- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Logistic growth and phase lines practice question 3
A beetle population satisfies the differential equation below. Write it in the logistic form and find the carrying capacity .
In the logistic form the bracket starts with 1. What must you factor out of the right-hand side to make that happen, and what then multiplies inside the bracket?
Factor out : . Then compare the bracket with .
A whole number.
Theory UBC Math 100 §12.1Key results
Worked solution
Strategy
Factor out rN so that the bracket starts with 1. The number multiplying N inside the bracket is then 1/K.
Working
Factor out rN = 0.6N so the bracket starts with 1.
The coefficient of N in the bracket is 1/K, so K is its reciprocal.
Check: the carrying capacity is the nonzero steady state, where growth stops.
Answer
The answer is the carrying capacity is 300 beetles.
Check
The growth rate 0.6N − 0.002N² peaks where 0.6 − 0.004N = 0, at N = 150, which is K/2 as logistic growth requires.
Where this goes wrong
Reporting 0.002/0.6 = 1/300. That is 1/K, the number in front of N inside the bracket; the carrying capacity is its reciprocal, 0.6/0.002 = 300.
Leans on
Factoring out a common factor, from Functions, graphs and inverses
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