Question

bug in a bottle
suppose at 11:00 there is one bug in a bottle. the number of bugs will double every minute. at the bottle is full of bugs. answer the following questions:
what time is the bottle ½ full? ____________
what time is the bottle ¼ full? ____________
what time did the bottle contain 16 bugs? ____________
11:55 how full was the bottle? ____________
suppose the first bug was very smart. at what time did he realize he was running out of room? ____________
the smart bug realized he was running out of room, he texted a friend to ask for 3 additional bottles.
what time did bottle number 2 fill up with bugs? ____________
what time were bottles 3 and 4 full of bugs? ____________
bottle tutorial: https://www.worldpopulationbalance.org/understanding-exponential-growth
using the data in the table below, predict the year that the human population may be equal to or exceed 10 billion. ____________. explain why you selected this year.

year	human population
1927	2 billion
1960	3 billion
1974	4 billion
1987	5 billion
1999	6 billion
2011	7 billion
2015	7.3 billion
	10 billion

compare the current human population situation to the bug in the bottle above. what conclusions can you draw?
chessboard/rice problem
what square are you on when you run out of room on a square? ____________
how many grains are on that square? ____________ (hint: use the formula n = 2^(n - 1) where n is the number of the square and n = the number of grains of rice on the square).
how many grains should be on the 10th square? ____________
the 32nd square is half way on the board; how many grains should be on it? ____________
using the graph paper provided, make a graph showing the number of grains on squares 1 - 32
does the graph represent constant or exponential growth? ____________
asim n5expgrow exponential population growth 5e student notes, revised 4/2020 page 3 of 6

Explanation:

- ## Step1: Since the number of bugs doubles every minute, if the bottle is full at 12:00, one - minute before it is half - full.
- If the full state is at a certain time \(t\), the half - full state is at \(t - 1\) minute.
- # Answer: 11:59
2. **What time is the bottle ¼ full?**
- # Explanation:
- ## Step1: Since the number of bugs doubles every minute, and the bottle is half - full at 11:59, one - minute before it is half - full, it is ¼ full.
- If the half - full state is at \(t_1\), the ¼ - full state is at \(t_1-1\) minute.
- # Answer: 11:58
3. **What time did the bottle contain 16 bugs?**
- # Explanation:
- ## Step1: We know the initial number of bugs \(N_0 = 1\), and the formula for the number of bugs \(N = N_0\times2^n\), where \(n\) is the number of minutes passed. We want to find \(n\) when \(N = 16\) and \(N_0 = 1\).
- Set \(16=1\times2^n\), then \(2^n = 16\), and \(n = 4\) (because \(2^4=16\)).
- ## Step2: Since the starting time is 11:00 and 4 minutes have passed.
- The time is 11:00 + 4 minutes.
- # Answer: 11:04
4. **At 11:55 how full was the bottle?**
- # Explanation:
- ## Step1: Let the full - bottle state be at 12:00 (60 minutes from 11:00). At 11:55, 55 minutes have passed from 11:00.
- ## Step2: The number of bugs at 11:55 is \(N_{55}=2^{55}\), and the number of bugs at 12:00 is \(N_{60}=2^{60}\). The fraction of the fullness \(f=\frac{2^{55}}{2^{60}}\).
- Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), we have \(\frac{2^{55}}{2^{60}}=2^{55 - 60}=2^{-5}=\frac{1}{2^5}=\frac{1}{32}\).
- # Answer: \(\frac{1}{32}\) full
5. **Suppose the first bug was very smart. At what time did he realize he was running out of room?**
- This is a bit of a subjective question, but a reasonable answer could be when the bottle is getting close to full. Let's say at 11:59 when the bottle is half - full, the bug might realize it's running out of room.
- # Answer: 11:59
6. **If the smart bug realized he was running out of room, he texted a friend to ask for 3 additional bottles. What time did bottle number 2 fill up with bugs?**
- # Explanation:
- ## Step1: Since the bugs double every minute, and when the first bottle is full at 12:00, the second bottle will start with the same number of bugs as the first bottle had at 11:59 (half - the full amount). So it will take 1 more minute to fill the second bottle.
- # Answer: 12:00
7. **What time were bottles 3 and 4 full of bugs?**
- # Explanation:
- ## Step1: Bottle 3 will start with the same number of bugs as bottle 2 had when it was half - full. Since bottle 2 is full at 12:00, it was half - full at 12:00 - 1 minute = 11:59. So bottle 3 will take 1 more minute to fill after bottle 2 is full, and bottle 4 will take 1 more minute after bottle 3 is full.
- Bottle 3 is full at 12:01 and bottle 4 is full at 12:02.
- # Answer: Bottle 3: 12:01; Bottle 4: 12:02
8. **Using the data in the table below, predict the year that the human population may be equal to or exceed 10 billion.**
- Looking at the table, the population reaches 10 billion in 2015.
- # Answer: 2015
9. **Compare the current human population situation to the bug in the bottle above. What conclusions can you draw?**
- # Brief Explanations:
- Both the bug population in the bottle and human population show exponential growth. Just as the bugs run out of space in the bottle, humans may face resource - limitation issues as the population grows, such as lack of living space, food shortages, and strain on natural resources.
- # Answer: Both show exponential growth and may face resource - limitation problems.
10. **For the chessboard/rice problem**:
- **At what square are you on when you run out of room on a square?**
- A standard chess - board has 64 squares. So the 64th square.
- # Answer: 64th square
- **How many grains are on that square?**
- # Explanation:
- ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 64\).
- \(N=2^{64 - 1}=2^{63}\).
- # Answer: \(2^{63}\)
- **How many grains should be on the 10th square?**
- # Explanation:
- ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 10\).
- \(N=2^{10 - 1}=2^9 = 512\).
- # Answer: 512
- **The 32nd square is half - way on the board; how many grains should be on it?**
- # Explanation:
- ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 32\).
- \(N=2^{32 - 1}=2^{31}\).
- # Answer: \(2^{31}\)
- **Using the graph paper provided, make a graph showing the number of grains on squares 1 - 32. Does the graph represent constant or exponential growth?**
- The graph represents exponential growth because the number of grains on each square is given by \(N = 2^{n - 1}\), and as \(n\) (the square number) increases, \(N\) increases exponentially.
- # Answer: Exponential growth

Answer:

1. What time is the bottle ½ full?

• # Explanation:
• ## Step1: Since the number of bugs doubles every minute, if the bottle is full at 12:00, one - minute before it is half - full.
• If the full state is at a certain time \(t\), the half - full state is at \(t - 1\) minute.
• # Answer: 11:59

1. What time is the bottle ¼ full?

• # Explanation:
• ## Step1: Since the number of bugs doubles every minute, and the bottle is half - full at 11:59, one - minute before it is half - full, it is ¼ full.
• If the half - full state is at \(t_1\), the ¼ - full state is at \(t_1-1\) minute.
• # Answer: 11:58

1. What time did the bottle contain 16 bugs?

• # Explanation:
• ## Step1: We know the initial number of bugs \(N_0 = 1\), and the formula for the number of bugs \(N = N_0\times2^n\), where \(n\) is the number of minutes passed. We want to find \(n\) when \(N = 16\) and \(N_0 = 1\).
• Set \(16=1\times2^n\), then \(2^n = 16\), and \(n = 4\) (because \(2^4=16\)).
• ## Step2: Since the starting time is 11:00 and 4 minutes have passed.
• The time is 11:00 + 4 minutes.
• # Answer: 11:04

1. At 11:55 how full was the bottle?

• # Explanation:
• ## Step1: Let the full - bottle state be at 12:00 (60 minutes from 11:00). At 11:55, 55 minutes have passed from 11:00.
• ## Step2: The number of bugs at 11:55 is \(N_{55}=2^{55}\), and the number of bugs at 12:00 is \(N_{60}=2^{60}\). The fraction of the fullness \(f=\frac{2^{55}}{2^{60}}\).
• Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), we have \(\frac{2^{55}}{2^{60}}=2^{55 - 60}=2^{-5}=\frac{1}{2^5}=\frac{1}{32}\).
• # Answer: \(\frac{1}{32}\) full

1. Suppose the first bug was very smart. At what time did he realize he was running out of room?

• This is a bit of a subjective question, but a reasonable answer could be when the bottle is getting close to full. Let's say at 11:59 when the bottle is half - full, the bug might realize it's running out of room.
• # Answer: 11:59

1. If the smart bug realized he was running out of room, he texted a friend to ask for 3 additional bottles. What time did bottle number 2 fill up with bugs?

• # Explanation:
• ## Step1: Since the bugs double every minute, and when the first bottle is full at 12:00, the second bottle will start with the same number of bugs as the first bottle had at 11:59 (half - the full amount). So it will take 1 more minute to fill the second bottle.
• # Answer: 12:00

1. What time were bottles 3 and 4 full of bugs?

• # Explanation:
• ## Step1: Bottle 3 will start with the same number of bugs as bottle 2 had when it was half - full. Since bottle 2 is full at 12:00, it was half - full at 12:00 - 1 minute = 11:59. So bottle 3 will take 1 more minute to fill after bottle 2 is full, and bottle 4 will take 1 more minute after bottle 3 is full.
• Bottle 3 is full at 12:01 and bottle 4 is full at 12:02.
• # Answer: Bottle 3: 12:01; Bottle 4: 12:02

1. Using the data in the table below, predict the year that the human population may be equal to or exceed 10 billion.

• Looking at the table, the population reaches 10 billion in 2015.
• # Answer: 2015

1. Compare the current human population situation to the bug in the bottle above. What conclusions can you draw?

• # Brief Explanations:
• Both the bug population in the bottle and human population show exponential growth. Just as the bugs run out of space in the bottle, humans may face resource - limitation issues as the population grows, such as lack of living space, food shortages, and strain on natural resources.
• # Answer: Both show exponential growth and may face resource - limitation problems.

1. For the chessboard/rice problem:

• At what square are you on when you run out of room on a square?
• A standard chess - board has 64 squares. So the 64th square.
• # Answer: 64th square
• How many grains are on that square?
• # Explanation:
• ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 64\).
• \(N=2^{64 - 1}=2^{63}\).
• # Answer: \(2^{63}\)
• How many grains should be on the 10th square?
• # Explanation:
• ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 10\).
• \(N=2^{10 - 1}=2^9 = 512\).
• # Answer: 512
• The 32nd square is half - way on the board; how many grains should be on it?
• # Explanation:
• ## Step1: Use the formula \(N = 2^{n - 1}\), where \(n = 32\).
• \(N=2^{32 - 1}=2^{31}\).
• # Answer: \(2^{31}\)
• Using the graph paper provided, make a graph showing the number of grains on squares 1 - 32. Does the graph represent constant or exponential growth?
• The graph represents exponential growth because the number of grains on each square is given by \(N = 2^{n - 1}\), and as \(n\) (the square number) increases, \(N\) increases exponentially.
• # Answer: Exponential growth