Problem 66 - Calculus and its applications

Question

Suppose 30 sparrows are released into a region where they have no natural predators. The growth of the region's sparrow population can be modeled by the uninhibited growth model \(d P / d t=k P\) where \(P(t)\) is the population of sparrows \(t\) years after their initial release. a) When the sparrow population is estimated at \(12,500,\) its rate of growth is about 1325 sparrows per year. Use this information to find \(k,\) and then find the particular solution of the differential equation. b) Find the number of sparrows after 70 yr. c) Without using a calculator, find \(P^{\prime}(70) / P(70)\).

Step-by-Step Solution

Answer

a) The growth constant \(k\) is approximately 0.106. b) After 70 years, the sparrow population is approximately 50,078. c) \( \frac{P'(70)}{P(70)} = 0.106 \).

Step 1: Setup the differential equation

The problem gives us a model for the sparrow population growth: \(\frac{dP}{dt} = kP\). We know that when the population \(P\) is 12,500, the rate of growth \(\frac{dP}{dt}\) is 1325 sparrows per year.

Step 2: Find the constant of proportionality \(k\)

Use the given rate of growth to find \(k\). Substitute \(P = 12500\) and \(\frac{dP}{dt} = 1325\) into the equation \(\frac{dP}{dt} = kP\):\[ 1325 = k \times 12500 \]Solve for \(k\):\[ k = \frac{1325}{12500} = 0.106 \] (approximately)

Step 3: Find the particular solution

The differential equation \(\frac{dP}{dt} = kP\) has the general solution \(P(t) = P_0 e^{kt}\), where \(P_0\) is the initial population. Here, \(P_0 = 30\) and we have already found \(k = 0.106\). Therefore, the particular solution is:\[ P(t) = 30 e^{0.106t} \]

Step 4: Calculate the population after 70 years

Use the particular solution to find \(P(70)\):\[ P(70) = 30 e^{0.106 \times 70} \]Calculate the exponent:\[ 0.106 \times 70 = 7.42 \]Then:\[ P(70) = 30 e^{7.42} \]Using the fact that \( e^{7.42} \approx 1669.270 \), we find:\[ P(70) = 30 \times 1669.270 \approx 50077.68 \]

Step 5: Calculate \( \frac{P'(70)}{P(70)} \)

The expression \( \frac{P'(t)}{P(t)} \) simplifies, by using \( P(t) = 30 e^{0.106t} \), to:\[ \frac{dP}{dt} = 0.106 P(t) \ldots\text{thus} \ldots \frac{P'(t)}{P(t)} = 0.106 \]So, \( \frac{P'(70)}{P(70)} = 0.106 \).

Key Concepts

Exponential Growth

Exponential growth refers to a situation where the growth rate of a population is proportional to its current size. This type of growth can be modeled by the equation \( \frac{dP}{dt} = kP \), where \( P(t) \) is the size of the population at time \( t \) and \( k \) is the growth constant. In the context of our sparrow example, this means that as the number of sparrows increases, the number of new sparrows born within the same time frame also increases proportionally. This concept is crucial for understanding how populations can rapidly increase when there are no limiting factors such as predators or limited food resources. When visualized on a graph, exponential growth appears as a curve that slopes upward steeply as time progresses. Understanding exponential growth helps in predicting future population sizes and making informed decisions in managing wildlife or comparing historical data to current trends.

Rate of Change

The rate of change in population dynamics refers to how quickly the population size changes over a certain period of time. For the sparrows, the rate of change is described mathematically by the derivative \( \frac{dP}{dt} \). In our exercise, when the sparrow population was 12,500, the rate at which the sparrows increased was 1325 sparrows per year. This information allowed us to determine the constant \( k \) in the exponential growth model. We found that \( k = 0.106 \), reflecting the proportional relationship between the growth rate and current population size. This measure is crucial because it helps in understanding the speed and nature of population changes, which is beneficial for developing strategies for population control or for further scientific modeling.

Initial Value Problem

An initial value problem in differential equations involves finding a particular solution that satisfies a given initial condition. In our sparrow problem, the initial value is the number of sparrows released, which is 30. This serves as \( P_0 \) in the general solution of the differential equation \( P(t) = P_0 e^{kt} \).By using the initial value in conjunction with the calculated growth constant \( k \), we derived a specific solution for our sparrow population: \( P(t) = 30 e^{0.106t} \). These problems are common in modeling real-world situations where we need to determine future outcomes based on starting conditions. Solving initial value problems allows us to tailor a general formula to reflect specific scenarios.

Population Dynamics

Population dynamics refers to the study of how and why population sizes change over time. Key factors include birth rates, death rates, immigration, and emigration. Our sparrow population example specifically examines the influence of exponential growth on population dynamics without the influence of predators, illustrating an idealized scenario where the population can grow uninhibitedly. In 70 years, our calculations show the sparrow population growing dramatically to approximately 50,078 sparrows from an initial 30 sparrows, underlining the power of exponential growth in population dynamics. This concept is integral for conservation biologists and ecologists as it helps predict trends, assess potential risks, and develop strategies for sustainable management of wildlife populations. It allows scientists to simulate different scenarios by adjusting various factors impacting population and observing their effects over time.