Population & Community Ecology

Question 1

(a) Separate variables: $\int \frac{dN}{N(1-N/K)} = \int r\,dt.$ Partial fractions: $\dfrac{1}{N(1-N/K)} = \dfrac{1}{N} + \dfrac{1/K}{1-N/K}$. (1) Integrate: $\ln N - \ln(1-N/K) = rt + C$, i.e. $\ln\dfrac{N}{1-N/K}=rt+C$. (2) At $t=0$: $C=\ln\dfrac{N_0}{1-N_0/K}$. (1) Solve for $N$: $\boxed{N(t) = \frac{K}{1 + \left(\frac{K-N_0}{N_0}\right)e^{-rt}}}.$ (2)

(b) Let $f(N)=rN(1-N/K)=rN - rN^2/K$. Then $f'(N)=r - 2rN/K = 0 \Rightarrow N=K/2$. (2) $f''(N) = -2r/K < 0$, so it's a maximum. (1) Maximum rate $f(K/2) = r(K/2)(1-1/2) = \dfrac{rK}{4}$. (1) Interpretation: population grows fastest at half carrying capacity; as $N\to K$ growth rate $\to 0$, so $K$ is the stable ceiling the density approaches (5.2.3). (1)

(c) Set $N(t^*)=K/2$: $1+\left(\frac{K-N_0}{N_0}\right)e^{-rt^*}=2$, so $e^{-rt^*}=\frac{N_0}{K-N_0}$. $\boxed{t^* = \frac{1}{r}\ln\!\left(\frac{K-N_0}{N_0}\right)}.$ (4) (2 for algebra, 2 for final form)

(d) Exponential: $N(t)=N_0 e^{rt}$. (1) As $N\ll K$, $1-N/K\to 1$, so $dN/dt\approx rN$ → early-phase logistic growth ≈ exponential (limit $K\to\infty$ or $N/K\to0$). (2) Condition: $N_0 \ll K$ / early times. (1)

(e)

import numpy as np, matplotlib.pyplot as plt
def logistic(r, K, N0, t):
    return K / (1 + ((K - N0)/N0)*np.exp(-r*t))
t = np.linspace(0, 30, 400)
r, K, N0 = 0.5, 1000, 10
N = logistic(r, K, N0, t)
tstar = (1/r)*np.log((K-N0)/N0)          # inflection time
plt.plot(t, N); plt.scatter([tstar],[K/2], color='red')
plt.xlabel('t'); plt.ylabel('N'); plt.show()

(3 for correct function, 2 for plot + inflection marking)

Question 2

(a) Set both derivatives to 0. $x(\alpha-\beta y)=0$, $y(\delta x-\gamma)=0$. Fixed points: $(0,0)$ and $\left(\dfrac{\gamma}{\delta}, \dfrac{\alpha}{\beta}\right)$. (4)

(b) Jacobian: $J=\begin{pmatrix}\alpha-\beta y & -\beta x\\ \delta y & \delta x-\gamma\end{pmatrix}.$ At $(x^*,y^*)=(\gamma/\delta,\ \alpha/\beta)$: $\alpha-\beta y^*=0$, $\delta x^*-\gamma=0$, so $J^*=\begin{pmatrix}0 & -\beta\gamma/\delta\\ \delta\alpha/\beta & 0\end{pmatrix}.$ (3) Characteristic: $\lambda^2 - \text{tr}\,\lambda + \det = 0$; $\text{tr}=0$, $\det = (\beta\gamma/\delta)(\delta\alpha/\beta)=\alpha\gamma>0$. So $\lambda^2 = -\alpha\gamma$, $\lambda = \pm i\sqrt{\alpha\gamma}$ — purely imaginary. (3) Classification: centre → neutrally stable closed orbits ⇒ sustained predator–prey oscillations, predator lagging prey (5.2.7). (1)

(c) $\dfrac{dV}{dt}=(\delta-\gamma/x)\dot x + (\beta-\alpha/y)\dot y$. (1) $\dot x = x(\alpha-\beta y)$, $\dot y=y(\delta x-\gamma)$. $(\delta-\gamma/x)x(\alpha-\beta y) = (\delta x-\gamma)(\alpha-\beta y)$. (1) $(\beta-\alpha/y)y(\delta x-\gamma) = (\beta y-\alpha)(\delta x-\gamma)$. (1) Sum $=(\delta x-\gamma)[(\alpha-\beta y)+(\beta y-\alpha)] = (\delta x-\gamma)(0)=0$. (2) ⇒ $V$ conserved.

(d) Modified prey eq: $\dot x = \alpha x(1-x/K) - \beta x y$. (1) The logistic term is density-dependent (self-limitation intensifies as $x$ rises), unlike density-independent factors (e.g. weather). It damps the neutral cycles into a stable spiral/focus converging to a fixed point rather than perpetual oscillation (5.2.4). (2)

(e)

import numpy as np
from scipy.integrate import solve_ivp
a,b,d,g = 1.1, 0.4, 0.1, 0.4
def f(t, z):
    x,y = z
    return [a*x - b*x*y, d*x*y - g*y]
sol = solve_ivp(f, [0,50], [10,5], t_eval=np.linspace(0,50,1000))
# sol.y[0]=prey, sol.y[1]=predator

(3 marks: correct RHS, params, integration call)

Question 3

(a) (4)

• Type I: convex, low early/mid mortality, sharp late drop → K-selected (large mammals, humans).
• Type II: straight line (log scale), constant mortality per age → intermediate (birds, rodents).
• Type III: concave, very high juvenile mortality, few survive to reproduce → r-selected (fish, insects, many plants). (1 each shape, 1 for r/K links)

(b) Constant survival $s=0.7$ ⇒ $l_x = s^x = 0.7^x$. (2) $l_3 = 0.7^3 = 0.343$. Survivors ≈ $343$ of 1000. (1)

(c) $p = (0.4,0.3,0.2,0.1)$. $\sum p_i^2 = 0.16+0.09+0.04+0.01 = 0.30$. Simpson $D = 1-0.30 = \boxed{0.70}$. (2) Shannon $H = -(0.4\ln0.4 + 0.3\ln0.3 + 0.2\ln0.2 + 0.1\ln0.1)$ $= -(-0.3665 -0.3612 -0.3219 -0.2303) = \boxed{1.2799}$. (2) Interpretation: high $D$ (≈0.7, near max 0.75 for 4 spp) and moderate–high $H$ indicate fairly even, diverse community; greater diversity confers stability/resilience and ecosystem function (5.2.11). (1)

(d) A keystone species exerts a disproportionately large effect on community structure relative to its abundance (5.2.10). Its removal can release a competitor or collapse a trophic interaction, causing secondary extinctions and a larger drop in $D/H$ than losing an abundant but functionally redundant dominant. (2)

[
  {"claim":"Logistic max growth rate at N=K/2 equals rK/4","code":"r,K,N=symbols('r K N',positive=True); f=r*N*(1-N/K); crit=solve(diff(f,N),N)[0]; result=(simplify(crit-K/2)==0) and (simplify(f.subs(N,K/2)-r*K/4)==0)"},
  {"claim":"Inflection time t*=(1/r)ln((K-N0)/N0) for N0=10,K=1000,r=0.5 gives ~9.19","code":"r,K,N0=0.5,1000,10; tstar=(1/r)*ln(Rational(K-N0,N0)); result=abs(float(tstar)-9.1901)<1e-3"},
  {"claim":"Simpson D=0.70 and l_3=0.343","code":"ps=[Rational(4,10),Rational(3,10),Rational(2,10),Rational(1,10)]; D=1-sum(p**2 for p in ps); l3=Rational(7,10)**3; result=(D==Rational(7,10)) and (l3==Rational(343,1000))"},
  {"claim":"Shannon H approx 1.2799","code":"ps=[0.4,0.3,0.2,0.1]; H=-sum(p*ln(p) for p in ps); result=abs(float(H)-1.27985)<1e-3"},
  {"claim":"LV coexistence eigenvalues purely imaginary: det=alpha*gamma, trace=0","code":"a,b,d,g=symbols('a b d g',positive=True); J=Matrix([[0,-b*g/d],[d*a/b,0]]); result=(J.trace()==0) and (simplify(J.det()-a*g)==0)"}
]