Problem 27 - Algebra and Trigonometry

Question

World Population The relative growth rate of world population has been decreasing steadily in recent years. On the basis of this, some population models predict that world population will eventually stabilize at a level that the planet can support. One such logistic model is $$ P(t)=\frac{73.2}{6.1+5.9 e^{-0.02 t}} $$ where \(t=0\) is the year 2000 and population is measured in billions. billions. (a) What world population does this model predict for the year 2200\(?\) For 2300\(?\) (b) Sketch a graph of the function \(P\) for the years 2000 to 2500 . (c) According to this model, what size does the world population seem to approach as time goes on?

Step-by-Step Solution

Answer

(a) 8.52 billion (2200), 11.14 billion (2300). (c) Approaches 12 billion.

Step 1: Calculate Population for Year 2200

To find the population in the year 2200, we substitute \( t = 200 \) into the equation, since 2200 is 200 years after the year 2000. Using the formula: \[P(200) = \frac{73.2}{6.1 + 5.9e^{-0.02 \times 200}}\] First, calculate the exponent: \(-0.02 \times 200 = -4\).Calculate \(e^{-4}\) and substitute into the denominator:\[ P(200) = \frac{73.2}{6.1 + 5.9 \times e^{-4}} \] Finally, compute to get \(P(200)\approx 8.52\) billion.

Step 2: Calculate Population for Year 2300

To find the population in the year 2300, let \( t = 300 \). Using the formula: \[P(300) = \frac{73.2}{6.1 + 5.9e^{-0.02 \times 300}}\]First, calculate the exponent: \(-0.02 \times 300 = -6\).Calculate \(e^{-6}\) and substitute into the denominator:\[ P(300) = \frac{73.2}{6.1 + 5.9 \times e^{-6}} \] Finally, compute to get \(P(300)\approx 11.14\) billion.

Step 3: Sketching the Graph of Function P(t)

To sketch the graph of \(P(t)\), consider the years \(t=0\) to \(t=500\), corresponding to the years 2000 to 2500. The key points are \( t=200 \) and \( t=300 \), where the populations are approximately calculated in previous steps. As \( t \rightarrow \infty \), observe that \( e^{-0.02t} \rightarrow 0 \), meaning \( P(t) \rightarrow \frac{73.2}{6.1} \approx 12\) billion. The graph should approach this horizontal asymptote.

Step 4: Calculating the Carrying Capacity

The carrying capacity is the maximum population that can be sustained, found as \( t \rightarrow \infty \), where the exponential term approaches zero. So, we calculate:\[P(\infty) = \frac{73.2}{6.1 + 5.9 \cdot 0} = \frac{73.2}{6.1} \approx 12\]Thus, the world population seems to stabilize around 12 billion.

Key Concepts

World Population Prediction

Population predictions are essential for understanding future global dynamics and planning for sustainable development. The logistic growth model is a powerful tool for making such predictions because it considers the natural limits to growth that exist. Unlike exponential models that suggest infinite growth, the logistic model predicts stabilization.In this exercise, the logistic growth model formula is given by:\[ P(t)=\frac{73.2}{6.1+5.9 e^{-0.02 t}} \]where \(t=0\) represents the year 2000, and the population is measured in billions. To predict the population for future years—such as 2200 and 2300—we substitute the corresponding \(t\) values into the equation.For example:

• For the year 2200, \( t = 200 \), resulting in a predicted population of approximately 8.52 billion.
• For the year 2300, \( t = 300 \), resulting in a predicted population of approximately 11.14 billion.

These predictions indicate an initial increase and eventual stabilization of the world population over time.

Exponential Decay

Exponential decay in this context refers to the decrease of the term \( e^{-0.02t} \) in the logistic growth function. As time \(t\) increases, the exponential decay causes \( e^{-0.02t} \) to get closer to zero. This happens because the exponent \(-0.02t\) becomes more negative as \(t\) increases.This property is essential for understanding how the logistic growth model functions. Unlike exponential growth, which leads to unbounded population growth, exponential decay helps to introduce a realistic approach. It limits the function by reducing the influence of certain terms as time progresses.In simpler terms, as time goes on:

• The influence of \( e^{-0.02t} \) diminishes.
• The population model approaches a stable value.

This behavior shows how populations may grow rapidly at first but will slow as they reach their sustainable limits.

Carrying Capacity

The carrying capacity is a core concept in the logistic growth model. It refers to the maximum population size that the environment can sustainably support. In mathematical terms, this is the value that the population approaches as time goes to infinity.For the logistic model given in the exercise, the carrying capacity is calculated by letting \( e^{-0.02t} \) approach zero as \( t \rightarrow \infty \):\[ P(\infty) = \frac{73.2}{6.1 + 5.9 \cdot 0} = \frac{73.2}{6.1} \approx 12 \text{ billion} \]This means that the predicted long-term stable world population, according to this model, is around 12 billion.Understanding carrying capacity is crucial for planning and managing resources. It helps make informed decisions about sustainability and supports policies that align with ecological limits.

Graph Sketching

Sketching the graph of the logistic growth function provides a visual understanding of how population grows and stabilizes over time. By plotting critical points and understanding the behavior of the function as \( t \) increases, we can create an informative graph.For the function \(P(t)\) from the year 2000 to 2500, key points include:

• Year 2000: Starting point \( P(0) = \frac{73.2}{6.1 + 5.9 \cdot 1} \).
• Year 2200: \(P(200) \approx 8.52\) billion.
• Year 2300: \(P(300) \approx 11.14\) billion.

As \( t \to \infty \), \( e^{-0.02t} \to 0 \), and the graph approaches a horizontal asymptote at around 12 billion. The overall graph will start with rapid growth, slow down as it nears the carrying capacity, and eventually level off. This demonstrates how real-world population dynamics stabilize over time.