Exponential Growth and Decay Section 3.5. Objectives Solve word problems requiring exponential...

Exponential Growth and Decay

Section 3.5

Objectives• Solve word problems requiring

exponential models.

Find the time required for an investment of $5000 to grow to $6800 at an interest rate of 7.5% compounded quarterly.

nt

nr

PA

1

Formula needed:

Find the time required for an investment of $5000 to grow to $6800 at an interest rate of 7.5% compounded quarterly.

138112586.4

)01875.1ln(436.1ln

)01875.1ln(436.1ln

01875.1ln36.1ln 4

t

t

t

t

t

t

t

4

4

4

01875.136.1

4075.

150006800

4075.

150006800

4

)01875.1log()36.1log(

4

)36.1(log 01875.1

t

Take the logarithm of both sides

Change to exponential form

t

t

4

)36.1(log

4)36.1(log

01875.1

01875.1

Change of base

It will take approximately 4.138112586 years.

The population of a certain city was 292000 in 1998, and the observed relative growth rate is 2% per year.

• Find a function that models the population after t years.

• Find the projected population in the year 2004.

• In what year will the population reach 365004?


teA 02.292000

• Find a function that models the population after t years.

We are going to let t = 0 represent the year 1998 and t is the number of years since 1998.

Populations are assumed to have continuous growth. Thus we use the formula rtPeA

This model will give us the population t years after 1998.


teA 02.292000

• Find the projected population in the year 2004.

We use the model from the first part of the question

Now we plug in for t. We need to know what t will represent the year 2004. We find that using a subtraction problem. 2004-1998 = 6 tells us that 2004 is 6 years since 1998 and thus our t is 6.

0807.329229

292000 6*02.

A

eA

This is not the answer. People must be whole, thus we must round this number. Either 329229 people or 329230 people would be considered correct.


teA 02.292000


We use the model from the first part of the question

Now we plug in 365004 for A and solve for t.

t

t

e

e

02.

02.

292000365004

292000365004

continued on next screen


15772551.1102.

292000365004

ln

)02(.292000365004

ln

ln)02(.292000365004

ln

ln292000365004

ln 02.

t

t

t

et

e t



15772551.1102.

292000365004

ln

02.292000365004

ln

t

t

t

change to logarithmic form

take a logarithm of both sides of the equation

OR



We still don’t have the answer to the question. We know that 11.15772551 years after 1998 the population will reach 365004. We need to know what year that is. We find this by adding

1998 + 11.15772551 = 2009.15772551

Now do we round up to 2010 or keep it at 2009. We must always keep it at the lower year (no matter what the decimal). Any decimal part of a year is still during that year. That the answer to the question is in the year 2009, the population will reach 365004.

The count in a bacteria culture was 600 after 15 minutes and 16054 after 35 minutes. Assume that growth can be modeled exponentially by a function of the form where t is in minutes.

• Find the relative growth rate.

We do not know a starting bacteria count, but the growth rate can be figured from any point in time to any other point in time as long as we use the passage of time as the t value.

We will use 600 as the starting population only for the purpose of this part of the problem. We then have 16054 bacteria 20 minutes later (the amount of time between 15 and 35 minutes). With these numbers we can solve for r.

rtPetA )(



r

r

e

e

r

r

20600

16054ln

20*600

16054ln

60016054

60016054

20*

20*• Find the relative growth rate.

We can now use this r for the other parts of this problem. It will be important tat we do not round our r value in future parts of the problem.

rtPetA )(


rtPetA )(

00083003.51

600

600

15*20600

16054ln

15*20600

16054ln

P

P

e

Pe

• What was the initial size of the culture? We know the relative growth rate r. We also have two different populations after two different periods of time. We can use either one for our population A after time t. We will need to solve for P.

I will use (15, 600) as the population after time t.

Since bacteria must be whole, the initial population must be 51 bacteria. This means that out model for future parts of the problem will be

t

etA

*20600

16054ln

51)(


rtPetA )(

t

t

t

e

e

e

*20600

16054ln

*20600

16054ln

*20600

16054ln

2

51102

51102

• Find the doubling period in minutes.

217784019.4

20600

16054ln

2ln

*20600

16054ln

2ln

t

t

t

Doubling time is how long it takes for the number of bacteria to get to be two times as many as the initial population. This means that we need to plug in an ending population of 2*51 or 102.

The doubling time is approximately 4.21778 minutes.


rtPetA )(

• Find the population after 110 minutes.

3617813887)(

51)(

110*20600

16054ln

tA

etA

For this one we just plug 110 into the model for t.


rtPetA )(

t

e

e

t

t

*20600

16054ln

5115000

ln

5115000

5115000

*20600

16054ln

*20600

16054ln

• When will the population reach 15000?

58688146.34

20600

16054ln

5115000

ln

t

t

For this one, we plug in 15000 for A(t) and solve the t.

The population will reach 15000 in approximately 34.58688 minutes.

• Find a function that models the mass m(t) remaining after t years.

In order to make a model, we need an initial amount and a decay rate. We are given the initial amount is 80 mg. We will have to calculate r. In order to do this, we will use the definition of half-life. Half-life is the number of years it takes to get from one amount to half of that amount. Thus for our problem, we know that in 28 years we will have 40 mg (half of 80). Now we plug all this into the continuous decay formula (same as continuous growth formula).

The half-life of strontium-90 is 28 years. Suppose we have a 80 mg sample.

28*

28*

28*

218040

8040

r

r

r

rt

e

e

e

PeA

r

r

2821

ln

28*21

ln

It will be important tat we do not round our r value in future parts of the problem.

t

eA*

28)5ln(.

80

Model:

• How much of the sample will remain after 100 years?


We will use the model found in the previous part of the problem and plug in 100 for the t.

729500963.6

80100*

28)5ln(.

A

eA

Model:

t

eA*

28)5ln(.

80

In 100 years, there will be 6.729500963 grams

• How long will it take the sample to decay to a mass of 20 mg?


Model:

t

e

e

e

t

t

t

*28

)5ln(.)25ln(.

25.

8020

8020

*28

)5ln(.

*28

)5ln(.

*28

)5ln(.

t

eA*

28)5ln(.

80For this part of the problem, we will use the model found in part 1. We will plug in 20 for A and solve for t.

56

28)5ln(.)25ln(.

t

t

It will take 56 years to get down to 20 mg. You could also have done this just using the definition of half-life.

A wooden artifact from an ancient tomb contains 35% of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.) For this problem we will need to find a model to represent the

decay of carbon-14. This will require that we find r. Once we have r and the model, we will be able to solve the problem.

It is assumed that a living tree has 100% of the carbon-14 in living plants. Once cut down, the amount of carbon-14 starts decreasing (decaying) based on the half-life. This means that we can use the half-life information with the staring point 100. The half-life information tells us that we have 50% of the carbon-14 after 5730 years.



A wooden artifact from an ancient tomb contains 35% of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.)

r

r

e

e

e

r

r

r

5730)5ln(.

5730*)5ln(.

5.10050

10050

5730*

5730*

5730*

t

etA*

5730)5ln(.

100)(

This means that our model is

Finding r using the half-life information

A wooden artifact from an ancient tomb contains 35% of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.)

50428.8678

5730)5ln(.)35ln(.

*5730

)5ln(.)35ln(.

35.

10035

10035

*5730

)5ln(.

*5730

)5ln(.

*5730

)5ln(.

t

t

t

e

e

e

t

t

t

This means that our wooden artifact is approximately 8678.50428 years old.

We use the model from the previous page. Since the ending amount of carbon-14 is 35%, we will plug 35 in for A(t) – the amount after time t. Now we solve for t.

An infectious strain of bacteria increases in number at a relative growth rate of 190% per hour. When a certain critical number of bacteria are present in the bloodstream, a person becomes ill. If a single bacterium infects a person, the critical level is reached in 24 hours. How long will it take for the critical level to be reached if the same person is infected with 10 bacteria?



An infectious strain of bacteria increases in number at a relative growth rate of 190% per hour. When a certain critical number of bacteria are present in the bloodstream, a person becomes ill. If a single bacterium infects a person, the critical level is reached in 24 hours. How long will it take for the critical level to be reached if the same person is infected with 10 bacteria?In order to answer this question, we have to know the relative growth rate and the number of bacteria that is classified as the critical level. In the problem, we are give the relative growth rate (r = 1.9). We know that if you start with 1 bacteria, you will have the critical level number in 24 hours. Thus we can use a model to find the critical level which we need to answer the question.

00000000006365439207

1 24*9.1

A

eA

Thus we need to find out how long it will take to reach this critical number 63654392070000000000. We will plug in this number for A and 10 for P and solve for t

It will take 22.7881131 hours to reach the critical level or for a person to become ill.

An infectious strain of bacteria increases in number at a relative growth rate of 190% per hour. When a certain critical number of bacteria are present in the bloodstream, a person becomes ill. If a single bacterium infects a person, the critical level is reached in 24 hours. How long will it take for the critical level to be reached if the same person is infected with 10 bacteria?

78811311.229.11000000000006365439207

ln

9.11000000000006365439207

ln

1000000000006365439207

1000000000006365439207

*9.1

*9.1

t

t

t

e

e

t

t

Thus we need to find out how long it will take to reach this critical number 63654392070000000000. We will plug in this number for A We will plug in 10 for P since we are asked how long it takes to get to the critical level if we start with 10 bacteria. Now we solve for t.

Exponential Growth and Decay Section 3.5. Objectives Solve word problems requiring exponential...

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