4.3 continuous compound interests perta x

continuous compound interests perta

1. Periodic Compound Interest Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation Recall the period-compound PINA formula.

2. Periodic Compound Interest Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A Recall the period-compound PINA formula.

3. Periodic Compound Interest Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A Recall the period-compound PINA formula. We use the following time line to see what is happening. 1 2 3 Nth period N–1

4. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward Recall the period-compound PINA formula. Periodic Compound Interest

5. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) Recall the period-compound PINA formula. Periodic Compound Interest

6. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 Recall the period-compound PINA formula. Periodic Compound Interest

7. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 Recall the period-compound PINA formula. Periodic Compound Interest

8. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 P(1 + i ) N - 1 Recall the period-compound PINA formula. Periodic Compound Interest

9. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 P(1 + i ) N - 1 Recall the period-compound PINA formula. P(1 + i ) N Periodic Compound Interest

10. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 P(1 + i ) N - 1 P(1 + i ) N = A Example A. $1,000 is in an account that has a monthly interest rate of 1%. How much will be there after 60 years? Recall the period-compound PINA formula. P(1 + i ) N Periodic Compound Interest

11. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 P(1 + i ) N - 1 Example A. $1,000 is in an account that has a monthly interest rate of 1%. How much will be there after 60 years? We have P = $1,000, i = 1 % = 0.01, N = 60 *12 = 720 months so by PINA , there will be 1000(1 + 0.01) 720 Recall the period-compound PINA formula. P(1 + i ) N = A P(1 + i ) N Periodic Compound Interest

12. Let P = principal i = (periodic) interest rate, N = number of periods A = accumulation then P (1 + i ) N = A We use the following time line to see what is happening. P 1 2 3 Nth period N–1 Rule: Multiply (1 + i ) each period forward P(1 + i ) P(1 + i ) 2 P(1 + i ) 3 P(1 + i ) N - 1 Example A. $1,000 is in an account that has a monthly interest rate of 1%. How much will be there after 60 years? We have P = $1,000, i = 1 % = 0.01 , N = 60 *12 = 720 months so by PINA , there will be 1000(1 + 0.01) 720 = $1,292,376.71 after 60 years. Recall the period-compound PINA formula. P(1 + i ) N = A P(1 + i ) N Periodic Compound Interest

13. The graphs shown here are the different returns with r = 20% with different compounding frequencies. Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia) Periodic Compound Interest

14. The graphs shown here are the different returns with r = 20% with different compounding frequencies. We observe that I. the more frequently we compound, the bigger the return Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia) Periodic Compound Interest

15. The graphs shown here are the different returns with r = 20% with different compounding frequencies. We observe that I. the more frequently we compound, the bigger the return II. but the returns do not go above the blue-line the continuous compound return , which is the next topic. Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia) Periodic Compound Interest

16. Continuous Compound Interest

17. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? Continuous Compound Interest

18. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, Continuous Compound Interest

19. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, Continuous Compound Interest

20. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Continuous Compound Interest

21. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000 Continuous Compound Interest

22. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ Continuous Compound Interest

23. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, Continuous Compound Interest

24. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Continuous Compound Interest

25. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000 Continuous Compound Interest

26. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000  4952.72 $ Continuous Compound Interest

27. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000  4952.72 $ For 10000 times a year, 10000 0.08 i = = 0.000008, Continuous Compound Interest

28. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000  4952.72 $ For 10000 times a year, 10000 0.08 i = = 0.000008, N = (20 years)(10000 times per years) = 200000 Continuous Compound Interest

29. Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year? P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000  4952.72 $ For 10000 times a year, 10000 0.08 i = = 0.000008, N = (20 years)(10000 times per years) = 200000 Hence A = 1000(1 + 0.000008 ) 200000 Continuous Compound Interest

30. P = 1000, r = 0.08, T = 20, For 100 times a year, 100 0.08 i = = 0.0008, N = (20 years)(100 times per years) = 2000 Hence A = 1000(1 + 0.0008 ) 2000  4949.87 $ For 1000 times a year, 1000 0.08 i = = 0.00008, N = (20 years)(1000 times per years) = 20000 Hence A = 1000(1 + 0.00008 ) 20000  4952.72 $ For 10000 times a year, 10000 0.08 i = = 0.000008, N = (20 years)(10000 times per years) = 200000 Hence A = 1000(1 + 0.000008 ) 200000  4953.00 $ Continuous Compound Interest Example B. We deposited $1000 in an account with annual compound interest rate r = 8%. How much will be there after 20 years if it's compounded 100 times a year? 1000 times a year? 10000 times a year?

31. We list the results below as the number compounded per year f gets larger and larger. Continuous Compound Interest

32. We list the results below as the number compounded per year f gets larger and larger. 4 times a year 4875.44 $ Continuous Compound Interest

33. We list the results below as the number compounded per year f gets larger and larger. 100 times a year 4949.87 $ 4 times a year 4875.44 $ Continuous Compound Interest

34. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $ Continuous Compound Interest

35. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $  Continuous Compound Interest

36. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $  4953.03 $ Continuous Compound Interest

37. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $  4953.03 $ We call this amount the continuously compounded return. Continuous Compound Interest

38. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $  4953.03 $ We call this amount the continuously compounded return. This way of compounding is called compounded continuously . Continuous Compound Interest

39. We list the results below as the number compounded per year f gets larger and larger. 10000 times a year 4953.00 $ 1000 times a year 4952.72 $ 100 times a year 4949.87 $ 4 times a year 4875.44 $  4953.03 $ We call this amount the continuously compounded return. This way of compounding is called compounded continuously . The reason we want to compute interest this way is because the formula for computing continously compound return is easy to manipulate mathematically. Continuous Compound Interest

40. Formula for Continuously Compounded Return ( Perta ) Continuous Compound Interest

41. Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

42. Continuous Compound Interest There is no “ f ” because it’s compounded continuously Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

43. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

44. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

45. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

46. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

47. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

48. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

49. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

50. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

51. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4  11023.18$ Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

52. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4  11023.18$ c. If r = 16%, how much will be there after 20 years? Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

53. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4  11023.18$ c. If r = 16%, how much will be there after 20 years? r = 16%, A = 1000 * e 0.16*20 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

54. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4  11023.18$ c. If r = 16%, how much will be there after 20 years? r = 16%, A = 1000 * e 0.16*20 = 1000 * e 3.2 Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

55. Example C. We deposited $1000 in an account compounded continuously. a. if r = 8%, how much will be there after 20 years? P = 1000, r = 0.08, t = 20. So the continuously compounded return is A = 1000 * e 0.08*20 = 1000 * e 1.6  4953.03$ b. If r = 12%, how much will be there after 20 years? r = 12%, A = 1000 * e 0.12 * 20 = 1000e 2.4  11023.18$ c. If r = 16%, how much will be there after 20 years? r = 16%, A = 1000 * e 0.16*20 = 1000 * e 3.2  24532.53$ Continuous Compound Interest Formula for Continuously Compounded Return ( Perta ) Let P = principal r = annual interest rate (compound continuously) t = number of years A = accumulated value, then Pe r*t = A where e  2.71828…

56. Continuous Compound Interest About the Number e

57. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Continuous Compound Interest About the Number e

58. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Where as π  3.14156… is a geometric constant–the ratio of the circumference to the diameter of a circle, e is derived from calculations. Continuous Compound Interest About the Number e

59. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Where as π  3.14156… is a geometric constant–the ratio of the circumference to the diameter of a circle, e is derived from calculations. For example, the following sequence of numbers zoom–in on the number, ( ) 1 , 2 1 … ( ) 4 , 5 4 ( ) 3 , 4 3 ( ) 2 , 3 2  2.71828… Continuous Compound Interest About the Number e

60. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Where as π  3.14156… is a geometric constant–the ratio of the circumference to the diameter of a circle, e is derived from calculations. For example, the following sequence of numbers zoom–in on the number, ( 2.71828…) the same as ( ) 1 , 2 1 … ( ) 4 , 5 4 ( ) 3 , 4 3 ( ) 2 , 3 2  2.71828…which is Continuous Compound Interest About the Number e

61. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Where as π  3.14156… is a geometric constant–the ratio of the circumference to the diameter of a circle, e is derived from calculations. For example, the following sequence of numbers zoom–in on the number, This number emerges often in the calculation of problems in physical science, natural science, finance and in mathematics. ( 2.71828…) the same as ( ) 1 , 2 1 … ( ) 4 , 5 4 ( ) 3 , 4 3 ( ) 2 , 3 2  2.71828…which is Continuous Compound Interest About the Number e

62. Just as the number π , the number e  2.71828… occupies a special place in mathematics. Where as π  3.14156… is a geometric constant–the ratio of the circumference to the diameter of a circle, e is derived from calculations. For example, the following sequence of numbers zoom–in on the number, http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 This number emerges often in the calculation of problems in physical science, natural science, finance and in mathematics. Because of its importance, the irrational number 2.71828… is named as “e” and it’s called the “natural” base number. ( 2.71828…) the same as http://www.ndt-ed.org/EducationResources/Math/Math-e.htm ( ) 1 , 2 1 … ( ) 4 , 5 4 ( ) 3 , 4 3 ( ) 2 , 3 2  2.71828…which is Continuous Compound Interest About the Number e

63. Continuous Compound Interest Growth and Decay

64. Continuous Compound Interest Growth and Decay In all the interest examples we have positive interest rate r, hence the return A = Pe rx grows larger as time x gets larger.

65. Continuous Compound Interest Growth and Decay In all the interest examples we have positive interest rate r, hence the return A = Pe rx grows larger as time x gets larger. We call an expansion that may be modeled by A = Pe rx with r > 0 as “ an exponential growth with growth rate r” .

66. Continuous Compound Interest y = e 1 x has the growth rate of r = 1 or 100%. Growth and Decay In all the interest examples we have positive interest rate r, hence the return A = Pe rx grows larger as time x gets larger. We call an expansion that may be modeled by A = Pe rx with r > 0 as “ an exponential growth with growth rate r” . For example,

67. Continuous Compound Interest y = e 1 x has the growth rate of r = 1 or 100%. Exponential growth are rapid expansions compared to other expansions as shown here by their graphs. y = x 3 y = 100x y = e x Growth and Decay In all the interest examples we have positive interest rate r, hence the return A = Pe rx grows larger as time x gets larger. We call an expansion that may be modeled by A = Pe rx with r > 0 as “ an exponential growth with growth rate r” . For example, An Exponential Growth

68. Continuous Compound Interest y = e 1 x has the growth rate of r = 1 or 100%. Exponential growth are rapid expansions compared to other expansions as shown here by their graphs. y = x 3 y = 100x y = e x The world population may be modeled with an exponential growth with r ≈ 1.1 % or 0.011 or that A ≈ 6.5e 0.011t in billions , with 2011as t = 0 . Growth and Decay In all the interest examples we have positive interest rate r, hence the return A = Pe rx grows larger as time x gets larger. We call an expansion that may be modeled by A = Pe rx with r > 0 as “ an exponential growth with growth rate r” . For example, An Exponential Growth

69. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger.

70. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” .

71. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” . For example, y = e –1 x has the decay or contraction rate of I r I = 1 or 100%. y = e –x An Exponential Decay

72. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” . For example, y = e –1 x has the decay or contraction rate of I r I = 1 or 100%. In finance, shrinking values is called “depreciation” or ”devaluation”. y = e –x An Exponential Decay

73. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” . For example, y = e –1 x has the decay or contraction rate of I r I = 1 or 100%. In finance, shrinking values is called “depreciation” or ”devaluation”. For example, a currency that is depreciating at a rate of 4% annually may be modeled by A = Pe – 0.04 x where x is the number of years elapsed. y = e –x An Exponential Decay

74. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” . For example, y = e –1 x has the decay or contraction rate of I r I = 1 or 100%. In finance, shrinking values is called “depreciation” or ”devaluation”. For example, a currency that is depreciating at a rate of 4% annually may be modeled by A = Pe – 0.04 x where x is the number of years elapsed. Hence if P = $1, after 5 years, its purchasing power is 1*e –0.04(5) = $0.82 or 82 cents. y = e –x An Exponential Decay

75. Continuous Compound Interest If the rate r is negative, or that r < 0 then the return A = Pe rx shrinks as time x gets larger. We call a contraction that may be modeled by A = Pe rx with r < 0 as “an exponential decay at the rate | r |” . For example, y = e –1 x has the decay or contraction rate of I r I = 1 or 100%. In finance, shrinking values is called “depreciation” or ”devaluation”. For example, a currency that is depreciating at a rate of 4% annually may be modeled by A = Pe – 0.04 x where x is the number of years elapsed. Hence if P = $1, after 5 years, its purchasing power is 1*e –0.04(5) = $0.82 or 82 cents. For more information: y = e –x An Exponential Decay http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm

76. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time .

77. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2

78. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2)

79. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r

80. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r ≈ 0.72/r

81. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r ≈ 0.72/r The last formula for the doubling time x is the 72-Rule, i.e. x ≈ 0.72/r

82. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r ≈ 0.72/r The last formula for the doubling time x is the 72-Rule, i.e. x ≈ 0.72/r This 72- Rule gives an easy estimation for the doubling time.

83. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r ≈ 0.72/r The last formula for the doubling time x is the 72-Rule, i.e. x ≈ 0.72/r This 72- Rule gives an easy estimation for the doubling time. For example, if r = 8% then the doubling time is ≈ 72/8 = 9 ( yrs ).

84. Doubling Time and the 72-Rule Given that r > 0, the return A = Pe rx grows as time x gets larger. The time it takes to double the principal P so that A = 2P. is called the doubling time . Solving for the doubling time we have Pe rx = 2P or that e rx = 2 so from the result of the next section, rx = In(2) and that the doubling time is x = In(2)/ r ≈ 0.72/r The last formula for the doubling time x is the 72-Rule, i.e. x ≈ 0.72/r This 72- Rule gives an easy estimation for the doubling time. For example, if r = 8% then the doubling time is ≈ 72/8 = 9 ( yrs ). if r = 12% then the doubling time is ≈ 72/12 = 6 ( yrs ). if r = 18 % then the doubling time is ≈ 72/18 = 4 ( yrs ).

85. Compound Interest B . Given the continuous compound annual rate r find the principal needed to obtain $ 1,000 after the given amount the time. 1. r = 1 %, time = 60 months. Exercise A. Given the continuous compound annual interest rate r and the amount the time, find the return with a principal of $1,000. 2. r = 1 %, time = 60 years. 3. r = 3 %, time = 60 years 4. r = 3½ %, time = 60 months. 5. r = 2¼ %, time = 6 months. 6. r = 1¼ %, time = 5½ years. 7. r = 3 3/8%, time = 52 months 8. r = 3/8%, time = 27 months. 1. r = 1 %, time = 60 months. 2. r = 1 %, time = 60 years. 3. r = 3 %, time = 60 years 4. r = 3½ %, time = 60 months. 5. r = 2¼ %, time = 6 months. 6. r = 1¼ %, time = 5½ years. 7. r = 3 3/8%, time = 52 months 8. r = 3/8%, time = 27 months.

86. Doubling Time and the 72-Rule 9 . For continuous growth with growth rate r > 0, the formula ≈ 3/2 x 0.72/r ≈ 1.08/r estimates the time that will take to triple the original amount. Given annual rate r = 6%, a. estimate how long it will take to have a return that’s 3 times the original amount? b. estimate how long it will take to have a return that’s 6 times the original amount? c. estimate how long it would take a have a return that’s 9 times the original amount ? a. estimate how long it will take to have a return that’s 4 times the original amount . How about 8 times the original amount ? b . estimate how long it will take to have a return that’s 12 times the original amount . 10. By the 72-formula, it will take ≈ 2 x 0.72/r = 1.44/r years to quadruple the original amount. Given r = 12 % , 11. How much off are the estimations in 10. a and b?

87. Doubling Time and the 72-Rule 12. Similar to 9 –11, given the continuous decay rate –r ( r > 0), by the half life 72-formula, it will take ≈ 3/2 x 0.72/r ≈ 1.08/r to decay to 1/3 of the initial amount, ≈ 2 x 0.72/r = 1.44/r to decay to ¼ of the original size. Given r = 4%, estimate how long it will take to decay to 1/6 of the initial amount? 1/9 of the initial amount? 1/12 of the initial amount? 13. Given r = 8%, estimate how long it will take to decay to 1/6 of the initial amount? 1/9 of the initial amount? 1/12 of the initial amount? 14. How much off are the estimations in 14?

88. Compound Interest Exercise B. 1. $ 1501.271 (Answers to the odd problems ) Exercise A. 3. $ 1161.834 5. $ 1011.3135 7. $1156.184 1. $ 951.229 3. $ 860.708 5. $ 988.813 7. $ 864.0289 9. a. 18 yrs b . 30 yrs c. 48 yrs 13. 22.3970 years, 31.0613 years

89. Continuous Compound Interest