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The rate of change
(dP)/(dt) of the number of people infected by a disease is modeled by a logistic differential equation. The maximum capacity of the village is 943 people. At 8 PM, the number of people infected by the disease is 106 and is increasing at a rate of 30 people per hour. Write a differential equation to describe the situation.

(dP)/(dt)=◻

The rate of change $\frac{d P}{d t}$ of the number of people infected by a disease is modeled by a logistic differential equation. The maximum capacity of the village is $943$ people. At $8$ PM, the number of people infected by the disease is $106$ and is increasing at a rate of $30$ people per hour. Write a differential equation to describe the situation. $\newline$ $\frac{d P}{d t}=\square$

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Write exponential functions: word problems

Full solution

Q. The rate of change

\frac{d P}{d t}

of the number of people infected by a disease is modeled by a logistic differential equation. The maximum capacity of the village is

943

people. At

8

PM, the number of people infected by the disease is

106

and is increasing at a rate of

30

people per hour. Write a differential equation to describe the situation.

\newline

\frac{d P}{d t}=\square

Logistic Differential Equation: The logistic differential equation is generally given by the formula: $\newline$ $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ $\newline$ where: $\newline$ - $P$ is the current population (number of people infected), $\newline$ - $r$ is the growth rate (rate of increase of the infection), $\newline$ - $K$ is the carrying capacity (maximum capacity of the village), $\newline$ - $\frac{dP}{dt}$ is the rate of change of the population (number of people infected) over time.
Given Carrying Capacity: We are given the carrying capacity $K = 943$ people.
Given Population and Rate of Change: We are also given that at a certain time ( $8$ PM), the number of people infected $P = 106$ and the rate of change of the number of people infected $\frac{dP}{dt} = 30$ people per hour.
Finding Growth Rate: To find the growth rate $r$ , we can use the given rate of change when $P = 106$ : $\newline$ $30 = r \cdot 106 \left(1 - \frac{106}{943}\right)$ $\newline$ Now we solve for $r$ : $\newline$ $r = \frac{30}{106 \left(1 - \frac{106}{943}\right)}$
Calculating Growth Rate: Calculate the value of $r$ : $\newline$ $r = \frac{30}{106 \left(1 - \frac{106}{943}\right)}$ $\newline$ $r = \frac{30}{106 \left(\frac{943 - 106}{943}\right)}$ $\newline$ $r = \frac{30}{106 \left(\frac{837}{943}\right)}$ $\newline$ $r = \frac{30 \cdot 943}{106 \cdot 837}$ $\newline$ $r \approx \frac{28329}{88962}$ $\newline$ $r \approx 0.3185$ (rounded to four decimal places)
Writing Logistic Differential Equation: Now we can write the logistic differential equation using the value of $r$ and the carrying capacity $K$ : $\newline$ $\frac{dP}{dt} = 0.3185P\left(1 - \frac{P}{943}\right)$

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