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EXPONENTIAL GROWTH MODEL WRITING EXPONENTIAL GROWTH MODELS A quantity is growing exponentially if it increases by the same percent in each time period. C is the initial amount. t is the time period. (1 + r) is the growth factor, r is the growth rate. The percent of increase is 100r. y = C (1 + r) t
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### Transcript of Exponential growth and decay EXPONENTIAL GROWTH MODEL

WRITING EXPONENTIAL GROWTH MODELS

A quantity is growing exponentially if it increases by the same percent in each time period.

C is the initial amount. t is the time period.

(1 + r) is the growth factor, r is the growth rate.

The percent of increase is 100r.

y = C (1 + r)t Finding the Balance in an Account

COMPOUND INTEREST You deposit \$500 in an account that pays 8% annual interest compounded yearly. What is the account balance after 6 years?

SOLUTION METHOD 1 SOLVE A SIMPLER PROBLEM

Find the account balance A1 after 1 year and multiply by the growth factor to

find the balance for each of the following years. The growth rate is 0.08, so the growth factor is 1 + 0.08 = 1.08.

•••

•••

A1 = 500(1.08) = 540 Balance after one year

A2 = 500(1.08)(1.08) = 583.20 Balance after two years

A3 = 500(1.08)(1.08)(1.08) = 629.856

A6 = 500(1.08) 6 793.437

Balance after three years

Balance after six years EXPONENTIAL GROWTH MODEL

C is the initial amount.

t is the time period.

(1 + r) is the growth factor, r is the growth rate.

The percent of increase is 100r.

y = C (1 + r)t

EXPONENTIAL GROWTH MODEL

500 is the initial amount. 6 is the time period.

(1 + 0.08) is the growth factor, 0.08 is the growth rate.

A6 = 500(1.08) 6 793.437 Balance after 6 years

A6 = 500 (1 + 0.08) 6

SOLUTION METHOD 2 USE A FORMULA

Finding the Balance in an Account

COMPOUND INTEREST You deposit \$500 in an account that pays 8% annual interest compounded yearly. What is the account balance after 6 years?

Use the exponential growth model to find the account balance A. The growthrate is 0.08. The initial value is 500. Writing an Exponential Growth Model

A population of 20 rabbits is released into a wildlife region. The population triples each year for 5 years. So, the growth rate r is 2 and the percent of increase each year is 200%.So, the growth rate r is 2 and the percent of increase each year is 200%.So, the growth rate r is 2 and the percent of increase each year is 200%.

1 + r = 31 + r = 3

Writing an Exponential Growth Model

A population of 20 rabbits is released into a wildlife region. The population triples each year for 5 years.

a. What is the percent of increase each year?

SOLUTION

The population triples each year, so the growth factor is 3.

1 + r = 3

The population triples each year, so the growth factor is 3.

Reminder: percent increase is 100r. A population of 20 rabbits is released into a wildlife region. The population triples each year for 5 years.

b. What is the population after 5 years?

Writing an Exponential Growth Model

SOLUTION

After 5 years, the population is

P = C(1 + r) t Exponential growth model

= 20(1 + 2) 5

= 20 • 3 5

= 4860

Help

Substitute C, r, and t.

Simplify.

Evaluate.

There will be about 4860 rabbits after 5 years. A Model with a Large Growth Factor

GRAPHING EXPONENTIAL GROWTH MODELS

Graph the growth of the rabbit population.

SOLUTION Make a table of values, plot the points in a coordinate plane, and draw a smooth curve through the points.

t

P 486060 180 540 162020

51 2 3 40

0

1000

2000

3000

4000

5000

6000

1 72 3 4 5 6Time (years)

Po

pu

lati

on

P = 20 ( 3 ) t Here, the large

growth factor of 3 corresponds to a rapid increase

Here, the large growth factor of 3 corresponds to a rapid increase WRITING EXPONENTIAL DECAY MODELS

A quantity is decreasing exponentially if it decreases by the same percent in each time period.

EXPONENTIAL DECAY MODEL

C is the initial amount.t is the time period.

(1 – r ) is the decay factor, r is the decay rate.

The percent of decrease is 100r.

y = C (1 – r)t Writing an Exponential Decay Model

COMPOUND INTEREST From 1982 through 1997, the purchasing powerof a dollar decreased by about 3.5% per year. Using 1982 as the base for comparison, what was the purchasing power of a dollar in 1997?

SOLUTION Let y represent the purchasing power and let t = 0 represent the year 1982. The initial amount is \$1. Use an exponential decay model.

= (1)(1 – 0.035) t

= 0.965 t

y = C (1 – r) t

y = 0.96515

Exponential decay model

Substitute 1 for C, 0.035 for r.

Simplify.

Because 1997 is 15 years after 1982, substitute 15 for t.

Substitute 15 for t.

The purchasing power of a dollar in 1997 compared to 1982 was \$0.59.

0.59 Graphing the Decay of Purchasing Power

GRAPHING EXPONENTIAL DECAY MODELS

Graph the exponential decay model in the previous example. Use the graph to estimate the value of a dollar in ten years.

SOLUTION Make a table of values, plot the points in a coordinate plane, and draw a smooth curve through the points.

0

0.2

0.4

0.6

0.8

1.0

1 123 5 7 9 11Years From Now

Pu

rch

asin

g P

ow

er

(do

lla

rs)

2 4 6 8 10

t

y 0.8370.965 0.931 0.899 0.8671.00

51 2 3 40

0.70.808 0.779 0.752 0.726

106 7 8 9

Your dollar of today will be worth about 70 cents in ten years.

Your dollar of today will be worth about 70 cents in ten years.

y = 0.965t

Help GRAPHING EXPONENTIAL DECAY MODELS

EXPONENTIAL GROWTH AND DECAY MODELS

y = C (1 – r)ty = C (1 + r)t

EXPONENTIAL GROWTH MODEL EXPONENTIAL DECAY MODEL

1 + r > 1 0 < 1 – r < 1

CONCEPT

SUMMARY

An exponential model y = a • b t represents exponential

growth if b > 1 and exponential decay if 0 < b < 1.C is the initial amount.t is the time period.

(1 – r) is the decay factor, r is the decay rate.

(1 + r) is the growth factor, r is the growth rate. (0, C)(0, C) Exponential Growth & Decay Models

• A0 is the amount you start with, t is the time, and k=constant of growth (or decay)

• f k>0, the amount is GROWING (getting larger), as in the money in a savings account that is having interest compounded over time

• If k<0, the amount is SHRINKING (getting smaller), as in the amount of radioactive substance remaining after the substance decays over time

ktoA A t A e Graphs

00

k

eAA kt

A0A0

0

0

ktA A e

k Example• Population Growth of the United States. In

1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau)– Find the exponential growth function.– What will the population be in 2020?– After how long will the population be double

what it was in 1990? Solution• At t = 0 (1990), the population was about 249 million. We

substitute 249 for A0 and 0.08 for k to obtain the exponential growth function.

A(t) = 249e0.08t

• In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:

A(3) = 249e0.08(3) = 249e0.24 317.

The population will be approximately 317 million in 2020. Solution continued• We are looking for the doubling time T.

498 = 249e0.08T

2 = e0.08T

ln 2 = ln e0.08T

ln 2 = 0.08T (ln ex = x)

= T

8.7 T

The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

ln 2

0.08 Exponential Decay• Decay, or decline, is represented by the function A(t) =

A0ekt, k < 0.

In this function:

A0 = initial amount of the substance, A = amount of the substance left after time, t = time, k = decay rate.

• The half-life is the amount of time it takes for half of an amount of substance to decay. Example Carbon Dating. The radioactive element carbon-14 has a

half-life of 5715 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?

Solution: The function for carbon dating is

A(t) = A0e-0.00012t.

If the charcoal has lost 7.3% of its carbon-14 from its initial amount A0, then 92.7%A0 is the amount present. Example continuedTo find the age of the charcoal, we solve the equation for t :

The charcoal was about 632 years old.

0.000120 092.7% tA A e

0.000120.927 te0.00012ln 0.927 ln te

ln 0.927 0.00012t

ln 0.927

0.00012632

t

t