Problem 17 - Applied Calculus

Question

The amount of land in use for growing crops increases as the world's population increases. Suppose \(A(t)\) represents the total number of hectares of land in use in year t. (A hectare is about \(2 \frac{1}{2}\) acres.) (a) Explain why it is plausible that \(A(t)\) satisfies the equation \(A^{\prime}(t)=k A(t) .\) What assumptions are you making about the world's population and its relation to the amount of land used? (b) In 1950 about \(1 \cdot 10^{9}\) hectares of land were in use; in 1980 the figure was \(2 \cdot 10^{9} .\) If the total amount of land available for growing crops is thought to be \(3.2 \cdot 10^{9}\) hectares, when does this model predict it is exhausted? (Let \(t=0\) in \(1950 .\) )

Step-by-Step Solution

Answer

The land will be exhausted around the year 2000, based on the model.

Step 1: Understanding the Model

The equation \(A'(t) = kA(t)\) suggests an exponential growth model for land usage. It assumes that the rate of change in land usage is proportional to the current land in use. This implies that as the population grows, the land usage scales up proportionally to accommodate the increase in population.

Step 2: Setting Initial Conditions

We know from the problem that in 1950, \(A(0) = 1 \times 10^9\) hectares and in 1980, \(A(30) = 2 \times 10^9\) hectares. We will use these initial conditions to solve for the constant \(k\).

Step 3: Solving the Differential Equation

The solution to the differential equation \(A'(t) = kA(t)\) is given by \(A(t) = A(0) e^{kt}\). Substituting the initial condition \(A(0) = 1 \times 10^9\), we have \(A(t) = 1 \times 10^9 e^{kt}\).

Step 4: Finding the Growth Rate \(k\)

Substitute \(A(30) = 2 \times 10^9\) into the equation \(1 \times 10^9 e^{30k} = 2 \times 10^9\). Simplifying gives \(e^{30k} = 2\), leading to \(30k = \ln(2)\). Thus, \(k = \frac{\ln(2)}{30}\).

Step 5: Predicting Land Exhaustion

To find when land usage will reach \(3.2 \times 10^9\) hectares, we set \(1 \times 10^9 e^{kt} = 3.2 \times 10^9\). This gives \(e^{kt} = 3.2\), or \(kt = \ln(3.2)\). Substitute \(k = \frac{\ln(2)}{30}\) to solve for \(t\), obtaining \(t = \frac{30 \ln(3.2)}{\ln(2)}\).

Step 6: Calculating \(t\) and Converting to Year

Calculate \(t\), which represents the number of years after 1950. \(t \approx 1950 + \frac{30(1.163)}{0.693} \approx 1950 + 50.3\), rounding down gives \(t \approx 2000\).

Key Concepts

Differential Equations

A differential equation is a mathematical tool used to describe how a quantity changes with respect to another. In the context of exponential growth, it helps us model how variables increase over time. Here's an equation we often see:\[ A'(t) = kA(t) \]This equation tells us that the growth rate of a variable, depicted as \( A'(t) \) (the derivative of \( A(t) \)), is proportional to the current size of that variable \( A(t) \). The constant \( k \) indicates the rate of growth.
The implication here is that as the land used for agriculture grows, it spurs more growth. Thus, the more land in use, the faster land usage increases. This is typical of exponential growth where resources like land expand significantly as populations grow.
The beauty of differential equations in this case is they allow us to predict future scenarios based on present data. Engineers, economists, and scientists use them regularly for modeling and forecasting.

Initial Conditions

Initial conditions are crucial in solving differential equations. They provide the starting point or baseline that helps plot out the entire scenario. In our problem, initial conditions were known:

• In 1950: \( A(0) = 1 \times 10^9 \) hectares
• In 1980: \( A(30) = 2 \times 10^9 \) hectares

These conditions provide us with critical snapshots of land usage at specific times, letting us calculate the growth rate \( k \). From the model \( A(t) = A(0) e^{kt} \), we know the function spreads over time like how brushfire spreads over grassland.
Once \( k \) is found, we can predict the amount of land used at any other time \( t \). Such models are very powerful, allowing us to see and understand how changes over time fit within practical, real-world limits.

Land Usage Modeling

Land usage modeling is pivotal in understanding how human activities impact available arable land. As populations grow, the demand for food increases, leading to more land cultivated for crops. The model we use assumes that land usage grows exponentially with population, which has profound implications.- It predicts a future point where a maximum land capacity might be reached if trends continue.- Decision-makers, like urban planners and agronomists, use it to forecast and prepare against potential shortages.The solution to our differential equation tells us when land might become fully used based on past data—aiming for a year when \( 3.2 \times 10^9 \) hectares may be needed.
These models are not perfect but can stimulate smarter land-use policies, ensuring sustainability. In a world of finite resources, they nudge us towards balance, innovation, and efficiency.