13.4 Compound Interest

Locker LESSON 13.4 Compound Interest Name Class Date 13.4 Compound Interest Common Core Math Standards The student is expected to: F-LE.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). Also F.IF.7e, F-IF.7, A-REI.11, F-IF.8, F-BF.1 Mathematical Practices MP.4 Modeling Language Objective Explain to a partner what simple interest is and what compound interest is. ENGAGE Essential Question: How do you model the value of an investment that earns compound interest? Possible answer: You must know the principal (amount invested) P, the annual interest rate r, and the number of times per year, n, that the interest is compounded. Then, the value V of the investment at time t (in years) is V (t) = P (1 + n ) nt. This model becomes V (t) = Pe rt when the interest is compounded continuously. Image Credits: Treasury Department/AP Images Essential Question: How do you model the value of an investment that earns compound interest? Explore Comparing Simple and Compound Interest A bond is a type of investment that you buy with cash and for which you receive interest either as the bond matures or when it matures. A conventional bond generates an interest payment, sometimes called a coupon payment, on a regular basis, typically twice a year. The interest payments end when the bond matures and the amount you paid up-front is returned to you. A zero-coupon bond, on the other hand, requires you to invest less money up-front and pays you its maturity value, which includes all accumulated interest, when the bond matures. Over the life of the bond, the interest that is earned each period itself earns interest until the bond matures. The basic difference between a conventional bond and a zero-coupon bond is the type of interest earned. A conventional bond pays simple interest, whereas a zero-coupon bond pays compound interest, which is interest that earns interest. A For $1000, you can buy a conventional bond that has a maturity value of $1000, has a maturity date of 4 years, and pays 5% annual interest. Calculate the interest that the investment earns annually by completing the table. (Bear in mind that interest earned is paid to you and not reinvested.) Conventional Bond Year Value of Investment at Beginning of Year Interest Earned for Year (Paid to the Investor) Value of Investment at End of Year 0 $1000 1 $1000 $1000 (0.05) = $ 50 $ 1000 2 $1000 $1000 (0.05) = $ 50 $ 3 $1000 $1000 (0.05) = $ 50 $ 1000 1000 Resource Locker PREVIEW: LESSON PERFORMANCE TASK View the online Engage. Discuss the photo and the concept of investing to meet a changing goal amount. Then preview the Lesson Performance Task. 4 $1000 $1000 (0.05) = $ 50 $ 1000 Module 13 681 Lesson 4 Name Class Date 13.4 Compound Interest Essential Question: How do you model the value of an investment that earns compound interest? F.LE.2 Construct... exponential functions,... given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). Also F.IF.7e, F.IF.7, A.REI.11 Houghton Mifflin Houghton Harcourt Mifflin Publishing Harcourt Company Publishing Image Company Credits: Treasury Department/AP Images Explore Comparing Simple and Compound Interest A bond is a type of investment that you buy with cash and for which you receive interest either as the bond matures or when it matures. A conventional bond generates an interest payment, sometimes called a coupon payment, on a regular basis, typically twice a year. The interest payments end when the bond matures and the amount you paid up-front is returned to you. A zero-coupon bond, on the other hand, requires you to invest less money up-front and pays you its maturity value, which includes all accumulated interest, when the bond matures. Over the life of the bond, the interest that is earned each period itself earns interest until the bond matures. The basic difference between a conventional bond and a zero-coupon bond is the type of interest earned. A conventional bond pays simple interest, whereas a zero-coupon bond pays compound interest, which is interest that earns interest. For $1000, you can buy a conventional bond that has a maturity value of $1000, has a maturity date of 4 years, and pays 5% annual interest. Calculate the interest that the investment earns annually by completing the table. (Bear in mind that interest earned is paid to you and not reinvested.) Year Value of Investment at Beginning of Year Conventional Bond Interest Earned for Year Value of Investment at (Paid to the Investor) End of Year 0 $1000 1 $1000 $1000 (0.05) = $ $ 2 $1000 $1000 (0.05) = $ $ Resource 50 1000 3 $1000 $1000 (0.05) = $ $ 4 $1000 $1000 (0.05) = $ $ 50 50 50 1000 1000 1000 HARDCOVER PAGES 495 504 Turn to these pages to find this lesson in the hardcover student edition. Module 13 681 Lesson 4 681 Lesson 13.4

B For $822.70, you can buy a zero-coupon bond that has a maturity value of $1000, has a maturity date of 4 years, and pays 5% annual interest. Calculate the interest that the investment earns annually by completing the table. (Bear in mind that interest earned is reinvested and not paid to you until the bond matures.) Zero-Coupon Bond Value of Investment Year at Beginning of Year Interest Earned for Year (Reinvested) Value of Investment at End of Year 0 $822.70 1 $822.70 $822.70 (0.05) = $ 41.14 $822.70 + $ 41.14 = $ 863.84 EXPLORE Comparing Simple and Compound Interest INTEGRATE TECHNOLOGY Students have the option of completing the Explore activity either in the book or online. 2 $ 863.84 $ 863.84 (0.05 ) = $ 43.19 $ 863.84 + $ 43.19 = $ 907.03 3 $ 907.03 $ 907.03 (0.05) = $ 45.35 $ 907.03 + $ 45.35 = $ 952.38 4 $ 952.38 $ 952.38 (0.05) = $ 47.62 $ 952.38 + $ 47.62 = $1000 QUESTIONING STRATEGIES Which type of bond generates the most interest? How much more interest is generated by this type of bond? The conventional bond generates more interest than the zero-coupon bond; $22.70 Reflect 1. Describe the difference between how simple interest is calculated and how compound interest is calculated. Simple interest is paid back to the investor, and only calculated from the initial amount invested. So, the interest amount is a constant value. Compound interest is reinvested. It is first calculated from the initial amount invested. Then, the interest is added to the initial amount invested and calculated again from this new value. So, the interest amount grows exponentially over time. 2. If V (t) represents the value of an investment at time t, in whole numbers of years starting with t = 0 and ending with t = 4, write a constant function for the last column in the first table and an exponential function for the last column in the second table. In the last column of the first table, the value of the investment is a constant $1000. So, V (t) = 1000. In the last column of the second table, the value of the investment grows exponentially with an initial investment of $822.70 and an annual growth rate of 5%. So, V (t) = 822.70 (1.05) t. If you are an investor, what is the advantage of investing in the zero-coupon bond? You may invest less money. Module 13 682 Lesson 4 PROFESSIONAL DEVELOPMENT Math Background The compound interest formula V (t) = P (1 + n ) nt gives the n V(1) total amount of money V (t) at the end of t years for an initial 2 2.25 investment P with an interest rate r that is compounded n 5 2.4883 times per year. When $1 is invested at 100% interest for 1 10 2.5937 n year, the formula becomes V (1) = (1 + 1_ n ). The table gives 100 2.7048 the value of V (1) for various values of n. Th e table suggests 1000 2.7169 that as n grows larger, the value of V (1) approaches a limiting value. This value, which is defined to be the constant e, is about 2.718281828459045. Written using the limit notation of calculus, n e = lim n (1 + 1_ n ). Compound Interest 682

EXPLAIN 1 Modeling Interest Compounded Annually QUESTIONING STRATEGIES What is the difference between simple interest and interest compounded annually? Simple interest is interest earned on the principal. Interest compounded annually is interest earned on the principal and on the interest. When is the simple interest earned on an investment equal to interest that is compounded annually? At the end of the first year, assuming the interest rates are the same. Explain 1 Modeling Interest Compounded Annually Recall that the general exponential growth function is ƒ (t) = a (1 + r) t. When this function is applied to an investment where interest is compounded annually at a rate r, the function is written as V (t) = P (1 + r) t where V (t) is the value V of the investment at time t and P is the principal (the amount invested). A person invests $3500 in an account that earns 3% annual interest. Find when the value of the investment reaches $10,000. Let P = 3500 and r = 0.03. Then V (t) = P (1 + r) t = 3500 (1.03) t. x Graph y = 3500 (1.03) on a graphing calculator. Also graph the line y = 10,000 and find the point where the graphs intersect. The value of the function is 10,000 at x 35.5. So, the investment reaches a value of $10,000 in approximately 35.5 years. A person invests $200 in an account that earns 6.25% annual interest. Find when the value of the investment reaches $5000. t Let P = 200 and r = 0.0625. Then V (t) = P (1 + r) t = 200 ( 1.0625 ). Graph the function on a graphing calculator. Also graph the line COMMUNICATE MATH The term compound has many meanings. It means one thing in science, another in the everyday world (for example, an army compound), and still another in mathematics. Talk with students about the word s different meanings, specifying its particular meaning in math class. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Discuss with students how compound interest is an example of exponential growth. Help them to see that the formula for compound interest is an exponential function. y = and find the point of intersection. 5000 The value of the function is at x 53.1. So, the investment reaches a value of $5000 in 53.1 approximately years. Your Turn 5000 3. A person invests $1000 in an account that earns 5.5% annual interest. Find when the value of the investment has doubled. Let P = 1000 and r = 0.055. Then V (t) = P (1 + r) t = 1000 (1.055) t. Graphing y = 1000 (1.055) x along with the line y = 2000 shows that the value of the function is 2000 at x 12.9. So, the investment reaches a value of $2000 in approximately 12.9 years. Module 13 683 Lesson 4 COLLABORATIVE LEARNING Small Group Activity Have students work in groups of 3 to 4. Instruct students to use the Internet to research various types of investments and the interest programs they offer. Have them create a scenario for each type of investment they identify, and show how the investment grows over time. Have them create posters showing the results of their research and present the posters to the class. 683 Lesson 13.4

Explain 2 Modeling Interest Compounded More than Once a Year Interest may be earned more frequently than once a year, such as semiannually (every 6 months), quarterly (every 3 months), monthly, and even daily. If the number of times that interest is compounded in a year is n, then the interest rate per compounding period is n (where r is the annual interest rate), and the number of times that interest is compounded in t years is nt. So, the exponential growth function becomes V (t) = P (1 + n) nt. A person invests $1200 in an account that earns 2% annual interest compounded quarterly. Find when the value of the investment reaches $1500. Let P = 1200, r = 0.02, and n = 4. Then V (t) = P (1 + n) nt 4t = 1200 ( 1 + _ 0.02 4 ) = 1200 (1.005) 4t. Graph the function y = 1200 (1.005) 4x on a graphing calculator. Also graph the line y = 1500 and find the point where the graphs intersect. EXPLAIN 2 Modeling Interest Compounded More Than Once a Year AVOID COMMON ERRORS Some students may mistake the number of months in a compounding period for the number of compounding periods per year. For example, for interest compounded quarterly, students may let n = 3, since quarterly means every 3 months. Clarify that n represents the number of times the interest is compounded each year. The value of the function is 1500 at x 11.2. So, the investment reaches a value of $1500 in approximately 11.2 years. A person invests $600 in an account that earns 6.25% annual interest compounded semiannually. Find when the value of the investment reaches $1700. Let P = 600, r = 0.0625, and n = 2. Then V (t) = P (1 + n ) nt = ) 2 0.0625 600 (1 + _ 2 t = 600 ( 1.03125 ) Graph the function on a graphing calculator. Also graph the line y = the graphs intersect. t. and find the point where The value of the function is 1700 at x 16.9. So, the investment reaches a value of $1700 in approximately 16.9 years. 1700 Module 13 684 Lesson 4 2 QUESTIONING STRATEGIES Why does the amount of interest earned increase as the frequency of compounding increases? The more often the interest is compounded, the sooner the new interest is added to the balance on which the interest is earned. INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Use a graphing calculator to illustrate how increasing the number of compounding periods per year increases the rate of growth of an investment. Have students graph and compare several functions, each based on the same initial investment, interest rate, and time period, but for varying values of n. DIFFERENTIATE INSTRUCTION Communicating Math Some students may feel overwhelmed by the complexity of the functions presented in the lesson. These students may benefit from rewriting the formulas using words in place of the various variables. Students can then use their verbal formulas to solve the problems. After achieving some success, students will likely feel less anxious about working with the formulas in their traditional forms. Compound Interest 684

EXPLAIN 3 Modeling Interest Compounded Continuously QUESTIONING STRATEGIES How is interest that is compounded continuously different from interest that is compounded daily? Interest that is compounded daily has 365 (or 366) compounding periods per year. Interest that is compounded continuously has, theoretically, infinitely many compounding periods per year. AVOID COMMON ERRORS Some students may, in error, raise the product of P and e to the exponent rt. Reinforce that the order of operations dictates that e be raised to the exponent, with the result multiplied by P. Your Turn 4. A person invests $8000 in an account that earns 6.5% annual interest compounded daily. Find when the value of the investment reaches $20,000. Let P = 8000, r = 0.065, and n = 365. Then V (t) = P (1 + n) nt = 8000 ( 1 + _ 0.065 365t 365 ) 8000 (1.000178) 365t. Graphing along y = 8000 (1.000178) 365x along with the y = 20,000 Explain 3 By letting m = Modeling Interest Compounded Continuously n) nt mrt as V (t) = P (1 + 1_ m) because mrt 1_ m) as V(t) = P rt m (1 + 1_ m) and m 1_ m) n_ r, you can rewrite the model V (t) = P (1 + n = 1_ m and nt = mrt. Then, rewriting V (t) = P (1 + letting n increase without bound, which causes m to increase without bound, you see that (1 + approaches e, and the model simply becomes V (t) = Pe rt. This model gives the value of an investment with principal P and annual interest rate r when interest is compounded continuously. shows that the value of the function is 20,000 at x = 14.1. So, the investment reaches a value of $20,000 in approximately 14.1 years. A person invests $5000 in an account that earns 3.5% annual interest compounded continuously. Find when the value of the investment reaches $12,000. Let P = 5000 and r = 0.035. Then V (t) = 5000e 0.035t. Graph y = 5000e 0.035x. on a graphing calculator. Also graph the line y = 12,000 and find the point where the graphs intersect. The value of the function is 12,000 at x 25. So, the investment reaches a value of $12,000 in approximately 25 years. Module 13 685 Lesson 4 LANGUAGE SUPPORT Communicate Math Have students work in pairs. Ask the first student to explain what simple interest is and how to find it, while the second student takes notes. The students switch roles; the second student explains what compound interest is and how to find it, while the first student takes notes. The first student should then do the same for constant compounding. Ask the pair to work together to read their notes and make any edits needed to complete their explanations. 685 Lesson 13.4

The principal amount, $350, earns 6% annual interest compounded continuously. Find when the value reaches $1800. 0.06t Let P = 350 and r = 0.06. Then V(t) = 350 e. Graph the function on a graphing calculator. Also graph the line y = 1800 and find the point where the graphs intersect. EXPLAIN 4 Finding and Comparing Effective Annual Interest Rates The value of the function is 1800 at x 27.3. So, the investment reaches a value of $1800 in 27.3 approximately years. Your Turn QUESTIONING STRATEGIES Which is greater, the nominal annual interest rate or its effective annual interest rate? Explain. The effective annual interest rate is greater because it takes into account the interest that is generated on the interest. 5. A person invests $1550 in an account that earns 4% annual interest compounded continuously. Find when the value of the investment reaches $2000. Let P = 1550 and r = 0.04. Then V (t) = 1550e 0.04t. Graphing y = 1550e 0.04x along with the line y = 2000 shows that the value of the function is 2000 at x 6.4. So, the investment reaches a value of $2000 in approximately 6.4 years. Explain 4 Finding and Comparing Effective Annual Interest Rates n) nt = P n) n The value-of-an-investment function V (t) = P (1 + ( 1 + ^t, where interest is compounded n times per year, is an exponential function of the form ƒ (t) = ab t where a = P and b = (1 + n) n. When the base of an exponential function is greater than 1, the function is an exponential growth function where the base is the growth factor and 1 less than the base is the growth rate. So, the growth rate for V (t) = P (1 + n) nt is b - 1 = (1 + n) n - 1. This growth rate is called the investment s effective annual interest rate R, whereas r is called the investment s nominal annual interest rate. Similarly, for the value-of-an-investment function V (t) = Pe rt =P e r t, where interest is compounded continuously, the growth factor is e r, and the growth rate is e r - 1. So, in this case, the effective annual interest rate is R = e r - 1. For an account that earns interest compounded more than once a year, the effective annual interest rate is the rate that would produce the same amount of interest if interest were compounded annually instead. The effective rate allows you to compare two accounts that have different nominal rates and different compounding periods. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 The concept of compound interest provides an opportunity to informally discuss the idea of a limiting process. As the frequency of compounding increases, the process gets closer and closer to continuous compounding. Tell students that the study of limiting processes is important in more advanced mathematics, especially calculus. Module 13 686 Lesson 4 Compound Interest 686

A Arturo plans to make a deposit in one of the accounts shown in the table. To maximize the interest that the account earns, which account should he choose? Account X Account Y Nominal Annual Interest Rate Compounding Period 2.5% 2.48% Quarterly Monthly B For Account X, interest is compounded quarterly, so n = 4. The nominal rate is 2.5%, so r = 0.025. R X = (1 + _ n) r - 1 Use the formula for the effective rate. = ( 1 + _ 0.025 4 4 ) - 1 Substitute. 0.02524 Simplify. For Account Y, interest is compounded monthly, so n = 12. The nominal rate is 2.48%, so r = 0.0248. R Y = (1 + _ n) r - 1 Use the formula for the effective rate. = ( 1 + _ 0.0248 12 12 ) - 1 Substitute. 0.02508 Simplify. Account X has an effective rate of 2.524%, and Account Y has an effective rate of 2.508%, so Account X has a greater effective rate, and Arturo should choose Account X. Harriet plans to make a deposit in one of two accounts. Account A has a 3.24% nominal rate with interest compounded continuously, and Account B has a 3.25% nominal rate with interest compounded semiannually. To maximize the interest that the account earns, which account should she choose? For Account A, interest is compounded continuously. The nominal rate is 3.24%, so r = 0.0324. R A = e r - 1 Use the formula for the effective rate. 0.0324 = e - 1 Substitute. 0.03293 Simplify. Module 13 687 Lesson 4 687 Lesson 13.4

For Account B, interest is compounded semiannually, so so n = 2. The nominal rate is 3.25%, so r = 0.0325 R B = (1 + _ n) r n - 1 Use the formula for the effective rate. ) 2 = (1 + _ 0.0325-1 Substitute. 2 0.03276 Simplify. Account A has an effective rate of 3.293 3.276 %, and Account B has an effective rate of %, so Account A has a greater effective rate, and Harriet should choose Account A. Your Turn 6. Jaclyn plans to make a deposit in one of the accounts shown in the table. To maximize the interest that the account earns, which account should she choose? Account X Account Y Nominal Annual Interest Rate 4.24% 4.18% Compounding Period Annually Daily For Account X, interest is compounded annually, so n = 1. The nominal rate is 4.24%, so r = 0.0424. R X = (1 + n - 1 n) = ( 1 + _ 0.0424 1 1 ) - 1 = 0.0424 For Account Y, interest is compounded daily, so n = 365. The nominal rate is 4.18%, so r = 0.0418. R Y = (1 + n - 1 n) = ( 1 + _ 0.0418 365 365 ) - 1 0.04268 Account X has an effective rate of 4.24%, and Account Y has an effective rate of 4.268%, so Account Y has a greater effective rate, and Jaclyn should choose Account Y. Image Credits: David R. Frazier Photolibrary, Inc./Alamy Module 13 688 Lesson 4 Compound Interest 688

ELABORATE INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Ask students to describe how the three functions used to model investments earning compound interest are related. Students should point out that the functions are all exponential growth functions, but that the growth factors differ in each case. Ensure that students use correct language in describing the mathematical relationships that exist among the functions. Elaborate 7. Explain the difference between an investment s nominal annual interest rate and its effective annual interest rate. An investment s nominal annual interest rate is the interest rate r that is used in the value function V (t) = P (1 + n ) nt or V (t) = Pe rt. The investment s effective annual interest rate is the interest rate R that would need to be used in the value function V (t) = P (1 + R) t, where interest is compounded annually, to produce the same amount of annual interest as the investment actually earns. 8. Essential Question Check-In List the three functions used to model an investment that earns compound interest at an annual rate r. Identify when each function is used. The three functions used to model an investment that earns compound interest at an annual rate r are V (t) = P (1 + r) t, V (t) = P (1 + n ) nt, and V (t) = Pe rt. The function V (t) = P (1 + r) t is used when interest is compounded annually. The function V (t) = P (1 + n ) nt is used when interest is compounded more than once a year but not continuously. The function V (t) = Pe rt is used when interest is compounded continuously. SUMMARIZE THE LESSON What functions can you use to model investments that earn compound interest? You can use V (t) = P (1 + r) t to model interest that is compounded annually. You can use V (t) = P (1 + n ) nt to model interest that is compounded n times per year. You can use V (t) = Pe rt to model interest that is compounded continuously. Evaluate: Homework and Practice 1. A person invests $2560 in an account that earns 5.2% annual interest. Find when the value of the investment reaches $6000. Let P = 2560 and r = 0.052. Then V (t) = P (1 + r) t t = 2560 (1.052). Graphing y = 2560 (1.052) x along with the line y = 6000 shows that the value of the function is 6000 at x 16.8. So, the investment reaches a value of $6000 in approximately 16.8 years. 2. A person invests $1800 in an account that earns 2.46% annual interest. Find when the value of the investment reaches $3500. Let P = 1800 and r = 0.0246. Then V (t) = P (1 + r) t = 1800 (1.0246) t. Graphing y = 1800 (1.0246) x along with the line y = 3500 shows that the value of the function is 3500 at x 27.4. So, the investment reaches a value of $3500 in approximately 27.4 years. Online Homework Hints and Help Extra Practice Module 13 689 Lesson 4 689 Lesson 13.4

3. Emmanuel invests $3600 and Kelsey invests $2400. Both investments earn 3.8% annual interest. How much longer will it take Kelsey s investment to reach $10,000 than Emmanuel s investment? Emmanuel s investment: Let P = 3600 and r = 0.038. Then V (t) = P (1 + r) t = 3600 (1.038) t. Graphing y = 3600 (1.038) x along with the line y = 10,000 shows that the value of the function is 10,000 at x 27.4. So, the investment reaches a value of $10,000 in approximately 27.4 years. Kelsey s investment: Let P = 2400 and r = 0.038. Then V (t) = P (1 + r) t = 2400 (1.038) t. Graphing y = 2400 (1.038) x along with the line y = 10,000 shows that the value of the function is 10,000 at x 38.3. So, the investment reaches a value of $10,000 in approximately 38.3 years. So, it takes Kelsey s investment approximately 38.3-27.4 = 10.9 years longer to reach a value of $10,000. 4. Jocelyn invests $1200 in an account that earns 2.4% annual interest. Marcus invests $400 in an account that earns 5.2% annual interest. Find when the value of Marcus s investment equals the value of Jocelyn s investment and find the common value of the investments at that time. For Jocelyn s investment, let P = 1200 and r = 0.024. Then V (t) = P (1 + r) t = 1200 (1.024) t. For Marcus s investment, let P = 400 and r = 0.052. Then V (t) = P (1 + r) t = 400 (1.052) t. The graphs of y = 1200 (1.024) x and y = 400 (1.052) x intersect at approximately the point (40.7, 3152.42). So, the value of Marcus s investment equals the value of Jocelyn s investment after approximately 40.7 years. The common value of the investments is approximately $3152.42. 5. A person invests $350 in an account that earns 3.65% annual interest compounded semiannually. Find when the value of the investment reaches $5675. Let P = 350, r = 0.0365, and n = 2. Then V (t) = P (1 + n ) nt = 350 (1 + _ 0.0365 2 ) 2t = 350 (1.01825) 2t. Graphing y = 350 (1.01825) 2x along with the line y = 5675, shows that the value of the function is 5675 at x 77. So, the investment reaches a value of $5675 in approximately 77 years. EVALUATE ASSIGNMENT GUIDE Concepts and Skills Explore Comparing Simple and Compound Interest Example 1 Modeling Interest Compounded Annually Example 2 Modeling Interest Compounded More Than Once a Year Example 3 Modeling Interest Compounded Continuously Example 4 Finding and Comparing Effective Annual Interest Rates VISUAL CUES Practice Exercises 1 4, 18 Exercises 5 8, 15 17 Exercises 9 11 Exercises 12 14 Suggest that students circle the numbers in the problem and, next to each number, write the variable for which the number will be substituted. Module 13 690 Lesson 4 Exercise Depth of Knowledge (D.O.K.) Mathematical Practices 1 2 1 Recall of Information MP.6 Precision 3 4 2 Skills/Concepts MP.5 Using Tools AVOID COMMON ERRORS The formulas presented in this lesson each contain numerous variables. Because of this, students may mistake e, in the formula for investments that earn interest that is compounded continuously, as a variable, and look to make a substitution when applying this formula to a problem. To avoid this, suggest that students underline e when using this formula, to remind them that e is a number, not a variable, and that no number is substituted for e. 5 7 1 Recall of Information MP.6 Precision 8 2 Skills/Concepts MP.5 Using Tools 9 10 1 Recall of Information MP.6 Precision 11 2 Skills/Concepts MP.5 Using Tools 12 13 1 Recall of Information MP.4 Modeling Compound Interest 690

QUESTIONING STRATEGIES If I (t) is a function that represents the amount of interest earned on an account in which interest is compounded continuously, what rule can you write for I (t)? I (t) = Pe rt - P 6. Molly invests $8700 into her son s college fund, which earns 2% annual interest compounded daily. Find when the value of the fund reaches $12,000. Let P = 8700, r = 0.02, and n = 365. Then V (t) = P (1 + n ) nt = 8700 (1 + 0.02 365 ) 365t 8700 (1.0000548) 365t. Graphing y = 8700 (1.0000548) 365x along with the line y = 12,000 shows that the value of the function is 12,000 at x 16.1. So, the fund reaches a value of $12,000 in approximately 16.1 years. Image Credits: Rubberball/Alamy 7. A person invests $200 in an account that earns 1.98% annual interest compounded quarterly. Find when the value of the investment reaches $500. Let P = 200, r = 0.0198, and n = 4. Then V (t) = P (1 + n ) nt = 200 (1 + _ 0.0198 4 ) 4t 200 (1.00495) 4t. Graphing y = 200 (1.00495) 4x along with the line y = 500, shows that the value of the function is 500 at x 46.4. So, the investment reaches a value of $500 in approximately 46.4 years. 8. Hector invests $800 in an account that earns 6.98% annual interest compounded semiannually. Rebecca invests $1000 in an account that earns 5.43% annual interest compounded monthly. Find when the value of Hector s investment equals the value of Rebecca s investment and find the common value of the investments at that time. For Hector s investment, let P = 800, r = 0.0698, and n = 2. Then V (t) = P (1 + n ) nt = 800 (1 + _ 0.0698 2 ) 2t 800 (1.0349) 2t. For Rebecca s investment, let P = 1000, r = 0.0543, and n = 12. Then V (t) = P (1 + n ) nt = 1000 (1 + _ 0.0543 12 ) 12t = 1000 (1.004525) 12t. The graphs of y = 800 (1.0349) 2x and y = 1000 (1.004525) 12x intersect at approximately the point (15.5, 2310.98). So, the value of Hector s investment equals the value of Rebecca s investment after approximately 15.5 years. The common value of the investments is approximately $2310.98. Module 13 691 Lesson 4 Exercise Depth of Knowledge (D.O.K.) Mathematical Practices 14 15 2 Skills/Concepts MP.5 Using Tools 16 3 Strategic Thinking MP.5 Using Tools 17 3 Strategic Thinking MP.3 Logic 18 3 Strategic Thinking MP.2 Reasoning 691 Lesson 13.4

9. A person invests $6750 in an account that earns 6.23% annual interest compounded continuously. Find when the value of the investment reaches $15,000. Let P = 6750 and r = 0.0623. Then V (t) = P e rt = 6570 e 0.0623t. Graphing y = 6750 e 0.0623x along with the line y = 15,000 shows that the value of the function is 15,000 at x 12.8. So, the investment reaches a value of $15,000 in approximately 12.8 years. 10. A person invests $465 in an account that earns 3.1% annual interest compounded continuously. Find when the value of the investment reaches $2400. Let P = 465 and r = 0.031. Then V (t) = P e rt = 465 e 0.031t. Graphing y = 465 e 0.031x along with the line y = 2400 shows that the value of the function is 2400 at x 52.9. So, the investment reaches a value of $2400 in approximately 52.9 years. COGNITIVE STRATEGIES As a memory device, students may find it helpful to remember that the effective annual interest rate is a result of the effect that the compounding of the interest has on the nominal interest rate. 11. Lucy invests $800 in an account that earns 6.12% annual interest compounded continuously. Juan invests $1600 in an account that earns 3.9% annual interest compounded continuously. Find when the value of Lucy s investment equals the value of Juan s investment and find the common value of the investments at that time. For Lucy s investment, let P = 800 and r = 0.0612. Then V (t) = P e rt = 800 e 0.0612t. For Juan s investment, let P = 1600 and r = 0.039. Then V (t) = P e rt = 1600 e 0.039t. The graphs of y = 800 e 0.0612x and y = 1600 e 0.039x intersect at approximately the point (31.2, 5407.00). So, the value of Lucy s investment equals the value of Juan s investment after approximately 31.2 years. The common value of the investments is approximately $5407.00. 12. Paula plans to make a deposit in one of the accounts shown in the table. To maximize the interest that the account earns, which account should she choose? Nominal Annual Interest rate Compounding Period Account X Account Y 2.83% 2.86% Continuously Annually For Account X, interest is compounded continuously. The nominal rate is 2.83%, so r = 0.0283. R X = e r - 1 = e 0.0283-1 0.0287 For Account Y, interest is compounded annually, so n = 1. The nominal rate is 2.86%, so r = 0.0286. R Y = (1 + n ) n - 1 = (1 + 0.0286 1 ) 1-1 0.0286 Account X has an effective rate of 2.87%, and Account Y has an effective rate of 2.86%, so Account X has a greater effective rate, and Paula should choose Account X. Module 13 692 Lesson 4 Compound Interest 692

INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Students may be interested to learn that five of the most famous numbers in mathematics (e, i, π, 1, and 0) are related to each other by the seemingly simple equation e iπ + 1 = 0. Students will learn the derivation of this relationship in more advanced courses. 13. Tanika plans to make a deposit to one of two accounts. Account A has a 3.78% nominal rate with interest compounded daily, and Account B has a 3.8% nominal rate with interest compounded monthly. To maximize the interest that the account earns, which account should she choose? For Account A, interest is compounded daily, so n = 365. The nominal rate is 3.78%, so r = 0.0378. R A = (1 + _ n ) n - 1 = (1 + 0.0378 365 ) 365-1 0.0385 For Account B, interest is compounded monthly, so n = 12. The nominal rate is 3.8%, so r = 0.038. R B = (1 + n ) n - 1 = (1 + 0.038 12 ) 12-1 0.0387 Account A has an effective rate of 3.85%, and Account B has an effective rate of 3.87%, so Account B has a greater effective rate, and Tanika should choose Account B. 14. Kylie plans to deposit $650 in one of the accounts shown in the table. She chooses the account with the greater effective rate. How much money will she have in her account after 10 years? Nominal Annual Interest Rate Compounding Period Account X Account Y 4.13% 4.12% Semiannually Quarterly For Account X, interest is compounded semiannually, so n = 2. The nominal rate is 4.13%, so r = 0.0413. R X = (1 + n ) n - 1 _ = (1 + 0.0413 2 ) 2-1 0.0417 For Account Y, interest is compounded quarterly, so n = 4. The nominal rate is 4.12%, so r = 0.0412. R Y = (1 + n ) n - 1 _ = (1 + 0.0412 4 ) 4-1 0.0418 Account Y has the greater effective rate, so Kylie chooses Account Y. The value of Kylie s investment after 10 years is: V (10) = 650 (1 + _ 0.0412 4 ) 4 (10) 979.33 So, Kylie will have $979.33 in her account after 10 years. Module 13 693 Lesson 4 693 Lesson 13.4

15. A person invests $2860 for 15 years in an account that earns 4.6% annual interest. Match each description of a difference in interest earned on the left with the actual difference listed on the right. A. Difference between compounding interest C $14.89 semiannually and annually B. Difference between compounding interest B $21.98 quarterly and semiannually C. Difference between compounding interest E $0.25 monthly and quarterly D. Difference between compounding interest A $42.75 daily and monthly E. Difference between compounding interest D $7.27 continuously and daily A. V S (15) - V A (15) = 2860 (1 + 0.046 2 ) 2 (15) - 2860 (1 + 0.046) 15 5657.64-5614.89 = 42.75 B. V Q (15) - V S (15) = 2860 (1 + 0.046 4 ) 4 (15) - 2860 (1 + 0.046 2 ) 2 (15) 5679.62-5657.64 = 21.98 C. V M (15) - V Q (15) = 2860 (1 + 0.046 12 ) 12 (15) - 2860 (1 + 0.046 4 ) 4 (15) 5694.51-5679.62 = 14.89 D. V D (15) - V M (15) = 2860 (1 + 0.046 365 ) 365 (15) - 2860 (1 + 0.046 12 ) 12 (15) 5701.78-5694.51 = 7.27 E. V C (15) - V D (15) = 2860 e 0.046 (15) - 2860 (1 + 0.046 365 ) 365 (15) 5702.03-5701.78 = 0.25 H.O.T. Focus on Higher Order Thinking 16. Multi-Step Ingrid and Harry are saving to buy a house. Ingrid invests $5000 in an account that earns 3.6% interest compounded quarterly. Harry invests $7500 in an account that earns 2.8% interest compounded semiannually. a. Find a model for each investment. Ingrid s investment: Let P = 5000, r = 0.036, and n = 4. Then V (t) = P (1 + n ) nt = 5000 (1 + 0.036 4 ) 4t = 5000 (1.009) 4t. Harry s investment: Let P = 7500, r = 0.028, and n = 2. Then V (t) = P (1 + n ) n = 7500 (1 + _ 0.028 2 ) 2t = 7500 (1.014) 2t. Image Credits: Visual Ideas/Camillo Morales/Blend Images/Corbis Module 13 694 Lesson 4 Compound Interest 694

PEER-TO-PEER DISCUSSION Ask students to work with a partner to make a conjecture as to how much more money an investment of $10,000 would make at a rate of 4% compounded annually for 5 years versus 4% compounded monthly for 5 years. Then have them apply the formulas from this lesson to check their conjectures. Students should find that when the interest is compounded monthly, the investment earns $43.44 more interest ($12,209.97 - $12,166.53). JOURNAL Have students describe the concept of compound interest, citing the different types of compound interest, and explaining why the functions that model investments that earn compound interest are exponential growth functions. b. Use a graphing calculator to find when the combined value of their investments reaches $15,000. The combined value of the investments is V (t) = 5000 (1.009) 4t + 7500 (1.014) 2t. Graphing the function y = 5000 (1.009) 4x + 7500 (1.014) 2x along with the line y = 15,000 shows that the value of the function is 15,000 at x 5.9. So, the combined investment reaches a value of $15,000 after approximately 5.9 years. 17. Explain the Error A student is asked to find when the value of an investment of $5200 in an account that earns 4.2% annual interest compounded quarterly reaches $16,500. The student uses the model V (t) = 5200 (1.014) 3t and finds that the investment reaches a value of $16,500 after approximately 27.7 years. Find and correct the student s error. The interest is compounded quarterly. This means that the interest is compounded every 3 months, or 4 times per year. The value of n in the function V (t) = P (1 + n ) nt is the number of times the interest is compounded per year, so the student should have used 4 instead of 3. Let P = 5200, r = 0.042, and n = 4. Then V (t) = P (1 + n ) nt = 5200 (1 + 0.042 4 ) 4t = 5200 (1.0105) 4t. Graphing y = 5200 (1.0105) 4x along with the line y = 16,500 shows that the value of the function is 16,500 at x 27.6. So, the investment reaches a value of $16,500 after approximately 27.6 years. 18. Communicate Mathematical Ideas For a certain price, you can buy a zero-coupon bond that has a maturity value of $1000, has a maturity date of 4 years, and pays 5% annual interest. How can the present value (the amount you pay for the bond) be determined from the future value (the amount you get when the bond matures)? The future value of the investment is $1000. The interest is compounded annually, so use the value function V (t) = P (1 + r) t to find the present value of the investment. Let r = 0.05, t = 4, and V (4) = 1000. Solve for the present value P. P (1 + 0.05) 4 = 1000 P (1.05) 4 = 1000 _ 1000 P = (1.05) 4 P 822.70 So, the present value of the investment is $822.70. Module 13 695 Lesson 4 695 Lesson 13.4

Lesson Performance Task The grandparents of a newborn child decide to establish a college fund for her. They invest $10,000 into a fund that pays 4.5% interest compounded continuously. a. Write a model for the value of the investment over time and find the value of the investment when the child enters college at 18 years old. b. In 2013, the average annual public in-state college tuition was $8893, which was 2.9% above the 2012 cost. Use these figures to write a model to project the amount of money needed to pay for one year s college tuition 18 years into the future. What amount must be invested in the child s college fund to generate enough money in 18 years to pay for the first year s college tuition? a. Use the formula for continuously compounded interest, with P = 10,000, r = 0.045, and t = 18. V (t) = P e rt (0.045)(18) = 10,000 e = 10,000 e 0.81 22,479.08 The value of the investment when the child enters college is $22,479.08. AVOID COMMON ERRORS Students should compare the answers in Parts A and B of the Lesson Performance Task and re-read the original question in each part to determine whether the answers are reasonable. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Students may consider alternative ways to approach the problem. For example, the growth during years 18 21 could also be considered, which would require a larger investment. Encourage students to pursue additional possibilities based on different constraints. Note that the growth rate will certainly change over time, so projections are unlikely to be very accurate. b. To estimate the cost in 18 years, use the formula for compound annual interest with P = 8893, r = 0.029, and t = 18. V (t) = P (1 + r) t = 8893 (1 + 0.029) 18 = 8893 (1.029) 18 14,877.33 The model predicts that tuition will cost $14,877.33 eighteen years into the future. Now, use the continuously compounded interest formula to find what initial principal P yields a value in 18 years of $14,877.33 when r = 0.045. 14,877.33 = P e rt (0.045)(18) 14,877.33 = P e 14,877.33 = P e 0.81 14,877.33 = P e 0.81 6618.30 The amount that must be invested is $6618.30. Module 13 696 Lesson 4 EXTENSION ACTIVITY Have students research the historical tuition costs of different four-year colleges and predict the savings needed to cover the projected tuitions for 4 years beginning 18 21 years from now. Have them consider different investment amounts and interest rates that would meet those costs, and present their findings to the class. Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. Compound Interest 696