Problem 1 - Applied Calculus

Question

If \(t\) is in years since \(1990,\) one model for the population of the world, \(P,\) in billions, is $$P=\frac{40}{1+11 e^{-0.08 t}}$$ (a) What does this model predict for the maximum sustainable population of the world? (b) Graph \(P\) against \(t\). (c) According to this model, when will the earth's population reach 20 billion? 39.9 billion?

Step-by-Step Solution

Answer

(a) 40 billion. (b) Use a graphing tool. (c) Year 2022 for 20 billion, 2055 for 39.9 billion.

Step 1: Analyze the Model for Maximum Population

The model given is \( P = \frac{40}{1+11e^{-0.08t}} \). As \( t \to \infty \), the exponential term \( e^{-0.08t} \to 0 \), simplifying the equation to \( P = \frac{40}{1+0} = 40 \). Therefore, the maximum sustainable population predicted by this model is 40 billion.

Step 2: Set Up the Graph of P against t

To graph \( P \) against \( t \), you will want to consider various values of \( t \) to compute corresponding \( P \) values. This helps visualize the change in population over time. Consider using a graphing software or calculator to plot this relationship effectively from \( t = 0 \) onwards.

Step 3: Solve for When Population Reaches 20 Billion

To find when \( P = 20 \), set the equation to 20: \( \frac{40}{1+11e^{-0.08t}} = 20 \). Multiply both sides by \( 1 + 11e^{-0.08t} \), then solve: \( 20 + 220e^{-0.08t} = 40 \), leading to \( 220e^{-0.08t} = 20 \). Solving gives \( e^{-0.08t} = \frac{1}{11} \), yielding \( t \approx \frac{-\ln(1/11)}{0.08} \). Solving this gives approximately \( t = 32 \). Hence, around the year 2022.

Step 4: Solve for When Population Reaches 39.9 Billion

For \( P = 39.9 \), set the equation to 39.9: \( \frac{40}{1+11e^{-0.08t}} = 39.9 \). Simplifying, \( 39.9(1+11e^{-0.08t}) = 40 \), leads to \( 39.9 + 438.9e^{-0.08t} = 40 \), thus \( 438.9e^{-0.08t} = 0.1 \). Solving gives \( e^{-0.08t} = \frac{0.1}{438.9} \), \( t \approx \frac{-\ln(0.1/438.9)}{0.08} \). This simplifies to \( t \approx 65 \), around the year 2055.

Key Concepts

Exponential Functions

Exponential functions often describe processes that change at rates proportional to their current value, making them essential for modeling phenomena like population growth. In population models, the function \(P = \frac{40}{1+11e^{-0.08t}}\) is an example of how exponential growth can be contained by some other limiting factor, represented here by the denominator.
When you see terms like \(e^{-0.08t}\) in the equation, the exponential function \(e^{x}\) tells us how the population changes over time. The constant \(-0.08\) represents how quickly population grows or shrinks, and as time \(t\) increases, this exponential term shrinks because negative exponentials tend to zero. This means that while growth may be quick initially, it slows as more time passes.
Studying these kinds of problems helps us understand that exponential functions are not always unlimited and, in our equation, the population approaches a "carrying capacity" or maximum sustainable level as \(t\) becomes very large.

Logarithmic Calculations

Logarithmic calculations are essential when you need to solve for time \(t\) given a specific population size \(P\). In our step-by-step solutions, logarithms help us make sense of when exactly certain population milestones are hit.
When finding out when the population will reach 20 billion, we have the equation \(\frac{40}{1+11e^{-0.08t}} = 20\). To solve for \(t\), isolating the exponential term requires us to rearrange terms and use logarithms. Solving \(e^{-0.08t} = \frac{1}{11}\) requires a logarithmic calculation: \(t = \frac{-\ln{(1/11)}}{0.08}\).
Logarithms invert exponential functions. Essentially, they tell us what power we need to raise \(e\) (approximately 2.718) to get a specific value. With logarithms like \(\ln \), we decode the timeline of growth, understanding not just when, but also how exponential processes take shape over time.

Graphing Population Data

Graphing population data visually represents how the population changes with time. It's a practical way to interpret the exponential model given by \(P = \frac{40}{1+11e^{-0.08t}}\). In such graphs, time \(t\) is plotted along the x-axis, while population \(P\) in billions goes on the y-axis.
When you graph this model, you see the characteristic S-shaped curve of logistic growth, where:

• The initial section rises steeply showing rapid growth.
• It then slows and flattens as it approaches the maximum population limit (40 billion).
• This section of the curve represents the slowing down due to limiting factors such as resources or space.

Graphing helps us visualize not only the current population status but also predict future scenarios. By seeing where the line flattens, we can easily observe when the population growth plateaus, indicating the maximum sustainability according to the model. Use tools like graphing calculators or software to make these plots more detailed, which enhances our understanding of population trends over time.