The document discusses equations to model growth and decay over time by accounting for limiting conditions. It presents an equation for population change over time and its solution. It then gives an example using Newton's law of cooling, showing that an equation in the form T = A + Ce^kt satisfies the law, where T is temperature, A is the constant outside temperature, and C and k are constants. It then applies this to calculate that given conditions, the temperature inside a room will drop to 10°C after 2 hours.

1. Further Growth & Decay

2. Further Growth & Decay
In order to take into account other conditions (e.g. the temperature of
an object will never be lower than the temperature of its
surroundings), we need to change the equation.

3. Further Growth & Decay
In order to take into account other conditions (e.g. the temperature of
an object will never be lower than the temperature of its
surroundings), we need to change the equation.
dP
 k P  N 
dt

4. Further Growth & Decay
In order to take into account other conditions (e.g. the temperature of
an object will never be lower than the temperature of its
surroundings), we need to change the equation.
dP
 k P  N 
dt
the solution to the equation is;
PN (trivial case)

5. Further Growth & Decay
In order to take into account other conditions (e.g. the temperature of
an object will never be lower than the temperature of its
surroundings), we need to change the equation.
dP
 k P  N 
dt
the solution to the equation is;
PN (trivial case)
OR
P  N  Ae kt
P = population at time t k = growth(or decay) constant
N + A = initial population t = time

6. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).

7. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.

8. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.
T  A  Ce kt

9. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.
T  A  Ce kt
dT
 kCe kt
dt

10. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.
T  A  Ce kt
dT
 kCe kt
dt
but T  A  Ce kt

11. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.
T  A  Ce kt
dT
 kCe kt
dt
but T  A  Ce kt
dT
 k T  A 
dt

12. e.g.(1993)
Let T be the temperature inside a room at time t and let A be the
constant outside air temperature.
Newton’s law of cooling states that the rate of change of the
temperature is proportional to (T – A).
a) Show that T  A  Ce kt (where C and k are constants) satisfies
Newton' s law of cooling.
T  A  Ce kt
dT
 kCe kt
dt
but T  A  Ce kt
dT
 k T  A 
dt
dT
T  A  Ce kt satisfies  k T  A
dt

13. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?

14. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?
T  5  Ce kt

15. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?
T  5  Ce kt
when t  0, T  20
 C  15
T  5  15e kt

16. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?
T  5  Ce kt when t  0.5, T  17
when t  0, T  20 i.e. 17  5  15e 0.5 k
 C  15
T  5  15e kt

17. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?
T  5  Ce kt when t  0.5, T  17
when t  0, T  20 i.e. 17  5  15e 0.5 k
 C  15 15e 0.5 k  12
T  5  15e kt

18. b) The outside air temperature is 5 C and a heating system breakdown
causes the inside room temperature to drop from 20  C to 17  C in
half an hour.
After how many hours is the inside room temperature equal to 10  C ?
T  5  Ce kt when t  0.5, T  17
when t  0, T  20 i.e. 17  5  15e 0.5 k
 C  15 15e 0.5 k  12
T  5  15e kt 12
e 0.5 k 
15
 12 
0.5k  log 
 15 
 12 
k  2 log 
 15 

19. when T  10, 10  5  15e kt

20. when T  10, 10  5  15e kt
15e kt  5
1
e 
kt
3
1
kt  log
3
1 1
t  log
k 3

21. when T  10, 10  5  15e kt
15e kt  5
1
e 
kt
3
1
kt  log
3
1 1
t  log
k 3
t  2.46167

22. when T  10, 10  5  15e kt
15e kt  5
1
e 
kt
3
1
kt  log
3
1 1
t  log
k 3
t  2.46167
1
 After 2 hours the temperature has dropped to 10 C
2

23. when T  10, 10  5  15e kt
15e kt  5
1
e 
kt
3
1
kt  log
3
1 1
t  log
k 3
t  2.46167
1
 After 2 hours the temperature has dropped to 10 C
2
Exercise 7H; 2, 4, 5, 8, 9, 10