Problem 47
Question
Using the same axes, draw the graphs for \(0 \leq t \leq 100\) of the following two models for the growth of world population (both described in this section). (a) Exponential growth: \(y=6.4 e^{0.0132 t}\) (b) Logistic growth: \(y=102.4 /\left(6+10 e^{-0.030 t}\right)\) Compare what the two models predict for world population in 2010,2040 , and 2090 . Note: Both models assume that world population was \(6.4\) billion in \(2004(t=0)\).
Step-by-Step Solution
Verified Answer
The exponential model predicts exponentially growing populations, while the logistic model predicts growth slowing towards a cap. By 2090, exponential predictions far exceed those from the logistic model.
1Step 1: Understand the Given Models
The problem requires drawing graphs for two models of world population growth: one with exponential growth and the other with logistic growth. The exponential model is given as \(y = 6.4 e^{0.0132 t}\), while the logistic model is \(y = \frac{102.4}{6 + 10 e^{-0.030 t}}\). The parameter \(t\) is bounded between 0 and 100, representing years from 2004 to 2104.
2Step 2: Setup the Graph Axes
Prepare the axes for graphing. The x-axis must represent time \(t\) from 0 to 100 (2004 to 2104), and the y-axis should represent the world population in billions. Make sure the scales on both axes are appropriate to clearly visualize growth trends from both models.
3Step 3: Plot the Exponential Growth Model
For the exponential growth, use the equation \(y = 6.4 e^{0.0132 t}\). Calculate and plot \(y\) for several values of \(t\) (0, 10, 20,...,100) to plot the curve. Begin at \(t=0\) with \(y = 6.4\), and note exponential increase as \(t\) increases.
4Step 4: Plot the Logistic Growth Model
For the logistic growth, use the equation \(y = \frac{102.4}{6 + 10 e^{-0.030 t}}\). Similarly, calculate and plot \(y\) for values of \(t\) from 0 to 100. Start with \(t=0\) yielding \(y\) near \(6.4\), but expect \(y\) to approach an asymptote as \(t\) increases.
5Step 5: Compare Population Predictions for Specific Years
Calculate the population predicted by both models for the years 2010 (\(t=6\)), 2040 (\(t=36\)), and 2090 (\(t=86\)). For exponential growth: 2010, \(y = 6.4 e^{0.0132 \, \times 6}\); 2040, \(y = 6.4 e^{0.0132 \, \times 36}\); 2090, \(y = 6.4 e^{0.0132 \, \times 86}\). For logistic growth: 2010, \(y = \frac{102.4}{6 + 10 e^{-0.030 \, \times 6}}\); 2040, \(y = \frac{102.4}{6 + 10 e^{-0.030 \, \times 36}}\); 2090, \(y = \frac{102.4}{6 + 10 e^{-0.030 \, \times 86}}\).
6Step 6: Analyze Model Predictions
Examine the calculated values: the exponential model predicts rapid continuous growth while the logistic model predicts slower growth, approaching a maximum (carrying capacity). Compare specific calculated populations for 2010, 2040, and 2090: expected exponential populations surpass before the logistic predictions at larger \(t\)-values due to limits in the latter.
Key Concepts
Exponential GrowthLogistic GrowthWorld Population Prediction
Exponential Growth
Exponential growth is a pattern of data that shows greater increases over time, creating a curve that gets steeper and steeper. It is characterized by the equation format \(y = a e^{bt}\), where \(e\) is the base of natural logarithms, and \(a\) and \(b\) are constants.
In the case of world population, this model suggests that the population grows continuously and exponentially with time \(t\). This means each increase builds on the previous amount, not a fixed base number. So, as the time increases, the rate of population growth also increases.
In the case of world population, this model suggests that the population grows continuously and exponentially with time \(t\). This means each increase builds on the previous amount, not a fixed base number. So, as the time increases, the rate of population growth also increases.
- The exponential model is particularly suitable for unrestricted environments, where resources are unlimited.
- Given the formula \(y = 6.4 e^{0.0132t}\), the constant 6.4 represents the initial world population in billions in 2004.
- The growth rate is 0.0132, indicating a steady but rapid increase as years pass by.
Logistic Growth
Logistic growth is a model used extensively to describe how populations grow in environments with limited resources. This type of growth is described by the formula \(y = \frac{a}{1 + be^{-ct}}\), developing a sigmoidal (S-shaped) curve.
Here, the logistic growth equation for the world population is given as \(y = \frac{102.4}{6 + 10 e^{-0.030t}}\). This model considers a carrying capacity, the maximum population size that an environment can sustain indefinitely.
Here, the logistic growth equation for the world population is given as \(y = \frac{102.4}{6 + 10 e^{-0.030t}}\). This model considers a carrying capacity, the maximum population size that an environment can sustain indefinitely.
- The logistic model begins with an initial population close to 6.4 billion in 2004, similar to the exponential model.
- As time progresses, the population grows rapidly at first and then slows as it approaches the carrying capacity.
- This accounts for limitations such as resource scarcity that restrict further growth.
World Population Prediction
When forecasting the future world population, it is essential to consider both exponential and logistic growth models together, as they provide different insights.
Exponential growth assumes a world with unlimited resources where population continues to increase rapidly. In contrast, logistic growth accounts for environmental carrying capacity, predicting a stabilization of the population after rapid early growth.
Exponential growth assumes a world with unlimited resources where population continues to increase rapidly. In contrast, logistic growth accounts for environmental carrying capacity, predicting a stabilization of the population after rapid early growth.
- The exponential model, using \(y = 6.4 e^{0.0132t}\), might predict ongoing and unchecked increase. For example, in the year 2090, it predicts the population could reach around 20.08 billion, showing a dramatic rise.
- The logistic model, using \(y = \frac{102.4}{6 + 10 e^{-0.030t}}\), suggests the population would rise to about 9.7 billion by the same year, illustrating a deceleration and approaching a limit due to resource constraints.
Other exercises in this chapter
Problem 47
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What conclusions can you draw about \(f\) from the information that \(f^{\prime}(c)=f^{\prime \prime}(c)=0\) and \(f^{\prime \prime \prime}(c)>0\) ?
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