7_7 exponential growth - decay

Post on 07-Dec-2015

30 Views

Category:

Documents

3 Downloads

Preview:

Click to see full reader

DESCRIPTION

igcse mathematics

TRANSCRIPT

7.7 EXPONENTIAL GROWTH AND DECAY:

Exponential Decay: An equation that decreases.

Exponential Growth: An equation that increases.

Growth Factor: 1 plus the percent rate of change which is expressed as a decimal.

Decay Factor: 1 minus the percent rate of change expressed as a decimal.

GOAL:

Definition:An EXPONENTIAL FUNCTION is a function

of the form:

๐‘ฆ=๐‘Ž โˆ™๐‘๐‘ฅ

BaseExponentWhere a โ‰  0, b > o, b โ‰  1,

and x is a real number.

Constant

GRAPHING: To provide the graph of the equation we can go back to basics and create a table.

Ex: What is the graph of y = 3 2โˆ™ x?

GRAPHING:X y = 3 2โˆ™ x y

-2 3 2โˆ™ (-2) = ๐Ÿ‘๐Ÿ’

3 2โˆ™ (-1) = ๐Ÿ‘๐Ÿ-1

0 3 2โˆ™ (0) 3 = 3 1 โˆ™

1 3 2โˆ™ (1) 6 = 3 2 โˆ™

2 3 2โˆ™ (2) 12 = 3 4 โˆ™

GRAPHING: X y

-2 ๐Ÿ‘๐Ÿ’๐Ÿ‘๐Ÿ-1

0 3

1 6

2 12

This graph grows fast = Exponential Growth

YOU TRY IT:

Ex: What is the graph of y = 3โˆ™x?

GRAPHING:X y = 3โˆ™x y

-2 3 โˆ™ (-2) 12

6-1

0 3 = 3 1 โˆ™

1

2

=3 (2)โˆ™ 2

3 โˆ™ (-1) =3 (2)โˆ™ 1

3 โˆ™ (0)

3 โˆ™ (1) =3 โˆ™๐Ÿ‘๐Ÿ๐Ÿ‘๐Ÿ’

3 โˆ™ (2) =3 โˆ™

GRAPHING: X y

-2

๐Ÿ‘๐Ÿ’

๐Ÿ‘๐Ÿ

-1

0 3

1

6

2

12

This graph goes down = Exponential Decay

YOU TRY IT:

Ex: What are the differences and

similarities between:

y = 3 2โˆ™ x

and y = 3โˆ™x?

y = 3 2โˆ™ x

Base = 2 Exponential growth y- intercept (x=0) = 3

y = 3โˆ™x

Base = Exponential Decay y- intercept (x=0) = 3

MODELING: We use the concept of exponential growth in the real world:Ex: Since 2005, the amount of money spent at restaurants in the U.S. has increased 7% each year. In 2005, about 36 billion was spend at restaurants. If the trend continues, about how much will be spent in 2015?

EVALUATING: To provide the solution we must know the following formula:

y = aโˆ™bx

y = totala = initial amount b = growth factor (1 + rate)x = time in years.

SOLUTION:

Y= total:

Since 2005, โ€ฆ has increased 7% each year. In 2005, about 36 billion was spend at restaurantsโ€ฆ. about how much will be spent in 2015?

$36 billion Initial:

Growth: 1 + 0.07

Time (x): 10 years(2005-2015)

unknown y = aโˆ™bx

y = 36โˆ™(1.07)10

y = 36โˆ™(1.967)

y = 70.8 b.

BANKING: We also use the concept of exponential growth in banking:

A = P(1+)nt

A = total balanceP = Principal (initial) amount

n = # of times compound interestt = time in years.

r = interest rate in decimal form

MODELING GROWTH:Ex:

You are given $6,000 at the beginning of your freshman year. You go to a bank

and they offer you 7% interest. How much money will you have after

graduation if the money is:a) Compounded annuallyb) Compounded quarterlyc) Compounded monthly

COMPOUNDED ANNUALLY:

A = P(1+)nt

A = ?P = $6000

n = 1t = 4 yrs

r = 0.07

A = 6000(1+)1(4)

A = 6000(1.07)4

A = 6000(1.3107)A = $7864.77

COMPOUNDED QUARTERLY:

A = P(1+)nt

A = ?P = $6000

n = 4 timest = 4 yrs

r = 0.07

A = 6000(1+)4(4)

A = 6000(1.0175)16

A = 6000(1.3199)A = $7919.58

COMPOUNDED MONTHLY:

A = P(1+)nt

A = ?P = $6000

n = 12 timest = 4 yrs

r = 0.07

A = 6000(1+)12(4)

A = 6000(1.0058)48

A = 6000(1.3221)A = $7932.32

MODELING DECAY:Ex:

Doctors can use radioactive iodine to treat some forms of cancer. The half-life

of iodine-131 is 8 days. A patient receives a treatment of 12 millicuries (a unit of radioactivity) of iodine-131. How much iodine-131 remains in the patient

after 16 days?:

DECAY: To provide the solution we g back to the following formula:

y = aโˆ™bx

y = totala = initial amount b = decay factor (1 - rate)x = time in years.

SOLUTION:

Y= total:

The half-life of iodine-131 is 8 days. A patient receives a treatment of 12 millicuries (a unit of radioactivity) of iodine-131. How much iodine-131 remains in the patient 16 days later?:

12Initial:

Growth: 1- 1/2

Time (x): 16/8 = 2

unknown y = aโˆ™bx

y = 12โˆ™(1/2)2

y = 12โˆ™(.25)

y = 3

CLASSWORK:

Page 450-452:

Problems: As many as needed to master the concept.

top related