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Chapter 2: Problem 9

In Problems , give a formula for \(N(t), t=0,1,2, \ldots\), on the basis of the information provided. Suppose \(N_{t}=100 \cdot 2^{t}, t=0,1,2, \ldots\), and one unit of time corresponds to 2 hours. Determine the amount of time it takes the population to triple in size.

Short Answer

Expert verified
It takes approximately 3.17 hours for the population to triple.

Step by step solution

01

Understand the Problem

We are given the formula for a population growth over time: \( N(t) = 100 \cdot 2^{t} \). Our goal is to determine how long it takes for this population to triple in size.
02

Set Up the Equation for Tripled Population

Since the initial population is \( N(0)=100 \), tripling the population means we want \( N(t) = 300 \). Use the formula: \( N(t) = 100 \cdot 2^{t} = 300 \).
03

Simplify the Equation

Divide both sides of the equation \( 100 \cdot 2^{t} = 300 \) by 100 to isolate \( 2^{t} \). This gives us: \[ 2^{t} = 3 \].
04

Solve for \( t \) Using Logarithms

Take the logarithm of both sides of the equation \( 2^{t} = 3 \). Use base 2 logarithm for simplicity: \[ \log_{2}(2^{t}) = \log_{2}(3) \]. Thus \( t = \frac{\log_{10}(3)}{\log_{10}(2)} \) which approximately equals 1.585.
05

Convert \( t \) to Time in Hours

Since \( t \) is in units where one unit equals 2 hours, multiply \( t \) by 2 to convert to hours: \[ 1.585 \times 2 = 3.17 \text{ hours} \].

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Key Concepts

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

Exponential Growth
Exponential growth is a powerful concept where something increases at a rate proportional to its current value. This is common in biological populations, investments, and even some physical processes. In exponential growth, the quantity grows by a constant factor over equal time intervals.

A basic formula for exponential growth is:
  • \[ N(t) = N_0 \cdot b^{t} \]
where:
  • \( N(t) \) is the quantity at time \( t \)
  • \( N_0 \) is the initial quantity
  • \( b \) is the base of the exponential, which represents the growth factor
  • \( t \) is the time elapsed
In our specific example, the growth factor is 2, meaning the population doubles every time period \( t \). This leads to rapid growth, which can sometimes outpace other processes like resource availability or space.
Logarithms
Logarithms are the inverses of exponential functions, helping us solve equations where the variable is an exponent. In simpler terms, if you know the result of an exponential equation, logarithms help find the time it took for the growth to reach that size.

For equation solving, we shift from an exponential form to a logarithmic form:
  • If \( b^x = y \), then \( x = \log_b(y) \)
In the given problem, we used logarithms to solve \( 2^{t} = 3 \). This can be expressed with base-10 logarithms:
  • \[ t = \frac{\log_{10}(3)}{\log_{10}(2)} \]
Understanding logs is crucial for converting between exponential growth rates and finding the actual time \( t \) needed for certain growth benchmarks, like tripling a population.
Time Calculation
Once we've determined \( t \) in the growth equation, understanding the actual time in real-world units is crucial. In this problem, a single unit of \( t \) corresponds to 2 hours. Once we've calculated \( t \) as 1.585, we need to convert this into hours.

To transform \( t \) to a real-world timing, multiply \( t \) by the time per unit:
  • \[ 1.585 \times 2 = 3.17 \text{ hours} \]
Thus, it takes approximately 3.17 hours for the population to triple. This conversion ensures we can apply the mathematical solution to actual scenarios, assisting in planning or further analysis of the growth scenario.

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Mountain Gorilla Conservation You are trying to build a mathematical model for the size of the population of mountain gorillas in a national park in Uganda. The data in this question are taken from Robbins et al. (2009). (a) You start by writing a word equation relating the population of gorillas \(t\) years after the study begins, \(N_{t}\), to the population \(N_{t+1}\) in the next year: \(N_{t+1}=N_{t}+\begin{array}{l}\text { number of gorillas } \\ \text { born in one year }\end{array}-\begin{array}{l}\text { number of gorillas } \\\ \text { that die in one yeai }\end{array}\) We will derive together formulas for the number of births and the number of deaths. (i) Around half of gorillas are female, \(75 \%\) of females are of reproductive age, and in a given year \(22 \%\) of the females of reproductive age will give birth. Explain why the number of births is equal to: \(\quad 0.5 \cdot 0.75 \cdot 0.22 \cdot N_{t}=0.0825 \cdot N_{t}\) (ii) In a given year \(4.5 \%\) of gorillas will die. Write down a formula for the number of deaths. (iii) Write down a recurrence equation for the number of gorillas in the national park. Assuming that there are 300 gorillas initially (that is \(N_{0}=300\) ), derive an explicit formula for the number of gorillas after \(t\) years. (iv) Calculate the population size after \(1,2,5\), and 10 years. (v) According to the model, how long will it take for population size to double to 600 gorillas? (b) In reality the population size is almost totally stagnant (i.e., \(N_{t}\) changes very little from year to year). Robbins et al. (2009) consider three different explanations for this effect: (i) Increased mortality: Gorillas are dying sooner than was thought. What percentage of gorillas would have to die each year for the population size to not change from year to year? (ii) Decreased female fecundity: Gorillas are having fewer offspring than was thought. Calculate the female birth rate (percentage of reproductive age females that give birth) that would lead to the population size not changing from year to year. Assume that all other values used in part (a) are correct. (iii) Emigration: Gorillas are leaving the national park. What number of gorillas would have to leave the national park each year for the population to not change from year to year?
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