(a) The rate of increase of x is proportional to the ratio of the area infected to the area not yet infected, which gives \(\frac{dx}{dt} = k \frac{x}{20-x}\). Given \(\frac{dx}{dt} = 1\) when x = 1, we find \(k = 19\).

(b) Separate variables and integrate: \(\int \frac{20-x}{x} \, dx = \int 19 \, dt\). This gives \(20 \ln x - x = 19t + C\). Using \(t = 0\) and \(x = 1\), find \(C = -19\). Substitute \(t = 1\) to get \(20 \ln x - x = 0\), leading to \(x = e^{0.9 + 0.05x}\).

(c) Use the iterative formula \(x_{n+1} = e^{0.9 + 0.05x_n}\) starting with \(x_0 = 2\). Iterate to find \(x = 2.83\) correct to 2 decimal places.

(d) Set \(x = 20\) in the equation \(20 \ln x - x = 19t - 19\) to find \(t = 2.15\).

(i) The rate of change of volume \(V\) with respect to time \(t\) is given by:

\(\frac{dV}{dt} = 1 - 0.2\sqrt{h}.\)

The volume \(V\) is related to the depth \(h\) by \(V = 2h\), so:

\(\frac{dV}{dt} = 2 \frac{dh}{dt}.\)

Equating the two expressions for \(\frac{dV}{dt}\), we have:

\(2 \frac{dh}{dt} = 1 - 0.2\sqrt{h}.\)

Rearranging gives:

\(\frac{dh}{dt} = \frac{1 - 0.2\sqrt{h}}{2}.\)

To find \(T\), integrate with respect to \(h\) from 0 to 4:

\(T = \int_0^4 \frac{10}{5 - \sqrt{h}} \, dh.\)

(ii) Using the substitution \(u = 5 - \sqrt{h}\), we have:

\(\frac{du}{dh} = -\frac{1}{2\sqrt{h}} \Rightarrow du = -\frac{1}{2\sqrt{h}} \, dh.\)

Substitute for \(h\) and \(dh\):

\(T = \int_3^5 \frac{20(5-u)}{u} \, du.\)

Integrate to obtain:

\(T = 100\ln u - 20u \bigg|_3^5.\)

Substitute the limits:

\(T = (100\ln 5 - 100\ln 3) - (20 \times 5 - 20 \times 3).\)

Calculate to find \(T = 11.1\).

(i) The area of triangle \(OPN\) is \(\frac{1}{2}xy\). The gradient of the curve is \(\frac{dy}{dx}\). Therefore, \(\frac{dy}{dx} = \frac{1}{2}xy\).

(ii) Separate variables: \(\frac{1}{y} dy = \frac{1}{2}x dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{1}{2}x dx\).

This gives \(\ln y = \frac{1}{4}x^2 + C\).

Using the point \((0, 2)\), \(\ln 2 = C\).

Thus, \(\ln y = \frac{1}{4}x^2 + \ln 2\).

Exponentiate both sides: \(y = e^{\frac{1}{4}x^2 + \ln 2} = 2e^{\frac{1}{4}x^2}\).

(iii) The curve passes through \((0, 2)\) and has a rapidly increasing positive gradient for \(x > 0\).

(i) The rate of increase of N is proportional to (N - 150), so we write the differential equation as \(\frac{dN}{dt} = k(N - 150)\).

(ii) Given \(\frac{dN}{dt} = 60\) and \(N = 900\), substitute into the differential equation to find \(k\):

\(60 = k(900 - 150)\)

\(60 = 750k\)

\(k = 0.08\)

Separate variables and integrate:

\(\int \frac{1}{N - 150} \, dN = \int 0.08 \, dt\)

\(\ln(N - 150) = 0.08t + c\)

When \(t = 0\), \(N = 650\):

\(\ln(650 - 150) = c\)

\(\ln 500 = c\)

Thus, \(\ln(N - 150) = 0.08t + \ln 500\)

\(N - 150 = 500e^{0.08t}\)

\(N = 500e^{0.08t} + 150\)

(iii) Substitute \(t = 15\) into the expression for \(N\):

\(N = 500e^{0.08 \times 15} + 150\)

\(N \approx 1810\)

Alternatively, solve for \(t\) when \(N = 2000\):

\(2000 = 500e^{0.08t} + 150\)

\(1850 = 500e^{0.08t}\)

\(e^{0.08t} = 3.7\)

\(0.08t = \ln 3.7\)

\(t \approx 16.4\)

Since \(t = 16.4\) is greater than 15, the target is not met.

(i) The rate of change of the population is given by \(\frac{dN}{dt} = kN(1 - 0.01N)\). Given \(\frac{dN}{dt} = 0.32\) when \(N = 20\), we have:

\(0.32 = k \times 20 \times (1 - 0.01 \times 20)\)

\(0.32 = 20k \times 0.8\)

\(0.32 = 16k\)

\(k = 0.02\)

(ii) Separate variables and integrate:

\(\int \frac{1}{N(1 - 0.01N)} \, dN = \int 0.02 \, dt\)

Use partial fraction decomposition:

\(\frac{1}{N(1 - 0.01N)} = \frac{A}{N} + \frac{B}{1 - 0.01N}\)

Solving gives \(A = 1\) and \(B = 0.01\).

Integrate:

\(\int \left( \frac{1}{N} + \frac{0.01}{1 - 0.01N} \right) \, dN = \int 0.02 \, dt\)

\(\ln N - \ln(1 - 0.01N) = 0.02t + C\)

At \(t = 0, N = 20\):

\(\ln 20 - \ln(1 - 0.2) = C\)

\(C = \ln 25\)

Thus, \(\ln N - \ln(1 - 0.01N) = 0.02t + \ln 25\)

Rearrange to find \(t\):

\(t = 50 \ln \left( \frac{4N}{100 - N} \right)\)

(iii) When the population doubles, \(N = 40\):

\(t = 50 \ln \left( \frac{4 \times 40}{100 - 40} \right)\)

\(t = 50 \ln 4\)

\(t = 49.0\)