# MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §9.6 Exponential Growth & Decay

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §9.5 → Exponential Equations  Any QUESTIONS About HomeWork §9.5 → HW-47 9.5 MTH 55

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 3 Bruce Mayer, PE Chabot College Mathematics Exponential Growth or Decay  Math Model for “Natural” Growth/Decay:  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

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 4 Bruce Mayer, PE Chabot College Mathematics Exponential Growth  An exponential GROWTH model is a function of the form  where A 0 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 A0A0 A(t)A(t) t 2A02A0 Doubling time

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 5 Bruce Mayer, PE Chabot College Mathematics Exponential Decay  An exponential DECAY model is a function of the form  where A 0 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 A0A0 A(t) t ½A 0 Half-life

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 6 Bruce Mayer, PE Chabot College Mathematics 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?

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Carbon Emissions  Solution: Assume the Exponential Growth Model Applies. Let t = 0 correspond to 1995, then  Find t so that  i.e., solve for t:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Carbon Emissions  Solution cont.  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.

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 9 Bruce Mayer, PE Chabot College Mathematics 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?

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Carbon Dating  Solution: Assume that the Exponential Decay Model Applies  The half-life of 14 C is approximately 5700 years and that means

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Carbon Dating  Soln cont.  Substitute this value for k:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Carbon Dating  Solution cont.  Since the bone contains 30% of the original carbon-14, have  Thus by RadioActive 14 C dating estimate that The person died about 9900 years ago

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 13 Bruce Mayer, PE Chabot College Mathematics 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?

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  King Tut’s Tomb  Solution: The half-life of carbon-14 is approximately 5700 years and that means  Solving the HalfLife Eqn Solving for k yields k = −0.0001216/year.  Subbing this Value of k into the Decay Eqn gives:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  King Tut’s Tomb  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 Decay Eqn  And The time t that elapsed between King Tut’s death and 1960 is t = 1960 + 1346 = 3306.

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  King Tut’s Tomb  The percent x 1 of the original amount of carbon-14 remaining after 3306 years is  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.

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  King Tut’s Tomb  The percent x 2 of the original amount of carbon-14 remaining after 3316 years is  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%

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 18 Bruce Mayer, PE Chabot College Mathematics 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, can be modeled by the Decay Eqn  Where T ≡ Constant Temperature of the surrounding medium u 0 ≡ Initial Temperature of the warm object

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 19 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  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

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 20 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  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

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 21 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  In Cooling Law Eqn Sub for T, u(t), u 0  Now Solve for the Time-Constant k

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 22 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  Divide both Sides by 72 °c  Next take Natural Log of Both Sides  Solve for k

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 23 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  Thus the Newton Model for cooling of a cup of hot water in a 21 °C room

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 24 Bruce Mayer, PE Chabot College Mathematics Newton’s Law of Cooling  The Students then graph the model

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 25 Bruce Mayer, PE Chabot College Mathematics 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

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 26 Bruce Mayer, PE Chabot College Mathematics 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) = 45,000e 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:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  Compound Interest  Soln: 60,743.65 = 45,000e r(5) 60,743.65/45,000 = e r(5) ln(1.349858889) = ln(e r(5) ) 1.349858889 = e r(5) ln(1.349858889) = 5r ln(1.349858889)/5 = r 0.06 ≈ r

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example  Compound Interest  The interest rate is about 0.06, or 6%, compounded continuously.  Thus the exponential growth function:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 29 Bruce Mayer, PE Chabot College Mathematics 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 →

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 30 Bruce Mayer, PE Chabot College Mathematics The Logistic Growth Model  Solution: in This Case We have t = 0 (1987), P(t) = 5 and M = 35.  Sub M & a into Eqn:

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 31 Bruce Mayer, PE Chabot College Mathematics The Logistic Growth Model  Now Solve for k given t = 16 (for 2003) and P(t) = 6  The growth rate was about 1.35%

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 32 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §9.6 Exercise Set 18, 20, 28, 30  The Heat Transfer Behind Newton’s Law of Cooling

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 33 Bruce Mayer, PE Chabot College Mathematics All Done for Today Carbon-14 ( 14 C) Dating

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 34 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

BMayer@ChabotCollege.edu MTH55_Lec-64_Fa08_sec_9-5b_Logarithmic_Eqns.ppt 35 Bruce Mayer, PE Chabot College Mathematics ReCall Logarithmic Laws  Solving Logarithmic Equations Often Requires the Use of the Properties of Logarithms