Individual Written Homework Assignment 3 Solutions

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1 Individual Written Homework Assignment 3 Solutions February 1, 2011 Assignment: pp 37-48, problems 1, 2, 3, 5, 15, 20, 21, 24, 28. And Section 1.4; problems 1, 2, 3, page 59. Note: All graphs from the homework solutions have been omitted. Use graphing calculator for solutions. 1. Sketch the graph of each of the following functions. Label each axis, and mark a scale of units on it. For each line that you draw, indicate: (i) its slope; (ii) its y-intercept; (iii) its x-intercept (where it crosses the x-axis). Solution: (a) y = 1 2x 7 x + 3 (b) y = 2 3 (c) 5x + 3y = 12 (i) m = 1 2 (i) m = 2 3 (i) m = 5 3 (ii) y int : (0, 3) (ii) y int : (0, 7 ) (ii) y int : (0, 4) 3 (iii) x int : (6, 0) (iii) x int : ( 7 2, 0) (iii) x int : (12 5, 0) 2. Graph the following functions. Put labels and scales on the axes. (a) V = 0.3Z 1; Solution: The graph of this function is a line with slope m = 0.3, y int : (0, 1) and x int : ( 10 3, 0). (b) W = 600 P 2. Solution: The graph of this function is a downward pointing parabola with vertex (0, 600) and x int : (± 600, 0). 3. Sketch the graph of each of the following functions. Put labels and scales on the axes. For each graph that you draw, indicate: i) its y-intercept; ii) its x-intercept(s). 1

2 For part (d) you will need the quadratic formula: x = b ± b 2 4ac 2a for the roots of the quadratic equation ax 2 + bx + c = 0. Solutions: (a) y = x 2 (c) y = (x + 1) 2 (i) y int : (0, 0) (i) y int : (0, 1) (ii) x int : (0, 0) (ii) x int : ( 1, 0) (b) y = x (d) y = 3x 2 + x 1 (i) y int : (0, 1) (i) y int : (0, 1) (ii) Graph has no x int (ii) x int : ( 1 ± 13, 0) 6 The next four questions refer to these functions: c(x, y) = 17 (1) j(z) = z (2) r(u) = 1 u (3) D(p, q) = p q (4) S(y) = y 2 (5) Q(v) = 2v + 1 (6) 3v 6 5 if x > 0 H(x) = x 2 if 0 x < 6 (7) 29 x if 6 x T (x, y) = r(x) + Q(y) (8) 5. True or false. Give reasons for your answers: if you say true, explain why; if you say false, give an example that shows why it is false. (a) For every non-zero number x, r(r(x)) = j(x). Solution: true. In order to show this is true, we will start at the left hand side, write down true statements, and eventually end up on the right hand side. Note that we want to consider an arbitrary x since the statement we re going to prove is for every non-zero x. As listed above, r(x) = 1 x Proof: r(r(x)) = 1 r(x) = 1 1 x and j(x) = x. = x = j(x). 2

3 (b) If a > 1,then S(a) > 1. Solution: true. We approach this problem with the same method we used in (a). Proof: Suppose a > 1. We can square both sides to get a 2 > 1 2 hence a 2 > 1. But s(a) = a 2 (plug a for y into equation 5). Therefore if a > 1 then s(a) > 1. (c) If a > b,then S(a) > S(b). Solution: false. In order to show that this statement is false we need to provide a counter example. Counter example: Let a = 4 and b = 2. Certainly 4 < 2 but ( 4) 2 = 16, ( 2) 2 = 4 and (d) For all real numbers a and b, S(a + b) = S(a) + S(b). Solution: false. Again, we need to provide a counter example. Counter example: Let a = 2 and b = 3. But 25 13, so S(2 + 3) S(2) + S(3). S(2 + 3) = (2 + 3) 2 = 5 2 = 25 S(2) + S(3) = = = 13 (e) For all real numbers a, b, and c, D((D(a, b)), c) = D(a, (D(b, c))). Solution: false. Again, we need to provide a counter example. Counter example: Let a = 5, b = 3, and c = 1. D((D(5, 3)), 1) = D((5 3), 1) = D(2, 1) = 2 1 = 1 D(5, (D(3, 1))) = D(5, (3 1)) = D(5, 2) = 5 2 = 3 But 1 3 so D((D(5, 3)), 1) D(5, (D(3, 1))). 15. Go back to the three functions given in problem 1. For each function, choose an initial value x 0 for x, find the corresponding value y 0 for y, and express the function in the form y y 0 = m(x x 0 ). Solution: for each function in problem 1 plug a point in for x 0, y 0 and plug in the slope for m. 20. Thermometers. There are two scales in common use to measure the temperature, called Fahrenheit degrees and the Celsius degrees. Let F and C, respectively, be the temperature on each of these scales. Each of these quantities is a linear function of the other; the relation between them in determined by the following table: physical measurement C F freezing point of water 0 32 boiling point of water

4 (a) Which represents a larger change in temperature, a Celsius degree or a Fahrenheit degree? Solution: celsius. (b) How many Fahrenheit degrees does it take to make the temperature go up one Celsius degree? How many Celsius degrees does it take to make it go up one Fahrenheit degree? Solution: this problem has a variety of correct solutions. (c) What is the multiplier m in the equation F = m C? What is the multiplier µ in the equation C = µ F? (The symbol µ is the Greek letter mu.) What is the relation between µ and m? Solution: m = 9 5 and µ = 5. Note that m is the multiplicative inverse of µ. 9 (d) Express F as a linear function of C. Graph this function. Put scales and labels on the axes. Indicate clearly the slope of the graph and its vertical intercept. Solution: F = 9 5 C + 32 with m = 9 5 and vertical intercept:(0, 32). (e) Express C as a linear function of F and graph this function. How are the graphs in parts (d) and (e) related? Give a clear and detailed explanation. Solution: C = 5 9 (F 32) with m = 5 9 F and C are inverses of each other. and vertical intercept:(0, 17.78). Note that (f) Is there any temperature that has the same reading on the two temperature scales? What is it? Does the temperature of the air ever reach this value? Where? Solution: This occurs at 40. The air does sometimes reach this value, but rarely and in very cold climates. 21. The Greenhouse Effect. Note: This problem was not graded, but solutions have been included. The concentration of carbon dioxide (CO 2 ) in the atmosphere is increasing. The concentration is measured in parts per million (PPM). Records kept at the South Pole show an increase of.8 PPM per year during the 1960s. (a) At that rate, how many years does it take for the concentration to increase by 5 PPM; by 15 PPM? Solution: We know the concentration increases 0.8 PPM per year, so for it to increase by 5 PPM it will take = 6.25 years. Similarly, for it to 5 PPM 0.8 PPM per year 15 PPM increase by 15 PPM it will take = years. 0.8 PPM per year (b) At the beginning of 1960 the concentration was about 316 PPM. What would it be at the beginning of 1970; at the beginning of 1980? PPM Solution: At the beginning of 1970 it would be 316 PPM+(0.8 ) (10 years) = PPM per year 4

5 324 PPM. Similarly for 1980 it would be 332 PPM. (c) Draw a graph that shows CO 2 concentration as a function of the time since Put scales on the axes and label everything clearly. Solution: Graph omitted. Graph has m = 0.8 and y-int: (0, 320). (d) The actual CO 2 concentration at the South Pole was 324 PPM at the beginning of 1970 and 338 PPM at the beginning of Plot these values on your graph, and compare them to your calculated values. Solution: = 14 PPM in 10 years, so the line between these two points has slope m = 14 PPM = year. (e) Using the actual concentrations in1970 and1980, calculate a new rate of increase in concentration. Using that rate, estimate what the increase in CO 2 concentration was between 1970 and Estimate the CO2 concentration at the beginning of Solution: from (d) we get that the new rate of increase in concentration is 1.4 PPM year. Thus the concentration in 1990 = concentration in (1.4) (20) = (1.4) (20) = 352 PPM. (f) Using the rate of.8 PPM per year that held during the 1960s, determine how many years before 1960 there would have been no carbon dioxide at all in the atmosphere. Solution: Since the concentration was 316 PPM in 1960 and it is increasing at a rate of 0.8 PPM per year, we know that = 395 years ago, or in year 1565 there would have been no carbon dioxide at all in the atmosphere. 24. In the equation P persons per year = k P, above, explain why the units for k are. The person number k is called the per capita growth rate. ( Per capit means per perso - per headó, literally.) Solution: By knowing the units for P and P, we can solve for k to see that the units are persons per year in fact person. This makes sense since P is proportional to the number of people in the existing population. 28. Note: 28 was not graded and solutions have been omitted because this problem is very similar in nature to others in this solutions set. A Model of an Orchard If an apple orchard occupies one acre of land, how many trees should it contain so as to produce the largest apple crop? This is an example of an optimization problem. The word optimum means best possible especially, the best under a given set of conditions. These exercises seek an optimum by analyzing a simple mathematical model of the orchard. The model is the function that describes how the total yield depends on the number of trees. 5

6 An immediate impulse is just to plant a lot of trees, on the principle: more trees, more apples. But there is a catch: if there are too many trees in a single acre, they crowd together. Each tree then gets less sunlight and nutrients, so it produces fewer apples. For example, the relation between the yield per tree, Y, and the number of trees, N, may be like that shown in the graph drawn on the left, below. Note: Graphs omitted. When there are only a few trees, they don t get in each other s way, and they produce at the maximum level-say, 750 pounds per tree. Hence the graph starts off level. At some point, the trees become too crowded to produce anything! In between, the yield per tree drops off as shown by the curved middle part of the graph. We want to choose N so that the total yield, T, will be as large as possible. We have T (N) = Y (N) N, but since we don t know Y (N) very precisely, it is difficult to analyze T (N). To help, let s replace Y (N) by the approximation shown in the graph on the right. Now carry out an analysis using this graph to represent Y (N). 1. Find a formula for the straight segment of the new graph of Y (N) on the interval 40 N 180. What is the formula for T (N) on the same interval? Solution: From the graphs given we see that the slope of the graph of Y (N) between 0 and 40 is = = 5.4. So we can write an equation for Y (N) on the inverval 0 N 40 using m = 5.4 and the point (40, 750). Thus we get Y (N) 750 = 5.4(N 40) in point slope form and Y (N) = 5.4N in slope-intercept form. Then since T (N) = Y (N) N, we get that T (N) = 5.3N N. 2. What are the formulas for T (N) when0 N 40 and when 180 N? Graph T as a function of N. Describe the graph in words. Solution: From question 1 we see that for 0 N 40, T (N) = 5.3N N. From the graphs given in the text, it s easy to see that for N 180, T (N) = 0 since Y (N) = 0 on that interval. If we graph T as a function of N, we get a downwards pointing parabola with its vertex at roughly (90, 43000). Then at N = 180, we have the horizontal line T = What is the maximum possible total yield T? For which N is this maximum attained? Solution: If we re only considering T on the interval [0, 40] and [180], then T obtains its maximum at N = 40 where T (40) = 5.3(40) (40) = 30, 000. If we broaden our perspective and consider T on the interval [0,180] then T obtains its maximum at the vertex of the parabola at (90, 43000) which is evident from the graph. 6