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In Problems , give a formula for \(N(t), t=0,1,2, \ldots\), on the basis of the information provided. \(N_{0}=1\); population doubles every 40 minutes; one unit of time is 80 minutes

Short Answer

Expert verified
The formula for the population at time \(t\) is \(N(t) = 4^t\).

Step by step solution

01

Define Variables and Initial Conditions

The problem states that the initial population, denoted as \(N_0\), is 1. The population is known to double every 40 minutes.
02

Determine the Growth Rate per Unit of Time

Since the population doubles every 40 minutes, it can be expressed using a growth rate. We are told that a unit of time equals 80 minutes. Thus, in one unit of time (80 minutes), the population will double twice: once at 40 minutes and once at 80 minutes.
03

Calculate the Effective Growth Factor per Unit of Time

In each 40-minute period, the population doubles, so the growth factor is 2 for each doubling period. Since a unit of time (80 minutes) includes two periods where the population doubles, the growth factor per unit time is \(2 \times 2 = 4\).
04

Develop the Formula

To form the explicit formula for \(N(t)\) which describes the population at time \(t\), where \(t\) is the number of 80-minute units, use the formula for exponential growth:\[ N(t) = N_0 \cdot 4^t \]This reflects that every unit of time the population quadruples.
05

Formulate Final Expression

Given that \(N_0 = 1\), the formula for the population at any time \(t\) becomes:\[ N(t) = 4^t \]

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Key Concepts

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

Understanding Population Doubling
Population doubling is a term that helps us understand how quickly a population grows over time. It means that after a specific period, the population will increase to twice its initial size. This rate of doubling is crucial in many biological and environmental contexts, such as estimating human population growth or analyzing the spread of species in a new habitat.
To determine how often a population doubles, we first need to know the doubling time. In our example, the doubling time is 40 minutes. This means that every 40 minutes, our population will be twice as large as it was at the start of that period.
Understanding this concept helps us see the power of exponential growth, where populations can increase rapidly if they keep doubling at regular intervals. The idea of doubling helps in setting plans for resource management and understanding potential future challenges.
Exploring the Growth Factor
The growth factor is an essential concept when dealing with exponential growth scenarios. It represents how much the population will multiply over a certain period. In our exercise, the growth factor is calculated based on the population doubling every 40 minutes.
Since we define one unit of time as 80 minutes, and in that span, the population doubles twice, the effective growth factor becomes the result of two consecutive doublings.
  • Every 40-minute period: Population doubles, growth factor of 2
  • In 80 minutes (one unit of time): Population doubles twice, leading to a growth factor of 4
This means that for every unit of time, the population grows by a factor of 4, reflecting the mathematical beauty of exponential growth models where quantities increase significantly over each period.
Delving into the Exponential Formula
The exponential formula allows us to predict the size of a population at any future point in time, assuming the growth conditions remain consistent. For our problem, the formula to find the population at time \( t \) is: \[ N(t) = N_0 \cdot 4^t \]Here, \( N_0 \) is the initial population, 1 in our case, and \( 4^t \) reflects the exponential growth resulting from the growth factor of 4.

This formula shows how, with each unit of time, the population is multiplied by the growth factor. It encapsulates the entire process from past to future populations, providing invaluable insights in both academic and real-world scenarios. By analyzing this formula, students can deeply understand how exponential growth functions in nature and technology.
Identifying Initial Conditions
Initial conditions lay the groundwork for understanding any mathematical model predicting future instances. In this exercise, our initial condition is the starting population, \( N_0 = 1 \). This value is essential as it acts as the baseline for all projections.
Initial conditions tell us about the present state, from which we calculate all future developments. Without clear initial conditions, projections and models can become unreliable. Knowing the starting population ensures that our exponential formula accurately represents reality.
Establishing initial conditions is not limited to population modeling. It’s a crucial step in fields like physics, chemistry, and engineering, where the initial state determines the system's behavior over time. Understanding these helps students navigate through complex problems with clarity.

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Most popular questions from this chapter

Investigate the behavior of the discrete logistic equation $$ x_{t+1}=R_{0} x_{t}\left(1-x_{t}\right) $$ Compute \(x_{t}\) for \(t=0,1,2, \ldots, 20\) for the given values of \(r\) and \(x_{0}\), and graph \(x_{t}\) as a function of \(t .\) \(R_{0}=3.1, x_{0}=0.9\)

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\(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=\ln \left(1+\frac{1}{n}\right), \epsilon=0.1 $$

Tylenol in the Body A patient is taking Tylenol (a painkiller that contains acetaminophen) to treat a fever. The data in this question is taken from Rawlins, Henderson, and Hijab (1977). At \(t=0\) the patient takes their first pill. One hour later the drug has been completely absorbed and the blood concentration, measured in \(\mu \mathrm{g} / \mathrm{ml}\), is 15 . Acetaminophen has first order elimination kinetics; in one hour, \(23 \%\) of the acetaminophen present in the blood is eliminated. (a) Write a recursion relation for the concentration \(c_{t}\) of drug in the patient's blood. For \(t \geq 1\) you may assume for now that no other pills are taken after the first one. (b) Find an explicit formula for \(c_{t}\) as a function of \(t\). (c) Suppose that the patient follows the directions on the pill box and takes another Tylenol pill 4 hours after the first (at time \(t=4\) ). What is the concentration at the time at which the second pill is taken? In others words, what is \(c_{4}\) ? (d) Over the next hour \(15 \mathrm{\mug} / \mathrm{ml}\) of drug enter the patient's bloodstream. So, \(c_{5}\) can be calculated from \(c_{4}\) using the word equation: $$ c_{5}=c_{4}+ $$ nt added \(\quad\) amount eliminated blo Given that the amount added is \(15 \mu \mathrm{g} / \mathrm{ml}\), and the amount eliminated is \(0.23 \cdot c_{4}\), calculate \(c_{5} .\) (e) For \(t=5,6,7,8\) the drug continues to be eliminated at a rate of \(23 \%\) per hour. No pills are taken and no extra drug enters the patient's blood. Compute \(c_{8}\). (f) At time \(t=8\), the patient takes another pill. Calculate \(c_{9} .\) Do not forget to include elimination of drug between \(t=8\) and \(t=9\). (g) We want to calculate the maximum concentration of drug in the patient's blood. We know that concentrations are highest in the hour after a pill is taken, namely at time \(t=1, t=5, t=\) \(9, \ldots\) Define a sequence \(C_{n}\) representing the concentration of the drug one hour after the \(n\) th pill is taken. (h) What terms of the original sequence \(\left\\{c_{r}: t=1,2, \ldots\right\\}\) are \(C_{1}\), \(C_{2}\), and \(C_{3} ?\) (i) Explain why $$ C_{n+1}=(0.77)^{4} \cdot C_{n}+15 $$ and \(c_{1}=15\) (j) From the recursion relation, assuming that the patient continues to take Tylenol pills at 4 -hour intervals, calculate \(C_{1}, C_{2}\), \(C_{3}, C_{4}, C_{5}\), and \(C_{6}\) (k) Does \(C_{n}\) increase indefinitely, or do you think that it converges? (1) By looking for fixing point of the recursion relation in (h), find the limit of \(C_{n}\) as \(n \rightarrow \infty\).

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