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EXPONENTIAL FUNCTIONS

EXPONENTIAL FUNCTIONS. 1 + r > 1. A quantity is growing exponentially if it increases by the same percent in each unit of time. This is called exponential growth. C is the initial amount. (1 + r ) is the growth factor. y = C (1 + r ) t. r is the growth rate.

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EXPONENTIAL FUNCTIONS

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  1. EXPONENTIAL FUNCTIONS

  2. 1 +r > 1 A quantity is growing exponentially if it increases by the same percent in each unit of time. This is called exponential growth. C is the initial amount (1 + r) is the growth factor y = C(1 + r)t r is the growth rate t is the time period

  3. You deposit $500 in an account that pays 8% annual interest compounded yearly. What is the account balance after 6 years? y = C(1 + r)t y = 500(1 + 0.08)6 Note: always convert the percentage into a decimal. = 500(1.08)6  793.437 The balance after 6 years will be about $793.44.

  4. Graphically 793.437 t The x-axis usually represents time

  5. A savings account certificate of $1000 pays 6.5% annual interest compounded yearly. What is the balance when the certificate matures after 5 years? y = C(1 + r)t y = 1000(1 + 0.065)5 = 1000(1.065)5  1370.09 The balance of the certificate after 5 years is about $1370.09.

  6. A newly hatched channel catfish typically weighs about 0.3 grams. During the first six weeks of life, its growth is approximately exponential, increasing by about 10% each day. Find the weight at the end of six weeks. y = C(1 + r)t y = 0.3(1 + .10)42 Note: since the problem stated that the rate of increase was daily, we must let t be represented in days. y = 0.3(1.1)42 y  16.42910977 The weight is about 16.4 grams.

  7. An experiment started with 100 bacteria. They double in number every hour. Find the number of bacteria after 8 hours. y = C(1 + r)t y = 100(1 + 1)8 Note: since the bacteria doubled every hour, they rate of increase was 100% or 1. y = 100(2)8 y = 25,600 The number of bacteria after 8 hours is 25,600.

  8. A population of 20 rabbits is released into a wild-life region. The population triples each year for 5 years. • What is the percent increase each year? • What is the population after 5 years? Solution a. The population triples each year, so the growth factor is 3. 1 + r = 3 • So, the growth factor rate, r, is 2 and the percent increase each year is 200%.

  9. A population of 20 rabbits is released into a wild-life region. The population triples each year for 5 years. • What is the percent increase each year? • What is the population after 5 years? Solution There will be about 4860 rabbits after 5 years. • P = C(1 + r)t • = 20(1 + 2)5 • = 20(3)5 • = 4860

  10. Graphically 4860 t

  11. A company starts with 50 employees and after one year has 75. It increases the employees at the same rate every year for 4 years. • What is the percent of increase each year? • Find the number of employees after 4 years? solution • After one year you have 75 employees which means there • are 25 more employees than the company started with. • Since there were 50 employees and 25 were added within • one year, the percent increase must be 50%. Try this example on a sheet of paper before clicking.

  12. A company starts with 50 employees and after one year has 75. It increases the employees at the same rate every year for 4 years. • What is the percent of increase each year? • Find the number of employees after 4 years? solution There will be 253 employees at the end of 4 years. • P = C(1 + r)t • = 50(1 + 0.5)4 • = 50(1.5)4 •  253

  13. 0 < 1 – r < 1 A quantity is decaying exponentially if it decreases by the same percent in each unit of time. This is called exponential decay. C is the initial amount (1 – r) is the decay factor y = C(1 r)t r is the decay rate t is the time period

  14. From 1982 through 1997, the purchasing power of a dollar decreased by about 3.5% per year. Using 1982 as the base for comparison, what was the purchasing power of a dollar in 1997? solution Let y represent the purchasing power and let t = 0 represent the year 1982. The initial amount is $1. Use an exponential decay model (Notice the dollar decreased, therefore, we use a decay model). y = C(1 –r)t Because 1997 is 15 years after 1982, substitute 15 for t. y = 1(1 – 0.035)15  0.59

  15. You bought a used car for $18,000. The value of the car will be less each year because of depreciation. The car depreciates (loses value) at the rate of 12% per year. • Write an exponential decay model to represent this situation. • Estimate the value of your car in 8 years. solution • The initial value C is $18,000. The decay rate r is 0.12. Let • tbe the age of the car in years. y = 18,000(1 – 0.12)t y = 18,000(0.88)t

  16. You bought a used car for $18,000. The value of the car will be less each year because of depreciation. The car depreciates (loses value) at the rate of 12% per year. • Write an exponential decay model to represent this situation. • Estimate the value of your car in 8 years. solution b. To find the value in 8 years, substitute 8 for t. y = 18,000(0.88)8 6473 The value of your car in 8 years will be about $6473.

  17. Checkpoint A house costs $140,000 in the year 2002. It appreciates 4.5% annually. • Write a model to represent the value of the house • after t years. • b. Find the value of the house after 20 years. The amount of a particular medicine in the body decreases at a rate of 40% an hour. You are given 250mg. • Write a model to represent the amount of medicine • in the body after t hours. • b. Find the amount of medicine after 5 hours.

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