# 8.3 Applications of Exponential Functions

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1 Section 83 Applications of Exponential Functions Applications of Exponential Functions In the preceding section, we examined a population growth problem in which the population grew at a fixed percentage each year In that case, we found that the population can be described by an exponential function A similar analysis will show that any process in which a quantity grows by a fixed percentage each year or each day, hour, etc) can be modeled by an exponential function Compound interest is a good example of such a process Discrete Compound Interest If you put money in a savings account, then the bank will pay you interest a percentage of your account balance) at the end of each time period, typically one month or one day For example, if the time period is one month, this process is called monthly compounding The term compounding refers to the fact that interest is added to your account each month and then in subsequent months you earn interest on the interest If the time period is one day, it s called daily compounding Let s look at monthly compounding in more detail Suppose that you deposit \$100 in your account, and the bank pays interest at an annual rate of 5% Let the function P t) represent the amount of money that you have in your account at time t, where we measure t in years We will start time at t = 0 when the initial amount, called the principal, is \$100 In other words, P 0) = 100 In the discussion that follows, we will compute the account balance at the end of each month Since one month is 1/ of a year, P 1/) represents the balance at the end of the first month, P 2/) represents the balance at the end of the second month, etc At the end of the first month, interest is added to the account balance Since the annual interest rate 5%, the monthly interest rate is 5%/, or 05/ in decimal form Although we could approximate 05/ by a decimal, it will be more useful, as well as more accurate, to leave it in this form Therefore, at the end of the first month, the interest earned will be 10005/), so the total amount will be ) 05 P 1/) = = ) 1) Now at the end of the second month, you will have the amount that you started that month with, namely P 1/), plus another month s worth of interest on that amount Therefore, the total amount will be ) 05 P 2/) = P 1/) + P 1/) = P 1/) ) 2) If we replace P 1/) in equation 2) with the result found in equation 1), then 1 Copyrighted material See:

2 790 Chapter 8 Exponential and Logarithmic Functions P 2/) = ) ) = ) 2 3) Let s iterate one more month At the end of the third month, you will have the amount that you started that month with, namely P 2/), plus another month s worth of interest on that amount Therefore, the total amount will be P 3/) = P 2/) + P 2/) 05 ) = P 2/) ) 4) However, if we replace P 2/) in equation 4) with the result found in equation 3), then P 3/) = ) ) = ) 3 5) The pattern should now be clear The amount of money you will have in the account at the end of m months is given by the function P m/) = ) m We can rewrite this formula in terms of years t by replacing m/ by t Then m = t, so the formula becomes P t) = ) t 6) What would be different if you had started with a principal of 200? By tracing over our previous steps, it should be easy to see that the new formula would be P t) = ) t Similarly, if the interest rate had been 4% per year instead of 5%, then we would have ended up with the formula P t) = ) t Thus, if we let P 0 represent the principal, and r represent the annual interest rate in decimal form), then we can generalize the formula to P t) = P r ) t 7) Example 8 If the principal is \$100, the annual interest rate is 5%, and interest is compounded monthly, how much money will you have after ten years? In formula 7), let P 0 = 100, r = 05, and t = 10: P 10) = ) 10 We can use our graphing calculator to approximate this solution, as shown in Figure 1

3 Section 83 Applications of Exponential Functions 791 Figure 1 Computing the amount after compounding monthly for 10 years Thus, you would have \$16470 after ten years Example 9 If the principal is \$10 000, the annual interest rate is 5%, and interest is compounded monthly, how much money will you have after forty years? In formula 7), let P 0 = , r = 05, and t = 40: P 40) = ) Thus, you would have \$73,58417 after forty years These examples illustrate the miracle of compound interest In the last example, your account is more than seven times as large as the original, and your total profit the amount of interest you ve received) is \$ Compare this to the amount you would have received if you had withdrawn the interest each month ie, no compounding) In that case, your profit would only be \$20 000: years months year interest month = 40 [ ) 05 )] = The large difference can be attributed to the shape of the graph of the function P t) Recall from the preceding section that this is an exponential growth function, so as t gets large, the graph will eventually rise steeply Thus, if you can leave your money in the bank long enough, it will eventually grow dramatically What about daily compounding? Let s again analyze the situation in which the principal is \$100 and the annual interest rate is 5% In this case, the time period over which interest is paid is one day, or 1/ of a year, and the daily interest rate is 5%/, or 05/ in decimal form Since we are measuring time in years, P 1/) represents the balance at the end of the first day, P 2/) represents the balance at the end of the second day, etc We ll follow the same steps as in the earlier analysis for monthly compounding At the end of the first day, you will have ) 05 P 1/) = = ) 10) At the end of the second day, you will have the amount that you started that day with, namely P 1/), plus another day s worth of interest on that amount Therefore, the total amount will be

4 792 Chapter 8 Exponential and Logarithmic Functions P 2/) = P 1/) + P 1/) ) 05 = P 1/) ) 11) If we replace P 1/) in equation 11) with the result found in equation 10), then P 2/) = ) ) = ) 2 ) At the end of the third day, you will have the amount that you started that day with, namely P 2/), plus another day s worth of interest on that amount Therefore, the total amount will be ) 05 P 3/) = P 2/) + P 2/) = P 2/) ) 13) Again, replacing P 2/) in equation 13) with the result found in equation ) yields P 3/) = ) ) = ) 3 14) Continuing this pattern shows that the amount of money you will have in the account at the end of d days is given by the function P d/) = ) d We can rewrite this formula in terms of years t by replacing d/ by t Then d = t, so the formula becomes P t) = ) t 15) More generally, if you had started with a principal of P 0 and an annual interest rate of r in decimal form), then the formula would be P t) = P r ) t 16) Comparing formulas 7) and 16) for monthly and daily compounding, it should be apparent that the only difference is that the number is used in the monthly compounding formula and the number is used in the daily compounding formula Looking at the respective analyses shows that this number arises from the portion of the year that interest is paid 1/ in the case of monthly compounding, and 1/ in the case of daily compounding) Thus, in each case, this number or ) also equals the number of times that interest is compounded per year It follows that if interest is compounded quarterly every three months, or 4 times per year), the formula would be P t) = P ) r 4t

5 Section 83 Applications of Exponential Functions 793 Similarly, if interest is compounded hourly 8760 times per year), the formula would be P t) = P r ) 8760t 8760 Summarizing, we have one final generalization: Discrete Compound Interest If P 0 is the principal, r is the annual interest rate, and n is the number of times that interest is compounded per year, then the balance at time t years is P t) = P r n) nt 17) Example 18 If the principal is \$100, the annual interest rate is 5%, and interest is compounded daily, what will be the balance after ten years? In formula 17), let P 0 = 100, r = 05, n =, and t = 10: P 10) = ) Thus, you would have \$16487 after ten years Example 19 If the principal is \$10 000, the annual interest rate is 5%, and interest is compounded daily, what will be the balance after forty years? In formula 17), let P 0 = , r = 05, n =, and t = 40: P 40) = ) Thus, you would have \$ after forty years As you can see from comparing Examples 8 and 18, and Examples 9 and 19, the difference between monthly and daily compounding is generally small However, the difference can be substantial for large principals and/or large time periods Example 20 If the principal is \$500, the annual interest rate is 8%, and interest is compounded quarterly, what will be the balance after 42 months? 42 months is 35 years, so let P 0 = 500, r = 08, n = 4, and t = 35 in formula 17): P 5) = ) Thus, you would have \$65974 after 42 months

6 794 Chapter 8 Exponential and Logarithmic Functions Continuous Compound Interest and the Number e Using formula 17), it is a simple matter to calculate the total amount for any type of compounding Although most banks compound interest either daily or monthly, it could be done every hour, or every minute, or every second, etc What happens to the total amount as the time period shortens? Equivalently, what happens as n increases in formula 17)? Table 1 shows the amount after one year with a principal of P 0 = 100, r = 05, and various values of n: compounding n P 1) monthly daily hourly every minute every second Table 1 Comparison of discrete compounding with P 0 = 100, r = 05, and t = 1 year Even if we carry out our computations to eight digits, it appears that the amounts in the right hand column of Table 1 are stabilizing In fact, using calculus, it can be shown that these amounts do indeed get closer and closer to a particular number, and we can calculate that number Starting with formula 17), we will let n approach In other words, we will let n get larger and larger without bound, as we started to do in Table 1 The first step is to use the Laws of Exponents to write P r ) [ nt = P0 1 + r ] rt r n n)n In the next step, replace n/r by m Since n/r = m, it follows that r/n = 1/m, and we have [ P r ) n ] rt [ r = P m ] rt n m) Now let n approach Since m = n/r and r is fixed, it follows that m also approaches We can use the TABLE feature of the graphing calculator to investigate the convergence of the expression in brackets as m approaches infinity Load 1+1/m) m into the Y= menu of the graphing calculator, as shown in Figure 2a) Of course, you must use x instead of m and enter 1+1/X)^X Use TBLSET and set Indepnt to Ask, select TABLE, then enter the numbers 10, 100, 1 000, , , and to produce the result shown in Figure 2b) Note that 1+1/X)^X appears to converge to the number If you move the cursor over the last result in the Y1 column, you can see more precision,

7 Section 83 Applications of Exponential Functions 795 a) b) c) Figure 2 Illustration of the convergence of 1 + 1/m) m to e as m increases to infinity Note that the numbers in the second column in Figure 2b) appear to stabilize Indeed, it can be shown by using calculus that the expression in brackets above gets closer and closer to a single number, which is called e To represent this convergence, we write m) m e 21) e is an irrational number, approximately 27183, as shown by the computations in Figure 2b) It follows that [ P m) 1 m ] rt P 0 e rt Because we took the discrete compound interest formula 17) and let the number of times compounded per year n) approach, this process is known as continuous compounding Continuous Compound Interest If P 0 is the principal, r is the annual interest rate, and interest is compounded continuously, then the balance at time t years is P t) = P 0 e rt 22) Before working the next examples, find the buttons on your calculator for the number e and for the exponential function e x Typing either e or e^1) using the e x button) will yield an approximation to the number e, as shown in Figure 2c) Compare this approximation with the one you obtained earlier in Figure 2b) Example 23 If the principal is \$100, the annual interest rate is 5%, and interest is compounded continuously, what will be the balance after ten years? In formula 22), let P 0 = 100, r = 005, and t = 10: P 10) = 100e 005)10) Use your calculator to approximate this result, as shown in Figure 3

8 796 Chapter 8 Exponential and Logarithmic Functions Figure 3 Computing the amount after compounding continuously for 10 years Thus, you would have \$16487 after ten years Example 24 If the principal is \$10,000, the annual interest rate is 5%, and interest is compounded continuously, what will be the balance after forty years? In formula 22), let P 0 = , r = 005, and t = 40: P 10) = e 005)40) Thus, you would have \$ after forty years Notice that the continuous compounding formula 22) is much simpler than the discrete compounding formula 17) Unless the principal is very large or the time period is very long, the preceding examples show that continuous compounding is also a close approximation to daily compounding In Example 23, the amount \$16487 is the same rounded to the nearest cent) as the amount for daily compounding found in Example 18 With a larger principal and longer time period, the amount \$ in Example 24 using continuous compounding is still only about \$10 more than the amount \$ for daily compounding found in Example 19 Remarks 25 1 The number e may strike you as a mere curiosity If so, that would be a big misconception The number e is actually one of the most important numbers in mathematics it s probably the second most famous number, following π), and it arises naturally as the limit described in 21) above Using notation from calculus, we write lim m = e ) m m) Although in our discussion above this limit arose in a man-made process, compound interest, it shows up in a similar manner in studies of many natural phenomena We ll look at some of these applications later in this chapter 2 Likewise, the exponential function e x is one of the most important functions used in mathematics, statistics, and many fields of science For a variety of reasons, the base e turns out to be the most natural base to use for an exponential function Consequently, the function fx) = e x is known as the natural exponential function

9 Section 83 Applications of Exponential Functions 797 Future Value and Present Value In this section we have derived two formulas, one for discrete compound interest, and the other for continuous compound interest However, in the examples presented so far, we ve only used these formulas to calculate future value: given a principal P 0 and interest rate r, how much will you have in your account in t years? Another type of question we can solve is known as a present value problem: how much money would you have to invest at interest r in order to have Q dollars in t years? Here are a couple of examples: Example 27 How much money would you have to invest at 4% interest compounded daily in order to have \$8000 dollars in 6 years? In this case, the principal P 0 is unknown, and we substitute r = 004, n =, and t = 6, into the discrete compounding formula 17) Since P 6) = 8000, we have the equation 8000 = P 6) = P ) )6) This equation can be solved by division: 8000 ) )6) = P 0 Figure 4 shows a calculator approximation for this result Figure 4 The present value of \$8000, compounded daily for six years Thus, the present value is approximately P 0 \$ If this amount is invested now at 4% compounded daily, then its future value in 6 years will be \$8000 Example 28 How much money would you have to invest at 7% interest compounded continuously in order to have \$5000 dollars in 4 years? As in the last example, the principal P 0 is unknown, and this time r = 007 and t = 4 in the continuous compounding formula 22) Then P 4) = 5000 yields the equation 5000 = P 4) = P 0 e 007)4) As in the last example, this equation can also be solved by division: 5000 e 007)4) = P 0 A calculator approximation for this result is shown in Figure 5

10 798 Chapter 8 Exponential and Logarithmic Functions Figure 5 The present value of \$5000, compounded continuously for four years Thus, the present value is approximately P 0 \$ If this amount is invested now at 7% compounded continuously, then its future value in 4 years will be \$5000 Additional Questions In terms of practical applications, there are also other types of questions that would be interesting to consider Here are two examples: 1 If you deposit \$1000 in an account paying 6% compounded continuously, how long will it take for you to have \$1500 in your account? 2 If you deposit \$1000 in an account paying 5% compounded monthly, how long will it take for your money to double? Let s look at the first question the second is similar) In this case, P 0 = 1000 and r = 006 Inserting these values into the continuous compounding formula 22), we obtain P t) = 1000e 006t Now we want the future value P t) of the account at some time t to equal \$1500 Therefore, we must solve the equation 1500 = 1000e 006t However, now we have a problem, because the variable t is located in the exponent of the expression on the right side of the equation Although we could approximate a solution graphically, we currently have no algebraic method for solving an equation such as this, where the variable is in the exponent these types of equations are called exponential equations) Over the course of the next few sections, we will define another type of function, the logarithm function, which will in turn provide us with a method for solving exponential equations Then we will return to these questions, and also discuss additional applications

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