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Chapter 9: Problem 23

In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model $$ \frac{d P}{d t}=k P \cos (r t-\phi) \quad P(0)=P_{0} $$ where \(k, r,\) and \(\phi\) are positive constants. (b) By graphing the solution for several values of \(k, r,\) and \(\phi,\) explain how the values of \(k, r,\) and \(\phi\) affect the solution. What can you say about \(\lim _{t \rightarrow \infty} P(t) ?\)

Short Answer

Expert verified
The solution is \( P(t) = P_0 \exp \left( \frac{k}{r} \sin(rt - \phi) \right) \). The parameters affect oscillation amplitude, frequency, and phase shift. \( P(t) \) remains bounded as \( t \to \infty \).

Step by step solution

01

Understanding the Differential Equation

We are given a differential equation \( \frac{dP}{dt} = kP \cos(rt - \phi) \). This equation relates the rate of change of a population \( P \) with respect to time \( t \) to a function involving \( \cos \), which accounts for seasonal variations. The constants \( k, r, \) and \( \phi \) will define the characteristics of these variations.
02

Using Separation of Variables

To solve the differential equation, we use the method of separation of variables. We rearrange terms to get \( \frac{1}{P} dP = k \cos(rt - \phi) dt \). This allows us to separate the variables \( P \) and \( t \) so that we can integrate both sides independently.
03

Integrating Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{1}{P} dP = \ln |P| + C \). The right side becomes \( \int k \cos(rt - \phi) dt = \frac{k}{r} \sin(rt - \phi) + C_1 \). These integrals provide us with an expression for \( P(t) \).
04

Solving for P(t)

Since both sides are equal, we can equate them: \( P(t) = P_0 \exp \left( \frac{k}{r} \sin(rt - \phi) \right)\). This equation gives us the solution to the differential equation, satisfying the initial condition \(P(0) = P_0\).
05

Analyzing the effects of k, r, and \phi

The parameter \( k \) affects the amplitude of the oscillations in growth rate. Higher \( k \) results in greater fluctuations. The parameter \( r \) affects the frequency of oscillations; higher \( r \) leads to faster oscillations. The phase shift \( \phi \) affects the starting point of the oscillation in time. Graphically, these parameters adjust how quickly and intensely the population grows and declines over time.
06

Determining Limiting Behavior as t 鈫� 鈭�

We analyze \( \lim_{t \to \infty} P(t) \). Since \( \sin(rt - \phi) \) oscillates between -1 and 1, \( \exp(\frac{k}{r} \sin(rt - \phi)) \) oscillates but remains bounded above by \( e^{\frac{k}{r}} \). The population \( P(t) \) will not diverge to infinity but instead exhibit bounded oscillations indefinitely.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve unknown functions and their derivatives. In the context of population dynamics, they describe how the population changes over time.
In our seasonal-growth model, the differential equation is \( \frac{dP}{dt} = kP \cos(rt - \phi) \). Here, \( P \) is the population, \( t \) is time, and \( k, r, \phi \) are constants that affect growth.
The equation is structured to capture the effect of seasonal variations on growth, represented by the cosine function. The coefficient \( k \) scales this effect, while \( r \) and \( \phi \) adjust its periodic nature. This intricate relationship makes differential equations essential for modeling complex dynamical systems.
Separation of Variables
The method of separation of variables is a technique for solving differential equations. It involves rearranging an equation so that each variable is on a different side of the equation, making it easier to integrate.
For our model, we rearrange the equation \( \frac{dP}{dt} = kP \cos(rt - \phi) \) as \( \frac{1}{P} dP = k \cos(rt - \phi) dt \). By "separating" \( P \) and \( t \), we can integrate each side separately.
This method allows us to solve for \( P(t) \), giving a clear expression of how the population evolves over time. It's a powerful tool not just limited to population dynamics, but applicable to various differential equations encountered in science and engineering.
Periodic Functions
Periodic functions repeat their values at regular intervals, and they are central to modeling cyclical phenomena. In the seasonal-growth model, the function \( \cos(rt - \phi) \) is periodic, affected by the constants \( r \) and \( \phi \).
The constant \( r \) alters the frequency of cycles. A higher \( r \) means more cycles per unit time, while \( \phi \) introduces a phase shift, modifying when these cycles start.
This periodic feature is crucial in representing real-world scenarios where population growth does not occur at a constant rate but varies seasonally. As a result, the growth pattern can mirror natural rhythms, like mating seasons or food availability.
Population Dynamics
Population dynamics studies how and why the number of individuals in a population changes over time. The seasonal-growth model is a brilliant example, showing how populations are influenced by external factors, such as seasonal variations in resources or environmental conditions.
In the model, the growth rate changes periodically due to the cosine function, reflecting realistic scenarios of boom and bust cycles in natural populations. By analyzing this model, we can predict how populations might grow, decline, or stabilize over time.
Parameters like \( k \), \( r \), and \( \phi \) allow us to simulate different situations, adjusting how severe or frequent these changes are. Understanding these dynamics is vital in ecology, conservation, and resource management to make informed decisions about species and ecosystems.

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Most popular questions from this chapter

Lynx eat snowshoe hares and snowshoe hares eat woody plants like willows. Suppose that, in the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially. In the absence of lynx and willow the hare population will decay exponentially. If \(L(t), H(t),\) and \(W(t)\) represent the populations of these three species at time \(t\) write a system of differential equations as a model for their dynamics. If the constants in your equation are all positive, explain why you have used plus or minus signs.

Determine whether the differential equation is linear. $$ y^{\prime}-x=y \tan x $$

A tank contains \(1000 \mathrm{L}\) of pure water. Brine that contains \(0.05 \mathrm{kg}\) of salt per liter of water enters the tank at a rate of \(5 \mathrm{L} / \mathrm{min}\). Brine that contains \(0.04 \mathrm{kg}\) of salt per liter of water enters the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\). The solution is kept thoroughly mixed and drains from the tank at a rate of \(15 \mathrm{L} / \mathrm{min}\). How much salt is in the tank (a) after \(t\) minutes and (b) after one hour?

Solve the initial-value problem. $$ t^{3} \frac{d y}{d t}+3 t^{2} y=\cos t, \quad y(\pi)=0 $$

(a) Program your computer algebra system, using Euler's method with step size \(0.01,\) to calculate \(y(2),\) where \(y\) is the solution of the initial-value problem \(y^{\prime}=x^{3}-y^{3} \quad y(0)=1\) (b) Check your work by using the CAS to draw the solution curve.
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