Math Problem Statement

Solve

Solution

To solve the problems related to the logistic growth model provided, let's analyze the equation and calculate the required values step by step.

Given the logistic growth model: P(t)=24001+24e0.128tP(t) = \frac{2400}{1 + 24e^{-0.128t}} where P(t)P(t) represents the population at time tt in years.

1. Number of Caribou Initially Planted

To find the initial number of caribou, we need to evaluate P(t)P(t) at t=0t = 0: P(0)=24001+24e0.1280=24001+24e0=24001+24=240025=96P(0) = \frac{2400}{1 + 24e^{-0.128 \cdot 0}} = \frac{2400}{1 + 24e^0} = \frac{2400}{1 + 24} = \frac{2400}{25} = 96 So, the initial number of caribou planted was 96 caribou.

2. Carrying Capacity for Caribou on the Island

The carrying capacity is the maximum population that the environment can sustain, which is the limit of P(t)P(t) as tt approaches infinity: limtP(t)=24001+24e0.128t\lim_{t \to \infty} P(t) = \frac{2400}{1 + 24e^{-0.128t}} As tt \to \infty, e0.128t0e^{-0.128t} \to 0: limtP(t)=24001+0=2400\lim_{t \to \infty} P(t) = \frac{2400}{1 + 0} = 2400 So, the carrying capacity for the caribou on the island is 2400 caribou.

3. Population After 8 Years

To find the population after 8 years, we substitute t=8t = 8 into the logistic growth model: P(8)=24001+24e0.1288=24001+24e1.024P(8) = \frac{2400}{1 + 24e^{-0.128 \cdot 8}} = \frac{2400}{1 + 24e^{-1.024}} First, we calculate e1.024e^{-1.024}: e1.0240.3597e^{-1.024} \approx 0.3597 Then, P(8)=24001+240.3597=24001+8.6328=24009.6328249P(8) = \frac{2400}{1 + 24 \cdot 0.3597} = \frac{2400}{1 + 8.6328} = \frac{2400}{9.6328} \approx 249 So, the population after 8 years is approximately 249 caribou.

4. Time When Population Reaches 1000

To find the time when the population reaches 1000, we set P(t)=1000P(t) = 1000 and solve for tt: 1000=24001+24e0.128t1000 = \frac{2400}{1 + 24e^{-0.128t}} Rearranging and solving for e0.128te^{-0.128t}: 1000(1+24e0.128t)=24001000(1 + 24e^{-0.128t}) = 2400 1+24e0.128t=240010001 + 24e^{-0.128t} = \frac{2400}{1000} 1+24e0.128t=2.41 + 24e^{-0.128t} = 2.4 24e0.128t=2.4124e^{-0.128t} = 2.4 - 1 24e0.128t=1.424e^{-0.128t} = 1.4 e0.128t=1.424e^{-0.128t} = \frac{1.4}{24} e0.128t=0.0583e^{-0.128t} = 0.0583 Taking the natural logarithm on both sides: 0.128t=ln(0.0583)-0.128t = \ln(0.0583) 0.128t2.841-0.128t \approx -2.841 t2.8410.12822.2t \approx \frac{2.841}{0.128} \approx 22.2 So, the population reaches 1000 after approximately 22 years.

Summary

  1. Initial number of caribou planted: 96 caribou
  2. Carrying capacity: 2400 caribou
  3. Population after 8 years: 249 caribou
  4. Time to reach 1000 caribou: 22 years

Would you like further details or have any questions?

Here are five more questions you might be interested in:

  1. How do you derive the logistic growth equation?
  2. Can you explain the significance of the carrying capacity in logistic models?
  3. How does the rate constant affect the growth of the population in the logistic model?
  4. What are the applications of logistic growth models in real-life scenarios?
  5. How can the initial population size influence the overall growth pattern?

Tip: In logistic growth models, the population growth rate slows down as it approaches the carrying capacity due to limited resources. This is a key characteristic differentiating logistic growth from exponential growth.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth Model
Exponential Functions
Natural Logarithm

Formulas

Logistic growth model formula P(t) = 2400 / (1 + 24e^(-0.128t))

Theorems

Limit of a function as time approaches infinity

Suitable Grade Level

Advanced High School or College