1. Module 4 Section 4.1 Exponential Functions and Models

2. 10/26/2012 Section 4.3 v5.0.1 2 Exponential Functions Growth Rates Linear Growth Consider the arithmetic sequence: 3, 5, 7, 9, 11, 13,... a 1 = 3 a 2 = a 1 + 2 = 3 + 2 a 3 = a 2 + 2 = 3 + 2 + 2 a 4 = a 3 + 2 = 3 + 2 + 2 + 2 a n = a 1 + (n – 1)2 = 3 + 2 + 2 + … + 2 Define function f(n) by: Let k = n – 1 n – 1 Terms = 3 + k(2) = 2k + 3 k = n – 1 f(n) = a n = a 1 + (n – 1)2 g(k) = a 1 + 2k = 2k + a 1 so that A Linear Function Question: How do we know this is linear ?

3. 10/26/2012 Section 4.3 v5.0.1 3 Exponential Functions Growth Rates Exponential Growth Consider the geometric sequence: 3, 6, 12, 24, 48, 96,... a 1 = 3 a 2 = a 1  2 = 3  2 1 a 3 = a 2  2 = a 1  2  2 = 3  2 2 a n = a n–1  2  …  2 = 3  2 n–1 Define function f(n) by: f(n) = a n = a 1 r n–1 = 3  2 n–1 Letting k = n – 1 n – 1 Factors k = n – 1 this becomes g(k) = a 1 r k = 3  2 k An Exponential Function

4. 10/26/2012 Section 4.3 v5.0.1 4 Exponential and Power Functions What’s the difference ? For any real number x, and rational number a we write the a th power of x as : Function f( x ) = x a is called a power function For any real numbers x and a, with a ≠ 1 and a > 0, the function f( x ) = a x is called an exponential function The general form is : where C is the constant coefficient, C > 0, and base is a x a Base x Exponent a f( x ) = C a x

5. 10/26/2012 Section 4.3 v5.0.1 5 Function Comparisons x y x y x y f(x) = 2x f(x) = x 2 Linear Function Exponential Function Power Function –2 –4 –1 –2 0 1212 2424 3636 –3 –6 x 2x –2 ¼ –1 ½ 0101 1212 2424 3838 –3 ⅛ x 2 x –2 4 –1 1 0 1 2424 3939 –3 9 x x 2 f = { (x, 2x)  x  R } f = { (x, x 2 )  x  R } Question: What is f(5) ?... and f(10) ?... and f(20) ? Which function grows fastest as x ? 

6. 10/26/2012 Section 4.3 v5.0.1 6 Exponential Functions Increasing/Decreasing Exponential Functions Exponential growth function : f(x) = C a x, a > 1 Exponential decay function g(x) = f(–x) = C a –x, 1 < a... a reflection of f(x) h(x) = Cb x, 0 < b < 1, x y f(x) = C2 x –2 ¼ 4 –1 ½ 2 011011 12½12½ 24¼24¼ 38⅛38⅛ –3 ⅛ 8 f = { (x, 2 x )  x  R } g(x) = f(–x) = C2 –x x 2 x 2 –x ● (0, C) g = { (x, 2 –x )  x  R } As ordered pairs (C = 1) : In tabular form (C = 1) : Questions: Intercepts ? Asymptotes ? Domain = R Range = { x  x > 0 }... OR 1 a b = Effects of larger/smaller a ? Growth factor a ? Decay factor a ? a > 1 0 < a < 1

7. 10/26/2012 Section 4.3 v5.0.1 7 Exponential Function Basics Let f(x) = a x with a > 0, a ≠ 1 f(0) = 1 Domain-of-f = R Range-of-f = { y  x > 0 } Graph is increasing for a > 1 and decreasing for a < 1 f is 1–1 For a > b > 1 : a x > b x for x > 0 and a x < b x for x < 0 Graphs of a x and b x intersect at (0, 1) If a x = a y then x = y = ( –, ) ∞ ∞ = ( 0, ) ∞ WHY ?

8. 10/26/2012 Section 4.3 v5.0.1 8 Exponential Equations Solve 1. 25 x = 125 (5 2 ) x = 5 3 5 2x = 5 3 2x = 3 x = 3/2 Solution set is 2. 9 x – 2 = 27 x (3 2 ) x – 2 = (3 3 ) x 3 2x – 4 = 3 3x 2x – 4 = 3x x = –4 Solution set is { –4 } { } 3 2 WHY ?

9. 10/26/2012 Section 4.3 v5.0.1 9 Exponential Decay Radioactive Decay Radioactive isotopes of some elements such as 14 C, 16 N, 238 U, etc decay spontaneously into more stable forms ( 12 C, 14 N, 236 U, 232 U, etc) Decay times range from a few microseconds to thousands of years Decay measurement Often can’t measure whole decay time Can measure limited decay, then calculate half-life Half-life = time for decay to half original measured amount Model with exponential functions Facts: Decay rate proportional to amount present Same proportion decays in equal time

10. 10/26/2012 Section 4.3 v5.0.1 10 Exponential Decay Radioactive Decay (continued) Let initial amount of radioactive of substance Q be A 0 and A( x ) the amount after x years of decay After half-life of k years, A( k ) = ½A 0 A( 2k ) = (½)A( k ) After n half-lives, x = nk so the amount left is A( x ) = A( nk ) = A 0  (½) n Since x = nk, then n = x/k and A( x ) = A 0  (½) x/k A( 3k ) = (½)A (2k), …, A( nk ) = A 0  (½) n, … = ( A 0 (½) ) (½) = A 0  (½) 2 … or just A 0  (½) = ( A 0 (½) 2 ) (½) = A 0  (½) 3 A( 4k ) = A 0 (½) 4

11. 10/26/2012 Section 4.3 v5.0.1 11 Exponential Decay Radioactive Decay (continued) Alternative view: just recognize that half-life can be modeled by an exponential function f(x) = C a x Then initial amount is f(0) = C and, for half-life k, f( k ) = (½)C = C a k Dividing out the constant C gives: a k = ½ and solving for a, we get Hence f( x ) = C a x = C ( (½) 1/k ) x = C(½) x/k Question: What does this look like graphically ? a = (½) 1/k

12. 10/26/2012 Section 4.3 v5.0.1 12 A(t) t Exponential Decay Radioactive Decay Graph Let :        A0A0 A0A0 1 2 A0A0 1 4 A0A0 1 8 A0A0 1 16 A0A0 1 64 k2k3k 4k5k6k 0 A0A0 1 32 A(t) = amount at time t A 0 = initial amount k = half-life A(t) = A 0 (½) t/k Since n = t/k A(t) = A 0  (½) n A 0 = A(0) Amount is reduced by half in each half-life After n half-lives t = nk A(t) = A 0  (½) t/k

13. 10/26/2012 Section 4.3 v5.0.1 13 Compounding $200 is deposited and earns 5% interest compounded annually. How much is in the account after three years ? After 1 year: Balance = 200 + (.05)(200) = 200(1 +.05) After 2 years: Balance = 200(1 +.05) + 200(1 +.05)(.05) = 200(1 +.05)(1 +.05) = 200(1 +.05) 2 After 3 years: Balance = 200(1 +.05) 2 + 200(1 +.05) 2 (.05) = 200(1 +.05) 2 (1 +.05) = 200(1 +.05) 3 = 231.525 ≈ $231.53 Question: What if interest is simple interest? Balance = $230.00

14. 10/26/2012 Section 4.3 v5.0.1 14 Suppose amount P draws r per cent interest (expressed as a decimal fraction) compounded annually for t years What is the amount A accumulated after t years? 0 year: 1 year: 2 years: 3 years: t years:... well... Is this obvious? Compound Interest in General A = P A = P + Pr = P(1 + r) A = P(1 + r) + P(1 + r)r = P(1 + r)(1 + r) = P(1 + r) 2 A = P(1 + r) t A = P(1 + r) 2 + P(1 + r) 2 r = P(1 + r) 2 (1 + r) = P(1 + r) 3 = P(1 + r) 1 Question: What if compounding is quarterly ? What if compounding is n times per year ? Now is this obvious?

15. 10/26/2012 Section 4.3 v5.0.1 15 Compound Interest in General Compounding n times per year we annualize interest to r/n 0 period: 1 period: 2 periods: 3 periods: k periods: In t years k = nt t years: A = P A = P + P(r/n) A = P(1 + r/n) + P(1 + r/n)(r/n) = P(1 + r/n) 2 A = 1000(1 +.05/4) 4(20) A = P(1 + r/n) 2 + P(1 + r/n) 2 (r/n)= P(1 + r/n) 3 = P(1 + r/n) 1 Example:$1000 for 20 years at 5% compounded quarterly Here P = 1000, r =.05, n = 4 and t = 20 A = P(1 + r/n) k = 1000(1.0125) 80 = 1000(2.701484941) A = P(1 + r/n) nt ≈ $2,701.48 Is this obvious?... consider... Ah, now it’s obvious !

16. 10/26/2012 Section 4.3 v5.0.1 16 Natural Exponential Function Compute the first few terms of the sequence anan n = 1 + n 1 ( ) nannan 12.000000000 22.250000000 32.370370370 42.441406250 52.488320000 62.521626372 102.593742460 1002.704813829 2002.711517123 4002.714891744 20002.717602569 100002.717942121 1000002.718268237 ? Does a n approach a value as n ∞ ? Question: 10000002.718280469 In fact, anan 2.7182 81828 45904 52353 60287.... We call this number e e is irrational (in fact transcendental) and is the base for natural exponential functions Natural exponential functions are of form f(x) = e x... and natural logarithms

17. 10/26/2012 Section 4.3 v5.0.1 17 Natural Exponential Function Graph of x e x 1 2.7182 81828 2 7.3890 56098 3 20.0855 36923 4 54.5981 50033 5 148.4131 59105 6 403.4287 93492 7 1096.6331 58428 8 2980.9579 87041 9 8103.0839 27575 10 22026.4657 94806 11 59874.1417 15197 12 162754.7914 19003 13 442413.3920 08920 ? 14 1202604.2841 64776 f(x) = e x x exex 1000 800 600 400 200 1200 1234567         x –10123 exex 1 2 3 4 5        f(x) = e x

18. 10/26/2012 Section 4.3 v5.0.1 18 We have shown that Thus Recall that amount P compounded n times per year at annual interest rate r for t years is given by Then As the number of compounding periods per year (n) increases periodic compounding approaches continuous compounding Thus an amount P compounded continuously for t years at annualized interest rate r yields amount A given by Continuous Compounding n 1 + n 1 ( ) e nx ) 1 + n 1 ( e x A = P(1 + r/n) nt x = n 1 + ( ) ( ) n 1 as n ∞ n 1 and0 as n ∞ n 1 and0 A = P(1 + r/n) (n/r)rt = P ( (1 + r/n) (n/r) ) rt What does this mean ? A = Pe rt Pe rt

19. 10/26/2012 Section 4.3 v5.0.1 19 Example: $1000 compounds continuously at 5% interest for 10 years What is the accumulated amount ? A = Pe rt = 1000e (.05)10 = 1000(1.648721271) ≈ 1648.72 The accumulated amount is $1,648.72 For 20 years this would be: $2,718.28 For 30 years this would be: $4,481.69 For 40 years this would be: $7,389.06 Continuous Compounding With simple annual interest $1,628.89 $2,653.30 $4,321.94 $7,039.99

20. 10/26/2012 Section 4.3 v5.0.1 20 Example: $25 is deposited at the end of each month in an account paying 5% annualized interest compounded continuously How much is in the account after 10 years? Let A n be the amount in the account at the end of month n and A 0 be the initial deposit A 1 = A 0 + A 0 (e.05/12 ) = A 0 (1 + (e.05/12 )) A 2 = A 0 + A 1 (e.05/12 ) = A 0 + A 0 (1 + (e.05/12 ))e.05/12 = A 0 (1 + e.05/12 + (e.05/12 ) 2 ) A k = A 0 (1 + e.05/12 + (e.05/12 ) 2 +... + (e.05/12 ) k ) Regular Saving 20 years?30 years? = A0A0 1 – ( e.05/12 ) k+1 1 – e.05/12 At 20 years, k = 240, A 240 = 10,356.18 At 10 years, k = 120, A 120 = 3,925.44 At 30 years, k = 360, A 360 = 20,958.68 At 40 years, k = 360, A 480 = 38,439.26 Geometric Series ( A k = S k+1 )

21. 10/26/2012 Section 4.3 v5.0.1 21 Example: The population of a certain country doubles every 50 years In 1950 the population was 150 M (million) What was the population in 1975 ? When will the population reach 600 M ? Solution: Let P(t) be the population at time t years Let t = 0 represent 1950 and P(0) = P 0 = 150 M P(t) = P 0 2 kt where k is a growth control In 2000, t = 50 the population is doubled: P(50) = 2P 0 = 300 = 150(2 50k ) 2 50k = 300/150 = 2 1 50k = 1 Population Growth How do we know this ?

22. 10/26/2012 Section 4.3 v5.0.1 22 Thus In 1975, t = 25 so P(25) = P 0 2 kt = (150)(2 25/50) ) When the population is 600 M we have P(t) = 600 = 150(2 t/50 ) 2 t/50 = 600/150 = 4 = 2 2 Thus t/50 = 2 Hence the population will be 600 M in the year 2050 Population Growth P(t) = P 0 2 kt 50k = 1 P(t) = 150(2 t/50 ) = 150  2 ≈ 212.1 M k = 1/50 and t = 100 years after 1950 Sound familiar ?

23. 10/26/2012 Section 4.3 v5.0.1 23 Think about it !