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)=◻

Question

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)=◻

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Logistic Differential Equation: The logistic differential equation is generally given by the formula: $$ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) $$ 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: $$ 37 = r \cdot 233 \left(1 - \frac{233}{942}\right) $$Solving for r: Now we solve for $ r $: $$ 37 = r \cdot 233 \left(\frac{942 - 233}{942}\right) $$ $$ 37 = r \cdot 233 \left(\frac{709}{942}\right) $$ $$ 37 = r \cdot 233 \cdot \frac{709}{942} $$ $$ r = \frac{37 \cdot 942}{233 \cdot 709} $$Calculating Growth Rate: Performing the calculation for $ r $: $$ r = \frac{37 \cdot 942}{233 \cdot 709} \approx \frac{34854}{165017} \approx 0.2112 \text{ per hour} $$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: $$ \frac{dP}{dt} = 0.2112P\left(1 - \frac{P}{942}\right) $$

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