F I t P P ( 1 r ) nt G J r t H K A P 1 n - - PowerPoint PPT Presentation

Lesson 5.1: Part 3 Applications

P P r where r rate and t years

t t

    ( ) , 1

Continuous Compound Interest

Annual Interest More frequent compounding

A P r n n

• f compoundings

per year r n rate per comp

nt

 

F H G I K J

  1 # / .

For Continuous Compounding

A Pert 

Ex 8: A total of $12,000 is invested for 5 years at an annual interest rate of 9%. Find the balance if it is compounded a.Quarterly b.Monthly c.Continuously

P = $12,000 r = 9% t = 5 years Quarterly Monthly

P

5 4 5

12000 1 09 4  

F H I K

.

( )

P

5 20

12000 1 0225  .

a f

P

5 20

12000 10225  .

a f

P

5

11  $18726. P

5 12 5

12000 1 09 12  

F H I K

.

( )

P

5 60

12000 1 0075  .

a f

P

5 60

12000 10075  .

a f

P

5

17  $18788.

A P r n

nt

 

F H I K

1

A P r n

nt

 

F H I K

1

P = $12,000 r = 9% t = 5 years Continuously

A Pert  A e 12000

09 5 . ( )

A e 12000

45 .

A  $18819.75

Ex 9: The half-life of the radioactive decay rate of the plutonium that spread across Chernobyl, in 1986, in the former Soviet Union, can be modeled using the function

Radioactive Decay

P

t

 F

H G I K J

10 1 2

24 360 / ,

where P is the amount of an original 10 lbs of plutonium remaining after t years. If t = 0 represents 1986, how much plutonium is left in 2005? How much plutonium would be left after 24,360 year? 100,000 years?

(extra writing space)

Homework: p373 #51-57 odd

2005 = 19 years

P  F

H I K

10 1 2

19 24360

P lbs  9 99 .

After 24,360 years

P  F

H I K

10 1 2

24360 24360

P  F

H I K

10 1 2

1

 5lbs

After 100,000 years

P  F

H I K

10 1 2

100000 24360  0 58

. lbs