Transcript Section 5.2

Applications of the Models
5.2
OBJECTIVE
• Perform computations involving interest
compounded continuously and continuous
money flow.
• Calculate the total consumption of a natural
resource.
• Find the present value of an investment.
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5.2 Applications of the Models
Growth Formula:
Decay Formula:

T

T
0
0
P0 kT
P0e dt   e  1.
k
kt
P0
 kt
P0
dt  1  e  kT .
k
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5.2 Applications of the Models
Definition
If P0 is invested for t years at interest rate k, compounded
continuously, then
P  t   P0e kt ,
where P  P0 at t  0. The value of P is called the future
value of P0 dollars invested at interest rate k, compounded
continuously, for t years.
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5.2 Applications of the Models
Quick Check 1
Find the future value of $10,000 invested for 3 years at an interest
rate of 6% compounded continuously.
We know that P  t   P0ekt , P0  $10,000, t  3, and k  0.06. So,
P  3  10,000e
0.06 3
 10,000e0.18  10,000 1.197217 
 $11,972.17
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5.2 Applications of the Models
FUTURE VALUE OF A CONTINUOUS
MONEY FLOW
If the yearly flow of money into an investment is given
by some function R(t), then the future value of the
continuous money flow at interest rate k, compounded
continuously over T years, is given by
A   R  t  ekt dt.
T
0
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5.2 Applications of the Models
Example 1: Find the future value of the continuous
money flow if $1000 per year flows at a constant rate
into an account paying 8%, compounded continuously,
for 15 yr.

15
0
15
$1000e
0.08 t
1000 0.08t 
dt  
e 
 0.08
0
1000 0.0815 0.08 0 

e
e
0.08
1000 1.2

e  1  $29,001.46
0.08


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
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5.2 Applications of the Models
Example 2: Consider a continuous flow of money
into an investment at the constant rate of P0 dollars per
year. What should P0 be so that the amount of a
continuous money flow over 20 yr, at an interest rate
of 8%, compounded continuously, will be $10,000?
20
10,000   P0e0.08t dt
0
20
 1 0.08t 
10,000  P0 
e 
 0.08
0
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5.2 Applications of the Models
Example 2 (concluded):
0.08 20 

10,000  P0 12.5e
 12.5


10,000  P0 12.51.6  12.5
10,000  49.4129053P0
$202.38  P0
A continuous money flow of $202.38 per year,
invested at 8%, compounded continuously for
20 years, will yield $10,000.
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5.2 Applications of the Models
DEFINITION:
The present value, P0, of an amount P, when P0 is
invested at interest rate k, compounded continuously,
and due t years later, is given by
P  P0ekt
P
P0  kt
e
P0  Pekt
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5.2 Applications of the Models
Example 3: Find the present value of $200,000 due
25 yr from now, at 8.7% compounded continuously.
P0  200,000e0.087(25)  $22, 721.63
Thus the present value is $22,721.63.
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5.2 Applications of the Models
Quick Check 2
Mira Bell, following the birth of a grandchild, wants to set up a trust
fund that will be worth $120,000 on the child’s 18th birthday. Mira
can get an interest rate of 5.6%, compounded continuously for the
time period. What amount will Mira have to deposit in the trust fund
to achieve her goal?
P0  Pe  kt
0.05618
P0  120,000e
P0  120,000e1.008
P0  120,000  0.364948146 
P0  43,793.78
So Mira must deposit $43,793.78 into the trust fund to achieve her goal.
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5.2 Applications of the Models
DEFINITION:
The accumulated present value, B, of a continuous
money flow into an investment at a rate of R(t) dollars
per year from now until T years in the future is given
T
by
B   R(t )e kt dt ,
o
where k is the interest rate, and interest is compounded
continuously.
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5.2 Applications of the Models
Example 4: Find the accumulated present value of
an investment over a 5-yr period if there is a
continuous money flow of $2400 per year and the
interest rate is 14%, compounded continuously.

5
0
5
2400e
0.14 t
dt

 2400 0.14 t 
 0.14 e

0

 17,142.86 e0.145  e0.140




17,142.86 e0.7  1
$8629.97
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5.2 Applications of the Models
Consumption of a Natural Resource
Suppose that P(t) is the amount of a natural resource
used at time t. If consumption of the resource is
growing exponentially at growth rate k, then the total
amount used during the interval [0, T ] is given by
T
P0 kT
kt
0 P0 e dt  k e  1 ,


where P0 represents the amount of the natural resource
used at time t = 0.
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5.2 Applications of the Models
The area under the graph
of
P t   P0 ekt
over the interval
[0, T] is is given
by

T
0
P0 kT
P0e dt   e  1 .
k
kt
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5.2 Applications of the Models
Example 5: In 2000 (t = 0), world gold production
was 2547 metric tons, and it was growing
exponentially at the rate of 0.6% per year. If the
growth continues at this rate, how many tons of gold
will be produced from 2000 to 2013?

13
0
2547e
0.006 t
dt

2547 0.00613 0.0060
e
e
0.006
0.078
424,500 e 1

34,437 metric tons



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5.2 Applications of the Models
Example 6: The world reserves of gold in 2000 were
estimated to be 77,000 metric tons. Assuming that the
growth rate for production given in Example 5
continues and that no new reserves are discovered,
when will the world reserves of gold be depleted?
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5.2 Applications of the Models
Example 6 (concluded):
77,000

77,000
0.1814
1.1814



ln1.1814 
ln1.1814 
28 yrs 
2547 0.006T
e
1
0.006
424,500 e0.006T  1
e0.006T  1
e0.006T




ln e0.006T
0.006T
T
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5.2 Applications of the Models
Quick Check 3
The movie Avatar is set in the year 2154 on the moon Pandora, of the
planet Polyphemus in the star system of Alpha Centauri. The conflict
in the movie is centered around a precious but scarce mineral,
Unobtanium.
a.) In 2010, the universe’s production of Unobtanium was
6800 metric tons and it was being used at a rate of 0.8% per year.
If Unobtanium continues to be used at this rate, how many tons of
Unobtanium will be used between 2010 and 2024?
b.) In 2010, the universe’s reserve of Unobtanium was 86,000 metric
tons. Assuming that the growth rate of 0.8% per year continues and
that no new reserves are discovered, when will the universe reserves
of Unobtanium be depleted?
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5.2 Applications of the Models
Quick Check 3 Continued
a.) In 2010, the universe’s production of Unobtanium was
6800 metric tons and it was being used at a rate of 0.8% per year.
If Unobtanium continues to be used at this rate, how many tons of
Unobtanium will be used between 2010 and 2024?


6800 0.00814
P0 kT

e
1
P   e  1
0.008
k
 850,000  e0.112  1
 850,000 .11851286 
 100,736
So 100,736 metric tons of Unobtanium will be used between 2010
and 2024.
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5.2 Applications of the Models
Quick Check 3 Concluded
b.) In 2010, the universe’s reserve of Unobtanium was 86,000 metric
tons. Assuming that the growth rate of 0.8% per year continues and
that no new reserves are discovered, when will the universe reserves
of Unobtanium be depleted?
P0 kT
P   e  1
k
6800 0.008T 
86000 
e
1
So the universe reserves of
0.008
Unobtanium will be depleted in
86000  850000  e0.112  1
12 years, or by the year 2022.
0.008 T
1.1012  e  

ln1.1012  0.008 T 
12  T
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5.2 Applications of the Models
Section Summary
• The future value of an investment is given by P  P0ekt , where P0
dollars are invested for t years at interest rate k, compounded
continuously.
• The accumulated future value of a continuous income stream is
given by
T
A   R  t  e kt dt ,
0
where R  t  represents the rate of the continuous income stream, k is
the interest rate, compounded continuously, at which the continuous
income stream is invested, and T is the number of years for which the
income stream is invested.
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5.2 Applications of the Models
Section Summary Continued
• If R  t  is a constant function, then
A
R t 
k
kT
e
 1.
• The present value is given by P0  Pe kt ,where the amount P is due
t years later is invested at interest rate k, compounded continuously.
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5.2 Applications of the Models
Section Summary Concluded
• The accumulated present value of a continuous income stream is
given by
T
B   R  t  e  kt dt ,
0
where R  t  represents the rate of the continuous income stream, k is
the interest rate, compounded continuously, at which the continuous
income stream is invested, and T is the number of years over which
the income stream is received.
• If R  t  is a constant function, then
R t 
B
1  e kT .
k


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