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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 858 students. At 6AM, the number of students who heard the rumor is 220 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 dPdt \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 858858 students. At 6AM 6 \mathrm{AM} , the number of students who heard the rumor is 220220 and is increasing at a rate of 3737 students per hour. Write a differential equation to describe the situation.\newlinedPdt=□ \frac{d P}{d t}=\square

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Q. The rate of change dPdt \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 858858 students. At 6AM 6 \mathrm{AM} , the number of students who heard the rumor is 220220 and is increasing at a rate of 3737 students per hour. Write a differential equation to describe the situation.\newlinedPdt=□ \frac{d P}{d t}=\square
  1. Logistic Differential Equation: The logistic differential equation is generally given by the formula: \newline(dPdt)=rP(1−PK)(\frac{dP}{dt}) = rP(1 - \frac{P}{K})\newlinewhere:\newline- PP is the current number of individuals,\newline- rr is the intrinsic growth rate,\newline- KK is the carrying capacity (maximum capacity), and\newline- (dPdt)(\frac{dP}{dt}) is the rate of change of the population with respect to time.\newlineWe need to find the value of rr, the intrinsic growth rate.
  2. Substitute Maximum Capacity: Given that the maximum capacity of the school, KK, is 858858 students, we can substitute this value into the equation.
  3. Find Intrinsic Growth Rate: At 6AM6\text{AM}, the number of students who heard the rumor, PP, is 220220. The rate of change of the number of students who heard the rumor, dPdt\frac{dP}{dt}, is 3737 students per hour. However, we cannot directly substitute this rate into the logistic equation because the rate of 3737 students per hour is not the intrinsic growth rate rr; it is the actual rate of change at P=220P = 220.
  4. Solve for rr: To find the intrinsic growth rate rr, we use the given rate of change when P=220P = 220:dPdt=rP(1−PK)\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)37=r×220×(1−220858)37 = r \times 220 \times \left(1 - \frac{220}{858}\right)
  5. Calculate r Value: Now we solve for rr:37=r×220×(1−220/858)37 = r \times 220 \times (1 - 220/858)37=r×220×(638/858)37 = r \times 220 \times (638/858)37=r×220×(319/429)37 = r \times 220 \times (319/429)37=r×220×0.743837 = r \times 220 \times 0.743837=r×163.63637 = r \times 163.636r=37/163.636r = 37 / 163.636r≈0.226r \approx 0.226
  6. Write Differential Equation: Now that we have the value of rr, we can write the logistic differential equation:\newlinedPdt=rP(1−PK)\frac{dP}{dt} = rP(1 - \frac{P}{K})\newlinedPdt=0.226P(1−P858)\frac{dP}{dt} = 0.226P(1 - \frac{P}{858})\newlineThis is the differential equation that describes the situation.

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