AP Calculus AB: 11.2.1 Exponential Growth
This content explains how population growth can be modeled using differential equations. It covers the exponential growth model—where growth rate is proportional to the current population—and introduces the logistic growth model that accounts for resource limitations by incorporating carrying capacity.
Exponential Growth
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.
Key Terms
Exponential Growth
Population growth can be modeled with a separable differential equation. Given the original size of the population as an initial condition and the ...
note
The beauty of mathematics is that it does such a good job modeling the events in nature.
One model of population growth assu...
note 2
Solving the logistic equation is challenging. The outline for the solution is presented here. The solution to the equation is simpler if we...
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
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| Term | Definition |
|---|---|
Exponential Growth | 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. |
note |
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note 2 |
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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? | 6 |
Suppose the fish population in a local lake increases at a yearly rate of 0.3 times the population at each moment. Which of the following differential equations describes the rate of change of the fish population in the lake? | dP/dt=0.3P |