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Chapter 5: Problem 4

What is the growth or decay factor for each given time period? a. Weight increases by \(0.2 \%\) every 5 days. b. Mass decreases by \(6.3 \%\) every year. c. Population increases \(23 \%\) per decade. d. Profit increases \(300 \%\) per year. e. Blood alcohol level decreases \(35 \%\) per hour.

Short Answer

Expert verified
(a) 1.002, (b) 0.937, (c) 1.23, (d) 4.0, (e) 0.65

Step by step solution

01

- Understanding Growth and Decay Factors

Growth or decay factors are calculated using the formula for exponential growth or decay: \[ \text{Factor} = 1 \text{ (+ or -) Rate} \text{ [expressed as a decimal]} \]For growth, use '+'. For decay, use '-'.
02

- Calculate Growth Factor for (a)

The weight increases by 0.2% every 5 days. Convert the percentage into a decimal:\[ \text{Rate} = 0.2 / 100 = 0.002 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 0.002 = 1.002 \]
03

- Calculate Decay Factor for (b)

The mass decreases by 6.3% every year. Convert the percentage into a decimal:\[ \text{Rate} = 6.3 / 100 = 0.063 \]Then apply the formula for decay:\[ \text{Decay Factor} = 1 - 0.063 = 0.937 \]
04

- Calculate Growth Factor for (c)

The population increases by 23% per decade. Convert the percentage into a decimal:\[ \text{Rate} = 23 / 100 = 0.23 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 0.23 = 1.23 \]
05

- Calculate Growth Factor for (d)

The profit increases by 300% per year. Convert the percentage into a decimal:\[ \text{Rate} = 300 / 100 = 3.0 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 3.0 = 4.0 \]
06

- Calculate Decay Factor for (e)

The blood alcohol level decreases by 35% per hour. Convert the percentage into a decimal:\[ \text{Rate} = 35 / 100 = 0.35 \]Then apply the formula for decay:\[ \text{Decay Factor} = 1 - 0.35 = 0.65 \]

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Key Concepts

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

growth factor
Growth factors are essential when dealing with exponential growth problems. They determine how much a quantity increases over a given period. The basic formula for finding a growth factor is: Growth Factor = 1 + Rate where the rate is expressed as a decimal. For instance, if a weight increases by 0.2% every 5 days, you first convert 0.2% to a decimal: Rate = 0.2 / 100 = 0.002 Then, apply the formula: Growth Factor = 1 + 0.002 = 1.002 This means the weight grows by a factor of 1.002 every 5 days. To make the conversion, always divide the percentage by 100 to turn it into a decimal.
decay factor
Decay factors are crucial when addressing exponential decay situations. They tell you how much a quantity decreases over time. The basic formula for finding a decay factor is: Decay Factor = 1 - Rate, where the rate is expressed as a decimal. For example, if a mass decreases by 6.3% every year, you convert 6.3% to a decimal: Rate = 6.3 / 100 = 0.063 Then, apply the formula: Decay Factor = 1 - 0.063 = 0.937 The mass decreases by a factor of 0.937 annually. Always remember to convert the percentage to a decimal by dividing by 100.
percentage conversion
Percentage conversion is the first step to solving growth and decay problems. This process involves turning a percentage into a decimal. To do this, simply divide the percentage by 100. For example: If you have a percentage of 23%, you'd convert it by calculating: 23 / 100 = 0.23 Similarly, for a percentage of 300%, you'd convert it by: 300 / 100 = 3.0 Once you have the decimal, you can use it in your growth or decay formulas. This conversion is a vital step for correct calculations.
exponential functions
Exponential functions describe growth or decay processes, where quantities change at a consistent rate. The general form of an exponential function is: y = a(1 + r)^t for growth or y = a(1 - r)^t for decay. where - y is the final amount - a is the initial amount - r is the growth (or decay) rate as a decimal - t represents time For example, if you start with a population of 1000 that grows by 23% per decade, the function is: y = 1000(1 + 0.23)^t Exponential functions help model real-world phenomena such as population growth, radioactive decay, and financial investments.

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Most popular questions from this chapter

Lead- 206 is not radioactive, so it does not spontaneously decay into lighter elements. Radioactive elements heavier than lead undergo a series of decays, each time changing from a heavier element into a lighter or more stable one. Eventually, the element decays into lead- 206 and the process stops. So, over billions of years, the amount of lead in the universe has increased because of the decay of numerous radioactive elements produced by supernova explosions. Radioactive uranium- 238 decays sequentially into thirteen other lighter elements until it stabilizes at lead-206. The half-lives of the fifteen different elements in this decay chain vary from 0.000164 seconds (from polonium- 214 to lead- 210 ) all the way up to 4.47 billion years (from uranium- 238 to thorium- 234 ). a. Find the decay rate per billion years for uranium- 238 to decay into thorium- 234 . b. Find the decay rate per second for polonium-214 to decay into lead-2.10.

Estimate the doubling time using the rule of 70 when: a. \(P=2.1(1.0475)^{t}\), where \(t\) is in years b. \(Q=2.1(1.00475)^{T}\), where \(T\) is in years

In a chain letter one person writes a letter to a number of other people, \(N,\) who are each requested to send the letter to \(N\) other people, and so on. In a simple case with \(N=2\), let's assume person Al starts the process. Al sends to \(\mathrm{B} 1\) and \(\mathrm{B} 2 ; \mathrm{B} 1\) sends to \(\mathrm{C} 1\) and \(\mathrm{C} 2 ; \mathrm{B} 2\) sends to \(\mathrm{C} 3\) and \(\mathrm{C} 4\); and so on. A typical letter has listed in order the chain of senders who sent the letters. So \(\mathrm{D} 7\) receives a letter that has \(\mathrm{A} 1, \mathrm{~B} 2\), and \(\mathrm{C} 4\) listed. If these letters request money, they are illegal. A typical request looks like this: \(\cdot\) When you receive this letter, send \(\$ 10\) to the person on the top of the list. \(\cdot\) Copy this letter, but add your name to the bottom of the list and leave off the name at the top of the list. \(\cdot\) Send a copy to two friends within 3 days. For this problem, assume that all of the above conditions hold. a. Construct a mathematical model for the number of new people receiving letters at each level \(L,\) assuming \(N=2\) as shown in the above tree. b. If the chain is not broken, how much money should an individual receive? c. Suppose A 1 sent out letters with two additional phony names on the list (say Ala and Alb) with P.O. box addresses she owns. So both \(\mathrm{B} 1\) and \(\mathrm{B} 2\) would receive a letter with the list \(\mathrm{A} 1, \mathrm{~A} 1 \mathrm{a},\) Alb. If the chain isn't broken, how much money would Al receive? d. If the chain continued as described in part (a), how many new people would receive letters at level \(25 ?\) e. Internet search: Chain letters are an example of a "pyramid growth" scheme. A similar business strategy is multilevel marketing. This marketing method uses the customers to sell the product by giving them a financial incentive to promote the product to potential customers or potential salespeople for the product. (See Exercise \(31 .)\) Sometimes the distinction between multilevel marketing and chain letters gets blurred. Search the U.S. Postal Service website (www.usps.gov) for "pyramid schemes" to find information about what is legal and what is not. Report what you find.

The following exponential functions represent population growth. Identify the initial population and the growth factor. a. \(Q=275 \cdot 3^{T}\) b. \(P=15,000 \cdot 1.04^{t}\) c. \(y=\left(6 \cdot 10^{8}\right) \cdot 5^{x}\) d. \(A=25(1.18)^{t}\) e. \(P(t)=8000(2.718)^{t}\) f. \(f(x)=4 \cdot 10^{5}(2.5)^{x}\)

(Graphing program recommended.) Cosmic ray bombardment of the atmosphere produces neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon-14 enters all living tissue through carbon dioxide (via plants). As long as a plant or animal is alive, carbon-14 is maintained in the organism at a constant level. Once the organism dies, however, carbon-14 decays exponentially into carbon-12. By comparing the amount of carbon- 14 to the amount of carbon-12, one can determine approximately how long ago the organism died. Willard Libby won a Nobel Prize for developing this technique for use in dating archaeological specimens. The half-life of carbon-14 is about 5730 years. In answering the following questions, assume that the initial quantity of carbon- 14 is 500 milligrams. a. Construct an exponential function that describes the relationship between \(A,\) the amount of carbon- 14 in milligrams, and \(t,\) the number of 5730 -year time periods. b. Generate a table of values and plot the function. Choose a reasonable set of values for the domain. Remember that the objects we are dating may be up to 50,000 years old. c. From your graph or table, estimate how many milligrams are left after 15,000 years and after 45,000 years. d. Now construct an exponential function that describes the relationship between \(A\) and \(T,\) where \(T\) is measured in years. What is the annual decay factor? The annual decay rate? e. Use your function in part (d) to calculate the number of milligrams that would be left after 15,000 years and after 45,000 years.
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