6/8/2016
Compound Interest: Non-Continuous
Section 5.2
Outline
Compound Interest: Non-Continuous
r

A  P 1  
 m
 Compound Interest: Non
Continuous
 Compound Interest:
mt
• P = principal amount invested
Continuous
• m = the number of times per year interest is compounded
 Applications of Interest
• r = the annual interest rate
• t = the number of years interest is being compounded
Compounded Continuously
• A = the compound amount, the balance after t years
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Compound Interest
Slide 2
Compound Interest: Continuous
Let P=1000 and r=6%
Compound Interest: Continuous
m
Growth Factor
A
1
1.06
1060
A  Pe rt
4
12
360
m
1.061363551 1.061677812 1.061831238 F=(1+r/m)^m
1061.36
1061.68
1061.83
P*F
 P = principal amount invested
Notice that as m increases, so does A. Therefore, the maximum amount of
interest can be acquired when m is being compounded all the time continuously.
 r = the interest rate
 t = the number of years interest is being compounded
 .06 
4
m  4 : 1 
  1.015  1.061364
4 

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 A = the compound amount, the balance after t years
Slide 3
Compound Interest: Continuous
EXAMPLE
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Slide 4
Compound Interest: Continuous
CONTINUED
(Continuous Compound) Ten thousand dollars is invested at 6.5% interest
compounded continuously. When will the investment be worth $41,787?
4.1787  e0.065t
ln 4.1787  0.065t
SOLUTION
We must first determine the formula for A(t). Since interest is being
compounded continuously, the basic formula to be used is A  Pe rt .
22  t
Divide by 10,000.
Rewrite the equation in
logarithmic form.
Divide by 0.065 and solve for t.
Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest
compounded continuously, in 22 years.
• The interest rate is r= 6.5%=0.065.
• Since ten thousand dollars is being invested, P = 10,000.
• And since the investment is to grow to become $41,787, A = 41,787.
• We will make the appropriate substitutions and then solve for t.
41,787  10,000e 0.065t
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Slide 5
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6/8/2016
Compound Interest: Present Value
If P dollars are invested today, the formula A=Pert gives the value of this investment after t years.
We say P is the present value of the amount A to be received in t years.
Compound Interest: Present Value
EXAMPLE
(Investment Analysis) An investment earns 5.1% interest compounded
continuously and is currently growing at the rate of $765 per year. What is the
present value of the investment?
SOLUTION
• Since the problem involves a rate of change, we will use the formula for the
derivative of A  Pe rt . That is, A΄ = rA.
 P = Present value of A to be received in t years
• The investment is growing at a rate of $765 per year means that A΄ = 765.
 r = the interest rate
• The interest rate is r=5.1%= 0.051. Thus, we have
 t = the number of years interest is being compounded continuously
 A = the amount to be received in t years
765  0.051A
A  15,000
Therefore, the value of A for this situation is 15,000. We can now use this, and
the present value formula, to determine P.
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Slide 7
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Slide 8
Compound Interest: Present Value
CONTINUED
P  Ae  rt
This is the present value
formula.
P  15,000e 0.0511
A = 15,000, r = 0.051 and t = 1
(since we were given the rate of
growth per year).
P  14,254.18
Simplify.
Therefore, the current value is $14,254.18.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
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