Problem 63 - Calculus and its applications

Question

The city of New River had a population of 17,000 in \(2002(t=0)\) with a continuous growth rate of \(1.75 \%\) per year. a) Write the differential equation that represents \(P(t)\) the population of New River after \(t\) years. b) Find the particular solution of the differential equation from part (a). c) Find \(P(10)\) and \(P^{\prime}(10)\). d) Find \(P^{\prime}(10) / P(10)\), and explain what this number represents.

Step-by-Step Solution

Answer

a) \(\frac{dP}{dt} = 0.0175P\); b) \(P(t) = 17000e^{0.0175t}\); c) \(P(10) \approx 20203\), \(P'(10) \approx 353.6\); d) \(\frac{P'(10)}{P(10)} = 0.0175\), the growth rate.

Step 1: Formulate the Differential Equation

The growth of a population with a continuous growth rate can be modeled by the differential equation \(\frac{dP}{dt} = rP\), where \(r\) is the growth rate. Given a growth rate of \(1.75\%\), we convert it to a decimal: \(r = 0.0175\). Therefore, the differential equation is \(\frac{dP}{dt} = 0.0175P\).

Step 2: Solve the Differential Equation

The differential equation \(\frac{dP}{dt} = 0.0175P\) is separable. Separate variables and integrate: \(\frac{dP}{P} = 0.0175 \, dt\). Integrating both sides, we get \(\ln |P| = 0.0175t + C\), where \(C\) is the integration constant. Taking the exponential of both sides, \(P(t) = Ce^{0.0175t}\).

Step 3: Determine the Particular Solution

Use the initial condition \(P(0) = 17000\) to find \(C\). Substitute into the equation: \(17000 = Ce^{0.0175 \times 0}\). This simplifies to \(17000 = C\). Thus, the particular solution is \(P(t) = 17000e^{0.0175t}\).

Step 4: Calculate \(P(10)\)

To find the population after 10 years, substitute \(t = 10\) into the solution: \(P(10) = 17000e^{0.0175 \times 10}\). Compute this to find \(P(10) \approx 20203\).

Step 5: Calculate \(P^{\prime}(t)\)

Differentiate the solution \(P(t) = 17000e^{0.0175t}\) with respect to \(t\): \(P^{\prime}(t) = 17000 \times 0.0175 \times e^{0.0175t} = 297.5e^{0.0175t}\).

Step 6: Calculate \(P^{\prime}(10)\)

Substitute \(t = 10\) into the derivative: \(P^{\prime}(10) = 297.5e^{0.0175 \times 10}\), which evaluates to \(P^{\prime}(10) \approx 353.6\).

Step 7: Compute \(\frac{P^{\prime}(10)}{P(10)}\)

Use the values found in Steps 4 and 6: \(\frac{P^{\prime}(10)}{P(10)} = \frac{353.6}{20203}\). This simplifies to approximately \(0.0175\). This number represents the continuous growth rate, confirming that the population is growing at \(1.75\%\) per year.

Key Concepts

Population Growth

Population growth is a fascinating phenomenon that can be seen in various organisms, including human populations. When studying population growth in mathematics, especially in calculus and differential equations, we often rely on functions and models to predict how a population will change over time based on given factors. In mathematical terms, population growth is commonly modeled using differential equations, which can calculate the growth rate at any given moment.

For the city of New River, we learned the population in 2002 was 17,000 and it grew continuously at a rate of 1.75% per year. By understanding how populations change with this steady growth, we can project future population sizes and the corresponding change rates.

A key takeaway is that population growth isn't always linear. Instead, it can be exponential, meaning the growth becomes faster as the population gets larger. This can have significant implications for planning and resource management in any growing community.

Initial Value Problem

An initial value problem is a type of differential equation accompanied by specific conditions at the start of the observation period. This initial condition allows us to find a unique solution, addressing situations where multiple potential solutions might exist.

In our New River example exercise, the initial value was the population in the year 2002, set at 17,000. This is crucial as it helps us determine the constant needed in the population growth function. By plugging this initial value into our differential equation solution, we refine the function to reflect the city's specific population growth dynamics.

Here's a quick look at the process:

• Setup the differential equation describing the change: \ \( \frac{dP}{dt} = rP \)
• Integrate and solve for the general solution.
• Apply the initial condition to find the constant, giving us a particular solution.

This straightforward approach helps us develop accurate, tailored models for different initial value problems.

Exponential Growth

Exponential growth is a key concept in understanding how populations can rapidly increase under certain conditions. This growth pattern is observed when the change rate of a population is proportional to its current size, resulting in each new generation adding more than the previous one.

Mathematically, it is expressed by the equation: \ \( P(t) = Ce^{rt} \). In this equation:

• \(P(t)\) is the population at time \(t\).
• \(C\) is a constant that includes the initial population.
• \(e\) is the base of the natural logarithm, approximately 2.718.
• \(r\) is the growth rate.

The example of the city of New River highlights this with a consistent 1.75% growth. This means the percentage growth remains the same, but the number of new individuals added each year increases. As a result, the population grows faster over time, following the exponential curve.

Understanding exponential growth is crucial for predicting future trends, and it underscores the importance of making strategic plans in response to rapid population increases.