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

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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 MTH 55

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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

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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

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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

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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?

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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:

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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.

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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?

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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

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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:

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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

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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?

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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 = − /year. Subbing this Value of k into the Decay Eqn gives:

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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 = = 3306.

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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 = = 3316.

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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 % and %

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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

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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

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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

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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

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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

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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

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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

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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

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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, 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:

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

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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:

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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 The Model →

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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:

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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%

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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

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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

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

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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

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