2026 HCI P1 Q11

2026 HCI P1 Q11

Junior College 2
12 marks

A botanist is monitoring the temperature, \(\theta\) (in degrees Celsius), of a specialised greenhouse at time \(t\) hours after noon. The temperature of the greenhouse is affected by the external environment and an automated internal cooling system. The rate of change of temperature of the greenhouse is modeled by the differential equation \[\frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac1{10}(\theta-20)-k(\theta-20)^2,\] where \(k\) is a positive constant. At noon, the temperature of the greenhouse is \(25^\circ\mathrm{C}\). It is known that if the temperature of the greenhouse is \(22^\circ\mathrm{C}\), it would decrease at a rate of \(0.26^\circ\mathrm{C}\) per hour.

  1. Show that \(\displaystyle k=\frac3{200}\).[2]
  2. By using the substitution \(u=\theta-20\), solve the differential equation to show that \(\displaystyle \theta=20+\frac{20}{7e^{0.1t}-3}\).[6]
  3. The cooling system triggers an alarm if the temperature of the greenhouse drops below \(21^\circ\mathrm{C}\). Find the time, to the nearest minute, when this alarm will be triggered.[2]
  4. Sketch the graph of \(\theta\) against \(t\), clearly labelling the coordinates of any intercepts and the equations of any asymptotes.[2]

Solution:

Solution locked

Sign in to view the step-by-step solution

Similar questions are unavailable for this question.
Answer:(a) \(k=\frac3{200}\) (b) \(\theta=20+\frac{20}{7e^{0.1t}-3}\) (c) \(11\text{ h}\hspace{0.5em}54\text{ min}\) after noon (d) decreasing from \((0,25)\) to asymptote \(\theta=20\)

Need help? Join our JC Math tuition classes.

Learn more