2026 ACJC P1 Q11

2026 ACJC P1 Q11

Junior College 2
12 marks

Two scientists are interested in studying the growth of a deer population after its introduction into a national park. They model the number of deer, \(P\), at time \(t\) years after the introduction of an initial population of 50 deer. Both \(P\) and \(t\) are assumed to be continuous variables.

Scientist A uses the Gompertz growth model. \[\frac{\mathrm{d}P}{\mathrm{d}t}=P(1-0.2\ln P).\]

  1. Using the substitution \(y=\ln P\), show that \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=1-0.2y\).[1]
  2. Solve the differential equation for \(y\) in terms of \(t\) and show that \(y\) can be expressed as \(y=5(1-Ke^{-0.2t})\), where \(K\) is a constant to be determined to 4 decimal places.[4]
  3. Hence find \(P\) in terms of \(t\).[1]
  4. Explain what will happen to the deer population in the long term.[1]
  5. Sketch the graph of \(P\) against \(t\).[2]
  6. Scientist B uses a model that assumes that, at any time \(t\), the growth rate of the deer population is \(\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}=0.2P\).
    Find \(P\) in terms of \(t\).[2]
  7. Comment on whether the model used by Scientist A or Scientist B is more appropriate.[1]

Solution:

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Answer:(a) \(\frac{\mathrm{d}y}{\mathrm{d}t}=1-0.2y\) (b) \(y=5(1-0.2176e^{-0.2t})\) (c) \(P=e^{5(1-0.2176e^{-0.2t})}\) (d) \(P\to e^5\approx148\) (e) increasing curve from \((0,50)\) with asymptote \(P=e^5\) (f) \(P=50e^{0.2t}\) (g) Scientist A

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