Transcript Exponential Growth Decay

Exponential
and
growth or decay constant
At   Ao e
amount at time t
For k > 0, k is the growth
constant or rate of growth
kt
time
amount initially
For k < 0, k is the decay
constant or rate of decline
k > 0 growth curves
k < 0 decay curves
Radiocarbon Dating
All living things
contain Carbon 14.
When they die, the
C-14 begins to
decay. We can
determine how long
something has been
dead by the amount
of C-14 left.
In 1940 a group of boys walking in the woods near the
village of Lascaux in France suddenly realized their dog
was missing. They soon found him in a hole too deep for
him to climb out. One of the boys was lowered into the
hole to rescue the dog and stumbled upon one of the
greatest archaeological discoveries of all time. What he
discovered was a cave whose walls were covered with
drawings of wild horses, cattle and a fierce-looking beast
resembling a modern bull. In addition, the cave contained
the charcoal remains of a small fire, and
from these remains scientists were able to
determine that the cave was occupied 15,000
years ago.
Charcoal left from the
logs contains Carbon-14
By chemical analysis it has been determined that
the amount of C-14 remaining in the samples of
the Lascaux charcoal was 15% of the amount
such trees would contain when living. The halflife of C-14 is approximately 5600 years.
At   Ao e
ln e
k 5600
kt
A5600  Ao e
1
 ln
2
Take the ln of both sides
k 5600
1
 Ao
2
1
ln this equation for k.
Solve
k  First2divide
 both
0.000123776
sides by Ao
5600
Divide by 5600 to find k
The half-life of a radioactive element is the time it takes
for 1/2 the initial amount to decay.
By chemical analysis it has been determined that
the amount of C-14 remaining in the samples of
the Lascaux charcoal was 15% of the amount
such trees would contain when living. The halflife of C-14 is approximately 5600 years.
k  0.000123776
Take the ln of both sides
0.15
AAt o  Ao e
Divide both sides by Ao
Divide both sides by k
kt
ln 0.15  ln e
kt
ln 0.15
t
 15,327
 0.000123776
So assuming the paintings were made at the time of the
fire in the cave, they are approximately 15,000 years old.
A culture of bacteria obeys the law of uninhibited
growth. If 500 bacteria are present initially and there are
800 after 1 hour, how many will be present in the culture
after 5 hours? How long is it until there are 20,000
bacteria?
You’ll need to determine the k
t   A e
A800
500
kt (1) value by using info that tells you
500o
after a certain time it was a certain
amount and subbing these values
500
in for t and A.
8
k
ln    ln e
5
k  0.47000363
Notice the k is positive this time
since this is exponential growth
instead of decay.
Now we know k we are ready to answer questions
about the bacteria.
A culture of bacteria obeys the law of uninhibited growth. If
500 bacteria are present initially and there are 800 after 1
hour, how many will be present in the culture after 5 hours?
How long is it until there are 20,000 bacteria?
k  0.47000363
A(5)
t   500
Ao e
kt (5)
Punch buttons in your
calculator using the k
value we already
found.
 5243
   500Ao e
20,000
At
500
kt
500
Solve for t
ln 40  ln e
kt
ln 40
t
 7.8 hrs
0.47000363
Another exponential equation that occurs
in nature and involves the natural base e is
called
Newton’s Law of Cooling
Temperature at time t
Initial temperature
u t   T  uo  T e
Surrounding Temperature
kt time
constant you determine
knowing a time and
temperature
A pizza baked at 450F is removed from the oven at 5:00
pm into a room that is a 70F. After 5 minutes the pizza is
at 300F. At what time can you begin eating the pizza if
you want its temperature to be 135F?
u t   70
T  450
uo  70
T e
u t5  70
T  450
uo  70
T e
380e
kt5
 300
We first
find k by
seeing
that after
5 min. it
was 300°
 230 
ln 

380

  .1004184
 230 k 
5
solve this for k
5k
kt
A pizza baked at 450F is removed from the oven at 5:00
pm into a room that is a 70F. After 5 minutes the pizza is
at 300F. At what time can you begin eating the pizza if
you want its temperature to be 135F?
Now that we have k we are ready to
answer the question asked.
k  .1004184
t   70
u135
T  450
uo  70
T e
solve this for t
65
kt
e
380
kt
So at 5:18 you
can be
munching away.
 65 
ln 

380 

t
 18 min
.1004184
One last type of equation we’ll look at is called the
Logistic Growth Model which is limited growth instead
of uninhibited growth.
carrying capacity - the limit or number the
population approaches as time approaches infinity
c
P t  
 bt
1  ae
population at time t
Since this is a negative
exponential, as t gets large it
gets small.
As t  , this term  0 and
the population  c.
The logistic growth equation below relates the
proportion of U.S. households that own a VCR to the
year. Let t = 0 represent 1984, t = 1 represent 1985,
and so on.
0 .9
P t  
 0.32t
1  6e
0.9 = 90%
Determine the maximum proportion of households that
will own a VCR. (Hint: this is the carrying capacity).
What proportion of the U.S. households owned a
VCR in 1984? (Hint: that is when t = 0)
0.9
0.9
P0  

 0.129  12.9%
 0.320 
1  6e
7
The logistic growth equation below relates the
proportion of U.S. households that own a VCR to the
year. Let t = 0 represent 1984, t = 1 represent 1985,
and so on.
need to solve for t by
0 .9
t  
P
0.8
 0.32t
1  6e
first multiplying both
sides by the
denominator
When will 0.8 (80%) of the households own a VCR?

0.8 1  6e
1  6e 0.32t
6e
 0.32t
0.32t
0.9

0.8
0.9

1
0.8
  0.9
e 0.32t
Isolate the exponential
0.125

6
ln e
 0.32t
0.125
 ln
6
0.125
ln
6  12.1 yrs
t
 0.32
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au