applications of exponential functions

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!**

nt

nrPA

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

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 2Quarterly 4Monthly 12

Daily 365Hourly 8760

Every Minute 525,600Every Second 31,536,000

Continuous Compounding and the Number e

Compounding Period n (1+1/n)n Annually 1 =2

Semiannually 2 =2.25Quarterly 4 ≈2.4414Monthly 12 ≈2.6130

Daily 365 ≈2.71457

Hourly 8760 ≈2.718127Every Minute 525,600 ≈2.7182792Every 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

Exponential Growth

trPxf )1()( Initial Amount = P Growth Factor = 1+ r Percent increase = r

Time = t

Exponential Decay

trPxf )1()( Initial Amount = P Decay Factor = 1+ r Percent decrease = r

Time = t

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 6: Solution Calculate the population in the year

2010.

f(10)= 26.9 million

Example 8: Chlorine Evaporation

Each day, 15% of the chlorine in a swimming pool evaporates. If the swimming pool started with 2.5 ppm of chlorine create an equation to model the change in chlorine level over time.

Example 8: Solution Since 15% of the chlorine evaporates

each day, and the initial amount of chlorine is 2.5 ppm

P = 2.5 r = .15

= 2.5(.85) x

• f(x) = 2.5( 1- . 15)x

Example 8: Solution

• How much chlorine will still be in the water after one week?

• .8 ppm