Chapter 7, Problem 59RE

a.

To determine

To Prove:

To prove that the function is a solution of a logistic differential equation. Identify $k$ and the carrying capacity.

Answer to Problem 59RE

The value of $k$ is $k=1$ and carrying capacity is $150$ .

Explanation of Solution

Given information:

The spread of flu in a certain school is given by the formula,

  $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

The general form is

  $\frac{dP}{dt}\propto P-P^2$

The given formula is $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

Differentiating with respect to t,

  $\frac{dP}{dt}=\frac{150e^{4.3-t}}{{\left(1+e^{4.3-t}\right)}^2}$

Now using the general form $\frac{dP}{dt}\propto P-P^2$ ,

  $\begin{array}{l}A\ast P-B\ast P^2=A\frac{150}{\left(1+e^{4.3-t}\right)}-B\frac{22500}{{\left(1+e^{4.3-t}\right)}^2}\\ P-P^2=A\frac{150\left(1+e^{4.3-t}\right)}{{\left(1+e^{4.3-t}\right)}^2}-B\frac{22500}{{\left(1+e^{4.3-t}\right)}^2}\end{array}$

For any given values of A and B the quantity $A\ast P-B\ast P^2$ will be same as $\frac{dP}{dt}$ .

Therefore, the value of $k$ is $k=1$ and carrying capacity is $150$ .

b.

To determine

To Estimate:

To estimate $P\left(0\right)$ and explain its meaning in the context of the problem.

Answer to Problem 59RE

The value of $P\left(0\right)$ is $P\left(0\right)\approx 2$ .

Explanation of Solution

Given information:

The spread of flu in a certain school is given by the formula,

  $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

The spread of flu in a certain school is given by the formula,

  $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

Here $P\left(0\right)$ represents the initial number of infected students.

Now substituting $t=0$ ,

  $\begin{array}{l}P\left(0\right)=\frac{150}{1+e^{4.3-0}}\\ P\left(0\right)=\frac{150}{1+e^{4.3}}\\ e^{4.3}=73.69979\dots \\ P\left(0\right)=\frac{150}{1+73.69979\dots }\\ P\left(0\right)=\frac{150}{74.69979\dots }\\ P\left(0\right)=2.00803\dots \end{array}$

Therefore, the value of $P\left(0\right)$ is $P\left(0\right)\approx 2$ .

c.

To determine

To Estimate:

To estimate the number of days it will take for a total of 125 students to become infected.

Answer to Problem 59RE

The number of days is appoximatey about 6 days

Explanation of Solution

Given information:

The spread of flu in a certain school is given by the formula,

  $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

The spread of flu in a certain school is given by the formula,

  $P\left(t\right)=\frac{150}{1+e^{4.3-t}}$

Here total of 125 students to become infected ,

Now substituting $P\left(t\right)=125$ ,

  $\begin{array}{l}125=\frac{150}{1+e^{4.3-t}}\\ 1+e^{4.3-t}=1.2\\ e^{4.3-t}=0.2\\ t=5.9\end{array}$

Therefore, the number of days is appoximatey about 6 days.