§9.6 Exponential Growth & Decay Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§9.6 Exponential

Growth & Decay

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot College Mathematics

1

Bruce Mayer, PE

BMayer@ChabotCollege.edu • MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt

Review § 9.5

MTH 55



Any QUESTIONS About

• §9.5 → Exponential Equations



Any QUESTIONS About HomeWork

• §9.5 → HW-47


2

Bruce Mayer, PE


Exponential Growth or Decay

 Math Model for “Natural” Growth/Decay:

   A

0 e kt

 A (t) = amount at time t

 A

0

= A (0), the initial amount

 k = relative rate of

• Growth ( k > 0)

• Decay ( k < 0)

 t = time


3

Bruce Mayer, PE


Exponential Growth

 An exponential GROWTH model is a function of the form

A

 



A

0 e

 kt k



0

A ( t )

2 A

0

A



A

0 e

 kt

A

0

 where A

0 is the population at time 0, A ( t ) is the population at time t ,

Doubling time t and k is the exponential growth rate

• The doubling time is the amount of time needed for the population to double in size


4

Bruce Mayer, PE


Exponential Decay

 An exponential DECAY model is a function of the form

A

 



A

0 e

 kt k



0

 where A

0 is the population at time 0, A ( t ) is the population at time t ,

A( t )

A

½A

0



A

0

A

0 e

 kt

Half-life and k is the exponential decay rate t

• The half-life is the amount of time needed for half of the quantity to decay


5

Bruce Mayer, PE


Example



Carbon Emissions

 In 1995, the United States emitted about 1400 million tons of carbon into the atmosphere. In the same year,

China emitted about 850 million tons.

 Suppose the annual rate of growth of the carbon emissions in the United

States and China are 1.5% and 4.5%, respectively.

 After how many years will China be emitting more carbon into Earth’s atmosphere than the United States?


6

Bruce Mayer, PE


Example




 Solution: Assume the

Exponential Growth

A

US

Model Applies. Let t = 0 correspond to 1995, then

 Find t so that

A

China

   A

US

A

China

 

.

 1400 e 0.015

t

 850 e 0.045

t

 i.e., solve for t : 850 e 0.045

t  1400 e 0.015

t

850 e 0.045

t  e  0.015

t  1400 e 0.015

t  e  0.015

t


7

850 e 0.03

t  1400

Bruce Mayer, PE


Example




8

 Solution cont.

e 0.03

t 

28

 So, in less than 17 years from 1995 ln e 0.03

t

17

 ln





(around 2012), under the present assumptions, China

0.03

t  ln





28

17 



28

17 

 will emitmore carbon into the Earth’s atmosphere than the U.S.

Chabot College Mathematics t  ln



 28

17 



0.03

 16.63

Bruce Mayer, PE


Example



Carbon Dating

 A human bone in the Gobi desert is found to contain 30% of the carbon-14 that was originally present. (There are several methods available to determine how much carbon-14 the artifact originally contained.)

 How long ago did the person die?


9

Bruce Mayer, PE


Example



Carbon Dating

 Solution: Assume that the Exponential

Decay Model Applies

 The half-life of 14 C is approximately

5700 years and that means

1

2

A

0

 A

0 e 5700 k


10

Bruce Mayer, PE


Example



Carbon Dating

 Soln cont.

ln





2

1

2 



1

 e 5700 k

 5700 k k 

 Substitute this value for k : ln



 1

2 



5700

  0.0001216

   A

0 e  0.0001216

t


11

Bruce Mayer, PE


Example



Carbon Dating

 Solution cont.

 Since the bone contains 30% of the original carbon-14, have

0.3

A

0

 A

0 e  0.0001216

t

0.3

 e  0.0001216

t t 

    0.0001216

t

 

 9901.09

 0.0001216

t

 Thus by RadioActive 14 C dating estimate that The person died about

9900 years ago


12

Bruce Mayer, PE


Example

 King Tut’s Tomb

 In 1960, a group of specialists from the British

Museum in London investigated whether a piece of art containing organic material found in

Tutankhamun’s tomb had been made during his reign or (as some historians claimed) whether it belonged to an earlier period.

 We know that King Tut died in 1346 B.C. and ruled Egypt for 10 years. What percent of the amount of carbon-14 originally contained in the object should be present in 1960 if the object was made during Tutankhamun’s reign?


13

Bruce Mayer, PE


Example


 Solution: The half-life of carbon-14 is approximately 5700 years and that means

1

2

A

0

 A

0 e 5700 k

 Solving the HalfLife Eqn Solving for k yields k = −0.0001216/year.

 Subbing this Value of k into the Decay

Eqn gives:    A

0 e  0.0001216

t .


14

Bruce Mayer, PE


Example


 Now Let x represent the percent of the original amount of 14 C in the antiquity object that remains after t yrs.

 Using x in the xA

0

 A

0 e  0.0001216

t

Decay Eqn x  e  0.0001216

t

 And The time t that elapsed between

King Tut’s death and 1960 is t = 1960 + 1346 = 3306.


15

Bruce Mayer, PE


Example


 The percent x

1 of the original amount of carbon-14 remaining after 3306 years is x

1

 e  0.0001216



3306

  0.66897

 66.897%

 King Tut ruled Egypt for 10 years, the time t

1 that elapsed from the beginning of his reign to 1960 is t

1

= 3306 + 10 = 3316.


16

Bruce Mayer, PE


Example


 The percent x

2 of the original amount of carbon-14 remaining after 3316 years is x

2

 e  0.0001216



3316

  0.66816

 66.816%

 Thus we conclude, that if the piece of art was made during King Tut’s reign, the amount of carbon-14 remaining in

1960 should be between 66.816% and

66.897%


17

Bruce Mayer, PE


Newton’s Law of Cooling

 The Famous Physicist Isaac Newton found that When a Warm object is placed in a cool convective environment that the temperature of the object, u , u can be modeled by the Decay Eqn

 



T



 u

0



T

 e kt and k



 Where

0

• T ≡ Constant Temperature of the surrounding medium

• u

0

≡ Initial Temperature of the warm object


18

Bruce Mayer, PE



 Some Chabot Engineering Students

Test Newton’s Law by Observing the cooling of Hot Coffee sitting on a table

 The Students Measure the coffee temperature over time, and graph the results


19

Bruce Mayer, PE



 During the course of the experiment the students find

• The Room Temperature, T = 21 °C

• The Initial Temperature, u

0

= 93 °C

• The Water Temperature is 55 °C after 32 minutes → in Fcn notation: u

0

(32min)= 55 °C

 Find Newton’s Cooling Law Model

Equation for this situation


20

Bruce Mayer, PE



 In Cooling Law Eqn Sub for T , u ( t ), u

0 u

 



T



 u

0



T

 e kt

55



C



21



C





93



C



21



C

 e k 32 min

21

 Now Solve for the Time-Constant k

55



C



21



C





72



C

 e k 32 min


34



C





72



C

 e k 32 min

Bruce Mayer, PE



 Divide both Sides by 72 °c

34



C 72



C

 e k 32 min ln

 Next take Natural Log of Both Sides



34 72



 ln

 e k 32 min



 ln



0 .

4722



 k 32 min

22 k

 Solve for k

 ln



0 .

4722

 k

 

0 .

7503


32 min

32 min

 

0 .

02345 min

Bruce Mayer, PE



 Thus the Newton Model for cooling of a cup of hot water in a 21 °C room u

 



T



 u

0



T

 e kt u



21



C





72



C

 e



0 .

02345 t min


23

Bruce Mayer, PE



 The Students then graph the model


24

Bruce Mayer, PE


ReCall Compound Interest

 When the “Principal” amount of money

P

0 is invested at interest rate r , compounded continuously, interest is computed every “instant” and added to the original amount. The balance

Amount A ( t ), after t years, is given by the exponential growth model

A t

 



P

0 e rt

Bruce Mayer, PE



25

Example



Compound Interest

 $45,000 is invested in a continously compounded saving account The $45k grows to $60,743.65 in 5 years. Find the exponential growth function

 We have P

0

= 45,000. Thus the exponential growth function is A (t) =

45,000 e rt , where r must be determined.

 Knowing that for t = 5 we have A (5) =

60,743.65, it is possible to solve for r :


26

Bruce Mayer, PE


Example




 Soln: 60,743.65 = 45,000 e r (5)

60,743.65/45,000 = e r (5)

1.349858889 = e r (5) ln(1.349858889) = ln( e r (5) ) ln(1.349858889) = 5 r ln(1.349858889)/5 = r


27

0.06 ≈ r

Bruce Mayer, PE


Example




 The interest rate is about 0.06, or 6%, compounded continuously.

 Thus the exponential growth function:

A

  

$ 45 k

 e

0 .

06 t


28

Bruce Mayer, PE


The Logistic Growth Model

 Suppose the carrying capacity M of the human population on Earth is 35 billion.

In 1987, the world population was about

5 billion. Use the logistic growth model of P. F. Verhulst to calculate the average rate, k , of growth of the population, given that the population was about 6 billion in 2003.

The Model →

  

M

1  ae  kt


29

Bruce Mayer, PE



 Solution: in This Case

We have t = 0 (1987), P ( t ) = 5 and M = 35 .

5 

35

1  ae  k

 



35

5 1  a

1  a

   35

1  a  7 a  6

 Sub M


30

& a into Eqn:

  

35

1  6 e  kt

.

Bruce Mayer, PE



 Now Solve for k given t = 16 (for 2003)

31

6

 and P ( t ) = 6



1



35

6 e

 k





16 e

 k



16 

29

36

6 e



16



16 k

 ln

29

6 1

  k



16



35

36

6



36 e

36 e



16

 k 

35 k

 

1

16 ln

29

 k 

29 36

 The growth rate was about 1.35%

Bruce Mayer, PE Chabot College Mathematics



0 .

0135


WhiteBoard Work

 Problems From §9.6 Exercise Set

• 18, 20, 28, 30

 The Heat Transfer Behind Newton’s

Law of Cooling

Q conv

 hA s



T s



T





Bruce Mayer, PE



32

All Done for Today

Carbon-14

( 14 C)

Dating


33

Bruce Mayer, PE


Chabot Mathematics

Appendix

r

2  s

2 

 r

 s

 r

 s



Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

–


34

Bruce Mayer, PE


ReCall Logarithmic Laws

 Solving Logarithmic Equations Often

Requires the Use of the Properties of

Logarithms


35

Bruce Mayer, PE


§9.6 Exponential Growth & Decay Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§9.6 Exponential

Growth & Decay

Any QUESTIONS About

Any QUESTIONS About HomeWork

All Done for Today

Chabot Mathematics

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§9.6 Exponential Growth &amp; Decay Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§9.6 Exponential

Growth & Decay

Any QUESTIONS About

Any QUESTIONS About HomeWork

All Done for Today

Chabot Mathematics

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§9.6 Exponential Growth & Decay Chabot Mathematics Bruce Mayer, PE