Logistic Growth Model: Estimating Population Growth and Carrying Capacity

Math Problem Statement

User responses cleared Homework:HW SECTION 9.5 Question 8, 9.5.38 Part 4 of 7 HW Score: 91.07%, 7.29 of 8 points Points: 0.29 of 1

Skip to Main content Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question content area top Part 1 A nation's population (to the nearest million) was 281 million in 2000 and 311 in 2010. It is projected that the population in 2050 will be 437 million. To construct a logistic model, both the growth rate and the carrying capacity must be estimated. There are several ways to estimate these parameters. Use parts (a) through (f) to use one approach. Question content area bottom Part 1 a. Assume that tequals0 corresponds to 2000 and that the population growth is exponential for the first ten years; that is, between 2000 and 2010, the population is given by Upper P left parenthesis t right parenthesis equals Upper P left parenthesis 0 right parenthesis e Superscript rt. Estimate the growth rate r using this assumption. requals 0.01014 (Round to five decimal places as needed.) Part 2 b. Write the solution of the logistic equation with the value of r found in part (a). Write any populations in the logistic equation in millions of people. P(t)equals StartFraction negative Upper K times 281 Over e Superscript negative 0.01014 t Baseline left parenthesis negative Upper K plus 281 right parenthesis minus 281 EndFraction Part 3 Use the projected value Upper P left parenthesis 50 right parenthesis equals 437 million to find a value of the carrying capacity K. Kequals 2744.39 (Type an integer or decimal rounded to the nearest hundredth as needed.) Part 4 c. According to the logistic model determined in parts (a) and (b), when will the country's population reach 95% carrying capacity? The population will reach 95% of the carrying capacity in the year enter your response here. (Type an integer or decimal rounded to the nearest hundredth as needed.) r(Round to five decimal places as needed.)P(t)K(Type an integer or decimal rounded to the nearest hundredth as needed.)The population will reach 95% of the carrying capacity in the year(Type an integer or decimal rounded to the nearest hundredth as needed.)K(Type an integer or decimal rounded to the nearest hundredth as needed.)K(Type an integer or decimal rounded to the nearest hundredth as needed.)Choose the correct response below.r(Round to five decimal places as needed.)P(t)K(Type an integer or decimal rounded to the nearest hundredth as needed.)The population will reach 95% of the carrying capacity in the year(Type an integer or decimal rounded to the nearest hundredth as needed.)K(Type an integer or decimal rounded to the nearest hundredth as needed.)K(Type an integer or decimal rounded to the nearest hundredth as needed.)Choose the correct response below.r(Round to five decimal places as needed.)P(t)K(Type an integer or decimal rounded to the nearest hundredth as needed.)The population will reach 95% of the carrying capacity in the year(Type an integer or decimal rounded to the nearest hundredth as needed.) Get more help

Solution

To solve the problem in Part 4, where we need to determine when the population will reach 95% of the carrying capacity using the logistic model, we need to follow these steps:

Step 1: Set Up the Logistic Model

We have the logistic equation from Part 2 as: $P(t) = \frac{K \cdot P(0)}{P(0) + (K - P(0))e^{-rt}}$ where:

$K = 2744.39$ million (carrying capacity from Part 3),
$P(0) = 281$ million (initial population in 2000),
$r = 0.01014$ (growth rate from Part 1).

Step 2: Set the Population to 95% of the Carrying Capacity

We want to find $t$ when the population $P(t)$ reaches 95% of $K$ : $P(t) = 0.95 \cdot K = 0.95 \cdot 2744.39 = 2607.17 \text{ million}$ So, we set up the equation: $2607.17 = \frac{2744.39 \cdot 281}{281 + (2744.39 - 281)e^{-0.01014t}}$

Step 3: Solve for $t$

To solve for $t$ , we rearrange the equation:

$2607.17 \cdot \left[ 281 + 2463.39e^{-0.01014t} \right] = 2744.39 \cdot 281$ $2607.17 \cdot 281 + 2607.17 \cdot 2463.39e^{-0.01014t} = 771148.39$ $642640.67 + 6420251.91e^{-0.01014t} = 771148.39$ $6420251.91e^{-0.01014t} = 128507.72$ $e^{-0.01014t} = \frac{128507.72}{6420251.91} \approx 0.020002$ Take the natural logarithm of both sides: $-0.01014t = \ln(0.020002)$ $t = \frac{\ln(0.020002)}{-0.01014} \approx 400.21$

Step 4: Find the Year

Since $t = 0$ corresponds to the year 2000, we add 400.21 years to 2000: $\text{Year} = 2000 + 400.21 \approx 2400.21$ Thus, the population will reach 95% of the carrying capacity around the year 2400.

Would you like any further details or have any questions? Here are some related questions to consider:

What is the general form of a logistic growth model?
How does the value of $r$ affect the shape of the logistic curve?
What happens to the population growth as it approaches the carrying capacity?
Why does the logistic model provide a more realistic prediction than an exponential model?
How would changes in initial conditions affect the time to reach 95% of the carrying capacity?

Tip: Remember that the logistic model accounts for the decreasing growth rate as the population approaches the carrying capacity, reflecting real-world limitations on growth.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth Model
Exponential Growth
Population Dynamics

Formulas

Logistic equation: P(t) = K * P(0) / (P(0) + (K - P(0))e^{-rt})
Exponential growth: P(t) = P(0)e^{rt}

Theorems

Logistic Growth Theorem

Suitable Grade Level

Grades 11-12

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