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The per capita growth rate of a population of cells varies over the course of a day. Assume that time \(t\) is measured in hours and $$ \frac{d N}{d t}=2\left(1-\cos \frac{2 \pi t}{24}\right) N $$ if \(N(0)=5\), find the number of cells after one day (that is, find \(N(24))\)

Short Answer

Expert verified
The number of cells after one day is extremely large: \(5e^{48}\).

Step by step solution

01

Understand the Problem

We are given a differential equation \(\frac{dN}{dt} = 2\left(1 - \cos \frac{2\pi t}{24}\right)N\) with an initial condition \(N(0) = 5\). Our task is to find \(N(24)\), or the number of cells after one day.
02

Recognize the Differential Equation Type

This equation is a first-order linear differential equation, where \(\frac{dN}{dt} = g(t)N\), with \(g(t) = 2\left(1 - \cos \frac{2\pi t}{24}\right)\). We can solve it by separating variables or using an integrating factor.
03

Use the Integrating Factor

The general solution of a linear differential equation \(\frac{dN}{dt} + P(t)N = 0\) is \(N(t) = N(0) e^{\int g(t) dt}\). Identify the integrating factor as \(e^{\int 2(1-\cos \frac{2\pi t}{24}) dt}\).
04

Compute the Integral

Calculate the integral \(\int 2(1 - \cos \frac{2\pi t}{24}) \, dt\). First, separate terms into \(\int 2\, dt - \int 2 \cos \frac{2\pi t}{24} \, dt\). This yields \(2t - 48\sin \frac{2\pi t}{24}\).
05

Substitute and Solve for N(t)

Substitute the result into the expression for \(N(t)\): \(N(t) = 5 e^{2t - 48 \sin \frac{2\pi t}{24}}\). Simplify for \(t = 24\): \(N(24) = 5 e^{48}\).
06

Compute Final Value

Calculate \(N(24) = 5 e^{48}\), which is a very large number indicating exponential growth. Without explicit computation of \(e^{48}\), understand that the number of cells is extremely large after a day.

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Key Concepts

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

Per Capita Growth Rate
The term "per capita growth rate" refers to the change in population size relative to the existing number of individuals. It's a measure of how the population's size changes over time.
In mathematical terms, the per capita growth rate is often expressed as \( \frac{1}{N} \frac{dN}{dt} \). This gives us the rate of change per individual in the population, allowing us to understand the growth dynamic on a per-individual basis rather than focusing on absolute numbers.
In the given problem, the rate \( 2(1-\cos \frac{2\pi t}{24}) \) affects how quickly the cell population increases during the day.
  • The component \( \cos \frac{2\pi t}{24} \) introduces periodicity, meaning that the growth rate changes at different times of the day.
  • Multiplying by \( 2 \) scales this periodic growth, modifying the intensity.
This approach allows for modeling a realistic environment where growth rates aren't static. Such insights are crucial in studies of ecology and other fields that involve dynamic systems.
Population Dynamics
Population dynamics is the study of how populations change over time under the influence of births, deaths, immigration, and emigration. It encompasses the actions and reactions of population components that lead to different growth patterns.
Understanding population dynamics is essential for managing resources, conserving species, and predicting how a population might respond to environmental changes.
  • The model given in the exercise introduces time-dependent changes in growth, i.e., the per capita growth rate is a function of time.
  • This showcases a dynamic system where parameters like growth are influenced by both internal biological factors and periodic external forces.
By incorporating periodic changes, like the cosine component seen in our differential equation, we simulate scenarios more closely aligned with real-world patterns—such as daily cycles in cellular activities or behaviors linked to environmental changes.
First-order Linear Differential Equation
First-order linear differential equations are equations that involve the first derivative of a function and are linear in the dependent variable. These equations are foundational in modeling natural processes because they often describe systems experiencing change over time.
The given differential equation \( \frac{dN}{dt} = 2(1-\cos \frac{2\pi t}{24})N \) falls under this category. Here's why it's crucial:
  • First-order implies the equation involves the first derivative—here, \( \frac{dN}{dt} \).
  • Linear means that \( N \) (the number of cells in this context) appears without exponents or other non-linear operations.
To solve this kind of differential equation, we frequently use methods like separation of variables or integrating factors.
In the solution outlined, an integrating factor is used. This approach involves multiplying through by a function derived from \( g(t) \), which transforms our equation so that it can be directly integrated.
This method provides a general solution that reflects how cell populations grow under the specified conditions, and these solutions help predict behaviors of the modeled systems in diverse fields, such as biology, economics, and engineering.

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Most popular questions from this chapter

A cell constantly gains or loses small molecules to its environment because the small molecules are able to diffuse through the cell membrane. We will build a model for this process. Suppose a molecule is present in the cell at a concentration \(C(t)\), and present in its environment at a concentration \(C_{\infty}\) (you may assume \(C_{\infty}\) is a constant). One model for the diffusion of molecules across the cell membrane is that the rate at which molecules travel through the membrane is proportional to the difference in concentration between the cell and its surroundings. That is: Rate at which $$ \text { molecules flow out }=k\left(C-C_{\infty}\right) $$ of cell The constant \(k\) is known as the permeability of the membrane: \(k>0\), and \(k\) depends on the surface area of the cell and the chemistry of the membrane, as well as the type of molecule. (a) Starting with a word equation for the amount of small molecules in the cell, show, if the cell volume is \(V\), then: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right) $$ (b) Find the equilibrium of \((8.53)\) and use a graphical analysis to determine whether it is stable or unstable. (c) Suppose that the molecule we are studying is produced within the cell. The cell produces the molecule at a rate \(r\); that is, a quantity \(r\) is produced (added to the cell) in unit time. Explain why the differential equation for the concentration of molecules in the cell should be modified to: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right)+\frac{r}{V} $$ (d) Analyze Equation (8.54) to find the equilibrium value of the cell concentration. Is this equilibrium stable or unstable? You may use a graphical argument or calculate the eigenvalue to determine the equilibrium's stability.

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d x}{d t}=x^{2}-h x $$

Suppose that \(N(t)\) denotes the size of a population at time \(t\). The population evolves according to the logistic equation, but in addition, predation reduces the size of the population so that the rate of change is given by $$ \frac{d N}{d t}=g(N) $$ where $$ g(N)=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N} $$ The first term on the right-hand side describes the logistic growth; the second term describes the effect of predation. (a) Make the vector field plot for this differential equation. (b) Find all equilibria of \((8.38)\). (c) Use your vector field plot in (a) to determine the stability of the equilibria you found in (b). (d) Repeat your analysis from part (c) but now use the method of eigenvalues to determine the stability of the equilibria you found in (b).

One of the key ideas for sketching solutions from vector field plots is that a solution curve must be monotonic, that is, \(x(t)\) is either increasing or decreasing or constant but cannot switch from one behavior to another. We showed that a solution \(x(t)\) could not start by increasing and then switch to decreasing. Suppose that \(x(t)\) is a solution of the differential equation \(\frac{d x}{d t}=g(x)\) and that \(x(t)\) starts off decreasing with time. Show that \(x(t)\) cannot switch to increasing.

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. \(\frac{d x}{d t}=h x-x^{2}\), where \(h\) is a constant and (a) \(h>0\), (b) \(h<0\)

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