5.3 Applications of Exponential Functions Objective: Create and use exponential models for a variety...

5.3 Applications of Exponential Functions

Objective:

•Create and use exponential models for a variety of exponential growth and decay application problems

Compound Interest When interest is paid on a balance that

includes interest accumulated from the previous time periods it is called compound interest.

Example 1:– If you invest $9000 at 4% interest,

compounded annually, how much is in the account at the end of 5 years?

Example 1: Solution After one year, the account balance is

– 9000 + .04(9000) Principal + Interest– 9000(1+0.04) Factor out 9000– 9000(1.04) Simplify (104% of Principal)– $9360 Evaluate

Note: The account balance changed by a factor of 1.04. If this amount is left in the account, that balance will change by a factor of 1.04 after the second year.– 9360(1.04) OR…– 9000(1.04)(1.04)= 9000(1.04)2

Example 1: Solution Continuing with this pattern shows

that the account balance at the end of t years can be modeled by the function B(t)=9000(1.04)t.

Therefore, after 5 years, an investment of $9000 at 4% interest will be:

– B(5)=9000(1.04)5=$10,949.88

Compound Interest Formula

If P dollars is invested at interest rate r (expressed as a decimal) per time period t, compounded n times per period, then A is the amount after t periods.

**NOTE: You are expected to know this formula!**

trPA 1

Example 2: Different Compounding Periods

Determine the amount that a $5000 investment over ten years at an annual interest rate of 4.8% is worth for each compounding period.– NOTE: Interest rate per period and the number of periods

may be changing!

A. annually

B. quarterly

C. monthly

D. daily

Example 2: Solution

Determine the amount that a $5000 investment over ten years at an annual interest rate of 4.8% is worth for each compounding period.

A. annually A = 5000(1+.048)10=$7990.66

B. quarterly A = 5000(1+.048/4)10(4)=$8057.32

C. monthly A = 5000(1+.048/12)10(12)=$8072.64

D. daily A = 5000(1+.048/365)10(365)=$8080.12

Example 3: Solving for the Time Period

If $7000 is invested at 5% annual interest, compounded monthly, when will the investment be worth $8500?

Example 3: Solution

If $7000 is invested at 5% annual interest, compounded monthly, when will the investment be worth $8500?

8500=7000(1+.05/12)t

Graphing each side of the equation allows us to use the Intersection Method to determine when they are equal.

After about 47 months, or 3.9 years, the investment will be worth $8500.

Continuous Compounding and the Number e

As the previous examples have shown, the more often interest is compounded, the larger the final amount will be. However, there is a limit that is reached.

Consider the following example:

– Example 4: Suppose you invest $1 for one year at 100% annual interest, compounded n times per year. Find the maximum value of the investment in one year.

Continuous Compounding and the Number e

The annual interest rate is 1, so the interest rate period is 1/n, and the number of periods is n.– A = (1+1/n)n

Now observe what happens to the final amount as n grows larger and larger…

Continuous Compounding and the Number e

Compounding Period n (1+1/n)n

Annually 1

Semiannually 2

Quarterly 4

Monthly 12

Daily 365

Hourly 8760

Every Minute 525,600

Every Second 31,536,000

Continuous Compounding and the Number e

Compounding Period n (1+1/n)n

Annually 1 =2

Semiannually 2 =2.25

Quarterly 4 ≈2.4414

Monthly 12 ≈2.6130

Daily 365 ≈2.71457

Hourly 8760 ≈2.718127

Every Minute 525,600 ≈2.7182792

Every Second 31,536,000 ≈2.7182825

The maximum amount of the $1 investment after one year is approximately $2.72, no matter how large n is.

Continuous Compounding and the Number e

When the number of compounding periods increases without

bound, the process is called continuous compounding.

(This suggests that n, the compounding period, approaches

infinity.) Note that the last entry in the preceding table is the

same as the number e to five decimal places. This example is

the case where P=1, r=100%, and t=1. A similar result occurs

in the general case and leads to the following formula:

A=Pert

**NOTE: You are expected to know this formula!**

Example 5: Continuous Compounding

If you invest $3500 at 3% annual interest compounded continuously, how much is in the account at the end of 4 years?

Example 5: Solution

If you invest $3500 at 3% annual interest compounded continuously, how much is in the account at the end of 4 years?

A=3500e(.03)(4)=$3946.24

Exponential Growth and Decay

Exponential growth or decay can be described by a

function of the form f(x)=Pax where f(x) is the quantity

at time x, P is the initial quantity, and a is the factor

by which the quantity changes (grows or decays)

when x increases by 1.

If the quantity f(x) is changing at a rate r per time

period, then a=1+r or a=1-r (depending on the type of

change) and f(x)=Pax can be written as

f(x)=P(1+r) x or f(x)=P(1-r) x

**NOTE: You are expected to know this formula!**

Example 6: Population Growth

The population of Tokyo, Japan, in the

year 2000 was about 26.4 million and

is projected to increase at a rate of

approximately 0.19% per year. Write

the function that gives the population

of Tokyo in year x, where x=0

corresponds to 2000.

Example 6: Solution The population of Tokyo, Japan, in the year

2000 was about 26.4 million and is

projected to increase at a rate of

approximately 0.19% per year. Write the

function that gives the population of Tokyo

in year x, where x=0 corresponds to 2000.

f(x)=26.4(1.0019)x

Example 7: Population Growth

A newly formed lake is stocked with

900 fish. After 6 months, biologists

estimate there are 1710 fish in the

lake. Assuming the fish population

grows exponentially, how many fish

will there be after 24 months?

Example 7: Solution Using the formula f(x)=Pax, we have f(x)=900ax,

which leaves us to determine a. Use the other given data about the population growth to determine a.

In 6 months, there were 1710 fish:– f(6)=1710 1710= 900a6 a=1.91/6

f(x)=900(1.91/6)x

f(24)= 900(1.91/6)24 = 11,728.89

There will be about 11,729 fish in the lake after 24 months.

Example 8: Chlorine Evaporation

Each day, 15% of the chlorine in a swimming pool evaporates. After how many days will 60% of the chlorine have evaporated?

Example 8: Solution Since 15% of the chlorine evaporates

each day, there is 85% remaining. This is the rate, or a value.

f(x)=(1-0.15)x=0.85x

• We want to know when 60% has evaporated; or, when the evaporation has left 40%.

f(x)=0.85x

0.40=0.85x

Example 8: Solution

• Solve for x by finding the point of intersection.

• x ≈ 5.638; After about 6 days, 60% of the chlorine has evaporated