Task 1:
a. Here is a table showing the number of bacteria present after each hour for the first six hours, based on a growth rate of 8:
| Hour | Number of Bacteria |
|------|-------------------|
| 1 | 8 |
| 2 | 64 |
| 3 | 512 |
| 4 | 4,096 |
| 5 | 32,768 |
| 6 | 262,144 |
To determine the number of bacteria after 24 hours, we need to continue the exponential growth. After 6 hours, we have 262,144 bacteria. If we continue multiplying by 8 for 18 more hours, we get:
262,144 * 8^18 ≈ 92,233,720,368,550,400 bacteria.
b. This table represents exponential growth because the number of bacteria is increasing at an exponential rate. Each hour, the number of bacteria is multiplied by the growth rate, resulting in a larger number of bacteria each time. The growth is not linear or proportional, but rather exponential.
c. In this example, any nonzero number raised to the power of zero is equal to one because it represents the initial starting point of the growth. When you raise a number to the power of zero, it means that there have been zero iterations of the growth, so the value remains at the starting point, which is one.
d. The rule for this table is that the number of bacteria after each hour is equal to the previous number of bacteria multiplied by the growth rate.
e. If we started with 100 bacteria instead of 1, the rule for the table would still be the same. The only difference is that the initial starting point is different. So instead of multiplying by 8^x, we would multiply by 100 * 8^x.
Task 2:
a. One city that has experienced population growth is London, England. On January 1st, 2000, the population of London was approximately 7,172,036.
An exponential function to represent London's population, y, based on the number of years that pass, x, after a period of exponential growth can be represented by the equation: y = 7,172,036 * (1.02)^x. In this equation, the variable y represents the population of London, and x represents the number of years since 2000. The growth rate is 1.02, indicating a 2% annual growth in population.
b. Another city whose population starts larger than London's in 2000, but experiences population decline during the same time is Detroit, Michigan. On January 1st, 2000, the population of Detroit was approximately 951,270.
An exponential function to represent Detroit's population, y, based on the number of years that pass, x, after a period of population decline can be represented by the equation: y = 951,270 * (0.98)^x. In this equation, the variable y represents the population of Detroit, and x represents the number of years since 2000. The decline rate is 0.98, indicating a 2% annual decline in population.
c. The equations in (a) and (b) are similar in that they both represent exponential growth or decline. However, the key difference is the growth or decline rate used in each equation. London's equation has a growth rate greater than 1 (1.02), indicating population growth, while Detroit's equation has a decline rate less than 1 (0.98), indicating population decline.
d. To determine the year when the population of city (a) (London) first exceeds that of city (b) (Detroit), we need to set the two equations equal to each other and solve for x:
7,172,036 * (1.02)^x = 951,270 * (0.98)^x.
Solving for x, we find:
x ≈ 23.12.
Therefore, the population of London will first exceed that of Detroit in the 24th year since 2000, which is 2024.
e. To determine the year when the population of city (a) (London) is at least twice the size of the population of city (b) (Detroit), we set the two equations equal to each other and solve for x:
7,172,036 * (1.02)^x ≥ 2 * 951,270 * (0.98)^x.
Solving for x, we find:
x ≈ 19.61.
Therefore, the population of London will be at least twice the size of the population of Detroit in the 20th year since 2000, which is 2020.
Task 3:
a. To model the population of Western Lowland Gorillas after 5, 10, and 20 years, with an annual decline of 2.7%, we can use the formula: y = 360,000 * (1 - 0.027)^x, where y represents the population of gorillas and x represents the number of years since 2022.
After 5 years (x = 5):
y = 360,000 * (1 - 0.027)^5
≈ 360,000 * (0.973)^5
≈ 360,000 * 0.875
≈ 315,000 gorillas.
After 10 years (x = 10):
y = 360,000 * (1 - 0.027)^10
≈ 360,000 * (0.973)^10
≈ 360,000 * 0.784
≈ 282,240 gorillas.
After 20 years (x = 20):
y = 360,000 * (1 - 0.027)^20
≈ 360,000 * (0.973)^20
≈ 360,000 * 0.608
≈ 218,880 gorillas.
b. The table showing the Gorilla population after 5, 10, and 20 years would be:
| Years | Population |
|-------|------------------|
| 5 | 315,000 |
| 10 | 282,240 |
| 20 | 218,880 |
c. The table shows exponential decay because the population of Western Lowland Gorillas is decreasing at an exponential rate. Each year, the population declines by 2.7%, and this decline is compounded over time. The percentage decline is constant, but the actual number of gorillas being lost each year increases as the starting population decreases. Scientists can use exponential decay to predict population changes in endangered species by studying the rate of decline and projecting it into the future. By understanding the growth or decline rate, they can estimate how quickly a population will decrease and take necessary conservation actions to prevent extinction.
Overall, the table and calculations demonstrate the concept of exponential decay and its application to understanding population changes in endangered species.