Chapter5 - Chapter 5 Applications of the Exponential and...

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 1 of 47 Chapter 5 Applications of the Exponential and Natural Logarithm Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 2 of 47 Exponential Growth and Decay Compound Interest Applications of the Natural Logarithm Function to Economics Further Exponential Models Chapter Outline
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 3 of 47 § 5.1 Exponential Growth and Decay
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 4 of 47 Exponential Growth The Exponential Growth and Decay Model Exponential Growth in Application Exponential Decay Exponential Decay in Application Section Outline
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 5 of 47 Exponential Growth Definition Example Exponential Growth : A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant ( 29 t e t P 4 3 =
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 6 of 47 Exponential Growth & Decay Model
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 7 of 47 Exponential Growth in Application EXAMPLE EXAMPLE SOLUTION SOLUTION ( World’s Population ) The world’s population was 5.51 billion on January 1, 1993 and 5.88 billion on January 1, 1998. Assume that at any time the population grows at a rate proportional to the population at that time. In what year will the world’s population reach 7 billion? Since the “oldest” information we’re given corresponds to 1993, that will serve as our initial time. Therefore the year 1993 will be the year t = 0 and the population at time t = 0 is 5.51 (measured in billions). Therefore, the year 1998 will be year t = 5 and the population at time t = 5 is 5.88 (measured in billions). Since the population grows at a rate proportional to the size of the population, we can use the exponential growth model P ( t ) = P 0 e kt to describe the population of the world. Since P 0 is the initial quantity, P 0 = 5.51. Therefore, our formula becomes
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 11e – Slide 8 of 47 Exponential Growth in Application Now we use the other given information (5.88 billion in 1998) to determine k . This is our function so far. CONTINUED CONTINUED ( 29 . 51 . 5 kt e t P = ( 29 kt e t P 51 . 5 = When t = 5, the population is 5.88 billion people. ( 29 5 51 . 5 88 . 5 5 k e P = = Divide. 5 07 . 1 k e = Rewrite in logarithmic form. k 5 07 . 1 ln = Solve for k . k
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Chapter5 - Chapter 5 Applications of the Exponential and...

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