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The rate of change
(dP)/(dt) of the number of students who heard a rumor is modeled by a logistic differential equation. The maximum capacity of the school is 942 students. At
5AM, the number of students who heard the rumor is 233 and is increasing at a rate of 37 students 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 students who heard a rumor is modeled by a logistic differential equation. The maximum capacity of the school is $942$ students. At $5 \mathrm{AM}$ , the number of students who heard the rumor is $233$ and is increasing at a rate of $37$ students 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 students who heard a rumor is modeled by a logistic differential equation. The maximum capacity of the school is

942

students. At

5 \mathrm{AM}

, the number of students who heard the rumor is

233

and is increasing at a rate of

37

students 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 $P$ is the population at time $t$ , $r$ is the growth rate, and $K$ is the carrying capacity (maximum capacity).
Given Carrying Capacity: We are given the carrying capacity $K = 942$ students. This is the maximum number of students that can hear the rumor according to the model.
Initial Population Data: We are also given that at $5$ AM, $P = 233$ students have heard the rumor, and the rate of change of $P$ at that time is $\frac{dP}{dt} = 37$ students per hour.
Finding Growth Rate: To find the growth rate $r$ , we can use the rate of change information when $P = 233$ . Plugging these values into the logistic equation, we get: $\newline$ $37 = r \cdot 233 \left(1 - \frac{233}{942}\right)$
Solving for r: Now we solve for $r$ : $\newline$ $37 = r \cdot 233 \left(\frac{942 - 233}{942}\right)$ $\newline$ $37 = r \cdot 233 \left(\frac{709}{942}\right)$ $\newline$ $37 = r \cdot 233 \cdot \frac{709}{942}$ $\newline$ $r = \frac{37 \cdot 942}{233 \cdot 709}$
Calculating Growth Rate: Performing the calculation for $r$ : $\newline$ $r = \frac{37 \cdot 942}{233 \cdot 709} \approx \frac{34854}{165017} \approx 0.2112 \text{ per hour}$
Final Logistic Differential Equation: Now we have the growth rate $r \approx 0.2112$ per hour and the carrying capacity $K = 942$ . We can write the logistic differential equation as: $\newline$ $\frac{dP}{dt} = 0.2112P\left(1 - \frac{P}{942}\right)$

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