6. Exponential Growth and Decay
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Chapter 6
6.
Exponential Growth and Decay
| Student Learning Objectives: |
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Why
This lesson explores the mathematical concepts of exponential growth and decay, emphasizing their real-world applications, particularly in finance and biology. For example, understanding exponential growth can help you appreciate how compound interest works in a savings account, allowing your money to grow over time. On the flip side, exponential decay is crucial in calculating the depreciation of assets like cars. These concepts are not just theoretical; they are tools that can help you make informed decisions in various aspects of life, from managing your finances to understanding biological phenomena like bacterial growth.
Lesson Settings & Tools
| | 12 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Exponential Growth and Decay
Slide of 12
Exponential functions are useful for modeling a wide variety of real-life scenarios. This lesson will introduce the concepts of exponential growth and exponential decay. How they are used to model real-life situations will also be understood.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Carbon-14 Decay and Archaeology
While watching a documentary about ancient civilizations, Ali wondered how scientists could determine the age of an object belonging to such ancient civilizations.
Ali recalled that his uncle, Mr. Jones, is an archaeologist! Ali called and asked him how are the ages of ancient objects determined.
Carbon-14 is a substance present in organisms that, once they expire, begin to be released from an object's body at a slow rate. To date an object this way consists of measuring the amount of carbon-14 in a sample and comparing it to known values of different ages.
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Explore
Exploring Exponential Functions
Consider the following exponential function. y=2 * b^x How does its graph change when the value of b changes? To visualize its effect, move the slider in the following applet.
- What is the behavior of the function when b>1?
- What is the behavior of the function when b<1?
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Discussion
Exponential Growth
For an exponential function y=a(b)^x, where a>0 and b>1, the y-values increases as the x-values increases. Therefore, it can be classified as an increasing function. y = ab^x ⇓ Increasing when a>0 and b >1 Such exponential functions are called exponential growth functions.
When a quantity increases by the same factor over equal intervals of time, it is said that such a quantity is in exponential growth. Exponential growth is modeled using exponential functions where a >0 and b>1. y = a(b)^t In this form, a is the initial amount, the base b is the growth factor, and t usually represents time. Like any other exponential function, a also represents the y-intercept.
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Example
Studying Bacterial Growth
Magdalena, excited for biology lab, is exploring about bacterial growth.
Bacteria are known to duplicate themselves within a certain amount of time. This means that after some time, there will be twice the amount of bacteria as before. Magdalena and her partners are studying E. coli, a bacteria responsible for many stomach related diseases. E. coli is known to duplicate about every 20 minutes.
a Write an equation that models this exponential growth in terms of the time elapsed since the start of the experiment, in minutes.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["t"],"constants":[]},"rendered":false},"text":""},"formTextBefore":"y=","formTextAfter":null,"answer":{"text":["2^{t\/20}"]}}
b How much bacteria will there be after two hours?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"bacteria","answer":{"text":["64"]}}
Hint
a The initial amount of bacteria is 1 and they duplicate each 20 minutes.
b Write 2 hours as minutes. Then substitute that into the equation obtained in Part A.
Solution
a Exponential growths are modeled using exponential functions of the following form.
y=a* b^t In the above equation, a represents the initial number of bacteria. Since the experiment starts with 1 bacteria, a equals 1. The bacteria duplicate themselves every 20 minutes. That means the base should be equal to 2. y= 1( 2)^t ⇒ y=2^t Notice that this function does not represent the situation exactly. The reason being that, for example, substituting 1 for t doubles the number of the bacteria. That is, the bacteria doubles itself every minute. For the equation to give the number of bacteria duplicating every 20 minutes, t must be divided by 20. y =2^t ⇒ y=2^(t 20) Therefore, y=2^(t20) correctly models the situation for E. coli. Comparing the number of bacteria on the applet and the y-values found by the equation, it can be checked if the equation is correct or not.
| Time Elapsed, t | y=2^(t20) | Number of Bacteria, y |
|---|---|---|
| 0 | 2^(020)=1 | 1 |
| 20 | 2^(2020)=2^1 | 2 |
| 40 | 2^(4020)=2^2 | 4 |
| 60 | 2^(6020)=2^3 | 8 |
| 80 | 2^(8020)=2^4 | 16 |
b Since the exponential growth function is a function of time in minutes, 2 hours must expressed in minutes as well. That can be done using a conversion factor.
2h* 60min/1h = 120min The converted time can now be substituted into the equation.
It has been found that there will be 64 bacteria after 2 hours. That is amazing, considering Magdalena began with only 1 bacteria of E. coli.
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Discussion
Compound Interest
Applications of exponential growth can also be encountered in the world of finance. Some people use the power of compound interest to grow their wealth exponentially.
Compound interest is the interest earned depending on both the initial investment and previously earned interest. To find the balance A of an account that earns compound interest, an exponential growth function can be used.
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Example
Investing Into a Savings Account
Kriz, determined and focused, won an online video game competition. The first place prize was $1000!
Kriz decides to not spend the prize money. Instead, their parent suggests placing all of it into a Certificate of Deposit. This is a type of savings account with compound interest. The catch is that the money cannot be taken out for a certain period of time. Ngân Hàng, a local bank, offers a Certificate of Deposit with the interest rate at 3 % compounded monthly.
a Write an equation that models the local bank's compound interest.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["t"],"constants":[]},"rendered":false},"text":""},"formTextBefore":"A=","formTextAfter":null,"answer":{"text":["1000*1.0025^{12*t}"]}}
b How much money will Kriz have 2 years after opening the account? Round the amount to 2 decimal places.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"$","formTextAfter":null,"answer":{"text":["1061.76"]}}
Hint
a The meaning of
compounded monthlyis that the interest is compounded each month of the year — meaning twelve times per year.
b Substitute 2 for t in the equation obtained in Part A.
Solution
a The local bank offers Kriz an interest rate of 3 % that is compounded monthly. Using the exponential growth function that models compound interest, first write the percentage as a decimal.
3 % ⇒ 0.03 It is given that the interest is compounded monthly. Therefore it will compound 12 times a year. Knowing that Kriz will store all of the $ 1000 in prize money, there is now enough information to write the equation that models this compound interest.
Using this equation, Kriz can calculate their earning after t years.
b In order to find how much money Kriz would have in their savings account after 2 years, 2 will be substituted for t into the equation solved for in Part A.
Kriz will have $1061.76 in their savings account after two years. What a good start to their gaming career!
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Discussion
Exponential Decay
When the base of an exponential function is a number greater than 0 and less than 1, the function is said to be decreasing. In such cases, the function represents what is known as exponential decay.
When a quantity decreases by the same factor over equal intervals of time it is said that such quantity is in exponential decay. Exponential decay is modeled using exponential functions with a>0 and a base b that is between 0 and 1. y = a(b)^t In this form, a is the initial amount, the base b is the decay factor, and t usually represents time. Like any other exponential function, a also represents the y-intercept.
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Example
Calculating the Devaluation of a Car
Diego has saved for the past few years dreaming of buying a car with a drop top so he can cruise the streets looking fly. Diego runs to the nearest car dealer and is met by Mr. Peterson, a car salesmen. They come to an agreement where Diego trades in his old car to help pay for the new car.
Diego bought his car five years ago at the same dealer for $20 000. Mr. Peterson states that the car depreciates at a rate of 15 % annually.
How much is Mr. Peterson going to give Diego for his old car? Round Mr. Peterson's offer to two decimal places.
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Hint
Since the value of the car depreciates, the situation can be modeled using an exponential decay function.
Solution
Previously, Diego bought a car with a value that has depreciated since the time of purchase. Therefore, this scenario can be modeled using an exponential decay function. The car was bought for $ 20 000, which is the initial amount.
y=a(1-r)^t ⇓ y= 20 000(1-r)^t
The car depreciates at a rate of 15 % annually. That value needs to be written as a decimal.
15 % ⇒ 0.15
There is now enough information to write an equation for the value of the old car.
It has been five years since Diego bought his car. Therefore, to find the current value of the car, 5 should be substituted for t into the function.
y=20 000(0.85)^t
Substitute
t= 5
y=20 000(0.85)^5
UseCalc
Use a calculator
y=8874.10625
RoundDec
Round to 2 decimal place(s)
y=8874.11
Considering the depreciation rate of 15 %, Mr.Peterson offers $8874.11 for Diego's 5-year-old car. If Diego accepts the offer, he will have a capital of $8874.11 to put toward the purchase of a new car from the dealership.
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Pop Quiz
Identifying Exponential Growth or Decay From a Table
Select the option that best describes the table of values given below.
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Pop Quiz
Identifying the Rate of Growth or Decay
Exponential functions y=ab^t can model exponential decay as well as exponential growth. Identify the rate of decay or growth r for the indicated function. Write the corresponding rate in decimal form.
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Closure
Modeling Carbon-14 Decay
This lesson introduced the interesting concepts of compound interest, exponential growth, and exponential decay. Using the knowledge gained from this lesson, the introductory challenge can be modeled using an exponential decay function. Recall what the archaeologist had to say.
Since the half-life of carbon-14 is 5730 years, an initial amount A of carbon-14 will decay by half that amount in 5730 years. Let y be the final amount of carbon-14 and write an equation that models this exponential decay.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["A","t"],"constants":[]},"rendered":false},"text":""},"formTextBefore":"y=","formTextAfter":null,"answer":{"text":["A*0.5^{\\dfrac{t}{5730}}","\\dfrac{A}{2}^{\\dfrac{t}{5730}}"]}}
Hint
Find the decay factor of this situation.
Solution
Exponential decay is modeled using exponential functions with a base b that is between 0 and 1.
y=a(b)^t
Since the given information is about the half-life of carbon-14, the base is 12, or 0.5. It is also given that the initial amount of carbon is represented by A.
y=a(b)^t ⇓ y= A( 0.5)^t
The half-life of carbon-14 is 5730 years, which means that after 5730 years, only half of the initial amount will remain. Therefore, t should be divided by 5730 so that for every 5730 years half the previous amount remains.
y=A(0.5)^(t 5730)
Wow! That is a great amount of time that carbon-14 takes in order to decay. That fact makes for a good reason to use it when determining the age of a fossil.
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Exponential Growth and Decay
Exercise 2.1
Select one of the following.
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
Select one of the following.
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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We are told that the population of a bacterial sample doubles each hour. This means that the population will increase with a growth rate of 2. This can be described using exponential growth.
Exponential Growth |- y=ab^x, b>1
Note that Graph B is a decreasing function, so it cannot properly describe this situation. We can also notice that the initial amount in graphs C and D is 0.
Furthermore they represent a quadratic and a linear function, respectively. The remaining option satisfies the description of the situation. Therefore, the answer is A.
We are told that the price of a stock decreases by half each year. This means that the price will decrease with a decay rate of 12. This can be described using exponential decay.
Exponential Decay
|-
y=ab^x, 0
Note that Graph B is the only decreasing function, so it is the only one that can properly describe this situation. Therefore, the answer is B.
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Hint & Answer
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Solution
{"type":"choice","form":{"alts":["A","B","C"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":""},"formTextBefore":"y=","formTextAfter":null,"answer":{"text":["50(2^x)"]}}
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We want to identify whether the given table of values represents an exponential growth, exponential decay, or neither. Let's begin by noticing that the t-values of the given table have a common difference of 1.
We will now focus on the y-values. As t increases by 1, the y-values become twice the previous amount. This means that the y-values have a common ratio of 2.
Since the t-values increase by the same factor over equal intervals of time, the table represents an exponential growth function. Furthermore, we can note that the common ratio is a number greater than 1. Therefore, the answer is A.
We found in Part A that the y-values increase by a factor of 2 as x increases by 1.
This means that the base of the exponential function is equal to 2. y=a* 2^x We can also note that when x is 0, we have that y is equal to 50. This means that the y-intercept or initial value is 50. With this information we have all we need to write the exponential function. Let's do it!
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Solution
{"type":"choice","form":{"alts":["A","B","C"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":""},"formTextBefore":"y=","formTextAfter":null,"answer":{"text":["5000(0.2)^x"]}}
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We can notice that the t-values of the given table have a common difference of 1. As t increases by 1, the y-values decrease. Let's see if we can find a common ratio!
The y-values have a common ratio of 0.2. The table represents an exponential decay function because the common ratio is a number less than 1. Therefore, the answer is B.
We found in Part A that the y-values decrease by a factor of 0.2 as the x-values increase by 1.
This means that the base of the exponential function is equal to 0.2. Let's write it as a fraction! 0.2 ⇔ 1/5 Knowing this, we can write part of the requested function. y=a*( 1/5)^x We can also note that when x is equal to 0, we have that y is equal to 5000. This means that the y-intercept or initial value is 5000. With this information we have all we need to write the exponential function. Let's do it!
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Solution
Exponential Growth and Decay
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