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Nov 2021 p33 q10
2294
A large plantation of area 20 km2 is becoming infected with a plant disease. At time t years the area infected is x km2 and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected.
When t = 0, x = 1 and \(\frac{dx}{dt} = 1\).
(a) Show that x and t satisfy the differential equation \(\frac{dx}{dt} = \frac{19x}{20-x}\).
(b) Solve the differential equation and show that when t = 1 the value of x satisfies the equation \(x = e^{0.9 + 0.05x}\).
(c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
(d) Calculate the value of t at which the entire plantation becomes infected.
Solution
(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 \(t = 2.15\).