chapter 4 additional derivative topics section 1 the constant e and continuous compound interest

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Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

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Page 1: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

Chapter 4

Additional Derivative Topics

Section 1

The Constant e and Continuous Compound

Interest

Page 2: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

2

Objectives for Section 4.1 e and Continuous Compound Interest

■ The student will be able to work with problems involving the irrational number e

■ The student will be able to solve problems involving continuous compound interest.

Page 3: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

3

The Constant e

Reminder:

By definition, e = 2.718 281 828 459 …

Do you remember how to find this on a calculator?

e is also defined as either one of the following limits:

s

s

n

ns

ne /1

01lim

11lim

Page 4: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

4

Compound Interest

Let P = principal, r = annual interest rate, t = time in years, n = number of compoundings per year, and A = amount realized at the end of the time period.

• Simple Interest: A = P (1 + r) t

• Compound interest:

•Continuous compounding: A = P ert.

A P 1r

n

nt

A P 1r

n

nt

Page 5: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

5

Compound Interest

Derivation of the Continuous Compound Formula: A = P ert.

limx

P 1r

n

nt

P limx

1r

n

n

r

rt

P lims 0

1 s 1

s

rt

P lims 0

1 s 1

s

rt

Pert

Page 6: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

6

ExampleGenerous Grandma

Your Grandma puts $1,000 in a bank for you, at 5% interest. Calculate the amount after 20 years.

Simple interest: A = 1000 (1 + 0.05•20) = $2,000.00

Compounded annually:A = 1000 (1 + .05)20 =$2,653.30

Compounded daily:

Compounded continuously: A = 1000 e(.05)(20) = $2,718.28

A 1000 1.05

360

360 20

$2,718.09

Page 7: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

7

Example IRA

After graduating from Barnett College, Sam Spartan landed a great job with Springettsbury Manufacturing, Inc. His first year he bought a $3,000 Roth IRA and invested it in a stock sensitive mutual fund that grows at 12% a year, compounded continuously. He plans to retire in 35 years.

a. What will be its value at the end of the time period?

b. The second year he repeated the purchase of an identical Roth IRA. What will be its value in 34 years?

Page 8: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

8

Example (continued)

After graduating from Barnett College, Sam Spartan landed a great job with Springettsbury Manufacturing, Inc. His first year he bought a $3,000 Roth IRA and invested it in a stock sensitive mutual fund that grows at 12% a year, compounded continuously. He plans to retire in 35 years.

a. What will be its value at the end of the time period?

A = Pert = 3000 e(.12)(35) =$200,058.99

b. The second year he repeated the purchase of an identical Roth IRA. What will be its value in 34 years?

$177,436.41

Page 9: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

9

Example Computing Growth Time

How long will it take an investment of $5,000 to grow to $8,000 if it is invested at 5% compounded continuously?

Page 10: Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

10

Example (continued)

How long will it take an investment of $5,000 to grow to $8,000 if it is invested at 5% compounded continuously?

Solution: Use A = Pert.

0.05

0.05

0.05

8,000 5,000

1.6

ln ln1.6

0.05 ln1.6

ln1.6

0.059.4 years

t

t

t

e

e

e

t

t

t