(a) The rate of water being added is 30 litres per minute, so \(a = 30\). The rate of water being lost is proportional to the volume, \(0.01V\), so \(b = 0.01\).

(b) The differential equation is \(\frac{dV}{dt} = 30 - 0.01V\). Separate variables and integrate:

\(\int \frac{1}{30 - 0.01V} \, dV = \int \, dt\)

\(-100 \ln(30 - 0.01V) = t + C\)

Using initial condition \(V = 0\) when \(t = 0\), find \(C\):

\(-100 \ln(30) = C\)

Thus, \(-100 \ln(30 - 0.01V) = t - 100 \ln(30)\)

\(100 \ln \left( \frac{30}{30 - 0.01V} \right) = t\)

For \(V = 1000\):

\(100 \ln \left( \frac{30}{30 - 10} \right) = t\)

\(t = 40.5\)

(c) Solving for \(V\) in terms of \(t\):

\(V = 3000(1 - e^{-0.01t})\)

As \(t \to \infty\), \(e^{-0.01t} \to 0\), so \(V \to 3000\).

(i) To find an expression for \(N\) in terms of \(t\), we start by separating variables and integrating both sides:

\(\int \frac{1}{N^{0.5}} \, dN = \int 1.2e^{-0.02t} \, dt\).

This gives:

\(2N^{0.5} = -60e^{-0.02t} + C\).

Using the initial condition \(N = 100\) when \(t = 0\), we find \(C\):

\(2(100)^{0.5} = -60e^{0} + C\)

\(20 = -60 + C\)

\(C = 80\).

Thus, the equation becomes:

\(2N^{0.5} = -60e^{-0.02t} + 80\).

Solving for \(N\):

\(N^{0.5} = -30e^{-0.02t} + 40\)

\(N = (40 - 30e^{-0.02t})^2\).

(ii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to 40^2 = 1600\).

(i) To solve the differential equation \(\frac{dN}{dt} = \frac{N(1800 - N)}{3600}\), we first separate variables:

\(\frac{1}{N(1800 - N)} = \frac{1}{3600} \frac{dN}{dt}.\)

We use partial fraction decomposition:

\(\frac{1}{N(1800 - N)} = \frac{A}{N} + \frac{B}{1800 - N}.\)

Solving for \(A\) and \(B\), we find:

\(\frac{2}{N} + \frac{2}{1800 - N}.\)

Integrating both sides gives:

\(2 \ln N - 2 \ln (1800 - N) = \frac{t}{2} + C.\)

Using the initial condition \(N = 300\) when \(t = 0\), we solve for \(C\):

\(2 \ln 300 - 2 \ln 1500 = C.\)

Thus, the solution is:

\(N = \frac{1800e^{t/2}}{5 + e^{t/2}}.\)

(ii) As \(t \to \infty\), \(e^{t/2} \to \infty\), so \(N \to 1800\).

(i) Separate variables: \(\frac{1}{\theta - A} d\theta = -k \, dt\).

Integrate both sides: \(\ln(\theta - A) = -kt + C\).

Use initial condition \(\theta = 4A\) when \(t = 0\):

\(\ln(4A - A) = C \Rightarrow C = \ln 3A\).

Thus, \(\ln(\theta - A) = -kt + \ln 3A\).

(ii) Substitute \(\theta = 3A\), \(t = 1\):

\(\ln(3A - A) = -k(1) + \ln 3A\).

\(\ln 2A = -k + \ln 3A\).

\(k = \ln 3A - \ln 2A = \ln \frac{3}{2}\).

(iii) Substitute \(t = 2\) into \(\ln(\theta - A) = -kt + \ln 3A\):

\(\ln(\theta - A) = -2\ln \frac{3}{2} + \ln 3A\).

\(\ln(\theta - A) = \ln \left(\frac{3A}{\left(\frac{3}{2}\right)^2}\right)\).

\(\theta - A = \frac{3A}{\frac{9}{4}} = \frac{4}{3}A\).

\(\theta = \frac{4}{3}A + A = \frac{7}{3}A\).

(i) Separate variables and integrate:

\(\int \frac{1}{N} \, dN = \int k \cos(0.02t) \, dt\)

\(\ln N = \frac{k}{0.02} \sin(0.02t) + C\)

Using \(t = 0, N = 125\):

\(\ln 125 = C\)

Thus, \(\ln N = \frac{k}{0.02} \sin(0.02t) + \ln 125\)

(ii) Given \(N = 166\) when \(t = 30\):

\(\ln 166 = \frac{k}{0.02} \sin(0.6) + \ln 125\)

Solving for \(k\), \(k \approx 0.0100479\) (accept \(k = 0.01\))

(iii) Expression for \(N\):

\(N = 125e^{0.502 \sin(0.02t)}\)

Least value occurs when \(\sin(0.02t) = -1\):

\(N = 125e^{-0.502} \approx 75.6\)

(i) Separate variables: \(\frac{1}{x} dx = -kt dt\).

Integrate both sides: \(\int \frac{1}{x} dx = \int -kt dt\).

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

Use initial condition \(t = 0, x = 100\) to find \(C\):

\(\ln 100 = C\).

Thus, \(\ln x = -\frac{1}{2}kt^2 + \ln 100\).

(ii) Use \(t = 20, x = 90\) to find \(k\):

\(\ln 90 = -\frac{1}{2}k(20)^2 + \ln 100\).

Solve for \(k\):

\(k = \frac{2(\ln 100 - \ln 90)}{400}\).

Substitute \(x = 50\) to find \(t\):

\(\ln 50 = -\frac{1}{2}k t^2 + \ln 100\).

\(t^2 = \frac{400(\ln 100 - \ln 50)}{\ln 100 - \ln 90}\).

Calculate \(t\):

\(t = 51.3\).

(i) Separate variables and integrate:

\\(\int \frac{1}{\sqrt{P-A}} \, dP = \int -k \, dt \\)

Integrating gives:

\\(2\sqrt{P-A} = -kt + c \\)

(ii) Use initial conditions to find constants:

When \\(t = 0, P = 5A \\):

\\(2\sqrt{5A-A} = c \\)

\\(c = 4\sqrt{A} \\)

When \\(t = 2, P = 2A \\):

\\(2\sqrt{2A-A} = -2k + 4\sqrt{A} \\)

Solve for \\(k \\):

\\(2\sqrt{A} = -2k + 4\sqrt{A} \\)

\\(k = \sqrt{A} \\)

(iii) Substitute \\(P = A \\) and solve for \\(t \\):

\\(2\sqrt{A-A} = -kt + 4\sqrt{A} \\)

\\(0 = -kt + 4\sqrt{A} \\)

\\(t = 4 \\)

(iv) Rearrange to express \\(P \\) in terms of \\(A \\) and \\(t \\):

\\(2\sqrt{P-A} = -kt + 4\sqrt{A} \\)

Square both sides:

\\(4(P-A) = (4\sqrt{A} - kt)^2 \\)

\\(P = \frac{1}{4}A(4 + (4-t)^2) \\)

(a) Separate variables: \(\int \frac{1}{x} \, dx = \int ke^{-0.1t} \, dt\).

Integrate both sides: \(\ln x = -10ke^{-0.1t} + C\).

Use initial condition \(x = 20\) when \(t = 0\): \(\ln 20 = C\).

Thus, \(\ln x = 10k(1 - e^{-0.1t}) + \ln 20\).

(b) Given \(x = 40\) when \(t = 10\), substitute into the equation:

\(\ln 40 = 10k(1 - e^{-1}) + \ln 20\).

Solve for \(10k\): \(10k = \frac{\ln 2}{1 - e^{-1}} = 1.09654\).

As \(t \to \infty\), \(e^{-0.1t} \to 0\), so \(x \to 20e^{10k}\).

Substitute \(10k = 1.09654\): \(x \to 59.9\).

(a) Separate variables: \(\frac{dN}{N^{\frac{3}{2}}} = k \cos 0.02t \, dt\).

Integrate both sides: \(-\frac{2}{\sqrt{N}} = 50k \sin 0.02t - 0.2\).

Use initial condition \(t = 0, N = 100\) to find constant: \(-\frac{2}{10} = -0.2\).

(b) Substitute \(N = 625\) and \(t = 50\) into the expression: \(-\frac{2}{25} = 50k \sin 1 - 0.2\).

Solve for \(k\): \(k = 0.00285\).

(c) Rearrange to find \(N\) in terms of \(t\): \(N = 4(0.2 - 0.142(607) \sin 0.02t)^{-2}\).

Find the greatest value of \(N\) by evaluating the expression at critical points.

(i) Separate variables: \(\frac{dx}{(20-x)(40-x)} = \frac{1}{5} dt\).

Use partial fractions: \(\frac{1}{(20-x)(40-x)} = \frac{A}{20-x} + \frac{B}{40-x}\).

Find \(A\) and \(B\): \(A = \frac{1}{20}\), \(B = -\frac{1}{20}\).

Integrate both sides: \(\int \left( \frac{1}{20} \ln(20-x) - \frac{1}{20} \ln(40-x) \right) dx = \int \frac{1}{5} dt\).

Integrate: \(-\frac{1}{20} \ln(20-x) + \frac{1}{20} \ln(40-x) = \frac{1}{5} t + C\).

Use initial condition \(x = 10\) when \(t = 0\) to find \(C\).

Solve for \(x\): \(x = \frac{60e^{4t} - 40}{3e^{4t} - 1}\).

(ii) As \(t\) becomes large, \(x\) approaches 20.

(i) Separate variables and integrate:

\(\int \frac{1}{x} \, dx = -k \int \frac{1}{\sqrt{t}} \, dt\)

\(\ln x = -2k\sqrt{t} + C\)

Use initial condition \(x = 100\) when \(t = 0\):

\(\ln 100 = C\)

Thus, \(\ln x = -2k\sqrt{t} + \ln 100\)

(ii) Given \(t = 25\) when \(x = 80\):

\(\ln 80 = -2k\sqrt{25} + \ln 100\)

\(\ln 80 = -10k + \ln 100\)

\(10k = \ln 100 - \ln 80\)

\(k = \frac{\ln 100 - \ln 80}{10}\)

Find \(t\) when \(x = 40\):

\(\ln 40 = -2k\sqrt{t} + \ln 100\)

\(2k\sqrt{t} = \ln 100 - \ln 40\)

\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2k}\)

Substitute \(k\):

\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2 \times \frac{\ln 100 - \ln 80}{10}}\)

\(\sqrt{t} = \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)}\)

\(t = \left( \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)} \right)^2\)

\(t \approx 64.1\)

(i) Separate variables: \(\frac{1}{x(4-x)} \frac{dx}{dt} = k\).

Integrate both sides: \(\int \frac{1}{x(4-x)} \, dx = \int k \, dt\).

Use partial fractions: \(\frac{1}{x(4-x)} = \frac{A}{x} + \frac{B}{4-x}\).

Find \(A = \frac{1}{4}\) and \(B = \frac{1}{4}\).

Integrate: \(\frac{1}{4} \ln x - \frac{1}{4} \ln(4-x) = kt + C\).

Use initial conditions \(t = 0, x = 0.4\) and \(t = 2, x = 2\) to find \(C\) and \(k\).

\(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Substitute \(t = 2, x = 2\): \(\ln 1 = 8k - 8k\).

Substitute \(t = 0, x = 0.4\): \(\ln \left( \frac{0.4}{3.6} \right) = -8k\).

Solve for \(k\): \(k = \frac{1}{4} \ln 3\).

(ii) When 90% of the area is infected, \(x = 3.6\).

Substitute \(x = 3.6\) into \(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Solve for \(t\): \(t = 4\).

(i) Separate variables and integrate:

\(\int \frac{1}{\sqrt{M}} \, dM = \int k \cos(0.02t) \, dt\)

Integrating both sides gives:

\(2\sqrt{M} = 50k \sin(0.02t) + C\)

Using initial condition \(M = 100\) when \(t = 0\), we find \(C = 20\).

Thus, \(2\sqrt{M} = 50k \sin(0.02t) + 20\).

(ii) Given \(M = 196\) when \(t = 50\), substitute to find \(k\):

\(2\sqrt{196} = 50k \sin(1) + 20\)

\(28 = 50k \sin(1) + 20\)

\(8 = 50k \sin(1)\)

\(k = \frac{8}{50 \sin(1)} \approx 0.190\)

(iii) Express \(M\) in terms of \(t\):

\(2\sqrt{M} = 50 \times 0.190 \sin(0.02t) + 20\)

\(\sqrt{M} = 4.75 \sin(0.02t) + 10\)

\(M = (4.75 \sin(0.02t) + 10)^2\)

The least possible number of micro-organisms occurs when \(\sin(0.02t) = -1\):

\(M = (4.75(-1) + 10)^2 = (5.25)^2 = 27.5625\)

Thus, the least possible number is approximately 28 or 27.5 or 27.6.

(i) Separate variables and integrate:

\(\int \frac{1}{x} \, dx = \int \frac{e^{-t}}{k + e^{-t}} \, dt\)

\(\ln x = -\ln(k + e^{-t}) + C\)

Given \(x = 10\) when \(t = 0\), substitute to find \(C\):

\(\ln 10 = -\ln(k + 1) + C\)

Thus, \(C = \ln 10 + \ln(k + 1)\).

Therefore, the solution is:

\(\ln x - \ln 10 = -\ln(k + e^{-t}) + \ln(k + 1)\)

(ii) Substitute \(x = 20\), \(t = 1\) into the solution:

\(\ln 20 - \ln 10 = -\ln(k + e^{-1}) + \ln(k + 1)\)

Simplify and solve for \(k\):

\(\ln 2 = \ln \left( \frac{k + 1}{k + \frac{1}{e}} \right)\)

\(2(k + \frac{1}{e}) = k + 1\)

\(2k + \frac{2}{e} = k + 1\)

\(k = 1 - \frac{2}{e}\)

(iii) As \(t \to \infty\), \(e^{-t} \to 0\):

\(\ln x = -\ln k + \ln(k + 1)\)

\(x = \frac{k + 1}{k}\)

Substitute \(k = 1 - \frac{2}{e}\):

\(x = \frac{1 - \frac{2}{e} + 1}{1 - \frac{2}{e}} = \frac{2 - \frac{2}{e}}{1 - \frac{2}{e}}\)

\(x \to 48\) as \(t \to \infty\).

(i) Separate variables and integrate:

\(\int \frac{1}{R} \, dR = \int \left( \frac{1}{x} - 0.57 \right) \, dx\)

\(\ln R = \ln x - 0.57x + C\)

Exponentiate both sides:

\(R = e^{\ln x - 0.57x + C} = e^C \cdot xe^{-0.57x}\)

When \(x = 0.5, R = 16.8\):

\(16.8 = e^C \cdot 0.5e^{-0.285}\)

\(e^C = \frac{16.8}{0.5e^{-0.285}} = 33.6\)

Thus, \(R = 33.6xe^{(0.285 - 0.57x)}\).

(ii) To find the maximum \(R\), set \(\frac{dR}{dx} = 0\):

\(\frac{1}{x} - 0.57 = 0\)

\(x = 0.57^{-1} \approx 1.75\)

Substitute back to find \(R\):

\(R = 33.6 \times 1.75 \times e^{(0.285 - 0.57 \times 1.75)} \approx 28.8\)

(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, \(N = 40\):

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

\(t = 50 \ln 4\)

\(t = 49.0\)

To solve the differential equation \(3y^2 \frac{dy}{dx} = 4(y^3 + 1) \cos^2 x\), we first use the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\) to rewrite the equation as:

\(3y^2 \frac{dy}{dx} = 2(y^3 + 1)(1 + \cos 2x)\).

Separate variables and integrate:

\(\int \frac{3y^2}{y^3 + 1} \, dy = \int 2(1 + \cos 2x) \, dx\).

The left side integrates to \(\ln(y^3 + 1)\), and the right side integrates to \(2x + \sin 2x + C\).

Thus, \(\ln(y^3 + 1) = 2x + \sin 2x + C\).

Using the initial condition \(y = 2\) when \(x = 0\), we find:

\(\ln(2^3 + 1) = \ln 9 = C\).

So, the equation becomes:

\(\ln(y^3 + 1) = 2x + \sin 2x + \ln 9\).

At stationary points, \(\frac{dy}{dx} = 0\), which implies \(\cos^2 x = 0\). Therefore, \(x = \frac{\pi}{2}\) or \(x = \frac{3\pi}{2}\).

For \(x = \frac{\pi}{2}\):

\(\ln(y^3 + 1) = 2\left(\frac{\pi}{2}\right) + \sin \pi + \ln 9 = \pi + \ln 9\).

\(y^3 + 1 = e^{\pi + \ln 9} = 9e^{\pi}\).

\(y^3 = 9e^{\pi} - 1\).

\(y = (9e^{\pi} - 1)^{1/3} \approx 5.9\).

For \(x = \frac{3\pi}{2}\):

\(\ln(y^3 + 1) = 2\left(\frac{3\pi}{2}\right) + \sin 3\pi + \ln 9 = 3\pi + \ln 9\).

\(y^3 + 1 = e^{3\pi + \ln 9} = 9e^{3\pi}\).

\(y^3 = 9e^{3\pi} - 1\).

\(y = (9e^{3\pi} - 1)^{1/3} \approx 48.1\).

(i) The volume of the cone is given by \(V = \frac{1}{3} \pi r^2 h\). Given the semi-vertical angle is 60°, \(r = h \tan(60^{\circ}) = \sqrt{3}h\). Thus, \(V = \pi h^3\). The rate of change of volume is \(\frac{dV}{dt} = -k\sqrt{h}\). Using \(V = \pi h^3\), \(\frac{dV}{dh} = 3\pi h^2\). Therefore, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 3\pi h^2 \cdot \frac{dh}{dt}\). Equating gives \(3\pi h^2 \cdot \frac{dh}{dt} = -k\sqrt{h}\), leading to \(\frac{dh}{dt} = -\frac{k}{3\pi} h^{-\frac{3}{2}}\). Let \(A = \frac{k}{3\pi}\), then \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\).

(ii) Separate variables: \(\int h^{\frac{3}{2}} \, dh = -A \int \, dt\). Integrating gives \(\frac{2}{5} h^{\frac{5}{2}} = -At + C\). At \(t = 0, h = H\), so \(C = \frac{2}{5} H^{\frac{5}{2}}\). At \(t = 60, h = 0\), so \(0 = -60A + \frac{2}{5} H^{\frac{5}{2}}\), giving \(A = \frac{1}{150} H^{\frac{5}{2}}\). Substitute back to find \(t = 60 \left( 1 - \left( \frac{h}{H} \right)^{\frac{5}{2}} \right)\).

(iii) Substitute \(h = \frac{1}{2}H\) into the expression for t: \(t = 60 \left( 1 - \left( \frac{1}{2} \right)^{\frac{5}{2}} \right) = 49.4\).

(i) The rate of change of volume \(\frac{dV}{dt}\) is given by the inflow minus the outflow: \(\frac{dV}{dt} = 80 - kV\).

Separate variables: \(\frac{dV}{80 - kV} = dt\).

Integrate both sides: \(-\frac{1}{k} \ln(80 - kV) = t + C\).

Using \(t = 0\) and \(V = 0\), find \(C\): \(-\frac{1}{k} \ln(80) = C\).

Substitute back: \(-\frac{1}{k} \ln(80 - kV) = t - \frac{1}{k} \ln(80)\).

Simplify to find \(V\): \(V = \frac{1}{k}(80 - 80e^{-kt})\).

(ii) Use the iterative formula \(k_{n+1} = \frac{4 - 4e^{-15k_n}}{25}\) with \(k_0 = 0.1\).

Perform iterations: \(k_1 = 0.1368\), \(k_2 = 0.1399\), \(k_3 = 0.1400\), \(k_4 = 0.1400\).

Final value of \(k\) is 0.14 to 2 significant figures.

(iii) Substitute \(t = 20\) into \(V = \frac{1}{k}(80 - 80e^{-kt})\) with \(k = 0.14\).

Calculate \(V \approx 535\) cm³.

As \(t \to \infty\), \(V \to \frac{80}{k} \approx 569\) cm³.

(i) Substitute \(x = 5e^{-3t}\) into the differential equation \(\frac{dy}{dt} = -0.6xy\) to get \(\frac{dy}{dt} = -0.6(5e^{-3t})y = -3e^{-3t}y\).

Separate variables: \(\frac{1}{y} dy = -3e^{-3t} dt\).

Integrate both sides: \(\ln y = e^{-3t} + C\).

Use the initial condition \(t = 0, y = 70\) to find \(C\): \(\ln 70 = 1 + C\), so \(C = \ln 70 - 1\).

Thus, \(\ln y = e^{-3t} + \ln 70 - 1\).

Rearrange to find \(y\): \(y = 70(e^{-3t} - 1)\).

(ii) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(y \to 70(-1) = -70\).

The percentage of the initial mass of B remaining is \(p = \frac{y}{70} \times 100\).

Thus, \(p \to \frac{100}{e}\) as \(t \to \infty\).

(i) The rate of formation of X is proportional to the product of the masses of Y and Z, which are \(10-x\) and \(20-x\) respectively. Therefore, \(\frac{dx}{dt} = k(10-x)(20-x)\). Given \(\frac{dx}{dt} = 2\) when t = 0 and x = 0, we have \(2 = k(10)(20)\), so \(k = 0.01\). Thus, \(\frac{dx}{dt} = 0.01(10-x)(20-x)\).

(ii) Separate variables: \(\frac{1}{(10-x)(20-x)} dx = 0.01 dt\).

Use partial fractions: \(\frac{1}{(10-x)(20-x)} = \frac{A}{10-x} + \frac{B}{20-x}\).

Solving gives \(A = \frac{1}{10}\) and \(B = -\frac{1}{10}\).

Integrate: \(-\frac{1}{10} \ln(10-x) + \frac{1}{10} \ln(20-x) = 0.01t + C\).

At t = 0, x = 0: \(C = \frac{1}{10} \ln 2\).

Rearrange: \(-\ln(10-x) + \ln(20-x) = 0.1t + \ln 2\).

\(\ln\left(\frac{20-x}{10-x}\right) = 0.1t + \ln 2\).

\(\frac{20-x}{10-x} = 2\exp(0.1t)\).

\(20-x = 2(10-x)\exp(0.1t)\).

\(x = \frac{20(\exp(0.1t)-1)}{2\exp(0.1t)-1}\).

(iii) As t becomes large, \(\exp(0.1t)\) becomes very large, so \(x\) approaches 10.

(i) The gradient of the curve is given by \(\frac{dy}{dx} = kxy\), where \(k\) is a constant. At \((1, 2)\), \(\frac{dy}{dx} = 4\), so \(4 = k \cdot 1 \cdot 2\), giving \(k = 2\). Thus, \(\frac{dy}{dx} = 2xy\).

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

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

This gives \(\ln |y| = x^2 + C\).

Exponentiate: \(y = e^{x^2 + C}\).

At \((1, 2)\), \(2 = e^{1^2 + C}\), so \(2 = e^{1 + C}\), giving \(C = \ln 2 - 1\).

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

(ii) The gradient at \((-1, 2)\) is \(\frac{dy}{dx} = 2(-1)(2) = -4\).

(a)(i) The gradient of the normal at \(P\) is \(-\frac{1}{\frac{dy}{dx}}\). The length \(MN\) is the horizontal distance from \(M\) to \(N\), which is \(y \cdot \frac{dy}{dx}\). Therefore, \(\frac{MN}{y} = \frac{dy}{dx}\).

(a)(ii) The area of triangle \(PMN\) is \(\frac{1}{2} \times MN \times y = \frac{1}{2} \times y \cdot \frac{dy}{dx} \times y = \frac{1}{2}y^2 \frac{dy}{dx}\). Given that this area equals \(\tan x\), we have \(\frac{1}{2}y^2 \frac{dy}{dx} = \tan x\).

(b) Separate variables: \(\frac{1}{2}y^2 dy = \tan x \, dx\).

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

This gives \(\frac{1}{6}y^3 = -\ln |\cos x| + C\).

Using the initial condition \(y = 1\) when \(x = 0\), we find \(C = \frac{1}{6}\).

Thus, \(\frac{1}{6}y^3 = -\ln |\cos x| + \frac{1}{6}\).

Rearranging gives \(y^3 = 1 - 6\ln |\cos x|\).

Therefore, \(y = \sqrt[3]{1 - 6\ln \cos x}\).

(i) The differential equation representing the biologist's claim is:

\(\frac{dA}{dt} = k\sqrt{2A - 5}\)

(ii) To solve the differential equation, separate variables and integrate:

\(\int \frac{1}{\sqrt{2A - 5}} \, dA = \int k \, dt\)

Integrating both sides, we obtain:

\((2A - 5)^{1/2} = kt + C\)

Using the initial condition \(t = 0\) and \(A = 7\), we find \(C\):

\((2 \times 7 - 5)^{1/2} = C\)

\(C = 3\)

Using \(t = 10\) and \(A = 27\) to find \(k\):

\((2 \times 27 - 5)^{1/2} = 10k + 3\)

\(7 = 10k + 3\)

\(k = 0.4\)

Substitute \(t = 20\), \(C = 3\), and \(k = 0.4\) to find \(A\):

\((2A - 5)^{1/2} = 0.4 \times 20 + 3\)

\((2A - 5)^{1/2} = 11\)

\(2A - 5 = 121\)

\(2A = 126\)

\(A = 63\)

(i) The rate of formation is proportional to \((20 - x)\), so \(\frac{dx}{dt} = k(20 - x)\). Given \(\frac{dx}{dt} = 1\) when \(t = 0\) and \(x = 0\), we have \(1 = k(20 - 0)\), thus \(k = 0.05\).

(ii) Separate variables: \(\frac{dx}{20 - x} = 0.05 \, dt\). Integrate both sides: \(-\ln|20 - x| = 0.05t + C\). Using \(t = 0\), \(x = 0\), we find \(C = -\ln 20\). Thus, \(\ln(20 - x) = \frac{1}{20}t\).

(iii) Substitute \(t = 10\) into \(\ln(20 - x) = \frac{1}{20}t\): \(\ln(20 - x) = \frac{1}{2}\). Solving gives \(x = 7.9\).

(iv) As \(t\) becomes very large, \(x\) approaches 20.

(i) The rate of increase of the surface area is proportional to the volume, so \(\frac{dA}{dt} = kV\). Given \(A = 4\pi r^2\) and \(V = \frac{4}{3}\pi r^3\), differentiate \(A\) with respect to \(t\):

\(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\).

Equating \(8\pi r \frac{dr}{dt} = k \frac{4}{3}\pi r^3\), solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{k}{6} r^2\).

Using \(\frac{dr}{dt} = 2\) and \(r = 5\), find \(k\):

\(2 = \frac{k}{6} \times 25 \Rightarrow k = 0.48\).

Thus, \(\frac{dr}{dt} = 0.08r^2\).

(ii) Separate variables and integrate:

\(\int \frac{1}{r^2} \, dr = \int 0.08 \, dt\).

\(-\frac{1}{r} = 0.08t + C\).

Using \(t = 0, r = 5\), find \(C\):

\(-\frac{1}{5} = C\).

Thus, \(-\frac{1}{r} = 0.08t - \frac{1}{5}\).

Solve for \(r\):

\(r = \frac{5}{1 - 0.4t}\).

(iii) For \(r\) to be positive, \(1 - 0.4t > 0\), so \(t < 2.5\). Therefore, \(0 \leq t < 2.5\).

(i) We start with the given \(V = \frac{4}{3}h^3\). Differentiating with respect to \(t\), we have \(\frac{dV}{dt} = 4h^2 \frac{dh}{dt}\). The rate of liquid poured in is 20 m\(^3\) per hour, and the rate of leakage is \(kh^2\). Thus, \(\frac{dV}{dt} = 20 - kh^2\). Equating the two expressions for \(\frac{dV}{dt}\), we have:

\(4h^2 \frac{dh}{dt} = 20 - kh^2\)

When \(h = 1\), \(\frac{dh}{dt} = 4.95\), so:

\(4(1)^2(4.95) = 20 - k(1)^2\)

\(19.8 = 20 - k\)

\(k = 0.2\)

Substituting \(k = 0.2\) back, we get:

\(4h^2 \frac{dh}{dt} = 20 - 0.2h^2\)

\(\frac{dh}{dt} = \frac{5}{h^2} - \frac{1}{20}\)

(ii) We need to verify:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Expanding the right side:

\(\frac{2000}{(10-h)(10+h)} = \frac{2000}{100-h^2}\)

Thus, \(-20 + \frac{2000}{100-h^2} = \frac{2000 - 20(100-h^2)}{100-h^2} = \frac{20h^2}{100-h^2}\)

(iii) Separating variables in the differential equation:

\(\frac{dh}{\frac{5}{h^2} - \frac{1}{20}} = dt\)

Integrating both sides, we have:

\(\int \frac{h^2}{5 - \frac{h^2}{20}} \, dh = \int dt\)

\(\int \frac{20h^2}{100-h^2} \, dh = t + C\)

Using partial fraction decomposition:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Integrating gives:

\(-20h + 100 \ln(10+h) + 100 \ln(10-h) = t + C\)

\(t = 100 \ln \left( \frac{10+h}{10-h} \right) + C\)

(i) The gradient of the curve at \(P\) is given by \(\frac{PN}{TN} = \frac{y}{TN}\). The area of triangle \(PTN\) is \(\frac{1}{2} \times TN \times y = \tan x\). Therefore, \(TN = \frac{2 \tan x}{y}\). The gradient \(\frac{dy}{dx} = \frac{y}{TN} = \frac{y^2}{2 \tan x} = \frac{1}{2}y^2 \cot x\).

(ii) Separate variables: \(\frac{dy}{y^2} = \frac{1}{2} \cot x \, dx\). Integrate both sides: \(-\frac{1}{y} = \frac{1}{2} \ln(\sin x) + C\). Given \(y = 2\) when \(x = \frac{1}{6}\pi\), substitute to find \(C\): \(-\frac{1}{2} = \frac{1}{2} \ln(\sin \frac{1}{6}\pi) + C\). Solve for \(C\) and substitute back to find \(y\): \(y = \frac{2}{1 - \ln(2 \sin x)}\).

(i) The rate of increase in height is proportional to \((9 - h)^{\frac{1}{3}}\), so we can write \(\frac{dh}{dt} = k(9 - h)^{\frac{1}{3}}\). Given \(\frac{dh}{dt} = 0.2\) when \(h = 1\), we have \(0.2 = k(9 - 1)^{\frac{1}{3}}\). Solving for \(k\), we find \(k = 0.1\). Thus, \(\frac{dh}{dt} = 0.1(9 - h)^{\frac{1}{3}}\).

(ii) Separate variables: \(\frac{dh}{(9 - h)^{\frac{1}{3}}} = 0.1 \, dt\). Integrate both sides: \(\int (9 - h)^{-\frac{1}{3}} \, dh = \int 0.1 \, dt\). This gives \(-\frac{3}{2}(9 - h)^{\frac{2}{3}} = 0.1t + C\). Use initial conditions \(t = 0, h = 1\) to find \(C\): \(-\frac{3}{2}(9 - 1)^{\frac{2}{3}} = C\). Solve for \(h\): \(h = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\).

(iii) The maximum height occurs when \(\frac{dh}{dt} = 0\), which implies \(9 - h = 0\), so \(h = 9\). The time taken is when \(h = 9\), which is \(t = 60\) years.

(iv) For half the maximum height, \(h = \frac{9}{2}\). Substitute into the expression for \(h\): \(\frac{9}{2} = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\). Solve for \(t\) to find \(t \approx 19.1\) years.

(i) The rate of formation of the substance is proportional to its mass, so we can write \(\frac{dx}{dt} = kx\) for some constant \(k\). The substance is also being removed at a constant rate of 25 grams per minute, so \(\frac{dx}{dt} = kx - 25\). Given \(\frac{dx}{dt} = 75\) when \(x = 1000\), we have:

\(75 = k(1000) - 25\)

\(75 = 1000k - 25\)

\(100 = 1000k\)

\(k = 0.1\)

Thus, \(\frac{dx}{dt} = 0.1x - 25\) can be rewritten as \(\frac{dx}{dt} = 0.1(x - 250)\).

(ii) To solve the differential equation \(\frac{dx}{dt} = 0.1(x - 250)\), we separate variables:

\(\frac{1}{x - 250} dx = 0.1 dt\)

Integrate both sides:

\(\int \frac{1}{x - 250} dx = \int 0.1 dt\)

\(\ln|x - 250| = 0.1t + C\)

Exponentiate both sides:

\(x - 250 = e^{0.1t + C}\)

\(x - 250 = e^C e^{0.1t}\)

Let \(e^C = A\), then \(x = A e^{0.1t} + 250\).

Using the initial condition \(x = 1000\) when \(t = 0\):

\(1000 = A e^{0} + 250\)

\(1000 = A + 250\)

\(A = 750\)

Thus, \(x = 750 e^{0.1t} + 250\).

Rearranging gives \(x = 250(3e^{0.1t} + 1)\).

(i) The rate of change of volume \(\frac{dV}{dt}\) is given by the inflow minus the outflow: \(\frac{dV}{dt} = 30 - k\sqrt{h}\). Since \(V = 1000h\), \(\frac{dV}{dt} = 1000 \frac{dh}{dt}\). Equating gives \(1000 \frac{dh}{dt} = 30 - k\sqrt{h}\). Given \(\frac{dh}{dt} = 0.02\) when \(h = 1\), we find \(k = 10\). Thus, \(\frac{dh}{dt} = 0.01(3 - \sqrt{h})\).

(ii) Substitute \(x = 3 - \sqrt{h}\), then \(\sqrt{h} = 3 - x\) and \(h = (3 - x)^2\). Differentiate to get \(\frac{dh}{dt} = -2(3-x) \frac{dx}{dt}\). Substitute into the differential equation to get \(-2(3-x) \frac{dx}{dt} = 0.01(3 - (3-x))\), simplifying to \((x-3) \frac{dx}{dt} = 0.005x\). Separate variables and integrate: \(\int \frac{x-3}{x} \, dx = \int 0.005 \, dt\). This gives \(x - 3 \ln x = 0.005t + C\). Using \(x = 3\) when \(t = 0\), find \(C = 0\). Thus, \(t = 200(x - 3 \ln x + 3 \ln 3)\).

(iii) When \(h = 4\), \(\sqrt{h} = 2\), so \(x = 3 - 2 = 1\). Substitute \(x = 1\) into the expression for \(t\): \(t = 200(1 - 3 \ln 1 + 3 \ln 3) = 200(1 + 3 \ln 3)\). Calculate to find \(t = 259\) seconds.

(i) Given that the rate of formation of X is proportional to the mass of Y, we have \(\frac{dx}{dt} = k(100 - x)\). Using the initial condition \(\frac{dx}{dt} = 1.9\) when \(x = 5\), we find \(1.9 = k(100 - 5)\), giving \(k = 0.02\). Thus, \(\frac{dx}{dt} = 0.02(100 - x)\).

(ii) To solve \(\frac{dx}{dt} = 0.02(100 - x)\), separate variables: \(\frac{1}{100 - x} dx = 0.02 dt\). Integrate both sides: \(-\ln|100 - x| = 0.02t + C\). Solve for x: \(x = 100 - Ce^{-0.02t}\). Use the initial condition \(x = 5\) when \(t = 0\) to find \(C = 95\). Thus, \(x = 100 - 95e^{-0.02t}\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(x \to 100\).

(i) The rate of growth of the infected area is proportional to the product of the infected area \(a\) and the uninfected area \(10 - a\). Therefore, \(\frac{da}{dt} = k a (10 - a)\). Given that initially \(a = 5\) and \(\frac{da}{dt} = 0.1\), we have:

\(0.1 = k \times 5 \times (10 - 5)\)

\(0.1 = 25k\)

\(k = 0.004\)

Thus, \(\frac{da}{dt} = 0.004a(10 - a)\).

(ii) Express \(\frac{1}{a(10-a)}\) in partial fractions:

\(\frac{1}{a(10-a)} = \frac{A}{a} + \frac{B}{10-a}\)

Solving, we find \(A = \frac{1}{10}\) and \(B = \frac{1}{10}\).

Separate variables and integrate:

\(\int \frac{da}{a(10-a)} = \int 0.004 \, dt\)

\(\frac{1}{10} \ln a - \frac{1}{10} \ln (10-a) = 0.004t + C\)

\(\ln \left( \frac{a}{10-a} \right) = 0.04t + C\)

Using initial conditions \(t = 0, a = 5\), solve for \(C\):

\(\ln 1 = C \Rightarrow C = 0\)

Thus, \(\ln \left( \frac{a}{10-a} \right) = 0.04t\)

\(t = 25 \ln \left( \frac{a}{10-a} \right)\)

(iii) For 90% infection, \(a = 9\):

\(t = 25 \ln \left( \frac{9}{1} \right)\)

\(t = 25 \ln 9 \approx 54.9\)

(a) The rate of change of volume is given by \(\frac{dV}{dt} = -k\sqrt{h}\). The volume of the hemisphere is \(V = \frac{1}{3} \pi (3rh^2 - h^3)\). Differentiating with respect to h, \(\frac{dV}{dh} = 2\pi rh - \pi h^2\).

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\), we have \(-k\sqrt{h} = (2\pi rh - \pi h^2) \cdot \frac{dh}{dt}\).

Rearranging gives \(\frac{dh}{dt} = -\frac{k\sqrt{h}}{2\pi rh - \pi h^2}\). Let \(B = k\pi\), then \(\frac{dh}{dt} = -\frac{B}{2rh^2 - h^3}\).

(b) Separate variables: \((2rh^2 - h^3) dh = -B dt\). Integrate both sides: \(\int (2rh^2 - h^3) dh = -\int B dt\).

The left side integrates to \(\frac{2}{3}rh^3 - \frac{1}{4}h^4\) and the right side to \(-Bt + C\).

At \(t = 0, h = r\), so \(\frac{2}{3}r^4 - \frac{1}{4}r^4 = C\).

At \(t = 14, h = 0\), so \(0 = -14B + C\).

Solving these gives \(C = \frac{5}{12}r^4\) and \(B = \frac{1}{15}r^2\).

Substitute back to find \(t\): \(t = 14 - 20\left(\frac{h}{r}\right)^{\frac{3}{2}} + 6\left(\frac{h}{r}\right)^{\frac{5}{2}}\).

(i) The rate of increase of m is proportional to \((50 - m)^2\), so \(\frac{dm}{dt} = k(50 - m)^2\). Given \(\frac{dm}{dt} = 5\) when \(m = 0\), we have \(5 = k(50)^2\). Solving for \(k\), we find \(k = \frac{5}{2500} = 0.002\). Thus, \(\frac{dm}{dt} = 0.002(50 - m)^2\).

(ii) Separate variables: \(\frac{1}{(50 - m)^2} dm = 0.002 dt\). Integrate both sides: \(-\frac{1}{50 - m} = 0.002t + C\). Using \(t = 0\), \(m = 0\), we find \(C = -\frac{1}{50}\). Thus, \(-\frac{1}{50 - m} = 0.002t - \frac{1}{50}\). Solving for m, we get \(m = 50 - \frac{500}{t + 10}\).

(iii) Substitute \(t = 10\) into the solution: \(m = 50 - \frac{500}{10 + 10} = 25\). For \(m = 45\), solve \(45 = 50 - \frac{500}{t + 10}\), giving \(t = 90\).

(iv) As \(t\) becomes very large, \(\frac{500}{t + 10}\) approaches 0, so \(m\) approaches 50.

(a) The differential equation is \(\frac{dy}{dx} = k \frac{y}{x\sqrt{x}}\). Separate variables: \(\frac{dy}{y} = k \frac{dx}{x\sqrt{x}}\). Integrate both sides: \(\ln y = -2k \frac{1}{\sqrt{x}} + C\). Use the point \((1, 1)\) to find \(C\): \(\ln 1 = -2k \cdot 1 + C\), so \(C = 2k\). Use the point \((4, e)\) to find \(k\): \(\ln e = -2k \cdot \frac{1}{2} + 2k\), solving gives \(k = 1\). Thus, \(y = \exp\left(-\frac{2}{\sqrt{x}} + 2\right)\).

(b) As \(x\) tends to infinity, \(-\frac{2}{\sqrt{x}}\) tends to 0, so \(y\) approaches \(e^2\).

(i) We are given \(\frac{dN}{dt} = -10\) when \(t = 0\) and \(N = 1000\). The differential equation is \(\frac{dN}{dt} = ke^{-0.02t}N\). Substituting the given values, \(-10 = k \times 1 \times 1000\), we find \(k = -0.01\).

(ii) Separate variables: \(\frac{1}{N} dN = -0.01 e^{-0.02t} dt\). Integrate both sides: \(\int \frac{1}{N} dN = \int -0.01 e^{-0.02t} dt\). This gives \(\ln N = 0.5 e^{-0.02t} + C\). Using \(N = 1000\) when \(t = 0\), \(\ln 1000 = 0.5 + C\), so \(C = \ln 1000 - 0.5\). Substitute \(N = 800\) to find \(t\): \(\ln 800 = 0.5 e^{-0.02t} + \ln 1000 - 0.5\). Solving gives \(t = 29.6\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to \frac{1000}{\sqrt{e}}\).

1. The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y^2}{x}\), where \(k\) is a constant.

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

3. Integrate both sides: \(-\frac{1}{y} = k \ln x + C\), where \(C\) is the constant of integration.

4. Use the point \((1, 1)\) to find \(C\): \(-1 = k \ln 1 + C \Rightarrow C = -1\).

5. Use the point \((e, 2)\) to find \(k\): \(-\frac{1}{2} = k \ln e - 1 \Rightarrow k = \frac{1}{2}\).

6. Substitute \(k\) and \(C\) back into the integrated equation: \(-\frac{1}{y} = \frac{1}{2} \ln x - 1\).

7. Solve for \(y\): \(y = \frac{2}{2 - \ln x}\).

(i) The gradient of the tangent at \(P\) is the same as the gradient of the curve at \(P\). Since \(TN = 2\), the horizontal distance from \(T\) to \(N\) is 2. The gradient of the tangent is \(\frac{y}{2}\), so \(\frac{dy}{dx} = \frac{1}{2}y\).

(ii) To solve the differential equation \(\frac{dy}{dx} = \frac{1}{2}y\), separate variables: \(\frac{1}{y} dy = \frac{1}{2} dx\).

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

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

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

Let \(e^C = A\), so \(y = Ae^{\frac{1}{2}x}\).

Using the point \((4, 3)\), substitute to find \(A\): \(3 = Ae^{2}\).

Thus, \(A = \frac{3}{e^2}\).

The equation of the curve is \(y = \frac{3}{e^2} e^{\frac{1}{2}x} = 3e^{\frac{1}{2}x - 2}\).

(i) The rate of increase of the mass of A, \(\frac{dx}{dt}\), is proportional to the mass of B. Since the total mass is constant at 50, the mass of B is \(50 - x\). Therefore, \(\frac{dx}{dt} = k(50 - x)\).

(ii) To solve \(\frac{dx}{dt} = k(50 - x)\), separate variables:

\(\frac{1}{50-x} dx = k \, dt\)

Integrate both sides:

\(-\ln|50-x| = kt + C\)

Using the initial condition \(x = 0\) when \(t = 0\), we find:

\(-\ln|50| = C\)

Thus, \(-\ln(50-x) = kt - \ln 50\).

Rearrange to find:

\(\ln\left(\frac{50}{50-x}\right) = kt\)

\(\frac{50}{50-x} = e^{kt}\)

\(50-x = 50e^{-kt}\)

\(x = 50(1 - e^{-kt})\)

Using \(x = 25\) when \(t = 10\), solve for k:

\(25 = 50(1 - e^{-10k})\)

\(0.5 = 1 - e^{-10k}\)

\(e^{-10k} = 0.5\)

\(-10k = \ln(0.5)\)

\(k = -\frac{\ln(0.5)}{10} \approx 0.0693\)

Thus, \(x = 50(1 - \exp(-0.0693t))\).

(i) Given \(\frac{dy}{dt} = -0.2xy\) and \(x = \frac{10}{(1+t)^2}\), substitute for x:

\(\frac{dy}{dt} = -0.2 \cdot \frac{10}{(1+t)^2} \cdot y = -\frac{2y}{(1+t)^2}\).

Separate variables: \(\frac{dy}{y} = -\frac{2}{(1+t)^2} dt\).

Integrate both sides: \(\int \frac{dy}{y} = \int -\frac{2}{(1+t)^2} dt\).

\(\ln y = \frac{2}{1+t} + C\).

Use initial condition \(y = 100\) when \(t = 0\):

\(\ln 100 = 2 + C\).

Thus, \(C = \ln 100 - 2\).

Final solution: \(\ln y = \frac{2}{1+t} - 2 + \ln 100\).

(ii) As \(t \to \infty\), \(\frac{2}{1+t} \to 0\), so \(\ln y \to \ln 100 - 2\).

Thus, \(y \to e^{\ln 100 - 2} = \frac{100}{e^2}\).

The mass of A, \(x = \frac{10}{(1+t)^2}\), tends to zero as \(t \to \infty\).