Exponential Growth & DecayModeling Data

• Objectives– Model exponential growth & decay– Model data with exponential & logarithmic

functions.

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Exponential Growth & Decay Models

• A0 is the amount you start with, t is the time, and k=constant of growth (or decay)

• f k>0, the amount is GROWING (getting larger), as in the money in a savings account that is having interest compounded over time

• If k<0, the amount is SHRINKING (getting smaller), as in the amount of radioactive substance remaining after the substance decays over time

ktoA A t A e

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Graphs

00

k

eAA kt

A0A0

0

0

ktA A e

k

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Could the following graph model exponential growth or decay?

• 1) Growth model.• 2) Decay model.

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Example• Population Growth of the United States.

In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau)– Find the exponential growth function.– What will the population be in 2020?– After how long will the population be double

what it was in 1990?

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Solution• At t = 0 (1990), the population was about 249 million. We

substitute 249 for A0 and 0.08 for k to obtain the exponential growth function.

A(t) = 249e0.08t

• In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:

A(3) = 249e0.08(3) = 249e0.24 317.

The population will be approximately 317 million in 2020.

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Solution continued• We are looking for the doubling time T.

498 = 249e0.08T

2 = e0.08T

ln 2 = ln e0.08T

ln 2 = 0.08T (ln ex = x)

= T

8.7 T

The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

ln 2

0.08

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Interest Compound Continuously

• The function A(t) = A0ekt can be used to calculate interest that is compounded continuously.

In this function:

A0 = amount of money invested,

A = balance of the account,

t = years,

k = interest rate compounded continuously.

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Example• Suppose that $2000 is deposited into an

IRA at an interest rate k, and grows to $5889.36 after 12 years.

– What is the interest rate?– Find the exponential growth function.– What will the balance be after the first 5

years?– How long did it take the $2000 to double?

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Solution• At t = 0, A(0) = A0 = $2000. Thus the exponential growth

function is A(t) = 2000ekt. We know that A(12) = $5889.36. We then

substitute and solve for k: $5889.36 = 2000e12k

125889.36

2000ke

125889.36ln ln

2000ke

5889.36ln 12

2000k

5889.36ln

2000120.095

k

k

The interest rate is approximately 9.5%10

Solution continued

• The exponential growth function is

A(t) = 2000e0.09t.

• The balance after 5 years is

A(5) = 2000e0.09(5)

= 2000e0.45

$3136.62

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Solution continued• To find the doubling time T, set A(T) = 2 A0 = $4000

and solve for T.

4000 = 2000e0.09T

2 = e0.09T

ln 2 = ln e0.09T

ln 2 = 0.09T

= T

7.7 T

The original investment of $2000 doubled in about 7.7 years.

ln 2

0.09

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Growth Rate and Doubling Time The growth rate k and the doubling time T are

related by

– kT = ln 2

or–

or–

* The relationship between k and T does not depend on A0.

ln 2k

T

ln 2T

k

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Example

• A certain town’s population is doubling every 37.4 years. What is the exponential growth rate?

Solution: ln 2 ln 2

1.9%37.4

kT

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Exponential Decay• Decay, or decline, is represented by the function

A(t) = A0ekt, k < 0.

In this function:

A0 = initial amount of the substance, A = amount of the substance left after time, t = time, k = decay rate.

• The half-life is the amount of time it takes for half of an amount of substance to decay.

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Example Carbon Dating. The radioactive element carbon-14

has a half-life of 5715 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?

Solution: The function for carbon dating is

A(t) = A0e-0.00012t.

If the charcoal has lost 7.3% of its carbon-14 from its initial amount A0, then 92.7%A0 is the amount present.

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Example continuedTo find the age of the charcoal, we solve the equation for t :

The charcoal was about 632 years old.

0.000120 092.7% tA A e

0.000120.927 te0.00012ln 0.927 ln te

ln 0.927 0.00012t

ln 0.927

0.00012632

t

t

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