Homework 2- Chapter 10. Problems and Solutions PDF

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Homework 2- Chapter 10. Math homework develops mental agility, combining both memory and logic towards solving problems. In regularly completing math homework, students strengthen their ability to retain large amounts of information to solve math problems....

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HOMEWORK 2

For your answers to questions, make sure to specify the units, when appropriate. (1) The number of species of coastal dune plants in Australia decreases as the latitude, in ◦ S, increases. There are 34 species at 11◦ S and 26 species at 44◦ S . (Note: “◦ S” abbreviates “degree South”. Australia is in the southern hemisphere.) (a) Find a formula for the number, N , of species of coastal dune plants in Australia as a linear function of latitude, l, in ◦ S . (b) Give units for and interpret the slope and vertical intercept of this function. (c) Roughly sketch the graph of this function between l = 11◦S and 44◦ S . (Australia lies between these latitudes.) (2) The following (four) functions give the populations of four towns with time t in years. For each of the following exponential functions, state the following: i. The initial population, i.e., the population in year t = 0. Specify the units. ii. The annual percentage growth rate, i.e., the percentage per year at which the population is change; also specify whether the population is growing or shrinking. Specify units. (a) P (t) = 1000(1.03)t. (b) P (t) = 900(0.90)t. (c) P (t) = 720 ∗ 3t . Hint for ii.: Recall from lecture that the base a = 1 + r, where r is the decimal representation of the change in percentage per unit time. (3) An air-freshener starts out with 30 grams and evaporates. In each of the following cases, write a formula for the quantity, Q, in grams, of air-freshener remaining t days after the start and sketch a graph of the function. (Remember to specify units!) The decrease is (a) 2 grams a day (b) 12% a day. (4) Each of the following functions P (t) = P0 at expresses a quantity as a function of time. For each of the following functions, do the following: i. Find P0 and a (and hence determine the exponential function). ii. Find the initial quantity (i.e., value at t = 0) and the percentage rate of change (i.e., percentage change per unit time). Make sure to specify whether the quantity is increasing (exponential growth) or decreasing (exponential decay). (a) P0 a4 = 18 and P0 a3 = 20. 1

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HOMEWORK 2

(b) P = 1600 when t = 3, and P = 1000 when t = 1. (5) The world’s population increased exponentially from 4.453 billion in 1980 to 5.937 billion in 1998, and continued to increase (approximately) at the same percentage rate between 1998 and 2008. (a) Give a function P (t) describing the world’s population in billions of people t years after 1980. Remember to specify units. (b) Use the function in (a) to calculate the world’s population in 2008. How does this compare with the actual population of 6.771 billion? Answers without work: 8 l + 110 (1) (a) N (l) = − 33 species at l ◦S. Or, using a calculator and ap3 proximating, N (l) = −0.24l + 36.7 species at l ◦ S. Both solutions are acceptable. (b) slope: −0.24 species/◦S. As we move southwards, 0.24 species disappear for each degree south that we pass. Since 4 ∗ 0.24 ≈ 1, another interpretation is that, as we move southwards, approximately one species disappears for each 4◦S that we pass. Either solution is acceptable.

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vertical intercept: (0◦ S, 36.7 species). If Australia were to stretch as far as the equator, then at the equator, there are 36.7 species. (Unfortunately, Australia does not reach the equator.) (a) i. 1000 people. ii. population increasing or growing by 3% per year. (b) i. 900 people. ii. population decreasing or shrinking by 10% per year. (c) i. 720 people. ii. population increasing or growing by 200% per year. This is a good example for the idea of looking for specific phrases in the word problem to figure out the form of the function (e.g., whether a straight line or an exponential function). (a) Q(t) = 30 − 2t grams left in t days after the start. (b) Q(t) = 30(0.88)t grams left in t days after the start. (a) i. P0 = 27.43, a = 0.9. ii. Initial quantity 27.43. Decaying (or decreasing) by 10% per unit time. √ (b) i. P0 = 793.7, a = 1.6 = 1.26. ii. Initial quantity 793.7. Growing (or increasing) by 26% per unit time. (a) P (t) = 4.453 ∗ (1.0161)t billion people t years since 1980. (b) P (28) ≈ 6.96 billion people. It’s close (just slightly bigger) to the actual population....