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Chapter 9: Problem 15

Justify the "Rule of \(69 "\) : If a quantity grows at a constant instantaneous rate of \(k \%\), then its doubling time is about \(69 / k .\) Example: In 2008 the population of Madagascar was about 20 million, growing at an annual rate of about \(3 \%,\) with a doubling time of about \(69 / 3=23\) years.

Short Answer

Expert verified
The doubling time for Madagascar's population, growing at an annual rate of 3%, is approximately 23 years, calculated using the Rule of 69 by dividing 69 by the growth rate (69 / 3).

Step by step solution

01

Understanding the Rule of 69

The Rule of 69 is a simplified way to estimate the doubling time of an investment or population growing at a constant rate. It states that if a quantity grows at a constant instantaneous rate of k%, then its doubling time is approximately 69 divided by k. This rule is an approximation and is rooted in the natural logarithm.
02

Applying the Rule of 69 to Madagascar's Population

To estimate the doubling time of Madagascar's population, which is growing at an annual rate of 3%, we apply the Rule of 69. This involves dividing 69 by the growth rate of 3%.
03

Calculating the Doubling Time

Perform the division: Doubling Time = 69 / k, where k is the growth rate. Therefore, Doubling Time = 69 / 3. Compute the result to find the estimated doubling time.

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Key Concepts

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

Exponential Growth
Exponential growth refers to an increase that occurs at a consistent relative growth rate over equal time periods. Imagine a scenario where you plant a single tree, and each year, that tree produces seeds that result in the growth of two new trees. Year after year, these trees also produce seeds, causing the forest to expand rapidly. That is exponential growth - a process where the quantity doubles or increases by a fixed percentage over time. In finance or population dynamics, this means the amount of money or the number of individuals can grow significantly over periods, provided that they grow at a fixed rate.
Doubling Time Estimation
The doubling time is an estimate of the period it takes for a quantity to double in size at a constant growth rate. Often used in financial contexts, this concept can also apply to population studies, like bacteria growth in a biology lab or human populations across the globe. To simplify these calculations, we use the 'Rule of 69.' This rule allows you to divide the number 69 by the growth rate to estimate doubling time. However, it's important to remember this is only an approximation, and the exactness can change based on the actual growth rate and time period considered. For our case with the population of Madagascar growing at a 3% rate, using this rule estimates that the population will double in roughly 23 years.
Instantaneous Growth Rate
The instantaneous growth rate is the rate at which a quantity is growing at any given instant. In contrast to average growth rates that measure growth over an interval, the instantaneous rate focuses on the growth at a precise moment, much like capturing a snapshot of growth. Mathematically, it is the derivative of the growth function with respect to time, indicating the slope of the tangent at any point on the growth curve. In the context of the Rule of 69, the instantaneous growth rate is the constant percentage rate at which a population or investment grows. As an approximation, the Rule of 69 simplifies complex logarithmic calculations into a more manageable formula to gauge how quickly a figure will double considering its instantaneous growth rate.

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Most popular questions from this chapter

Solve the systems in Exercises 31 through \(34 .\) Give the solution in real form. Sketch the solution. $$\frac{d \vec{x}}{d t}=\left[\begin{array}{rr} 0 & 1 \\ -4 & 0 \end{array}\right] \vec{x} \quad \text { with } \quad \vec{x}(0)=\left[\begin{array}{l} 1 \\ 0 \end{array}\right]$$

Solve the initial value problems. $$f^{\prime}(t)-5 f(t)=0, f(0)=3$$

Consider a linear system \(d \vec{x} / d t=A \vec{x},\) where \(A\) is a \(2 \times 2\) matrix that is diagonalizable over \(\mathbb{R}\). When is the zero state a stable equilibrium solution? Give your answer in terms of the determinant and the trace of \(A.\)

Consider a quadratic form \(q(\vec{x})=\vec{x} \cdot A \vec{x}\) of two variables, \(x_{1}\) and \(x_{2}\). Consider the system of differential equations $$\left|\begin{array}{l} \frac{d x_{1}}{d t}=\frac{\partial q}{\partial x_{1}} \\ \frac{d x_{2}}{d t}=\frac{\partial q}{\partial x_{2}} \end{array}\right|$$ or, more succinctly, $$\frac{d \vec{x}}{d t}=\operatorname{grad} q$$ a. Show that the system \(d \vec{x} / d t=\operatorname{grad} q\) is linear by finding a matrix \(B\) (in terms of the symmetric matrix \(A\) ) such that grad \(q=B \vec{x}\) b. When \(q\) is negative definite, draw a sketch showing possible level curves of \(q .\) On the same sketch, draw a few trajectories of the system \(d \vec{x} / d t=\operatorname{grad} q\) What does your sketch suggest about the stability of the system \(d \vec{x} / d t=\operatorname{grad} q ?\) c. Do the same as in part b for an indefinite quadratic form. d. Explain the relationship between the definiteness of the form \(q\) and the stability of the system \(d \vec{x} / d t\) \(=\operatorname{grad} q\)

Let \(A\) be an \(n \times n\) matrix and \(k\) a scalar. Consider the following two systems: $$\begin{array}{l}\frac{d \vec{x}}{d t}=A \vec{x} \quad (I)\\\\\frac{d \vec{c}}{d t}=k A \vec{c} \quad (II)\end{array}$$ Show that if \(\vec{x}(t)\) is a solution of system (I), then \(\vec{c}(t)=\) \(\vec{x}(k t)\) is a solution of system (II). Compare the vector fields of the two systems.
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