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# Notes 6-4

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e and Natural Logarithms

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### Notes 6-4

1. 1. Section 6-4 e and Natural Logarithms
2. 2. Begin by opening your books to page 389. Pair up with a partner to work on the In-Class Activity. Make sure to record your observations on your note sheet.
3. 3. e:
4. 4. e: Comes out to be approximately 2.7182818284590...
5. 5. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield
6. 6. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1 r
7. 7. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 −1
8. 8. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367
9. 9. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63%
10. 10. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year
11. 11. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year 10,000 + 10,000(e .0735 − 1)
12. 12. Example 1 If \$10,000 is put into bonds that pay 7.35% interest compounded continuously, ﬁnd: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year 10,000 + 10,000(e .0735 − 1) ≈ \$10,762.69
13. 13. Continuous Change Model
14. 14. Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is:
15. 15. Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is: A(t) = Pe rt
16. 16. Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is: A(t) = Pe rt This means we have a model we can work with for continuous compounding, just like the other types of compounding we saw earlier in the year.
17. 17. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period?
18. 18. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe rt
19. 19. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5)
20. 20. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20
21. 21. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20 b. How does this compare with the balance if the interest were compounded annually?
22. 22. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) t
23. 23. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) t 5
24. 24. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) ≈ \$14,256.41 t 5
25. 25. Example 2 Suppose \$10,000 is put into a 5-year certiﬁcate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ \$14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) ≈ \$14,256.41 t 5 About \$185 more
26. 26. The Exponential Function with Base e:
27. 27. The Exponential Function with Base e: A function of the form f(x) = ex
28. 28. The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm:
29. 29. The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x
30. 30. The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x In other words, ln x = loge x
31. 31. The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x In other words, ln x = loge x This means that the Exponential Function with Base e and the Natural Logarithm are inverses of each other.
32. 32. Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0?
33. 33. Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1
34. 34. Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1 ≈ 1.945910149
35. 35. Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1 ≈ 1.945910149 units 2
36. 36. Homework
37. 37. Homework p. 394 #1-20