Problem 1: Finding growth rate (dN/dt)

Given: N = 300, K = 600, r = 0.5/year. Find dN/dt.

Solution:

dN/dt = rN(K-N)/K
      = 0.5 × 300 × (600 - 300)/600
      = 0.5 × 300 × 300/600
      = 0.5 × 300 × 0.5
      = 0.5 × 150
      = 75 individuals/year

Note: N = 300 = K/2 = 600/2 → this is the maximum growth rate point.

Problem 2: Mark-Recapture Estimation

Given: First capture (n_{1}) = 200 frogs marked and released. Second sample (n_{2}) = 150. Marked in second sample (m) = 30.

Solution:

N = (n_{1} × n_{2}) / m
  = (200 × 150) / 30
  = 30,000 / 30
  = 1,000 frogs

Problem 3: Energy Flow Calculation

Given: Producer level fixes 5,000,000 kJ. Find energy at tertiary consumer ($T_{4}$) level.

Solution:

$T_{1}$ = 5,000,000 kJ
$T_{2}$ = 5,000,000 × 0.1 = 500,000 kJ
$T_{3}$ = 500,000 × 0.1 = 50,000 kJ
$T_{4}$ = 50,000 × 0.1 = 5,000 kJ

Or directly: $T_{4}$ = $T_{1}$ × (0.1)^{3} = 5,000,000 × 0.001 = 5,000 kJ ✓

Problem 4: NPP Calculation

Given: GPP = 12,000 g/$m^{2}$/year; Plant respiration = 4,800 g/$m^{2}$/year.

Solution:

NPP = GPP - Respiration = 12,000 - 4,800 = 7,200 g/$m^{2}$/year

If herbivores consume 1,440 g/$m^{2}$/year:

Herbivore consumption as % of NPP = (1,440/7,200) × 100 = 20%
Entering detritus food chain = 7,200 - 1,440 = 5,760 g/$m^{2}$/year (80%)

Problem 5: Maximum Growth Rate at K/2

Given: K = 1,000, r = 0.4/year. What is maximum growth rate (dN/dt_max)?

Solution:

Maximum growth rate occurs at N = K/2 = 500
dN/dt_max = r × (K/2) × (K - K/2)/K
           = r × (K/2) × (K/2)/K
           = r × $K^{2}$/4K
           = rK/4
           = 0.4 × 1,000/4
           = 100 individuals/year