Presentation on theme: "5.3 Applications of Exponential Functions"— Presentation transcript:
15.3 Applications of Exponential Functions Objective:Create and use exponential models for a variety of exponential growth and decay application problems
2Compound InterestWhen 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 $9000 at 4% interest, compounded annually, how much is in the account at the end of 5 years?
3Example 1: Solution After one year, the account balance is (9000) Principal + Interest9000(1+0.04) Factor out 90009000(1.04) Simplify (104% of Principal)$ EvaluateNote: The account balance changed by a factor of 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
4Example 1: SolutionContinuing 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
5Compound 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!**
6Example 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. annuallyB. quarterlyC. monthlyD. daily
7Example 2: SolutionDetermine the amount that a $5000 investment over ten years at an annual interest rate of 4.8% is worth for each compounding period.A. annually A = 5000(1+.048)10=$B. quarterly A = 5000(1+.048/4)10(4)=$C. monthly A = 5000(1+.048/12)10(12)=$D. daily A = 5000(1+.048/365)10(365)=$
8Example 3: Solving for the Time Period If $7000 is invested at 5% annual interest, compounded monthly, when will the investment be worth $8500?
9Example 3: SolutionIf $7000 is invested at 5% annual interest, compounded monthly, when will the investment be worth $8500?8500=7000(1+.05/12)tGraphing 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.
10Continuous 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.
11Continuous 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)nNow observe what happens to the final amount as n grows larger and larger…
12Continuous Compounding and the Number e Compounding Periodn(1+1/n)nAnnually1Semiannually2Quarterly4Monthly12Daily365Hourly8760Every Minute525,600Every Second31,536,000
13Continuous Compounding and the Number e Compounding Periodn(1+1/n)nAnnually1=2Semiannually2=2.25Quarterly4≈2.4414Monthly12≈2.6130Daily365≈Hourly8760≈Every Minute525,600≈Every Second31,536,000≈The maximum amount of the $1 investment after one year is approximately $2.72, no matter how large n is.
14Continuous 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!**
15Example 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?
16Example 5: SolutionIf you invest $3500 at 3% annual interest compounded continuously, how much is in the account at the end of 4 years?A=3500e(.03)(4)=$
17Exponential 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 asf(x)=P(1+r) x or f(x)=P(1-r) x**NOTE: You are expected to know this formula!**
18Example 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.
19Example 6: SolutionThe 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
20Example 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?
21Example 7: SolutionUsing 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)= = 900a6 a=1.91/6f(x)=900(1.91/6)xf(24)= 900(1.91/6)24 = 11,728.89There will be about 11,729 fish in the lake after 24 months.
22Example 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?
23Example 8: SolutionSince 15% of the chlorine evaporates each day, there is 85% remaining. This is the rate, or a value.f(x)=(1-0.15)x=0.85xWe want to know when 60% has evaporated; or, when the evaporation has left 40%.f(x)=0.85x0.40=0.85x
24Example 8: Solution Solve for x by finding the point of intersection. x ≈ 5.638; After about 6 days, 60% of the chlorine has evaporated