AQA C4 2006 June — Question 8

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2006
SessionJune
TopicDifferential equations

8 A disease is spreading through a colony of rabbits. There are 5000 rabbits in the colony. At time \(t\) hours, \(x\) is the number of rabbits infected. The rate of increase of the number of rabbits infected is proportional to the product of the number of rabbits infected and the number not yet infected.
    1. Formulate a differential equation for \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) in terms of the variables \(x\) and \(t\) and a constant of proportionality \(k\).
    2. Initially, 1000 rabbits are infected and the disease is spreading at a rate of 200 rabbits per hour. Find the value of the constant \(k\).
      (You are not required to solve your differential equation.)
  1. The solution of the differential equation in this model is $$t = 4 \ln \left( \frac { 4 x } { 5000 - x } \right)$$
    1. Find the time after which 2500 rabbits will be infected, giving your answer in hours to one decimal place.
    2. Find, according to this model, the number of rabbits infected after 30 hours.