of 14 /14
9.6 Notes Part I Exponential Growth and Decay
• Author

others
• Category

## Documents

• view

0

0

Embed Size (px)

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

• 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