Respuesta :
In Argentina, there is a super colony of ants that spans over 3,700 miles and is estimated to have 307 million ants in it in 2023. Suppose a virus attacks the colony and the ants start to die off at a rate of 22% every 6 years. Assuming the number of ants is changing exponentially, how many ants do you estimate there will be in 4 years? Demonstrate mathematically why it is not true that the number of ants in the colony decreases by 11% every 3 years. Mathematically demonstrate and explain what the 3 year percent change will actually be. Algebraically determine how long it will take for the number of ants in the colony to decrease by 50%. (Hint: use Desmos or Logs) Write a formula that will model the number of ants in the colony with respect to time. Be sure to well-define variables. Show each step of your process as you work through the problems and explain the purpose of each step. Be specific!
Show drafts
Problem Solving:
1. Number of ants in 4 years:
a) Formula: The number of ants at any given time can be modeled by an exponential decay function:
N(t) = N_0 * (1 - decay_rate)^t
where:
N(t) is the number of ants at time t
N_0 is the initial number of ants (307 million)
decay_rate is the percentage of decrease per interval (as a decimal)
t is the time interval (in years)
b) Calculation:
Decay rate: We need to convert the given 22% decrease every 6 years to a decimal. Decay rate = (1 - 22%) = 0.78.
Time interval: We are interested in the number of ants in 4 years, so t = 4 years.
Number of ants in 4 years:
N(4) = 307 million * (0.78)^4
≈ 130.2 million ants
2. Misconception of 11% decrease every 3 years:
It is not accurate to say the number of ants decreases by 11% every 3 years because exponential decay is compounding over time. This means the percentage decrease is applied to the remaining population after each interval, not the initial population. So, the decrease after 3 years is not simply half of the 6-year decrease (22%/2).
3. Actual 3-year percent change:
a) Calculation:
Year 1: Population decreases by 22% of the initial: N(1) = 307 million * (0.78) ≈ 240.6 million
Year 2: Population decreases by 22% of the new population: N(2) = 240.6 million * (0.78) ≈ 187.3 million
Year 3: Population decreases by 22% of the Year 2 population: N(3) = 187.3 million * (0.78) ≈ 146.7 million
Percent change in 3 years:
((N_0 - N(3)) / N_0) * 100%
= ((307 million - 146.7 million) / 307 million) * 100% ≈ 51.9% decrease
Therefore, the actual decrease in the ant population over 3 years is approximately 51.9%, not 11%.
4. Time to decrease by 50%:
a) Approach: We can use the formula and solve for the time (t) when the population (N(t)) is 50% of the initial population (N_0).
b) Calculation:
N(t) = N_0 * (1 - decay_rate)^t
0.5 * N_0 = N_0 * (1 - 0.78)^t
0.5 = (1 - 0.78)^t
Taking the logarithm of both sides (use Desmos or calculator):
log(0.5) = t * log(1 - 0.78)
t ≈ 9.4 years (rounded to two decimal places)
5. Formula for number of ants:
Therefore, the formula that models the number of ants (N(t)) in the colony with respect to time (t) is:
N(t) = 307 million * (1 - 0.78)^t
where:
N(t) is the number of ants at time t (in years)
307 million is the initial number of ants
0.78 is the decay rate (1 - 22% decrease)
This formula takes into account the initial population and the exponential decay rate, accurately predicting the number of ants at any given time.
Show drafts
Problem Solving:
1. Number of ants in 4 years:
a) Formula: The number of ants at any given time can be modeled by an exponential decay function:
N(t) = N_0 * (1 - decay_rate)^t
where:
N(t) is the number of ants at time t
N_0 is the initial number of ants (307 million)
decay_rate is the percentage of decrease per interval (as a decimal)
t is the time interval (in years)
b) Calculation:
Decay rate: We need to convert the given 22% decrease every 6 years to a decimal. Decay rate = (1 - 22%) = 0.78.
Time interval: We are interested in the number of ants in 4 years, so t = 4 years.
Number of ants in 4 years:
N(4) = 307 million * (0.78)^4
≈ 130.2 million ants
2. Misconception of 11% decrease every 3 years:
It is not accurate to say the number of ants decreases by 11% every 3 years because exponential decay is compounding over time. This means the percentage decrease is applied to the remaining population after each interval, not the initial population. So, the decrease after 3 years is not simply half of the 6-year decrease (22%/2).
3. Actual 3-year percent change:
a) Calculation:
Year 1: Population decreases by 22% of the initial: N(1) = 307 million * (0.78) ≈ 240.6 million
Year 2: Population decreases by 22% of the new population: N(2) = 240.6 million * (0.78) ≈ 187.3 million
Year 3: Population decreases by 22% of the Year 2 population: N(3) = 187.3 million * (0.78) ≈ 146.7 million
Percent change in 3 years:
((N_0 - N(3)) / N_0) * 100%
= ((307 million - 146.7 million) / 307 million) * 100% ≈ 51.9% decrease
Therefore, the actual decrease in the ant population over 3 years is approximately 51.9%, not 11%.
4. Time to decrease by 50%:
a) Approach: We can use the formula and solve for the time (t) when the population (N(t)) is 50% of the initial population (N_0).
b) Calculation:
N(t) = N_0 * (1 - decay_rate)^t
0.5 * N_0 = N_0 * (1 - 0.78)^t
0.5 = (1 - 0.78)^t
Taking the logarithm of both sides (use Desmos or calculator):
log(0.5) = t * log(1 - 0.78)
t ≈ 9.4 years (rounded to two decimal places)
5. Formula for number of ants:
Therefore, the formula that models the number of ants (N(t)) in the colony with respect to time (t) is:
N(t) = 307 million * (1 - 0.78)^t
where:
N(t) is the number of ants at time t (in years)
307 million is the initial number of ants
0.78 is the decay rate (1 - 22% decrease)
This formula takes into account the initial population and the exponential decay rate, accurately predicting the number of ants at any given time.
