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(a) There is a trick, called the Rule of 70, that can be used to get a quick estimate of the doubling time or half-life of an exponential model. According to this rule, the doubling time or half-life is roughly 70 divided by the percentage growth or decay rate. For example, we showed in Example 5 that with a continued growth rate of \(1.10 \%\) per year the world population would double every 63 years. This result agrees with the Rule of 70 , since \(70 / 1.10 \approx 63.6\). Explain why this rule works. (b) Use the Rule of 70 to estimate the doubling time of a population that grows exponentially at a rate of \(1 \%\) per year. (c) Use the Rule of 70 to estimate the half-life of a population that decreases exponentially at a rate of \(3.5 \%\) per hour. (d) Use the Rule of 70 to estimate the growth rate that would be required for a population growing exponentially to double every 10 years.

Short Answer

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(a) The Rule of 70 approximates the time for doubling or halving using natural logarithms. (b) 70 years for 1% growth. (c) 20 hours for 3.5% decay. (d) 7% growth rate for doubling in 10 years.

Step by step solution

01

Understanding the Rule of 70

The Rule of 70 is a way to estimate the time it takes for a quantity undergoing exponential growth to double, or for a quantity undergoing exponential decay to halve. This is approximated by dividing 70 by the percent growth rate for doubling, or by the percent decay rate for halving. This works because the natural logarithm of 2, which is used in the exact formula for doubling time or half-life, is approximately 0.693, which rounds to 0.7.
02

Estimating Doubling Time for 1% Growth Rate

To find the doubling time using the Rule of 70 for a growth rate of 1%, we use the formula: \( \text{Doubling time} = \frac{70}{\text{growth rate}} = \frac{70}{1} = 70 \text{ years} \). Thus, a population growing at 1% per year will take approximately 70 years to double.
03

Estimating Half-Life for 3.5% Decay Rate

Using the Rule of 70 to find the half-life for a decay rate of 3.5% involves the calculation: \( \text{Half-life} = \frac{70}{\text{decay rate}} = \frac{70}{3.5} = 20 \text{ hours} \). Therefore, a population decreasing at 3.5% per hour will halve in approximately 20 hours.
04

Finding Required Growth Rate for Doubling in 10 Years

To find the required growth rate for a population to double in 10 years, we rearrange the Rule of 70 to solve for growth rate: \( \text{growth rate} = \frac{70}{\text{time to double}} = \frac{70}{10} = 7\% \). Hence, a growth rate of 7% is needed for doubling every 10 years.

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Key Concepts

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

Rule of 70
The Rule of 70 is a simple yet powerful tool for estimating how fast something that's growing or shrinking will change in size. It's a quick trick to find out how long it will take for something growing at a steady rate to double, or for something shrinking at a steady rate to halve. The rule works by dividing the number 70 by the annual growth or decline rate (as a percentage). For example, if a quantity is growing at a rate of 1.1% per year, then by dividing 70 by 1.1, you get approximately 63.6 years, which is the doubling time. This approximation is useful because it stems from the natural logarithmic properties linked to exponential growth and decay. The natural logarithm of 2, which represents doubling, is roughly 0.693, rounding closely to 0.7, hence the use of 70 in the rule.
Doubling Time
When you hear "doubling time," think of how long it takes for something to become twice as big. This concept is key in understanding growth in scenarios where a constant growth rate is assumed over time. To calculate the doubling time using the Rule of 70, you simply take 70 and divide it by the growth rate. Let's say we have a growth rate of 1% per year. Using our formula, \[ \text{Doubling time} = \frac{70}{1\%} = 70 \text{ years} \]This means every 70 years the population size will double. This concept is widely applicable in fields such as finance, ecology, and demography, where predicting future size based on current growth rates is essential.
Half-Life
Half-life refers to the time taken for a quantity to reduce to half its initial value in a system experiencing exponential decay. Unlike doubling time, we use decay rates here, but the Rule of 70 still applies. Think of it as the opposite: instead of growing and doubling, something is shrinking and reaching half its size. For example, if a decay rate is 3.5% per hour, you can find the half-life by calculating:\[ \text{Half-life} = \frac{70}{3.5\%} = 20 \text{ hours} \]This means every 20 hours, the amount will have reduced to half its starting value. The half-life concept is crucial in fields like chemistry and physics (particularly nuclear physics), where it is used to understand how substances decompose over time.
Growth Rate Calculation
The growth rate calculation is the process of determining the rate required for a variable to change over time. If you want to know what rate is needed for a population to double in a specific period, rearrange the Rule of 70 formula to solve for the growth rate. For a population to double in 10 years, the calculation would be:\[ \text{Growth rate} = \frac{70}{10 \text{ years}} = 7\% \]This calculation tells us that a 7% annual growth rate is needed to double the population size every decade. This approach can be flipped to find rates for business profits, investments, or any system experiencing exponential growth, making it a versatile tool in economic and scientific forecasting.

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Most popular questions from this chapter

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Suppose that \(P\) dollars is invested at an annual interest rate of \(r \times 100 \%\). If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually; if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3 -month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest. (a) Show that if interest is compounded \(n\) times a year at equally spaced intervals, then the value \(A\) of the investment after \(t\) years is $$ A=P\left(1+\frac{r}{n}\right)^{n t} $$ (b) One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the limit one can conceive of interest compounded at each instant of time; this is called continuous compounding. Thus, from part (a), the value \(A\) of \(P\) dollars after \(t\) years when invested at an annual rate of \(r \times 100 \%\) compounded continuously, is $$ A=\lim _{n \rightarrow+\infty} P\left(1+\frac{r}{n}\right)^{n t} $$ Use the fact that \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\) to prove that \(A=P e^{r t}\) (c) Use the result in part (b) to show that money invested at continuous compound interest increases at a rate proportional to the amount present.

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