Transcript Sullivan College Algebra Section 6.7

Sullivan PreCalculus
Section 4.8
Exponential Growth and Decay;
Newton’s Law; Logistic Models
Objectives of this Section
• Find Equations of Populations That Obey the Law of
Uninhibited Growth
• Find Equations of Populations That Obey the Law of Decay
• Use Newton’s Law of Cooling
• Use Logistic Growth Models
Many natural phenomena have been found to
follow the law that an amount A varies with
time t according to the function:
A(t )  A0 e
kt
If k > 0, there is uninhibited growth.
If k < 0, there is decay.
A(0) = A0 represents the original amount
A culture of bacteria increases according to the
law of uninhibited growth. If 100 grams bacteria
are present initially, and there are 175 grams of
bacteria after 1 hour, how many will be present
after 4 hours?
N 0  100 N (1)  175
N(t)  N0e
kt
N (1)  175  100e k (1)
175
.  ek
ln175
. k
k  05596
.
N(t)  N0e
kt
k  05596
.
N (t )  100e
0.5596t
N (4)  100e
 937.8 grams
0.5596( 4 )
The half-life of Uranium-234 is 200,000 years. If 50
grams of Uranium-234 are present now, how much
will be present in 1000 years. NOTE: The half-life
of is the time required for half of radioactive
substance to decay.
A(t)  Ae
0
kt
1
A(t )  A0
2
1
kt
A0  A0e
2
1 k ( 200,000)
e
2
1

ln   200,000k
 2
k
 2
ln 1
200,000
 3.466  10
6
A(t)  A0e
kt
k  3.466  10
A ( t )  A0 e
6
 3.466 10 t
6
3.466106 (1000)
A(1000)  50e
 49.8 grams
Newton’s Law of Cooling
u(t )  T + u0  Te
kt
k <0
T : Temperature of surrounding medium
uo : Initial temperature of object
k : A negative constant
A cup of hot chocolate is 100 degrees
Celsius. It is allowed to cool in a room
whose air temperature is 22 degrees Celsius.
If the temperature of the hot chocolate is 85
degrees Celsius after 4 minutes, when will its
temperature be 60 degrees Celsius?
u(t )  T + u0  T e kt 85  22 + 100  22e k ( 4 )
63  78e
4k
63
 e4 k
78
63

ln   4 k
 78
63

ln 
 78
 0.0534
k
4
u( t )  22 + 78e
u(t )  T + u0  Te
kt
T  22, u0  100
 0.0534 t
60  22 + 78e  0.0534 t
38  78e  0.0534 t
38
 0.0534 t
e
78
19 

ln   0.0534t
 39 

ln 19

39
 13.5 minutes
t
 0.0534
Logistic Growth Model
c
P(t ) 
bt
1 + ae
where a, b, and c are constants with c > 0
and b > 0. The constant c is called the
carrying capacity.
The logistic growth model
500
P(t ) 
 0.2476 t
1 + 6.67e
represents the amount of a bacteria (in
grams) after t days.
What is the carrying capacity?
Graph the function using a graphing utility.
500
P(t ) 
 0.2476t
1 + 6.67e
What was the initial amount of bacteria?
500
500
P(0) 

 0.2476( 0)
1 + 6.67e
1 + 6.67
 65 grams
When will the amount of bacteria be 300
grams?
500
P(t ) 
1 + 6.67e
 0.2476 t
500
300 
 0.2476 t
1 + 6.67e
1 + 6.67e
6.67e
 0.2476 t
 0.2476 t
500

300
5
 1
3
6.67e
e
 0.2476 t
 0.2476 t
5
 1
3
5
1
3
 0.1
6.67
 0.2467t  ln 01
.
ln 01
.
t
 9.3 days
 0.2467