Problem 64 - Calculus Early Transcendentals

Question

Some species have growth rates that oscillate with an (approximately) constant period \(P .\) Consider the growth rate function $$N^{\prime}(t)=A \sin \left(\frac{2 \pi t}{P}\right)+r,$$ where \(A\) and \(r\) are constants with units of individuals/year. A species becomes extinct if its population ever reaches 0 after \(t=0\). a. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. b. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=100,\) does the population ever become extinct? Explain. c. Suppose \(P=10, A=50,\) and \(r=5 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. d. Suppose \(P=10, A=50,\) and \(r=-5 .\) Find the initial population \(N(0)\) needed to ensure that the population never becomes extinct.

Step-by-Step Solution

Answer

Answer: a. For \(P=10\), \(A=20\), \(r=0\), and \(N(0) = 10\): The population becomes extinct. b. For \(P=10\), \(A=20\), \(r=0\), and \(N(0) = 100\): The population does not become extinct. c. For \(P=10\), \(A=50\), \(r=5\), and \(N(0) = 10\): The population does not become extinct. d. For \(P=10\), \(A=50\), and \(r=-5\), we need to find \(N(0)\) such that the population never becomes extinct: \(N(0) = 25 - 5t\).

Step 1: Integrate the growth rate function

To find the population function \(N(t)\), we need to integrate the growth rate function \(N^{\prime}(t)\) with respect to time \(t\): $$ N(t) = \int N^{\prime}(t) dt = \int \left(A \sin\left(\frac{2 \pi t}{P}\right) + r\right) dt $$

Step 2: Solve the integral

Solving the integral gives the population function (Consider \(C\) to be the constant of integration): $$ N(t) = -\frac{A P}{2 \pi} \cos \left(\frac{2 \pi t}{P}\right) + rt + C $$

Step 3: Determine the constant C

Since we are given that the initial population is \(N(0)\), the constant of integration can be computed from the initial condition: $$ N(0) = -\frac{A P}{2 \pi} \cos \left(0\right) + r(0) + C $$ $$ N(0) = -\frac{A P}{2 \pi} + C $$ Thus, \(C = N(0) + \frac{A P}{2 \pi}\). Now, the population function is given by: $$ N(t) = -\frac{A P}{2 \pi} \cos \left(\frac{2 \pi t}{P}\right) + rt + N(0) + \frac{A P}{2 \pi} $$

Step 4: Analyze Population Extinction

To analyze the population extinction, we need to find the conditions when the population function \(N(t) = 0\). If this happens at any time after \(t=0\), the population becomes extinct. Now, we can apply the given values and check the conditions for each part. a. For \(P=10\), \(A=20\), \(r=0\), and \(N(0) = 10\): $$ N(t) = -10 \cos \left(\frac{2 \pi t}{10}\right) + 10 $$ The population oscillates between 0 and 20, thus, it becomes extinct. b. For \(P=10\), \(A=20\), \(r=0\), and \(N(0) = 100\): $$ N(t) = -10 \cos \left(\frac{2 \pi t}{10}\right) + 100 $$ The population oscillates between 90 and 110, thus, it does not become extinct. c. For \(P=10\), \(A=50\), \(r=5\), and \(N(0) = 10\): $$ N(t) = -25 \cos \left(\frac{2 \pi t}{10}\right) + 5t + 10 $$ As the population increases linearly with time, it does not become extinct. d. For \(P=10\), \(A=50\), and \(r=-5\), we need to find \(N(0)\) such that a population never becomes extinct: $$ N(t) = -25 \cos\left(\frac{2 \pi t}{10}\right) - 5t + N(0) $$ For the population never to become extinct, the minimum value of the cosine term should be equal to the maximum value of the linear term, \(5t\). Thus, \(N(0)\) should be equal to the difference between the maximum and minimum values: $$ N(0) = -25(1) - 5t + 25(1) $$ $$ N(0) = 25 - 5t $$ This ensures that the population never becomes extinct, as the term \(5t\) will always equal or exceed the minimum cosine term, keeping the population positive.

Key Concepts

Population Dynamics

Population dynamics studies how and why populations change over time. It provides insights into factors that control population size and growth. Populations can be affected by resources, space availability, and interaction with other species. Understanding these dynamics helps us predict future population sizes.
In the exercise provided, the concept of population dynamics is modeled using a differential equation. This equation includes parameters like oscillating growth rates that are affected by both natural and external factors. ##[Learn about core population dynamics](#)## to deepen your understanding of how different factors influence the rate of population growth and decay.

Oscillating Growth Rate

An oscillating growth rate describes populations that grow and shrink in a repetitive cycle over time. This cycle is influenced by various environmental factors such as seasonal changes or food availability.
The growth rate given by the equation \( N^{\prime}(t)=A \sin \left(\frac{2 \pi t}{P}\right)+r \) shows how the population's growth can oscillate between positive and negative values. Parameters \(A\) and \(r\) are crucial.

• \(A\) (amplitude) determines how drastically the population fluctuates.
• \(r\) (average rate) shifts the graph vertically.

Special attention is required when \(r = 0\), as it indicates the population's tendency to return to zero, leading to extinction. To avoid extinction, managing these oscillations by adjusting parameters is essential.

Calculus Integration

Integration is the process of finding the antiderivative of a function. In population dynamics, it transforms a growth rate function into a population function that predicts population size over time.
For the given differential equation of the growth rate \( N^{\prime}(t)=A \sin \left(\frac{2 \pi t}{P}\right)+r \), integrating it provides the population function \( N(t) \). Integration turns the complex oscillating rate into a usable formula.The result, \(-\frac{A P}{2 \pi} \cos \left(\frac{2 \pi t}{P}\right) + rt + C\), gives insights into:

• Long-term trends when the sine wave balances out over time.
• The real-world initial conditions affecting growth via constant \(C\).

Understanding calculus integration is key to solving these real-world problems.

Initial Value Problem

Initial value problems in differential equations deal with finding a specific solution given an initial condition. These problems help tailor the general solution to fit real-world situations, such as predicting future population based on starting data.
In the exercise, initial conditions like \(N(0)\) are used to determine the integration constant \(C\). This ensures that the solution satisfies the condition \(N(0)=\text{{Initial Population}}\).Given \( N(t) = -\frac{A P}{2 \pi} \cos \left(\frac{2 \pi t}{P}\right) + rt + N(0) + \frac{A P}{2 \pi} \), setting initial values allows us to predict whether populations will persist or go extinct. Tools like these are vital for scientists and researchers aiming to preserve species or understand ecosystem dynamics.