Math Problem Statement

The population P(t) of a species satisfies the logistic differential equation dP/dt = P(2 - P/5000), where the initial population is P(0) = 3000 and t is the time in years. a. What is the carrying capacity? b. Determine the rate when the population is growing the fastest. c. Determine the time when the population is growing the fastest. d. Given P(0) = 3000, identify the time interval when P(t) is increasing.

Solution

Let's analyze and solve the given logistic differential equation:

dPdt=P(2P5000)\frac{dP}{dt} = P \left(2 - \frac{P}{5000} \right)

with the initial condition P(0)=3000P(0) = 3000.

(a) Carrying Capacity

The carrying capacity KK is the population level where growth stops, which occurs when dPdt=0\frac{dP}{dt} = 0. Setting the growth term to zero:

2P5000=02 - \frac{P}{5000} = 0

Solving for PP:

P=5000P = 5000

Thus, the carrying capacity is 5000.


(b) Rate of Maximum Growth

The logistic growth model grows fastest at half of the carrying capacity, i.e., at:

P=50002=2500P = \frac{5000}{2} = 2500

To find the maximum growth rate, substitute P=2500P = 2500 into the differential equation:

dPdt=2500(225005000)\frac{dP}{dt} = 2500 \left(2 - \frac{2500}{5000} \right)

=2500(20.5)=2500×1.5=3750= 2500 \left(2 - 0.5\right) = 2500 \times 1.5 = 3750

Thus, the population grows the fastest at a rate of 3750 individuals per year.


(c) Time of Fastest Growth

For a logistic equation, the fastest growth occurs when P=K2P = \frac{K}{2}. The general solution to a logistic differential equation is:

P(t)=K1+CertP(t) = \frac{K}{1 + Ce^{-rt}}

where rr is the intrinsic growth rate. To determine the exact time, we need to solve for tt when P=2500P = 2500, given P(0)=3000P(0) = 3000. If needed, we can proceed with solving this equation explicitly.


(d) Interval Where P(t)P(t) is Increasing

The population is increasing when dPdt>0\frac{dP}{dt} > 0, which happens when:

P(2P5000)>0P \left(2 - \frac{P}{5000} \right) > 0

Since PP is always positive, the critical point is:

2P5000>02 - \frac{P}{5000} > 0

P<5000P < 5000

So, the population is increasing in the interval 0<P<50000 < P < 5000. Given P(0)=3000P(0) = 3000, the population will continue to increase until it reaches 5000.


Follow-up Questions

  1. What is the general solution to this logistic equation?
  2. How do we determine the value of the intrinsic growth rate?
  3. What happens to the population as tt \to \infty?
  4. How does the logistic model compare to an exponential growth model?
  5. What real-world examples fit this logistic growth scenario?

Tip: In logistic growth, the fastest population growth happens at half the carrying capacity because of the balance between available resources and reproduction rate.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth
Differential Equations
Population Dynamics

Formulas

dP/dt = P(2 - P/5000)
P(t) = K / (1 + Ce^(-rt))

Theorems

Logistic Growth Model

Suitable Grade Level

Grades 11-12