Section 6.8 Exponential Models; Newton's Law of Cooling; Logistic Models

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1 Section 6.8 Exponential Models; Newton's Law of Cooling; Logistic Models 197 Objective #1: Find Equations of Populations that Obey the Law of Uninhibited Growth. In the last section, we saw that when interest is compounded continuously, we can find the future value in the account using the formula A = Pe rt. This is an example of "uninhibited growth." Similarly, the formula for the present value is P = Ae rt is an example of "uninhibited decay." In such examples, the amount of growth or decay depends on the size of the account. There are many physical phenomena such as cell division, growth of a population, and the growth of bacteria that exhibit uninhibited growth and the amount of growth depends on the size of the population. The uninhibited growth model can be used to model the growth of the population before the limitations of the environment start restrict the growth. Uninhibited Growth Let N o be the initial population and k be a positive constant representing the growth rate. If the growth is uninhibited, the population N after N 0 time t will be N(t) = N 0 e kt. We can also say that the population grows exponentially or experiences exponential growth. Solve the following: Ex. 1 The population of a town increases according to the model P(t) = 3200e t where t is the time in years after a) Determine the initial population. b) What is the growth rate of the population? c) Predict the population in the town in the year d) How long will it take the population to reach 5000 people? e) What is the doubling time of the population? Solution: a) Since P(0) = 3200e (0) = 3200e 0 = 3200, the initial population is 3200 people in the year b) Since k = , then the growth rate is 3.21%.

2 198 Ex. 2 c) The year 2030 corresponds to t = 20 years (think: ). Thus, evaluating P at 20 yields: P(20) = 3200e (20) = 3200e ( ) 6081 people. So, the population will be 6081 people in d) We will need to solve the equation P(t) = 5000: 3200e t = 5000 (divide both sides by 3200) e t = (rewrite as a logarithm) t = ln(1.5625) (divide both sides by ) t = ln(1.5625) 13.9 years It will take 13.9 years for the population to reach e) When the population doubles, P(t) = Solving yields: 3200e t = 6400 (divide both sides by 3200) e t = 2 (rewrite as a logarithm) t = ln(2) (divide both sides by ) t = ln(2) 21.6 years The population doubles every 21.6 years. A colony of bacteria increases according to the law of uninhibited growth. a) Let N be the number of bacteria after time t in hours. Express N as a function of t. b) If the number of bacteria doubles in size in five hours, write the function using the appropriate growth constant. c) How long will it take for the number of bacteria to be 32 times its original size? d) How long will it take for the number of bacteria to triple? Solution: a) Since the growth is uninhibited, the function is N(t) = N 0 e kt. b) Since the population doubles in five hours, then N(5) = 2N 0. Thus, 2N 0 = N 0 e k(5). Solving for k yields: 2N 0 = N 0 e 5k (divide both sides by N 0 ) 2 = e 5k (rewrite as a logarithm) 5k = ln(2) (divide both sides by 5) k = ln(2) Hence, the equation becomes N(t) = N 0 e t.

3 199 c) Since the population doubles every 5 hours, then t = 0 N(0) = N 0 t = 5 N(5) = 2N 0 t = 10 N(10) = 2 2N 0 = 4N 0 t = 15 N(15) = 2 4N 0 = 8N 0 t = 20 N(20) = 2 8N 0 = 16N 0 t = 25 N(25) = 2 16N 0 = 32N 0 Thus, it will take 25 hours. d) When the population triple, N(t) = 3N 0. Thus, 3N 0 = N 0 e t. Solving for t yields: 3N 0 = N 0 e t (divide both sides by N 0 ) 3 = e t (rewrite as a logarithm) t = ln(3) (divide both sides by ) t = ln(3) hours It will take approximately 7.9 hours for the population to triple. Objective #2: Find Equations of Populations that Obey the Law of Decay. Finding the present value of an account that is expected to grow to amount A in t years when interest is compounded continuously is an example uninhibited decay. Again, the amount of decay depends on the size of the account will be in t years. Several phenomena such as the decomposition of chemical in a solution and radioactive decay exhibit uninhibited decay. The uninhibited decay model can be used to model the decay of the population before the limitations of the environment start restrict the decay. Uninhibited Decay Let N o be the initial population and k be a negative constant representing the decay rate. If the decay is uninhibited, the population N after N 0 time t will be N(t) = N 0 e kt. We can also say that the population decays exponentially or experiences exponential decay.

4 200 The half-life of a radioactive substance is the amount of time required for the substance to decay to half of its original size. For instance, live organisms contain two types of carbon, carbon 12 ( 12 C) which is stable and carbon 14 ( 14 C) which is radioactive. After the organism dies, it cannot accumulate any carbon so the carbon 14 in the dead organism begins to decay exponentially while the amount of carbon 12 does not change. Since carbon 14 has a half-life of about 5730 years, scientists then can compare the amount of carbon 14 to the amount of carbon 12 in a dead organism to determine how long ago the organism died. Solve the following: Ex. 3 The amount of a sample of radioactive substance remaining after t years is given by Q(t) = Q o e 0.005t. a) What is the decay rate? b) What is the half-life? c) At the end of years, 4,000 grams of the substance remained. How many grams were initially present? Solution: a) Since k = 0.005, the decay rate is 0.5%. b) When the substance is half it size, Q(t) = 0.5Q o. Solving yields: Q o e 0.005t = 0.5Q o (divide both sides by Q 0 ) e 0.005t = 0.5 (rewrite as a logarithm) 0.005t = ln(0.5) (divide both sides by 0.005) t = ln(0.5) years The half-life is approximately years. c) When t = years, Q(t) = 4,000 grams. Plugging into Q(t) = Q o e 0.005t yields: Q() = Q o e 0.005() = Q o e 1.5 = 4000 Now, we can solve for Q o : Q o e 1.5 = 4000 Q o ( ) = 4000 Q o 17, grams Initially, there was about 17, grams of the substance present. Ex. 4 How old is a wooden tool if the ratio of 14 C to 12 C is 1 the ratio 7 found in the atmosphere if the half-life of 14 C is 5,730 years?

5 Solution: When the substance is half it size, A(t) = 0.5A o and t = 5730 years (Note, the book uses 5600 years for the half-life of 14 C). Solving yields: A o e k(5730) = 0.5A o (divide both sides by A 0 ) e 5730k = 0.5 (rewrite as a logarithm) 5730k = ln(0.5) (divide both sides by 5730) k = ln(0.5) Thus, our formula becomes A(t) = A 0 e t Since the ratio of 14 C to 12 C is 1 7, then A(t) = 1 7 A 0. Plugging and solving yields: 1 7 A 0 = A 0 e t (divide both sides by A 0 ) 1 7 = e t (rewrite as a logarithm) t = ln( 1 7 ) t = ln( 1 7 ) ,100 years. The tool is approximately 16,100 years old. Objective #3: Newton's Law of Cooling. The temperature of a heated object will exponentially decrease to the temperature of the surrounding environment. This is known as Newton's Law of Cooling. It is often used to determine the time of death of a body found at a crime scene. Newton's Law of Cooling Let T e be the temperature of the environment, T 0 be the initial temperature of the heated object, k be a negative constant. The temperature of the object at time t is T(t) = T e + (T 0 T e )e kt Solve the following: Ex. 5 A pizza is removed from an oven that was set at 425 F at 5:30 pm. After 3 minutes, the temperature of the pizza is F. If the temperature of the room is 75, find the following: a) After 10 minutes, what will the temperature of the pizza be? b) How much time should elapse after the pizza is removed from the oven until it is safe to eat ( 110 F). c) What do you notice about the temperature as time passes? 201

6 Solution: Since T 0 = 425 F and T e = 75 F, our equation becomes: T(t) = T e + (T 0 T e )e kt = 75 + (425 75)e kt = e kt After the three minutes, the temperature is F, thus T(3) =. Plugging in and solving for k yields: = e k(3) (subtract 75 from both sides) 225 = 350e 3k (divide by 350) 9 14 = e3k (write as a logarithm) 3k = ln( 9 ) (divide both sides by 3) 14 k = ln( 9 14 ) = Thus, our equation is T(t) = e t a) To find the temperature after 10 minutes, replace t by 10: T(10) = e (10) = e After ten minutes, the temperature will be F. b) Replace T(t) by 110 F and solve for T: 110 = e t (subtract 75 from both sides) 35 = 350e t (divide by 350) 0.1 = e t (write as a logarithm) t = ln(0.1) (divide both sides by ) t = ln(0.1) = minutes It will take about minutes until the pizza is safe to eat. c) As t increases without bound, e t approaches 0 so T(t) approaches (0) = 75 which is the room temperature. 202 Objective #4: Use Logistic Models In the real world, most situations that experience exponential growth or decay actually do not grow or decay uninhibitedly. Eventually, the factors in the environment such as resources and space will limit the growth and decay. The logistic curve can be used in such situations to model the data. With the logistic model, the population will approach a fixed value c called the "carrying capacity" as time t increases without bound for growth models and as time t decreases without bound in the decay models.

7 Logistic Model Let a, b, and c be nonzero constant with a and c both being positive. Then, the population P after time t is given by c 203 P(t) = 1+ae bt where the model is a growth model if b > 0 and a decay model if b < 0. The value c is the carrying capacity. A point called an "inflection point" where the graph changes from curving up to curving down or vice-versa. For logistic models, the inflection point will occur when the population reaches half of the carrying capacity. Growth Model (b > 0) Decay Model (b < 0) P(t) P(t) y = c y = c (0, P(0)) 1 2 c 1 2 c (0, P(0)) t t Logistic curves have the following properties: Properties of Logistics Curves 1) The domain is all real numbers. The range is (0, c) where c is the carrying capacity. 2) The y-intercept is (0, P(0)). There are no x-intercepts. 3) y = 0 and y = c are the two horizontal asymptotes. 4) If b > 0, then P(t) is increasing on its domain. If b < 0, then P(t) is decreasing on its domain. 5) The inflection point is where P(t) = 1 2 c. 6) The function has a "continuous" smooth graph with no corners and no gaps.

8 Solve the following: 204 Ex. 6 In re-introducing a duck population, several female-male pairs are released into a habitat. The population of the ducks is expected to grow according to the model: P(t) = where t is measured in years. 1+29e 0.438t a) Find the carrying capacity and growth rate. b) Determine the number of ducks released. c) Find the population of ducks after 5 years. d) When will the duck population reach 200 ducks? e) How long will it take the duck population to reach half of its carrying capacity? Solution: a) Since b = and c =, the carrying capacity is ducks and the growth rate is 43.8%. b) P(0) = = 1+29e 0.438(0) 1+29 = 10 There were 10 ducks released. c) P(5) = = 1+29e 0.438(5) After five years. there will be approximately 71 ducks. d) Set P(t) = 200 and solving yields: = e 0.438t (multiply both sides by e 0.438t ) = 200(1 + 29e 0.438t ) (divide both sides by 200) 1.5 = e 0.438t (subtract 1 from both sides) 0.5 = 29e 0.438t (divide both sides by 29) = e 0.438t (replace by 1 58 and write as a logarithm) 0.438t = ln( 1 ) (divide both sides by 0.438) 58 t = ln( 1 58 ) = years It will take about 9.3 years for the population to reach 200 ducks.

9 Ex. 7 e) Set P(t) = 150 and solving yields: = e 0.438t (multiply both sides by e 0.438t ) = 150(1 + 29e 0.438t ) (divide both sides by 150) 2 = e 0.438t (subtract 1 from both sides) 1 = 29e 0.438t (divide both sides by 29) 1 29 = e 0.438t (write as a logarithm) t = ln( 1 ) (divide both sides by 0.438) 29 t = ln( 1 29 ) = years It will take about 7.7 years for the population to reach half of the carrying capacity. Due to unusually warm and wet spring, the monarch butterfly population exploded at the beginning of the summer. As the summer progresses and climate conditions return to more of a normal pattern, the monarch butterfly population (in thousands) is expected to decay according to the model: P(t) = e 0.358t where t is measured in days. a) What is the decay rate? b) Determine the initial population. c) What is the population after 30 days? d) How many days will it take the population to reach half of its size? Solution: a) Since b = 0.358, the decay rate is 35.8%. b) P(0) = = e 0.358(0) = thousand The initial population was about 35,200,000 butterflies. c) P(30) = = e 0.358(30) thousand After 30 days, the population is about 6,370 butterflies.

10 206 d) Setting P(t) = 35200/2 = and solving yields: = (multiply both sides by e 0.358t ) e 0.358t = 17600( e 0.358t ) (divide both sides by 17600) = e0.358t (subtract 1 from both sides) = 0.136e0.358t (divide both sides by 0.136) = e0.358t (write as a logarithm) t = ln ( (divide both sides by 0.358) 0.136) t = 14 ln = days It will take about 6.25 days for the butterfly population to reach half of its initial size.