Question

Mathmatical modelling!

Q 4. (10 MARKS) Consider T(t), the number of travellers in an airport, where t denotes the time in hours after midnight. The rate that travellers arrive at the airport varies periodically and the rate that travellers depart from the airport is a constant 600 per hour, so we model T(t) by the differential equation 0, dr = a (2 +cos dt where a is a positive constant. At midnight each day there are 2000 travellers in the airport (a) According to the model, what is the period of the variation in the rate that travellens arrive at the airport? (b) Solve the differential equation and determine the value of a (c) According to the model, what is the maximum value of T(t)? (1 mark) (6 marks) (3 marks)

Answer 1

The differential equation is

dT = a | 2 + cos dt -600 4

a) The period of variation is obtained ny comparing with $\frac{\mathrm{d} T}{\mathrm{d} t}=a\cos\left ( \frac{2\pi t}{T} \right )+C$ .

Thus,

T-8 hours 4

a) Integrating the differential equation,

T-600 4 4 4600t + c

It is given that at . Thus

t=0,T=2000

T(O)-2000-a(0+4sin(:) -600 x 0 + C C = 2000

Also the minimum value of
dT dt
, so $a\left ( 2-1\right ) \right )-600=0\Rightarrow a=600$

Thus the solution is

T-600 (2tt sin(+2000

c) The maximum value of occurs when
-600(2+cos傑))-600-0 cos( )--1+t-4 dT dt
.

Thus at

4

$T(4\pi )=600\left (8\pi +\frac{4}{\pi }\times 0\right )-600(4\pi )+2000\\ {\color{Blue} T(4\pi )=9,539.8224}$