Chapter 3.4, Problem 58E

(a)

To determine

To find: the initial number of students who have infected with the flu.

Answer to Problem 58E

The initial number of students who have infected with the flu is $1$ .

Explanation of Solution

Given information :

The spread of a flu is modeled by the equation $P\left(t\right)=\frac{200}{1+e^{5-t}}$ where $t$ days.

From initial number, plug $t=0$ in the equation $P\left(t\right)=\frac{200}{1+e^{5-t}}$ .

  $\begin{array}{c}P\left(0\right)=\frac{200}{1+e^{5-0}}\\ =\frac{200}{1+e^5}\\ =1.338570\\ \approx 1\end{array}$

The initial number of students who have infected with the flu is $1$ .

(b)

To determine

To find: the rate at which flu spreading after 4 days.

Answer to Problem 58E

The rate at which flu spreading after 4 days is $52$ students per day.

Explanation of Solution

Given information :

The spread of a flu is modeled by the equation $P\left(t\right)=\frac{200}{1+e^{5-t}}$ where $t$ days.

Formula needed :

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and $\frac{d}{dx}\left(a^x\right)=a^x\ln a$

Differentiate $P\left(t\right)=\frac{200}{1+e^{5-t}}$ with respect to $t$ .

  $\begin{array}{c}{P}^{\prime }\left(t\right)=200\cdot \frac{d}{dt}{\left(1+e^{5-t}\right)}^{-1}\\ =200\cdot \left[-{\left(1+e^{5-t}\right)}^{-2}\right]\cdot \frac{d}{dt}\left(1+e^{5-t}\right)\\ =200\cdot \left[-{\left(1+e^{5-t}\right)}^{-2}\right]\cdot e^{5-t}\frac{d}{dt}\left(5-t\right)\\ =200\cdot \left[-{\left(1+e^{5-t}\right)}^{-2}\right]\cdot e^{5-t}\left(-1\right)\end{array}$

Further simplifying,

  $P^{\prime }\left(t\right)=\frac{200}{{\left(1+e^{5-t}\right)}^2}{e}^{5-t}$

Substitute $t=4$ in the above function.

  $\begin{array}{c}{P}^{\prime }\left(4\right)=\frac{200}{{\left(1+e^{5-4}\right)}^2}{e}^{5-4}\\ =\frac{200}{{\left(1+e\right)}^2}{e}^1\\ =39.32238\\ \approx 39\end{array}$

The rate at which flu spreading after 4 days is $39$ students per day.

(c)

To determine

To find: the maximum rate and time at which flu spread.

Answer to Problem 58E

The maximum rate and time at which flu spreading is $50$ students per day and 5 days respectively.

Explanation of Solution

Given information :

The spread of a flu is modeled by the equation $P\left(t\right)=\frac{200}{1+e^{5-t}}$ where $t$ days.

Formula needed :

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and $\frac{d}{dx}\left(a^x\right)=a^x\ln a$

Differentiate $P^{\prime }\left(t\right)=\frac{200}{{\left(1+e^{5-t}\right)}^2}{e}^{5-t}$ with respect to $t$ .

  $\begin{array}{c}{P}^{″}\left(t\right)=\frac{200}{{\left(1+e^{5-t}\right)}^2}{e}^{5-t}\\ =200\left[e^{5-t}\cdot \frac{d}{dt}{\left(1+e^{5-t}\right)}^{-2}+{\left(1+e^{5-t}\right)}^{-2}\cdot \frac{d}{dt}\left(e^{5-t}\right)\right]\\ =200\left[e^{5-t}\cdot 2e^{5-t}{\left(1+e^{5-t}\right)}^{-3}+{\left(1+e^{5-t}\right)}^{-2}\cdot \left(-e^{5-t}\right)\right]\\ =200e^{5-t}{\left(1+e^{5-t}\right)}^{-2}\left[\frac{2e^{5-t}}{\left(1+e^{5-t}\right)}-1\right]\end{array}$

Take $P^{″}\left(t\right)=0$ .

  $\begin{array}{c}\frac{200e^{5-t}}{{\left(1+e^{5-t}\right)}^2}\left[\frac{2e^{5-t}-1-e^{5-t}}{\left(1+e^{5-t}\right)}\right]=0\\ \frac{200e^{5-t}}{{\left(1+e^{5-t}\right)}^2}\left[\frac{e^{5-t}-1}{\left(1+e^{5-t}\right)}\right]=0\\ e^{5-t}-1=0\\ t=5\end{array}$

After solving and simplifying,

  $t=5$ days

Substitute $t=5$ in the function $P^{\prime }\left(t\right)=\frac{200}{{\left(1+e^{5-t}\right)}^2}{e}^{5-t}$ .

  $\begin{array}{c}{P}^{\prime }\left(5\right)=\frac{200}{{\left(1+e^{5-5}\right)}^2}{e}^{5-5}\\ =\frac{200}{{\left(1+e^0\right)}^2}{e}^0\\ =\frac{200}{4}\cdot 1\\ =50\end{array}$

The maximum rate and time at which flu spreading is $50$ students per day and 5 days respectively.