Exponential models #2 - Questions and Answers

Question 1 What is the difference between linear growth and exponential growth? Answer 1 Linear growth is always at the same rate, and exponential growth increases in speed over time. Answer 2 Linear growth occurs by adding the same numbers, and exponential growth occurs by multiplying the same numbers. Linear growth is a slower form of growth, and exponential is faster form. Linear equations dont Answer 3 If growth is plotted in a diagram and it resembles a straight line, this is called linear growth. Additive processes produce linear growth. Exponential growth is much faster than linear growth. Exponential growth occurs in cases of unrestrained positive feedback. Change is multiplied during each time period. Question 2 Find the exponential model y = ae^bt that fits the points (0, 2) and (4, 3) Answer Use the first point to fina a. y = ae^bx 2 = ae^b*0 2 = a * e^0 2 = a * 1 a = 2 Use the second point to find b. y = 2e^bx 3 = 2e^4b 3/2 = e^4b ln(3/2) = ln(e^4b) 4b = ln(3/2) b = 1/4 * ln(3/2) Write the equation for the exponential model. y = 2 * e^(1/4 * ln(3/2)x) Result: y = 2e^(1/4 ln (3/2)x) Question 3 The number of users on a website has grown exponentially since its launch. After 2 months, there were 300 users. After 4 months there were 30000 users. Find the exponential function that models the number of users x months after the website was launched. Answer Consider the model y = a * b^x. After 2 months, there were 300 users and after 4 months there were 30000 users. 300 = a * b² (1) 30000 = a * b⁴ (2) a * b⁴ / a * b² = 30000 / 300 b² = 100 b = 10 300 = a * (10)² 300 = a * 100 a = 300/100 a = 3 The exponential function that models the number of users x months after the website was launched is y = 3 * (10)ˣ Question 4 The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially. (a) Find a function that models the population t years after 1990. (b) Find the time required for the population to double. (c) Use the function from part (a) to predict the population of California in the year 2010. Look up California's actual population in 2010, and compare. Answer The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially. Let A₀ and r be the initial population & growth rate of population respectively. Then population in California t years after 1990 is, P(t) = A₀e^rt (1) a. Since the population in 1990 is 29.76 million, therefore A₀ = 29.76 In 2000, the population is 33.87 million. That is, P(t) = 33.87 when t = 10 Therefore from equation (1), we get 33.87 = 29.76 * e^10r => e^10r = 1.13810483 => 10r = ln (1.13810483) => r = ln(1.13810483) / 10 => r = 0.012936 Therefore the population t years after 1990 is, P(t) = 29.76e^0.012936t b) The double of 29.76 million is 59.52 million. So put P(t) = 59.52 in P(t) = 29.76e^0.012936t, we get 59.52 = 29.76e^0.012936t => e^0.012936t = 2 => 0.012936t = ln (2) => t = ln (2) / 0.012936 => t = 53.5828 Hence the population will double in 53.5828 years. c) Since 2010-1990=20, therefore to predict the population in 2010, put t = 20 in P(t) = 29.76e^0.012936t, we get P(20) = 29.76e^0.012936(20) = 38.5472 million The population in California in 2010 is 37.3 million, therefore the actual population is approximately 1 million less than the obtained population. Question 5 Newton's law of cooling indicates that the temperature of a warn object will decrease exponentially with time and will approach the temperature of the surrounding air. The temperature T(t) is modeled by T(t) = Ta + (To - Ta)e^-kt. In this model, Ta represents the temperature of the surrounding air, To represents the initial temperature of the object and t is the time after the object starts cooling. The value of k is the cooling rate and is a constant related to the physical properties the object. A cake comes out of the oven at 335F and is placed on a cooling rack in a 70F kitchen. After checking the temperature several minutes later, it is determined that the cooling rate k is 0.050. Write a function that models the temperature T(t) of the cake t minutes after being removed from the oven. Answer Given: The initial temperature (TO) of the cake is 335°F. The tempearture (Ta) of the cake when place in cooling tray is 70°F. The cooling rate (k) is 0.050. Known fact: By the Newton's law of cooling, the temperature is modeled as T(t) = Ta + (To - Ta)e^-kt where, Ta is the temperature surronded by air. To is the initial temperature t is the time after the object starts cooling, k is the cooling rate. Calculation: The function that models the temperature of the cake t minutes after being removed from the oven is computed as follows. Substitute To = 335F, Ta = 70F and k = 0.050 ∈ T(t) = Ta + (To + Ta)e^-kt T(t) = 70 + (335 - 70)e^-(0.050)t = 70 + 265e^-(0.050)t Answer: The function that models the temperature of the cake t minutes after it is removed from the oven is T(t) = 70 + 265e^-(0.050)t