Flashcards in 11.2.1 Exponential Growth Deck (11):
• Population growth can be modeled with a separable differential equation. Given the original size of the population as an initial condition and the rate of growth, you can solve the equation and predict the size of the population at any time.
- The beauty of mathematics is that it does such a good job modeling the events in nature.
- One model of population growth assumes that the change in population over time (dP/dt) is directly proportional to the current population (P). Using the constant k as a constant of proportionality produces a separable differential equation.
- Integrate both sides to solve the differential equation.
- Raising e to the power of each side allows you to solve for P, which can be written as P(t).
- At time t = 0 the population is e c . This is often denoted P 0 .
- Now you have a function that models population growth.
- Use the population growth function to model the growth of a petri dish population.
- The initial population is 1000 and the population doubles every two hours.
- On the left, first replace P 0 with the initial population. Since the population doubles every two hours, the population at t = 2 is 2000. When you plug in two you get an exponential equation with only the unknown k.
- On the right, you can solve the equation for k and plug it into the population function.
- Now you have a function that models the given population.
- Check to see if your function works. You know that after six hours your population will have doubled three times: to 2000, then to 4000, and finally to 8000. The function agrees!
- The exponential model of population growth fits a population that is growing with unlimited resources. But if populations grew exponentially then we would long ago have been up to our ears in everything. In real populations, some resource (food, water, home sites) always limits growth, either by reducing births or increasing deaths. Growth with a limiting resource can be modeled with the logistic equation.
- The rate of change of the population depends on the
population size itself (the first P). It also depends on how
close the population is to the carrying capacity (the second expression). When the population is at the carrying capacity (P = k 2 ), then dP/dt = 0; the population is not growing.
- Solving the logistic equation is challenging. The outline for the solution is presented here. The solution to the equation is simpler if we define a new variable x = P/k 2 .
- The equation can now be solved through separation of variables, but it helps to know and apply this identity: 1/(y(1 − y)) = 1/y + 1/(1 − y).
- Evaluate the integral from time zero to some arbitrary time T. This arbitrary time can later be generalized to any time by replacing it with the variable t.
- The final form of the equation is obtained after some
rearrangement. This looks complicated, but remember the logic of the model behind the original differential equation. Given that relatively simple model, this is the equation that can tell us how many individuals are expected in a population with a carrying capacity after a given period of time.
- The behavior of the solution to the logistic growth equation can best be visualized graphically.
- There are three elements of this graph to note. First, the population goes through an initial exponential growth phase before the effect of the carrying capacity influences the population. Second, there is an inflection point at half the carrying capacity where the rate of population growth begins to decline. This corresponds to a zero value for the second-order differential. Finally, the population levels off at the carrying capacity of 1000 individuals (in this case). At this point the rate of growth of the population is zero.
Suppose the fish population in a local lake increases at a yearly rate of 0.4 times the population at each moment. If there are 100,000 fish in the lake immediately after 10 years, which of the following is the fish population now?
1.8 × 10^3
Suppose the fish population in a local lake increases at a rate proportional to the population each moment. There were 1000 fish 5 years ago and 2000 fish 4 years ago. How many fish are there right now?
3.2 × 10^4
Suppose the bacteria population in a specimen increases at a rate proportional to the population at each moment. There were 100 bacteria 4 days ago and 100,000 bacteria 2 days ago. How many bacteria will there be by tomorrow?
3.2 × 10^9
Suppose the fish population in a local lake increases at a yearly rate of 0.4 times the population at each moment. There are 2000 fish in the lake right now. Which of the following is the fish population right after 10 years?
1.1 × 10^5
Suppose the bacteria population in a specimen increases at the hourly rate of 0.2 times the population at each moment. There are 1000 bacteria in the specimen right now. Which of the following is the bacteria population immediately after one day?
1.2 × 10^5
Suppose the fish population in a local lake increases at a yearly rate of 0.3 times the population at each moment. There are 1000 fish in the lake right now. Which of the following is the fish population at time t ?
P = 1000 e ^0.3t
Suppose the bacteria population in a specimen increases at an hourly rate of 0.2 times the population at each moment. After three days, there are 10,000,000 bacteria in the specimen. How many bacteria were there at the beginning?