Number of People Contacted

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1 Section 7.2 Problem Solving Using Exponential Functions (Exponential Growth and Decay) Key Words: exponential, growth, decay, critical mass Key Concepts: exponential growth, exponential decay, simple interest, compound interest Goals. In this section, examples of math models that use the exponential function will be introduced. After completing the exercises, students should have a strong understanding of the importance of exponential growth and decay. There are two basic types of real world problems which can be modeled using the exponential function. Those in which there is a "fast and steady" increase after a critical mass has been achieved. This type of problem is referred to as an exponential growth problem. The basic function for this type of problem is an exponential of the form x y = a, a > 1. Two key examples are population growth problems and compound interest problems The second major class of applications is exponential decay. A typical example of this type is radioactive decay. Here there is a "fast and steady" decrease after a critical mass has been achieved. The basic function for this type of problem is an exponential of the form y = a = -x 1 x a. Example 1. (exponential growth) A local charity has decided to raise funds through the following scheme; Person A will contact two people and ask each of them to contribute $1.00. Each of these people will contact two people and ask each of them to contribute $1.00 and to contact two others. Assume that this process continues and that each transaction takes 1 week. The following table outlines this process: Week Number of People Contacted Contributions = 2 $ = $ = 8 $ = 1, 02 $1, = 67,108, 86 $67,108, 86 From the above, the money collected during week x is given by the function y = 2 x. The total amount of money collected is the sum of all money collected since the first week. For example, the total collected for the first 3 weeks (if we assume the first person gives $1.00) is = $15.00 (2-1). The astounding fact about exponential growth is how fast it grows. Look at the amount collected during the 26th week in the above table. It is $67,108,86. If we want the total collected for the first 26 weeks the sum is $ 13,217,727 (2 27-1). Example 2. (exponential growth) The following is typical of the so called population growth

2 problems. Assume that a certain bacteria is such that a single cell splits into two cells after one hour, and during the second hour each of these two cells splits to form 2 new cells, giving cells. Assume that this doubling process continues so that for any time t (in hours) the population of bacteria, P(t) is given by the function P(t) = 2 t. The function P(t) enables us to compute the number of bacteria grown at any given hour t. So after 10 hours P(10) = 2 10 = 1,02 bacteria are present. How many bacteria are present after 10 hours if 1000 were present at the start of an experiment. Let B 0 stand for the number of bacteria present at the beginning (at time t= 0) of the experiment, then the mathematical model of this situation is P(t) = B 0 2 t and P(10) = = = 1,02,000 Example 3. (exponential growth) A woman has three daughters and each daughter has 3 daughters and so on through successive generations. How many female descendants are born in the fourth generation? How many descendants are there in the first four generations? Let P(t) stand for the number of daughters born in generation t. Then we have P(t) = 3 t and P() = 3 = 81 (Note: this is not counting the initial woman as the first generation.) daughters born in the fourth generation. Note here the initial population was one woman so P 0 = 1 in the formula P(t) = P 0 3 t. 3 1, 3 2, 3 3, 3 are the number of daughters born in generations 1,2,3 and respectively so = 120 are born in the first four generations. Figures and illustrate two ways in which our results can be displayed. The first graph is called a bar graph. It displays the values of P(t) only for the whole numbers 1, 2, 3, and. The second graph is a continuous graph that also gives us the values of P(t) for non-integer values. In this case those values have no reasonable meaning since the variable is generations (what is a half generation?). However, there will be cases where we can interpret this data; and so we will display mostly continuous graphs from here on

3 Figure The graph of P(t) = 3 t, t = 1, 2, 3, Figure The graph of P(t) = 3 t, 0 t Population Formula (used for all exponential growth problems): P(t) = P 0. b t where P(t) stands for the population at time t P 0 stands for the initial population (population at time t = 0) b stands for the number of offspring produced by each individual Example. (exponential decay) Jim inherits $10,000. He gambles and loses one fourth of the money he has remaining each day. How much money will he have at the end of the fifth day? Let M(t) stand for the amount of money he has remaining at day t Let M 0 stand for his inheritance After day t, he will have

4 M(t) = $10000(3/) t Then at the end of day 5 he has M(5) = $10000(3/) 5 $2,373 left of his inheritance $8,000 $7,000 $6,000 $5,000 $,000 $3,000 $2,000 $1,000 $ Figure Bar graph of M(t) = 10000(3/) t, t = 1, 2,..., 7. Figure 7.2. Continuous graph of M(t) = 10000(3/) t 0 t 7 Other interesting exponential growth and decay applications come from finance. The Mathematics of Finance (A matter of interest) There are two major types of interest used today simple interest and compound interest. Definition: Simple interest is interest computed on the original principal (original amount) only. Example 5. A student borrows $1000 at 15% simple interest for one year. Let A stand for the amount owed at the end of the year

5 Let P stand for the principal, the amount borrowed. Let i stand for the interest rate Then A = P + Pi = P(1 + i) = 1000 (1 +.15) = 1000 (1.15) = $1, Note: This example is of the form A = 1000(1.15) n,another exponential growth problem. If n = the time in years and if i = annual interest rate, the balance after n years using simple interest is A = P(1 + i) n Example 6. A loan of $300 is made for a period of 3 years at a simple interest rate of 17% annually. Find the amount to be paid at the end of the loan period. Here P = 300, i =.17 and n = 3. If we designate the amount to be paid by A we have A = 300(1+0.17) 3 = 80.8 So, the amount to be paid is $80.8. Simple Interest Formula : Let A stand for the amount (principal plus interest) Let P stand for the principal Let i stand for the annual interest rate Let n stand for the time in years Then A = P(1 + i) n Today, simple interest is rarely used because of the constantly changing balance on our credit cards, and other loans. The balance on a loan is rarely constant long enough to apply a years worth of interest to it. When interest is added to the principal at specific times (periods) then the interest is called compound interest. In this situation the interest itself earns interest as the following example illustrates. For example if we compound interest quarterly, the interest rate is divided by four and at the end of each quarter year the quartered rate is applied to the principle. Example 7. $1,000 is placed in a bank at 7% (annual rate) compounded quarterly. How much money is in the bank at the end of 1 year? As above: Let A stand for the amount (principal plus interest) in the account Let P stand for the principal = $1,000 Let i stand for the annual interest rate = 7% =.07 Let n stand for the time in years = 1 Since the interest is compounded quarterly, times per year (or every 3 months) the interest rate per

6 quarter is.07 At the end of the first quarter 1 A = At the end of the second quarter A 2 = A1 1 + = Why? At the end of the third quarter A 3 = A2 1 + = Why? At the end of the fourth quarter, one year A = A3 1 + = Therefore, ( ) ( ) A = = = $ Notice that simple annual interest of 7% would yield $1070. The observant reader will note that the equation A ( ) = is of the same form as the Population Formula above. That is, it is an exponential growth function. A plot of A(t) for t between 0 and 8 might be somewhat of a surprise since the graph looks almost linear. This is due to the fact that the growth rate is really quite small. Regrettably, this seems to be the case when we watch our investments grow. Figure The graph of A(t) = 1000(1.0175) t, 0 t 8 Compound Interest Formula (exponential growth): Let A stand for the amount (principal plus interest) Let P stand for the principal Let i stand for the annual interest rate

7 Let n stand for the time in years Let m stand for the number of times per year interest is compounded Then mn A m = P 1 + i m Example 8. $1,000 is placed in a bank at 7% (annual rate) compounded quarterly. How much money is in the bank at the end of 2 years? 3 years? n years? From example 7 At the end of year 1 we have A = 1000(1.0175) At the end of year 2 we have A = 1000(1.0175) (1.0175) =1000((1.0175) ) 2 = 1000(1.0175) 8 At the end of year 3 we have A = 1000(1.0175) (1.0175) (1.0175) =1000((1.0175) ) 3 = 1000(1.0175) 12 At the end of year n we have A =1000((1.0175) ) n = 1000(1.0175) n Theorem If interest is compounded m times a year for n years the amount A is given by the formula: m n A = P 1 + i m = P 1 + i mn m