Problem 20 - Calculus Concepts

Question

Plow Patents The number of patents issued for plow sulkies between 1865 and 1925 was increasing with respect to time at a rate jointly proportional to the number of patents already obtained and to the difference between the number of patents already obtained and the carrying capacity of the system. The carrying capacity was approximately 2700 patents, and the constant of proportionality was about \(7.52 \cdot 10^{-5} .\) By 1883,980 patents had been obtained. (Source: Hamblin, Jacobsen, and Miller, \(A\) Mathematical Theory of Social Change, New York: Wiley, 1973 ) a. Write a differential equation describing the rate of change in the number of patents with respect to the number of years since 1865 b. Write a general solution for the differential equation. c. Write the particular solution for the differential equation. d. Estimate the number of patents obtained by 1900 .

Step-by-Step Solution

Answer

By 1900, approximately 1376 patents were obtained.

Step 1: Describe the Differential Equation

The problem states that the rate of change in the number of patents is jointly proportional to two factors: the number of patents already obtained \( P \), and the difference between the carrying capacity \( K \), which is 2700, and \( P \). This can be expressed as the differential equation \( \frac{dP}{dt} = k \cdot P \cdot (K - P) \), where \( k = 7.52 \times 10^{-5} \).

Step 2: Write the Differential Equation

Given the above understanding, the differential equation that describes the rate of change in the number of patents is: \[\frac{dP}{dt} = 7.52 \times 10^{-5} \cdot P \cdot (2700 - P)\]

Step 3: Find the General Solution

To solve this differential equation, we perform separation of variables and integrate both sides. The separated equation is:\[\int \frac{1}{P (2700 - P)} \, dP = 7.52 \times 10^{-5} \int \, dt\]After integration, we use partial fraction decomposition to find the general solution, yielding:\[\ln \left| \frac{P}{2700 - P} \right| = 7.52 \times 10^{-5} t + C\]Solving for \( P \), we get:\[P(t) = \frac{2700}{1 + C e^{-7.52 \times 10^{-5} t}}\]

Step 4: Solve for the Particular Solution

To find the particular solution, use the initial condition that by 1883, 980 patents had been obtained. This means that at \( t = 18 \) (since 1883 is 18 years after 1865), \( P = 980 \). Substituting in the values:\[980 = \frac{2700}{1 + C e^{-7.52 \times 10^{-5} \times 18}}\]Solving for \( C \), we find:\[C \approx 1.7551\]So the particular solution is:\[P(t) = \frac{2700}{1 + 1.7551 e^{-7.52 \times 10^{-5} t}}\]

Step 5: Estimate the Number of Patents by 1900

To estimate the number of patents by 1900, calculate \( P \) when \( t = 35 \) (1900 is 35 years after 1865) using the particular solution:\[P(35) = \frac{2700}{1 + 1.7551 e^{-7.52 \times 10^{-5} \times 35}}\]Calculating this gives approximately:\[P(35) \approx 1376\]

Key Concepts

Carrying Capacity

Carrying capacity, in the context of differential equations related to population dynamics, is a vital concept. When applied to systems like the number of patents being filed, it represents the maximum number that the system can sustain over time. In our example, the carrying capacity represents a theoretical limit of 2700 patents. It implies that once this number is reached, the growth in the number of patents will slow down and eventually stop. This happens because resources become limited or demand plateaus, making it harder to sustain further increases.

• Carrying capacity is symbolized by the letter \( K \).
• It affects the rate of growth by setting an upper limit.
• It acts as a soft barrier that impacts how a system grows over time.

Understanding carrying capacity helps model real-life situations where growth isn't infinite, such as populations, market saturation, or potential patents in a technological sector.

Initial Conditions

Initial conditions are essential for finding particular solutions to a differential equation. They give us the specific values that a system or function will take on at a starting point in time. These values help hone in on precise solutions rather than generic or possible ones. For the patent example, the initial condition was 980 patents by 1883, which corresponds to 18 years since the reference year of 1865.

• Initial conditions are necessary for making exact predictions.
• They transform a general mathematical solution into a particular one.
• In the patent example, they provide the year-dependent anchor for calculating future patent numbers.

Without considering initial conditions, models might become overly abstract, lacking practical applicability for real-world predictions or estimations.

Separation of Variables

Separation of variables is a technique used to solve differential equations, especially when dealing with functions that depend on multiple variables. It involves rearranging an equation so that each type of variable is isolated on opposite sides of the equation, allowing integration to solve it.
This method is particularly useful for linear equations where multiplication of the dependents is involved. For our example, this meant isolating \( P \) from \( t \), the time.

• It simplifies complex differential equations involving proportions.
• The process involves rearranging and integrating both sides of the equation separately.
• It's a foundational method that helps uncover the general and particular solutions of problems.

This technique is crucial in applied mathematics, where closed-form solutions can be derived for otherwise difficult-to-solve equations.

Proportionality Constants

Proportionality constants are values that help describe the relationship between changes in variables within a system. They provide a measure that relates how much of one quantity will change in relation to another quantity. In the patent example, the constant \( k = 7.52 \times 10^{-5} \) indicates how the rate at which patents are issued grows over time, dependent on both the current number of patents and the carrying capacity.

• This constant regulates the speed or steepness of growth in the model.
• It's essential for ensuring the model reflects reality; without it, predictions and estimations would lack accuracy.
• These constants provide insights into system behaviors and growth dynamics.

The value of \( k \) helps predict how quickly a system approaches its carrying capacity and generally plays a central role in controlling the behavior of solutions over time.