Math Problem Statement

do not round any of the numbers while evaluation

During the period from 1790 to 1920, a country's population P(t) (in years) grew from 3.9 million to 102.9 million. Throughout this period, P(t) remained close to the solution of the initial value problem dP/dt = 0.03133P - 0.0001492P^2, P(0) = 3.9. (a) What 1920 population does this logistic equation predict? (b) What limiting population does it predict? (c) The country's population in 2000 was 313 million. Has this logistic equation continued since 1920 to accurately model the country's population?

Solution

We are given a logistic growth equation:

dPdt=0.03133P0.0001492P2,P(0)=3.9.\frac{dP}{dt} = 0.03133P - 0.0001492P^2, \quad P(0) = 3.9.

Let's solve the questions step by step.


(a) Predicted Population in 1920

The time tt is measured in years, starting from t=0t=0 in 1790. Thus, 1920 corresponds to t=19201790=130t = 1920 - 1790 = 130. To find the population at t=130t=130, we solve the logistic growth equation.

The solution to the logistic equation is:

P(t)=K1+(KP01)ert,P(t) = \frac{K}{1 + \left(\frac{K}{P_0} - 1\right)e^{-rt}},

where:

  • r=0.03133r = 0.03133,
  • K=rc=0.031330.0001492210K = \frac{r}{c} = \frac{0.03133}{0.0001492} \approx 210 (carrying capacity),
  • P0=3.9P_0 = 3.9 (initial population).

Substitute t=130t = 130 into the equation. Let me compute this precisely.


(b) Limiting Population

The limiting population KK is the carrying capacity of the logistic model, which we compute as:

K=rc.K = \frac{r}{c}.


(c) Comparison with 2000 Data

For t=20001790=210t = 2000 - 1790 = 210, we calculate P(210)P(210) using the logistic model and compare it to the actual population in 2000, which was 313 million. This will indicate whether the model remains accurate after 1920.

Let me compute these values. sympy as sp

Parameters

r = 0.03133 c = 0.0001492 P0 = 3.9

Carrying capacity (K)

K = r / c

Time for part (a) and (c)

t_1920 = 130 t_2000 = 210

Logistic growth equation solution

t = sp.symbols('t') P = K / (1 + ((K / P0) - 1) * sp.exp(-r * t))

Predicted populations

P_1920 = P.subs(t, t_1920) P_2000 = P.subs(t, t_2000)

Display results

K, P_1920, P_2000### Results:

  1. Carrying capacity (KK) = 209.99 million (approximately 210 million).

  2. Predicted population in 1920 (t=130t = 130) = 110.53 million.

  3. Predicted population in 2000 (t=210t = 210) = 195.63 million.


Analysis for Part (c):

The logistic equation predicts a population of 195.63 million in 2000, which is significantly lower than the actual population of 313 million. This suggests that the logistic equation no longer accurately modeled the population after 1920.


Would you like a detailed explanation of any step? Here are related questions:

  1. How is the logistic equation derived from its differential form?
  2. Why is carrying capacity (KK) considered the limiting population?
  3. What factors might cause the logistic equation to fail after 1920?
  4. How do changes in growth rates or parameters affect predictions in logistic models?
  5. Can we modify the logistic model to fit the 2000 population more closely?

Tip: For a logistic equation, the initial conditions and parameters greatly influence long-term predictions. Always check if assumptions remain valid over time.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth
Differential Equations
Exponential Functions

Formulas

dP/dt = rP - cP^2
P(t) = K / [1 + (K/P_0 - 1)e^(-rt)]
K = r/c

Theorems

Logistic Growth Model

Suitable Grade Level

Grades 11-12