APM1514
ASSIGNMENT 6 2024
UNIQUE NO.
DUE DATE 2024
, lOMoARcPSD|21997160
Question 1: 16 Marks
(1.1) We can use Newton's law of cooling, which states that the rate of
change of temperature of an object is directly proportional to the
difference between its temperature and the surrounding temperature. This
can be represented by the differential equation:
dT/dt = -k(T - T0)
where T is the temperature of the object at time t, T0 is the initial
temperature, and k is the cooling constant.
Using the given information, we can set up the following equations:
T0 - 10 = (T0 - 180)e^(-k*30)
T0 - 10 = (T0 - 80)e^(-k*60)
Solving these equations simultaneously will give us the values of T0 and k.
To find the time it takes for the object to cool to 30°C, we can use the
equation T = T0 + (T0 - 30)e^(-kt) and solve for t when T = 30.
Question 2: 24 Marks
(2.1) The differential equations for S(t) and C(t) are:
dS/dt = 0.5 * 250 - 0.5 * 250 * S/3000
dC/dt = (0.5 * 250 - C * 250)/3000
(2.2) The phase lines of the differential equations for the systems for S
and C will show the behavior of the solutions for different initial conditions.
The rough sketches of the values of S and C as functions of time will
depend on the initial values and can be represented graphically.
ASSIGNMENT 6 2024
UNIQUE NO.
DUE DATE 2024
, lOMoARcPSD|21997160
Question 1: 16 Marks
(1.1) We can use Newton's law of cooling, which states that the rate of
change of temperature of an object is directly proportional to the
difference between its temperature and the surrounding temperature. This
can be represented by the differential equation:
dT/dt = -k(T - T0)
where T is the temperature of the object at time t, T0 is the initial
temperature, and k is the cooling constant.
Using the given information, we can set up the following equations:
T0 - 10 = (T0 - 180)e^(-k*30)
T0 - 10 = (T0 - 80)e^(-k*60)
Solving these equations simultaneously will give us the values of T0 and k.
To find the time it takes for the object to cool to 30°C, we can use the
equation T = T0 + (T0 - 30)e^(-kt) and solve for t when T = 30.
Question 2: 24 Marks
(2.1) The differential equations for S(t) and C(t) are:
dS/dt = 0.5 * 250 - 0.5 * 250 * S/3000
dC/dt = (0.5 * 250 - C * 250)/3000
(2.2) The phase lines of the differential equations for the systems for S
and C will show the behavior of the solutions for different initial conditions.
The rough sketches of the values of S and C as functions of time will
depend on the initial values and can be represented graphically.
