Question 1 of 7intermediate🔧 ApplyNumerical2 marks

If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?

Correct Answer

Given: Population doubles in 3 years, so N_t = 2N_0 and t = 3 years

Using the integral form of exponential growth equation: N_t = N_₀e^(rt)

Substituting the values: 2N_₀ = N_₀e^(r×3)

Simplifying: 2 = e^(3r)

Taking natural logarithm on both sides: ln 2 = 3r r = ln 2 / 3 r = 0.693 / 3

r = 0.231 per year

Exercise: EXERCISES | Q: 2 | (Page 239)
For More Understanding

Explanation

This question tests the application of the exponential growth equation N_t = N_₀e^(rt) provided in the textbook. When a population doubles, N_t becomes 2N_₀. By substituting t = 3 years and solving for r using natural logarithms, we get r = 0.231 per year. The value of r represents the intrinsic rate of natural increase, which is a measure of the inherent potential of a population to grow under unlimited resource conditions.

Solution Steps

  1. Step 1: Identify given values - population doubles (N_t = 2N_₀) in time t = 3 years

  2. Step 2: Write the exponential growth equation: N_t = N_₀e^(rt)

  3. Step 3: Substitute values: 2N_₀ = N_₀e^(3r)

  4. Step 4: Simplify to get: 2 = e^(3r)

  5. Step 5: Take natural log: ln 2 = 3r, therefore r = ln 2/3 = 0.693/3 = 0.231 per year

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