(a) Show that the general solution of \( \frac{\mathrm{d} P}{\mathrm{~d} t} = k P \) is \( P = A e^{k t} \):

Separate variables:

\[ \frac{1}{P} \mathrm{d} P = k \mathrm{d} t \quad (M1) \]

Integrate both sides:

\[ \int \frac{1}{P} \mathrm{d} P = \int k \mathrm{d} t \quad (A1) \]

\[ \ln P = k t + c \quad (A1)(A1) \]

Exponentiate:

\[ P = e^{k t + c} \quad (A1) \]

\[ P = A e^{k t} \text{, where } A = e^c \quad (AG) \]

[5 marks]

(b) (i) Population after 10 years:

Initial condition: \( t = 0 \), \( P = 1000 \):

\[ P = A e^{0.003 t} \]

\[ 1000 = A e^{0} \implies A = 1000 \quad (A1) \]

At \( t = 10 \):

\[ P(10) = 1000 e^{0.003 \cdot 10} = 1000 e^{0.03} \approx 1030 \]

(A1)

[2 marks]

(b) (ii) Number of years for population to triple:

Triple the initial population: \( P = 3000 \).

\[ 3000 = 1000 e^{0.003 t} \quad (M1) \]

\[ e^{0.003 t} = 3 \]

\[ 0.003 t = \ln 3 \]

\[ t = \frac{\ln 3}{0.003} \approx 366 \text{ years} \quad (A1) \]

[2 marks]

(b) (iii) \( \lim_{t \to \infty} P \):

\[ P = 1000 e^{0.003 t} \]

As \( t \to \infty \), \( e^{0.003 t} \to \infty \), so:

\[ \lim_{t \to \infty} P = \infty \quad (A1) \]

[1 mark]

(c) (i) Solution of \( \frac{\mathrm{d} P}{\mathrm{~d} t} = (0.003 + 0.002 t) P \):

Separate variables:

\[ \int \frac{1}{P} \mathrm{d} P = \int (0.003 + 0.002 t) \mathrm{d} t \quad (M1) \]

\[ \ln P = 0.003 t + 0.001 t^2 + c \quad (A1)(A1) \]

\[ P = e^{0.003 t + 0.001 t^2 + c} \quad (A1) \]

Initial condition: \( t = 0 \), \( P = 1000 \):

\[ 1000 = e^c \quad (M1) \]

\[ P = 1000 e^{0.003 t + 0.001 t^2} \]

[5 marks]

(c) (ii) Number of years for population to triple:

\[ 3000 = 1000 e^{0.003 t + 0.001 t^2} \quad (M1) \]

\[ e^{0.003 t + 0.001 t^2} = 3 \]

\[ 0.001 t^2 + 0.003 t = \ln 3 \quad (A1) \]

Solve the quadratic equation \( 0.001 t^2 + 0.003 t – \ln 3 = 0 \):

Use quadratic formula or GDC: \( t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 0.001 \), \( b = 0.003 \), \( c = -\ln 3 \approx -1.098612 \).

\[ t \approx 31.7 \text{ years} \quad (M1)(A1) \]

[4 marks]

(d) Show that \( \frac{\mathrm{d} P}{\mathrm{~d} t} = \frac{m}{L} P (L – P) \):

Given:

\[ k = m \left(1 – \frac{P}{L}\right) \quad (A1) \]

Substitute into \( \frac{\mathrm{d} P}{\mathrm{~d} t} = k P \):

\[ \frac{\mathrm{d} P}{\mathrm{~d} t} = m \left(1 – \frac{P}{L}\right) P \]

\[ = \frac{m}{L} P (L – P) \quad (A1)(AG) \]

[2 marks]

(e) Solve \( \frac{\mathrm{d} P}{\mathrm{~d} t} = \frac{m}{L} P (L – P) \):

Separate variables:

\[ \int \frac{1}{P (L – P)} \mathrm{d} P = \int \frac{m}{L} \mathrm{d} t \quad (M1) \]

Use partial fractions:

\[ \frac{1}{P (L – P)} = \frac{A}{P} + \frac{B}{L – P} \quad (M1) \]

\[ 1 \equiv A (L – P) + B P \quad (A1) \]

Solve for \( A \) and \( B \):

\[ A = \frac{1}{L}, \quad B = \frac{1}{L} \quad (A1) \]

\[ \frac{1}{L} \int \left( \frac{1}{P} + \frac{1}{L – P} \right) \mathrm{d} P = \int \frac{m}{L} \mathrm{d} t \]

\[ \frac{1}{L} \left( \ln P – \ln (L – P) \right) = \frac{m}{L} t + c \]

\[ \ln \left( \frac{P}{L – P} \right) = m t + d \text{, where } d = c L \]

\[ \frac{P}{L – P} = C e^{m t} \text{, where } C = e^d \quad (A1) \]

Solve for \( P \):

\[ P (1 + C e^{m t}) = C L e^{m t} \quad (M1) \]

\[ P = \frac{C L e^{m t}}{1 + C e^{m t}} \quad (A1) \]

Alternatively:

\[ P = \frac{L}{D e^{-m t} + 1} \text{, where } D = \frac{1}{C} \]

[10 marks]

(f) Number of years to triple with \( P_0 = 1000 \), \( L = 10000 \), \( m = 0.003 \):

Initial condition: \( t = 0 \), \( P = 1000 \):

\[ 1000 = \frac{10000}{D + 1} \quad (M1) \]

\[ D + 1 = 10 \implies D = 9 \quad (A1) \]

Triple: \( P = 3000 \):

\[ 3000 = \frac{10000}{9 e^{-0.003 t} + 1} \quad (M1) \]

\[ 9 e^{-0.003 t} + 1 = \frac{10000}{3000} = \frac{10}{3} \]

\[ 9 e^{-0.003 t} = \frac{7}{3} \]

\[ e^{-0.003 t} = \frac{7}{27} \]

\[ -0.003 t = \ln \left( \frac{7}{27} \right) \]

\[ t = \frac{\ln \left( \frac{27}{7} \right)}{0.003} \approx 450 \text{ years} \quad (A1) \]

[4 marks]