CHAPTER ELEVEN 1 Solutions for Section 11.1 1. (a) 80-90 F (b) 60-7 F (c) 60-100 F. predicted high temperature 100 90 80 70 south Topeka north - distance from Topeka Figure 11.1 3. 110 Boise 100 90 80 60 100 Boise 90 80 60 North South West East Figure 11. Figure 11.3 4. The amount of mone spent on beef equals the product of the unit price p and the quantit of beef consumed: M = pc = pf(;p): Thus, we multipl each entr in Table 11.1 on page 3 b the price at the top of the column. This ields Table 11.1. TABLE 11.1 Amount of mone spent on beef ($/household/week) incomenprice 3:00 3:50 4:00 4:50 0 7:95 9:07 10:04 10:94 40 1:4 14:18 15:76 17:46 60 15:33 17:50 19:88 1:78 80 16:05 18:5 0:76 :8 100 17:37 0:0 :40 4:89
CHAPTER ELEVEN /SOLUTONS 5. (a) Beef consumption b households making $0;000/ear is given b Row 1 of Table 11.1 TABLE 11. p 3:00 3:50 4:00 4:50 f(0;p) :65 :59 :51 :43 For households making $0;000/ear, beef consumption decreases as price goes up. (b) Beef consumption b households making $100; 000/ear is given b Row 5 of Table 11.1 TABLE 11.3 p 3:00 3:50 4:00 4:50 f(100;p) 5:79 5:77 5:60 5:53 For households making $100;000/ear, beef consumption also decreases as price goes up. (c) Beef consumption b households when the price of beef is $3:00/lb is given b Column 1 of Table 11.1. TABLE 11.4 0 40 60 80 100 f(;3:00) :65 4:14 5:11 5:35 5:79 When the price of beef is $3:00/lb, beef consumption increases as income increases. (d) Beef consumption b households when the price of beef is $4:00/lb is given b Column 3 of Table 11.1. TABLE 11.5 0 40 60 80 100 f(;4:00) :51 3:94 4:97 5:19 5:60 When the price of beef is $4:00/lb, beef consumption increases as income increases. 6. f the price of beef is held constant, beef consumption for households with various incomes can be read from a fied column in Table 11.1. For eample, the column corresponding to p = 3:00 gives the function h() =f(;3:00); it tells ou how much beef a household with income will bu at $3.00/lb. Looking at the column from the top down, ou can see that it is an increasing function of.thisistrueinevercolumn. This sas that at an fied price for beef, consumption goes up as household income goes up which makes sense. Thus, f is an increasing function of for each value of p.
11.1 SOLUTONS 3 7. Table 11.7 gives the amount M spent on beef per household per week. Thus, the amount the household spent on beef in a ear is 5M. Since the household s annual income is thousand dollars, the proportion of income spent on beef is P = 5M 1000 = 0:05M : Thus, we need to take each entr in Table 11.7, divide it b the income at the left, and multipl b 0.05. Table 11.6 shows the results. TABLE 11.6 spent on beef. Proportion of annual income TABLE 11.7 Amount of mone spent on beef ($/household/week) ncome Price of Beef ($) ($1,000) 3.00 3.50 4.00 4.50 0 0.01 0.04 0.06 0.08 40 0.016 0.018 0.00 0.03 60 0.013 0.015 0.017 0.019 80 0.010 0.01 0.013 0.015 100 0.009 0.011 0.01 0.013 incomenprice 3:00 3:50 4:00 4:50 0 7:95 9:07 10:04 10:94 40 1:4 14:18 15:76 17:46 60 15:33 17:50 19:88 1:78 80 16:05 18:5 0:76 :8 100 17:37 0:0 :40 4:89 8. n the answer to Problem 7 we saw that and in the answer to Problem 4 we saw that P = 0:05 M ; M = pf(;p): Putting the epression for M into the epression for P,gives: P = 0:05 pf(;p) : 9. We have M = f(b;t)=b(1:05) t. M 50 40 30 0 10 0 4 6 8 10 Figure 11.4 B = 30 f(30;t) B = 0 f(0;t) B = 10 f(10;t) t M 50 40 30 0 10 0 f(b;10) t = 10 f(b;5) t = 5 t = 0 f(b;0) 10 0 30 40 50 Figure 11.5 B Figure 11.4 gives the graphs of f as a function of t for B fied at 10, 0, and 30. For each fied B, the function f(b;t) is an increasing function of t. The larger the fied value of B, thelargerf(b;t) is. Figure 11.5 gives the graphs of f as a function of B for t fied at 0, 5, and 10. For each fied t, f(b;t) is an increasing (and in fact linear) function of B. Thelargert is, the larger the slope of the line.
4 CHAPTER ELEVEN /SOLUTONS 10. (a) The dail fuel cost is calculated: Cost = Price per gallon Number of gallons: TABLE 11.8 Cost vs. Price & Gallons Number of Price per gallon (dollars) Gallons 1.00 1.05 1.10 1.15 1.0 1.5 1.30 5 5.00 5.5 5.50 5.75 6.00 6.5 6.50 6 6.00 6.30 6.60 6.90 7.0 7.50 7.80 7 7.00 7.35 7.70 8.05 8.40 8.75 9.10 8 8.00 8.40 8.80 9.0 9.60 10.00 10.40 9 9.00 9.45 9.90 10.35 10.80 11.5 11.70 10 10.00 10.50 11.00 11.50 1.00 1.50 13.00 11 11.00 11.55 1.10 1.65 13.0 13.75 14.30 1 1.00 1.60 13.0 13.80 14.40 15.00 15.60 (b) Note: Table entries ma var depending on the price increase interval chosen. Number of gallons = (Distance in miles) = (30 miles per gallon) Cost = Price per gallon Number of gallons TABLE 11.9 Cost vs. Price & Distance Distance Price per gallon (dollars) (miles) 1.00 1.05 1.10 1.15 1.0 1.5 1.30 100 3.33 3.50 3.67 3.83 4.00 4.17 4.33 150 5.00 5.5 5.50 5.75 6.00 6.5 6.50 00 6.67 7.00 7.33 7.67 8.00 8.33 8.67 50 8.33 8.75 9.17 9.58 10.00 10.4 10.83 300 10.00 10.50 11.00 11.50 1.00 1.50 13.00 350 11.67 1.5 1.83 13.4 14.00 14.58 15.17 400 13.33 14.00 14.67 15.33 16.00 16.67 17.33 450 15.00 15.75 16.50 17.5 18.00 18.75 19.50 Note: Table entries ma var depending on the price increase interval chosen. 11. (a) The acceleration due to gravit decreases as h increases, because the gravitational force gets weaker the farther awa ou are from the planet. (n fact, g is inversel proportional to the square of the distance from the center of the planet.) (b) The acceleration due to gravit increases as m increases. The more massive the planet, the larger the gravitational force. (n fact, g is proportional to m.)
11.1 SOLUTONS 5 1. (a) According to the table in the problem, it feels like,31 F. (b) A wind of 10 mph, according to the table in the problem. (c) About 5:5 mph. Since at a temperature of 5 F, when the wind increases from 5 mph to 10 mph, the temperature adjusted for wind-chill decreases from 1 Fto10 F, we can sa that a 5 mph increase in wind speed causes an 11 F decrease in the temperature adjusted for wind-chill. Thus, each 0:5 mph increase in wind speed brings about a1 F drop in the temperature adjusted for wind-chill. (d) With a wind of 15 mph, approimatel 3:5 F would feel like 0 F. With a 15 mph wind speed, when air temperature drops five degrees from 5 Fto0 F, the temperature adjusted for wind-chill drops 7 degrees from Fto,5 F. We can sa that for ever 1 F decrease in temperature there is about a1:4 F (= 7=5) drop in the temperature ou feel. 13. TABLE 11.10 Temperature adjusted for wind-chill at 0 F Fnmph 5 10 15 0 5 0 F 16 3,5,10,15 TABLE 11.11 0 F Temperature adjusted for wind-chill at Fnmph 5 10 15 0 5 0 F,5,,31,39,44 14. TABLE 11.1 Temperature adjusted for wind-chill at 5 mph (mph)n F 35 30 5 0 15 10 5 0 5mph 33 7 1 16 1 7 0,5 TABLE 11.13 Temperature adjusted for wind-chill at 0 mph (mph)n F 35 30 5 0 15 10 5 0 0 mph 1 4,3,10,17,4,31,39 15. Distance-wise, this wave has half the period of the original wave, that is, the distance from crest to trough at an one moment is halved. Time-wise it also has half the period of the original wave, that is, the time it takes for an one person to complete one ccle is halved. One wa to find the speed of the wave is to compare the position of the crest at time t and at time t + 1. At an time, t, the crest of the wave is at an that makes the quantit So cos(, t)=1, t = k; where k is an integer such that t + k is a non negative integer. Thus at an time t the position of the crest is at = t + k where k is an integer such that t + k is a non negative integer. For t = 0, the crest is at = 0; for t = 1 the crest is at =. Thus the wave is moving at seats/second.
6 CHAPTER ELEVEN /SOLUTONS 16. Since the wave in the tet has formula h(; t)=5 + cos(0:5, t), a wave moving in the opposite direction has formula: h(; t)=5 + cos(0:5 + t): 17. (a) For t = 0, we have = f(; t)=sin, 0 1 = For t = =4, we have = f(; t)= p p Figure 11.6 sin, 0 = For t = =, we have = f(; t)=0 Figure 11.7 Figure 11.8 For t = 3=4, we have = f(; t)=,p sin, 0 = p - Figure 11.9 For t =, wehave = f(; t)=, sin, 0 =,1 Figure 11.10
11.1 SOLUTONS 7 (b) The graphs show an arch of a sine wave which is above the -ais, concave down at t = 0, is straight along the -ais at t = =, and below the -ais, concave up at t =, like a guitar string vibrating up and down. 18. The function = f(; 0)=cos 0 sin = sin gives the displacement of each point of the string when time is held fied at t = 0. The function f(; 1)=cos 1 sin = 0:54 sin gives the displacement of each point of the string at time t = 1. Graphing f(; 0) and f(; 1) gives in each case an arch of the sine curve, the first withamplitude1 and the second withamplitude0:54. For each different fied value of t, we get a different snapshot of the string, each one a sine curve with amplitude given b the value of cos t. The result looks like the sequence of snapshots shown in Figure 11.11. 1 0:54,0:54,1 f(; 0)=sin f(; 1)=0:54 sin Figure 11.11 19. The function f(0;t)=cos t sin 0 = 0 gives the displacement of the left end of the string as time varies. Since that point remains stationar, the displacement is zero. The function f(1;t)=cos t sin 1 = 0:84 cos t gives the displacement of the point at = 1 as time varies. Since cos t oscillates back and forthbetween 1 and,1, this point moves back and forth with maimum displacement of 0:84 in either direction. Notice the maimum displacements are greatest at = = where sin = 1. 0. (a) For g(; t) =cos t sin, our snapshots for fied values of t are still one arch of the sine curve. The amplitudes, which are governed b the cos t factor, now change twice as fast as before. That is, the string is vibrating twice as fast. (b) For = h(; t) =cos t sin, the vibration of the string is more complicated. f we hold t fied at an value, the snapshot now shows one full period, i.e. one crest and one trough, of the sine curve. The magnitude of the sine curve is time dependent, given b cos t. Now the center of the string, = =, remains stationar just like the end points. This is a vibrating string with the center held fied, as shown in Figure 11.1. t = t = 3 4 t = t = 0 t = 4 Figure 11.1: Another vibrating string: = h(; t)=cos t sin