[email protected] MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical

[email protected] • MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§9.6 Exponential§9.6 ExponentialGrowth & DecayGrowth & Decay



Review §Review §

Any QUESTIONS About• §9.5 → Exponential Equations

Any QUESTIONS About HomeWork• §9.5 → HW-47

9.5 MTH 55



Exponential Growth or DecayExponential Growth or Decay

Math Model for “Natural” Growth/Decay:

A t A0ekt

A(t) = amount at time t

A0 = A(0), the initial amount

k = relative rate of • Growth (k > 0)

• Decay (k < 0)

t = time



Exponential GrowthExponential Growth

An exponential GROWTH model is a function of the form

00 keAtA kt

where A0 is the population at time 0, A(t) is the population at 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

A0

A(t)

t

2A0

Doubling time

kteAA 0



Exponential DecayExponential Decay

An exponential DECAY model is a function of the form

00 keAtA kt

where A0 is the population at time 0, A(t) is the population at time t, and k is the exponential decay rate • The half-life is the amount of time needed for half of the quantity to decay

A0

A(t)

t½A0

Half-life

kteAA 0



Example Example Carbon Emissions 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?



Example Example Carbon Emissions Carbon Emissions

Solution: Assume the Exponential Growth Model Applies. Let t = 0 correspond to 1995, then

AUS 1400e0.015t

AChina 850e0.045t

Find t so that AChina t AUS t .

i.e., solve for t: 850e0.045t 1400e0.015t

850e0.045t e 0.015t 1400e0.015t e 0.015t

850e0.03t 1400



Example Example Carbon Emissions Carbon Emissions

Solution cont. e0.03t 28

17

lne0.03t ln28

17

0.03t ln28

17

t ln

2817

0.0316.63

So, in less than 17 years from 1995 (around 2012), under the present assumptions, China will emitmore carbon into the Earth’s atmosphere than the U.S.



Example Example Carbon Dating 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 didthe person die?




Solution: Assume that the Exponential Decay Model Applies

The half-life of 14C is approximately 5700 years and that means

1

2A0 A0e

5700k




Solncont.

1

2e5700k

ln1

2

5700k

k ln

12

5700 0.0001216

Substitute this value for k: A t A0e

0.0001216t




Solution cont. Since the bone

contains 30% of the original carbon-14, have

0.3A0 A0e 0.0001216t

0.3 e 0.0001216t

ln 0.3 0.0001216t

t ln 0.3

0.0001216t9901.09

Thus by RadioActive 14C dating estimate that The person died about 9900 years ago



Example Example King Tut’s Tomb 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?



Example Example King Tut’s Tomb King Tut’s Tomb

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

1

2A0 A0e

5700k

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

Subbing this Value of k into the Decay Eqn gives:




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

Using x in theDecay Eqn

xA0 A0e 0.0001216t

x e 0.0001216t

And The time t that elapsed between King Tut’s death and 1960 is t = 1960 + 1346 = 3306.




The percent x1 of the original amount of carbon-14 remaining after 3306 years is

x1 e 0.0001216 3306 0.66897 66.897%

King Tut ruled Egypt for 10 years, the time t1 that elapsed from the beginning of his reign to 1960 is t1 = 3306 + 10 = 3316.




The percent x2 of the original amount of carbon-14 remaining after 3316 years is

x2 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%



Newton’s Law of CoolingNewton’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, can be modeled by the Decay Eqn

0and0 keTuTtu kt

Where• T ≡ Constant Temperature of the surrounding medium

• u0 ≡ Initial Temperature of the warm object




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




During the course of the experiment the students find• The Room Temperature, T = 21 °C

• The Initial Temperature, u0 = 93 °C

• The Water Temperature is 55 °C after 32 minutes → in Fcn notation: u0(32min)= 55 °C

Find Newton’s Cooling Law Model Equation for this situation




In Cooling Law Eqn Sub for T, u(t), u0

kteTuTtu 0

min3221932155 keCCCC Now Solve for the Time-Constant k

min32722155 keCCC

min327234 keCC




Divide both Sides by 72 °c

min.lnlnln min 32472207234 32 kek

min327234 keCC Next take Natural Log of Both Sides

Solve for k

min.min.

min.ln

0234503275030

3247220

k

k




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

kteTuTtu 0

teCCtu min

02345.0

7221




The Students then graph the model



ReCall Compound InterestReCall Compound Interest

When the “Principal” amount of money P0 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

rtePtA 0



Example Example Compound Interest 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 P0 = 45,000. Thus the exponential growth function is A(t) = 45,000ert, where r must be determined.

Knowing that for t = 5 we have A(5) = 60,743.65, it is possible to solve for r:




Soln: 60,743.65 = 45,000er(5)

60,743.65/45,000 = er(5)

ln(1.349858889) = ln(er(5))

1.349858889 = er(5)

ln(1.349858889) = 5r

ln(1.349858889)/5 = r

0.06 ≈ r




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

Thus the exponential growth function:

tektA 06045 .$



The Logistic Growth ModelThe 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 →

P t M

1 ae kt




Solution: in This CaseWe have t = 0 (1987), P(t) = 5 and M = 35.

5 35

1 ae k 0 35

1 a5 1 a 35

1 a 7

a 6

P t 35

1 6e kt . Sub M & a into Eqn:




Now Solve for k given t = 16 (for 2003) and P(t) = 6

The growth rate was about 1.35%

2936

35366

35616

61

356

16

16

16

16

k

k

k

k

e

e

e

e

0135.036

29ln

16

1

36

29ln16

36

2916

k

k

e k



WhiteBoard WorkWhiteBoard Work

Problems From §9.6 Exercise Set• 18, 20, 28, 30

The Heat Transfer Behind Newton’sLaw of Cooling

TThAQ ssconv



All Done for TodayAll Done for Today

Carbon-14(14C)

Dating



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



ReCall Logarithmic LawsReCall Logarithmic Laws

Solving Logarithmic Equations Often Requires the Use of the Properties of Logarithms

Documents

[email protected] MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical