Problem 28 - Calculus

Question

According to an article in The New York Times,' ' pigweed has acquired resistance to the weedkiller Roundup. Let \(N\) be the number of acres, in millions, where Roundup-resistant pigweed is found. Suppose the relative growth rate, \((1 / N) d N / d t,\) was \(15 \%\) when \(N=5\) and \(14.5 \%\) when \(N=10 .\) Assuming the relative growth rate is a linear function of \(N,\) write a differential equation to model \(N\) as a function of time, and predict how many acres will eventually be afflicted before the spread of Roundup-resistant pigweed halts.

Step-by-Step Solution

Answer

The model predicts that 155 million acres will eventually be afflicted before the spread halts.

Step 1: Define the Growth Rate Function

We are given two relative growth rates at specific values of \(N\): 15% when \(N = 5\) and 14.5% when \(N = 10\). We assume the relative growth rate \(r\) is a linear function of \(N\), i.e., \(r(N) = aN + b\). Using the given data, we can set up two equations: \(0.15 = 5a + b\) and \(0.145 = 10a + b\).

Step 2: Solve for Coefficients a and b

We need to solve the system of equations for \(a\) and \(b\). Subtract the first equation from the second:\[(10a + b) - (5a + b) = 0.145 - 0.15\]Simplifying, \[5a = -0.005 \ \Rightarrow a = -0.001\]Substitute \(a\) back into the first equation to find \(b\):\[0.15 = 5(-0.001) + b \ \Rightarrow b = 0.155\]

Step 3: Write the Relative Growth Rate Function

The linear function for the relative growth rate is \(r(N) = -0.001N + 0.155\). This represents the growth rate as a function of acres \(N\).

Step 4: Develop the Differential Equation for N

We know the relative growth rate is \(\frac{1}{N} \frac{dN}{dt} = r(N)\). So, the differential equation becomes:\[\frac{1}{N} \frac{dN}{dt} = -0.001N + 0.155\]which we can rewrite as\[\frac{dN}{dt} = (-0.001N + 0.155)N\]Thus, the differential equation is \(\frac{dN}{dt} = -0.001N^2 + 0.155N\).

Step 5: Determine the Limiting Value for N

The spread of Roundup-resistant pigweed will halt when the growth rate is zero: \(-0.001N + 0.155 = 0\). Solving for \(N\), we get:\[-0.001N + 0.155 = 0 \ \Rightarrow N = \frac{0.155}{0.001} = 155\]Therefore, the maximum number of acres that will be affected is 155 million.

Key Concepts

Relative Growth Rate

The concept of the relative growth rate is a crucial aspect when studying population dynamics. In the context of our example, the relative growth rate measures how quickly pigweed acreage increases in relation to the current total acreage. To calculate this, we use the formula \( \frac{1}{N} \frac{dN}{dt} \), where \( N \) represents the current acreage. This value essentially tells us what portion of the existing acreage will be added as new acreage over a unit of time.
Let's break this down a bit:

• If the relative growth rate is 15%, it means an additional 15% of the current acreage is expected to grow over the next time period.
• The given exercise assumes this growth rate changes linearly with the acreage, meaning as \( N \) changes, the growth rate linearly declines.

For example, when \( N = 5 \), the rate is 15%, and when \( N = 10 \), it drops slightly to 14.5%. This decline indicates that as the pigweed spreads, the percentage of acreage growth relative to the current size decreases slightly.

Linear Function

A linear function is one of the simplest forms of any mathematical function, expressed as \( f(x) = ax + b \). Here, it forms a straight line when graphed. In the context of the problem provided, the linear function describes how the relative growth rate \( r(N) \) changes with respect to the acreage \(N\).
In our exercise, we establish the linear function \( r(N) = -0.001N + 0.155 \) using the given data points:

• 15% at \( N = 5 \)
• 14.5% at \( N = 10 \)

This tells us two things:

• The slope \(a = -0.001\) signifies a slight decrease in the growth rate as the acreage expands.
• The y-intercept \(b = 0.155\) indicates the initial growth rate when \(N\) is zero.

Thus, by fitting this linear equation to the data, we can predict how the growth rate decreases as the acreage becomes larger. It helps in modeling the scenario more realistically.

Modeling Population Growth

Modeling population growth often involves differential equations, allowing us to understand how populations change over time. In this context, the population we're interested in is the acreage of pigweed resistant to Roundup. We are tasked with forming a differential equation based on the relative growth rate, which is portrayed as a function of \( N \), the acreage.
From the solution, the differential equation is derived as \( \frac{dN}{dt} = (-0.001N + 0.155)N \). This equation models the rate of change of \( N \) over time, influenced by the current acreage.

• The term \(-0.001N^2\) shows that the rate of increase decreases as the acreage becomes larger.
• The term \(0.155N\) reflects the initial rate of increase without taking the limit into consideration.

The final step is to determine the limiting value of \( N \), which is done by setting the growth rate to zero: \(-0.001N + 0.155 = 0\). Solving this gives us a maximum acreage value, beyond which spread stops (i.e., 155 million acres). This type of model assists in predicting long-term outcomes, allowing for better resource and agricultural management.