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The rate of change 
(dP)/(dt) of the number of people in a park is modeled by the following differential equation:

(dP)/(dt)=(4085)/(36816)P(1-(P)/( 860))
At 
t=0, the number of people in the park is 236 and is increasing at a rate of 19 people per hour. Find 
lim_(t rarr oo)P^(')(t).
Answer:

The rate of change dPdt \frac{d P}{d t} of the number of people in a park is modeled by the following differential equation:\newlinedPdt=408536816P(1P860) \frac{d P}{d t}=\frac{4085}{36816} P\left(1-\frac{P}{860}\right) \newlineAt t=0 t=0 , the number of people in the park is 236236 and is increasing at a rate of 1919 people per hour. Find limtP(t) \lim _{t \rightarrow \infty} P^{\prime}(t) .\newlineAnswer:

Full solution

Q. The rate of change dPdt \frac{d P}{d t} of the number of people in a park is modeled by the following differential equation:\newlinedPdt=408536816P(1P860) \frac{d P}{d t}=\frac{4085}{36816} P\left(1-\frac{P}{860}\right) \newlineAt t=0 t=0 , the number of people in the park is 236236 and is increasing at a rate of 1919 people per hour. Find limtP(t) \lim _{t \rightarrow \infty} P^{\prime}(t) .\newlineAnswer:
  1. Analyze Differential Equation: Analyze the given differential equation.\newlineThe differential equation given is dPdt=408536816P(1P860)\frac{dP}{dt} = \frac{4085}{36816}P\left(1 - \frac{P}{860}\right). This is a logistic growth model where the growth rate of the population PP is proportional to both the current population and the difference between the current population and the carrying capacity, which in this case is 860860.
  2. Determine Steady State: Determine the steady state of the population.\newlineIn the logistic growth model, the steady state, or equilibrium, occurs when the rate of change of the population dPdt\frac{dP}{dt} is zero. This happens when PP is at its carrying capacity or when PP is zero. Since we are looking for the limit as tt approaches infinity, we are interested in the carrying capacity, which is P=860P = 860.
  3. Calculate Limit: Calculate the limit of P(t)P'(t) as tt approaches infinity.\newlineAs tt approaches infinity, the population PP approaches its carrying capacity. Therefore, the term (1P860)(1 - \frac{P}{860}) in the differential equation will approach zero, because PP will be very close to 860860. This means that the rate of change of the population dPdt\frac{dP}{dt} will also approach zero.
  4. Conclude Result: Conclude the limit of P(t)P'(t) as tt approaches infinity.\newlineSince the rate of change of the population dPdt\frac{dP}{dt} approaches zero as tt approaches infinity, we can conclude that limtP(t)=0\lim_{t \to \infty}P'(t) = 0.

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