have caught the measles to date.

Calculation:

  $P(t)=\frac{200}{1+e^{5-t}},$

To find the initial number of students that were infected put t = 0,

  $P(0)=\frac{200}{1-e^{5-0}}=\frac{200}{1-e^5}\approx 1$ .

b.

To determine

To find: about how many students in all will get the measles.

Answer to Problem 78RE

Nearly 200 students will be infected by measles over time.

Explanation of Solution

Given information: The spread of measles in a certain school is given by,

  $P(t)=\frac{200}{1+e^{5-t}},$ Where t is the number of days since the measles first appeared, and P(t) is total number of students who have caught the measles to date.

Calculation:

This function follows a logistic model; therefore the number on top is the upper limit. The total number that the function will reach approaches this number.

Nearly 200 students will be infected by measles over time.

c.

To determine

To find: when will the rate of spread of measles be greatest and what this rate.

Answer to Problem 78RE

5 days, 50 students per day.

Explanation of Solution

Given information: The spread of measles in a certain school is given by,

  $P(t)=\frac{200}{1+e^{5-t}},$ Where t is the number of days since the measles first appeared, and P(t) is total number of students who have caught the measles to date.

Calculation:

The rate of spread of the measles will be the greatest at the maximum of P’(t) so find P’(t) using the power rule and chain rule.

  $\begin{array}{c}P(t)=\frac{200}{1+e^{5-t}}=200{(}^1.\\ P\text{'}(t)=-200{(}^1(-e^{5-t}).\\ P\text{'}(t)=\frac{200e^{5-t}}{{(1+e^{5-t})}^2}.\\ P\text{'}\text{'}(t)=\frac{200e^{5-t}(e^{5-t}-1)}{{(1+e^{5-t})}^3}.\\ SetP\text{'}\text{'}(t)=0\text{and}\text{solve}\text{for}t:\\ \frac{200e^{5-t}(e^{5-t}-1)}{{(1+e^{5-t})}^3}=0.\\ 200e^{5-t}(e^{5-t}-1)=0.\\ \text{Since}200e^{5-t}\ne 0\text{for}\text{all}\text{values}\text{of}t.\\ (e^{5-t}-1)=0.\\ e^{5-t}=1.\\ \ln e^{5-t}=\ln 1.\\ 5-t=0.\\ t=5.\\ \end{array}$

To find the maximum rate , evaluate P’(t) at t = 5;

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

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