Digital SATMath: Advanced MathMedium
A biologist is studying a population of insects that grows according to the function P(t) = 2000 / (1 + 4e^(-0.1t)), where P(t) is the population after t days. What is the population of the insects as t approaches infinity?
- A2000
- B0
- C400
- D8000
Show answer & explanationAnswer & explanation
Correct answer: A. 2000
As t approaches infinity, the term e^(-0.1t) approaches 0 because the exponent becomes a very large negative number (e^-large_number approaches 0). Therefore, the denominator (1 + 4e^(-0.1t)) approaches (1 + 4*0) = 1. So, P(t) approaches 2000 / 1 = 2000. This is the carrying capacity of the logistic model.
Why the other options are wrong
- B. This would happen if the numerator were 0 or the denominator approached infinity very quickly without a limit.
- C. Incorrect calculation; this might be 2000 / (1 + 4), if e^(-0.1t) was incorrectly treated as 1.
- D. Incorrect calculation; this might come from multiplying 2000 by 4, or misunderstanding the limit.
Limits of Logistic Functions at Infinity
For a logistic growth function P(t) = K / (1 + Ce^(-rt)), as t approaches infinity, the term e^(-rt) approaches 0, and the function approaches the carrying capacity K.
- Represents the long-term stable population size.
- The limit is always the numerator K.
- Assumes r > 0 for growth towards K.
Memory trick: Time flies, 'e' shrinks, K capacity it thinks.