Chapter 3, Problem 76RE

Solution

(a)

To calculate: The initial number of infected students by flu.

  $5.4$

Given information: The function $P\left(t\right)=\frac{300}{1+e^{4-t}}$

Calculation:

Find the initial number of infected students as:

  $\begin{array}{c}P\left(t\right)=\frac{300}{1+e^{4-t}}\text{ Replace }\text{'}t\text{'}=0\\ P\left(t\right)=\frac{300}{1+e^{4-0}}\\ P\left(t\right)\approx 5.4\end{array}$

Therefore, the number of students infected with flu is $5.4$ .

(b)

To calculate: Number of students infected after $3$ days.

  $\text{81 students}$ was affected by flu.

Given information: function $P\left(t\right)=\frac{300}{1+e^{4-t}}$

Calculation:

Substitute $t=3$ in formula to get answer rounded to the nearest integer, means that $81$ students are infected after three days

  $\begin{array}{c}P\left(t\right)=\frac{300}{1+e^{4-t}}\\ =80.68\end{array}$

Or $\text{81 students}$

(c)

To calculate: why side must be positive of the triangle.

After $\text{3 days \& 7 hours}$

Given information: The function $P\left(t\right)=\frac{300}{1+e^{4-t}}$

Calculation:

We let $P\left(t\right)=100$ in original equation:

  $100=\frac{300}{1+e^{4-t}}$

Rearrange:

  $1+e^{4-t}=\frac{300}{100}$

Simplify and isolate exponential:

  $\begin{array}{c}1+e^{4-t}=3\\ e^{4-t}=2\end{array}$

Take logarithm of both side and apply logarithmic properties:

  $\begin{array}{l}\ln 1+e^{4-t}=3\\ \ln e^{4-t}=\ln 2\\ 4-t=\ln 2\\ t=4-\ln 2\\ =3.31\end{array}$

So $100$ students will be infected after about $3.31$ days, or about $\text{3 days \& 7 Houres}$ .

(d)

To calculate: Maximum number of infected students.

  $\text{300 students}$ will be infected.

Given information: function $P\left(t\right)=\frac{300}{1+e^{4-t}}$

Calculation:

Considering values of $t>0$ the maximum students infected is reached when denominator of formula approaches $1$ , so $P(t)=\frac{300}{1}$ when ${\text{e}}^{\text{4-t}}$ is very small, very large numbers for $t$ . Therefore, maximum number of infected students is $300$ .

  $\underset{x\to \infty }{\lim }\frac{300}{1+e^{4-t}}=300$