A population of a very rare insect in an area in a discrete rural region in ASEAN will grow at a rate that is proportional to their current population. In the absence of any outside factors, the population will triple in two weeks’ time. On any given day there is a net migration (into the area) of 15 insects and there are also an extinction cause by identified factor such as 16 are eaten by the local bird population and 7 die of natural causes. If there are initially 100 insects in the area, predict either the population of the rare insect will survive or not? If not, when do they will die out?
Hmmm. If the population is p, then we have
dp/dt = kp
dp/p = k dt
ln(p) = kt + c
p = c e^kt
Now we know that p triples in 14 days, so we can forgo using e, and just use 3 as our base:
p(t) = Po * 3^(t/14)
so, starting with 100, and gaining 15 per day and losing 23 each day, making a net loss of 8 per day.
p(t) = 100*3^(t/14) - 8t
so, take a look at the graph and decide:
http://www.wolframalpha.com/input/?i=100*3^%28t%2F14%29+-+8t
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Let's break down the information given:
1. The population of the rare insect will triple in two weeks without any outside factors affecting it.
2. On any given day, there is a net migration of 15 insects into the area.
3. There is also a daily extinction rate of 16 due to being eaten by birds and 7 due to natural causes.
4. Initially, there are 100 insects in the area.
To determine if the population will survive or die out, we need to calculate the population change over time.
Let's start by calculating the population change without any outside factors:
In two weeks (14 days), the population triples, so the growth rate per day can be calculated by taking the cube root of 3:
Growth rate = (3)^(1/14) ≈ 1.057
Now, let's calculate the population change on any given day:
Net population change = (Growth rate * Current population) + Net migration - Extinction rate
= (1.057 * 100) + 15 - (16 + 7)
= 105.7 - 23
= 82.7 (approximately)
Since the net population change is positive (82.7), the population will increase each day. Therefore, the population of the rare insect will survive.
Note: The growth rate may not exactly match the population growth in real life, but this is a simplified representation based on the given information.
To predict whether the population of the rare insect will survive or not, we can analyze the net growth rate of the population.
Given that the population will triple in two weeks' time in the absence of any outside factors, we can assume that the growth rate is proportional to the current population. Let's denote the population at time t as P(t).
We can express the growth rate as a differential equation:
dP(t)/dt = k * P(t)
Where k is a constant representing the growth rate.
Now let's consider the daily changes in the population.
On any given day, there is a net migration into the area of 15 insects. This means that the growth rate for migration is 15. So the differential equation becomes:
dP(t)/dt = k * P(t) + 15
Additionally, the population faces extinction causes, where 16 insects are eaten by the local bird population and 7 die of natural causes. This means that the net decrease in the population due to these factors is 23. So the differential equation becomes:
dP(t)/dt = k * P(t) + 15 - 23
Simplifying, we get:
dP(t)/dt = k * P(t) - 8
Given that the initial population is 100 insects, we have:
P(0) = 100
To determine whether the population survives or not, we need to solve this differential equation.
One way to solve this is by using an integrating factor. Let's assume the integrating factor is e^(kt):
e^(kt) * dP/dt = k * e^(kt) * P(t) - 8 * e^(kt)
Now, this equation is in a form that can be integrated:
∫ e^(kt) * dP = ∫ k * e^(kt) * P(t) - 8 * e^(kt) dt
Integrating both sides, we get:
e^(kt) * P(t) = (k * ∫ e^(kt) * P(t) dt) - (8 * ∫ e^(kt) dt) + C
where C is the constant of integration.
At this point, solving for P(t) to predict the population becomes complex. It would require additional information on the constant k, as well as numerical methods for solving differential equations.
In conclusion, based on the given information, it is not possible to accurately predict whether the population of the rare insect will survive or die out without additional information or assumptions regarding the growth rate constant.