Does modeling pay off badly?
Mathematics HTL 2, textbook
127 4.2 Modeling Growth Using First Order Linear Difference Equations 562 Herbert takes out a loan of € 20,000. The annual interest rate is 5.75%. At the end of each year, Herbert pays € 2500 back to the bank. How big is Herbert's debt level after t years? a. Model this problem using a difference equation. b. Report the debt level at the end of the first 10 years. 563 Anna has a savings account with which she receives 2% interest annually on her balance. At the end of each year she pays in € 1,000. How big is your balance after t years? Describe this task by a difference equation. Find the first 10 terms of your solution. 564 The number of fish in a fish pond would grow by 340% annually if the pond were not fished. At the beginning of the first year there are 100 fish in this pond. a. Under these assumptions, describe the growth of the fish stock using a difference equation and give the first 10 terms of its solution. b. Is such a model realistic in practice? Justify. What are the problems in assuming exponential growth? 565 (Continuation of exercise 564) How does the fish population change if one assumes that annually (after the closed season, i.e. after the fish have spawned) a. 200 fish, b. the double of the previous year's fish stock are fished out of the pond? Give the appropriate difference equation and calculate the first 10 terms of its solution. Modeling of limited growth and decline processes 566 60 goats currently live on a small island. The island's food supply is enough for a maximum of 100 goats. As the number of goats increases, the diet deteriorates and the goats reproduce more slowly. We therefore assume that the increase in the number of goats per year is proportional to the difference between 100 and the number of goats in the previous year. How many goats live on the island after t years? Also draw a diagram. We denote the number of goats after t years with y t. Then, according to our assumption about the increase in the number of goats, yt + 1 = yt + r (100 - yt) so yt + 1 = (1 - r) yt + r 100, where r is a real number between 0 and 1 that we have to choose. The solution of the difference equation yt + 1 = (1 - r) yt + r 100 with y 0 = 60 is the sequence k (1 - r) n 60 + 100r ((1 - r) n - 1) ___ (1 - r) - 1 l = k 100 - 40 (1 - r) nl. If r = 1 _ 3, then after n years there will be 100 - 40 · 2 2 _ 3 3 n goats. After one year it's about 73, after two years it's about 82 and after five years it's about 95 goats. We are talking about limited growth (the number of goats) in this case. If we describe a process by the difference equation yt + 1 = yt + r · (K - yt) with y 0 = a, where r, K and a are real numbers with 0 ≤ a ≤ K and 0
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