9.6 Notes Part I Exponential Growth and Decay...exponential decay function to model this situation....

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9.6 Notes Part I Exponential Growth and Decay

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  • 9.6 Notes Part I Exponential Growth

    and Decay

  • I. Exponential Growth

    (1 )ty C r

    Final Amount

    Initial Amount

    Rate of Change

    Time

  • Ex 1: The original value of a painting is $9000 and the value increases by 7% each year. Write an exponential growth function to model this situation. Then find the painting’s value in 15 years.

    Step 1 Write the exponential growth function for this situation.

    y = a(1 + r)t

    = 9000(1 + 0.07)t

    = 9000(1.07)t

    Substitute 9000 for a and 0.07 for r.

    Simplify.

    Write the formula.

    Step 2 Find the value in 15 years.

    y = 9000(1.07)t

    = 9000(1.07)15

    ≈ 24,831.28

    The value of the painting in 15 years is $24,831.28.

    Substitute 15 for t.

    Use a calculator and round to the

    nearest hundredth.

  • Ex 2: An investment is increasing in value at a rate of 8% per year, and its value in 2000 was $1200. Write an exponential growth function to model this situation. Then find the investment’s value in 2006.

    Step 1 Write the exponential growth function for this situation.

    y = a(1 + r)t

    = 1200(1.08)t

    Write the formula.

    Substitute 1200 for a and 0.08 for r.

    Simplify.

    = 1200(1 + 0.08)t

    Step 2 Find the value in 6 years.

    y = 1200(1.08)t

    = 1200(1.08)6

    ≈ 1,904.25

    The value of the painting in 6 years is $1904.25.

    Substitute 6 for t.

    Use a calculator and round to the

    nearest hundredth.

  • II. Exponential Decay

    (1 )ty C r

    Final Amount

    Initial Amount

    Rate of Change

    Time

  • Ex 3: The population of a town is decreasing at a rate of 3% per year. In 2000 there were 1700 people. Write an exponential decay function to model this situation. Then find the population in 2012.

    Step 1 Write the exponential decay function for this situation.

    y = a(1 – r)t

    = 1700(1 – 0.03)t

    = 1700(0.97)t

    Write the formula.

    Substitute 1700 for a and 0.03 for r.

    Simplify.

    Step 2 Find the population in 2012.

    ≈ 1180

    Substitute 12 for t. y = 1700(0.97)12

    Use a calculator and round to the

    nearest whole number.

    The population in 2012 will be approximately 1180 people.

  • Ex 4: The fish population in a local stream is decreasing at a rate of 3% per year. The original population was 48,000. Write an exponential decay function to model this situation. Then find the population after 7 years.

    Step 1 Write the exponential decay function for this situation.

    y = a(1 – r)t

    = 48,000(1 – 0.03)t

    = 48,000(0.97)t

    Write the formula.

    Substitute 48,000 for a and 0.03 for

    r.

    Simplify.

    Step 2 Find the population in 7 years.

    ≈ 38,783

    Substitute 7 for t. y = 48,000(0.97)7

    Use a calculator and round to the

    nearest whole number.

    The population after 7 years will be approximately 38,783 people.

  • A common application of exponential decay is half-life. The half-life of a substance is the time it takes for one-half of the substance to decay into another substance.

    III. Half-Life

  • Ex 1: Astatine-218 has a half-life of 2 seconds.

    Find the amount left from a 500 gram sample of astatine-218 after 10 seconds.

    Step 1 Find t, the number of half-lives in the given time period.

    Divide the time period by the half-life. The

    value of t is 5.

    Write the formula.

    = 500(0.5)5 Substitute 500 for P and 5 for t.

    = 15.625 Use a calculator.

    There are 15.625 grams of Astatine-218 remaining after 10 seconds.

    Step 2 A = P(0.5)t

  • Ex 2: Astatine-218 has a half-life of 2 seconds.

    Find the amount left from a 500-gram sample of Astatine-218 after 1 minute.

    Step 1 Find t, the number of half-lives in the given time period.

    Divide the time period by the half-life. The

    value of t is 30.

    1(60) = 60 Find the number of seconds in 1 minute.

    Step 2 A = P(0.5)t Write the formula.

    = 500(0.5)30 Substitute 500 for P and 30 for t.

    = 0.00000047 g Use a calculator.

    There are 0.00000047 grams of Astatine-218 remaining after 60 seconds.

  • Ex 3: Cesium-137 has a half-life of 30 years. Find the amount of Cesium-137 left from a 100 milligram sample after 180 years.

    Step 1 Find t, the number of half-lives in the given time period.

    Divide the time period by the half-life. The

    value of t is 6.

    Step 2 A = P(0.5)t Write the formula.

    = 100(0.5)6

    Use a calculator.

    There are 1.5625 milligrams of Cesium-137 remaining after 180 years.

    Substitute 100 for P and 6 for

    t.

    = 1.5625 mg

  • Ex 4: Bismuth-210 has a half-life of 5 days.

    Find the amount of Bismuth-210 left from a 100-gram sample after 5 weeks. (Hint: Change 5 weeks to days.)

    Step 1 Find t, the number of half-lives in the given time period.

    Divide the time period by the half-life. The

    value of t is 5.

    5 weeks = 35 days Find the number of days in 5 weeks.

    Step 2 A = P(0.5)t Write the formula.

    = 100(0.5)7

    = 0.78125 g Use a calculator.

    There are 0.78125 grams of Bismuth-210 remaining after 5 weeks.

    Substitute 100 for P and 7 for t.

  • Lesson Quiz: Part I

    1. The number of employees at a certain company is 1440 and is increasing at a rate of 1.5% per year. Write an exponential growth function to model this situation. Then find the number of employees in the company after 9 years.

    y = 1440(1.015)t; 1646

    Write a compound interest function to model each situation. Then find the balance after the given number of years.

    2. $12,000 invested at a rate of 6% compounded quarterly; 15 years

    A = 12,000(1.015)4t, $29,318.64

  • Lesson Quiz: Part II

    4. The deer population of a game preserve is decreasing by 2% per year. The original population was 1850. Write an exponential decay function to model the situation. Then find the population after 4 years.

    y = 1850(0.98)t; 1706

    5. Iodine-131 has a half-life of about 8 days. Find the amount left from a 30-gram sample of Iodine-131 after 40 days.

    0.9375 g