Week #9  The Integral Section 5.1


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1 Week #9  The Inegral Secion 5.1 From Calculus, Single Variable by HughesHalle, Gleason, McCallum e. al. Copyrigh 005 by John Wiley & Sons, Inc. This maerial is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS 7. For ime,, in hours, 0 1, a bug is crawling a a velociy, v, in meers/hour given by v = Use = 0. o esimae he disance ha he bug crawls during his hour. Find an overesimae and an underesimae. Then average he wo o ge a new esimae. Using = 0., we need he following values v() = /(1+0) = /(1+0.) = 5/ /(1+0.4) = 5/ /(1+0.6) = 5/ /(1+0.8) = 5/ /(1+1.0) = 5/10 The lefhand esimae is The righhand esimae is v(0) + v(0.) v(0.8) = (0.)[1 + 5/6 + 5/7 + 5/8 + 5/9] meers v(0.) + v(0.4) v(1.0) = (0.)[5/6 + 5/7 + 5/8 + 5/9 + 5/10] meers Because he bug is moving slower and slower (see he able), he lefhand sum is an overesimae of he disance raveled, while he righhand sum is an underesimae. The average of he wo esimaes is meers. This a reasonable esimae of he disance he bug has raveled. 1
2 9. The velociy of a car is f() = 5 meers/sec. Use a graph of f() o find he exac disance raveled by he car, in meers, from = 0 o = 10 seconds. A graph of he simple linear funcion is shown below, along wih he area (marked in grey) beneah he graph f() 0 10 The area under he recangle represens he disance raveled, and i is = 50 The car ravels 50 meers in he firs 10 seconds. 16. Roger runs a marahon. His friend Jeff rides behind him on a bicycle and clocks his speed every 15 minues. Roger sars ou srong, bu afer an hour and a half he is so exhaused ha he has o sop. Jeff s daa follow: Time since sar (min) Speed (mph) (c) How ofen would Jeff have needed o measure Roger s speed in order o find lower and upper esimaes wihin 0.1 mile of he acual disance he ran? (a,b) See soluions o Quiz Prep problems (c) The difference beween Roger s pace a he beginning and he end of his run is 1 mph. If he ime beween he measuremens is h, hen he difference beween he upper and lower esimaes is 1h. We wan 1h < 0.1, so h < hours = 30 seconds Thus Jeff would have o measure Roger s pace every 30 seconds. QUIZ PREPARATION PROBLEMS 15. A suden is speeding down Roue 11 in his fancy red Porsche when his radar sysem warns him of an obsacle 400 fee ahead. He immediaely applies he brakes, sars o slow down, and spos a skunk in he road direcly ahead of him. The black box in he Porsche records he car s speed every wo seconds, producing he following able. The speed decreases hroughou he 10 seconds i akes o sop, alhough no necessarily a a uniform rae. Time since brakes applied (sec) Speed (f/sec)
3 (a) Wha is your bes esimae of he oal disance he suden s car raveled before coming o res? (b) Which one of he following saemens can you jusify from he informaion given? (i). The car sopped before geing o he skunk. (ii). The black box daa is inconclusive. The skunk may or may no have been hi. (iii). The skunk was hi by he car. (a) Our bes esimae would be he average of he lef and righhand sums. Since he inerval beween measuremens is = seconds, he lefhand sum is [v(0) + v() + v() + v(6) + v(8)] = [ ] = 530 fee The righhand sum is [v() + v() + v(6) + v(8) + v(10)] = [ ] = 330 fee Since he driver was braking coninuously, he velociy should have been decreasing he whole ime. This means ha he lefhand sum is an overesimae of he sopping disance while he righhand sum is an underesimae. A more accurae esimae would be o average he wo numbers: 430 fee. (b) All we can be sure of is ha he disance raveled lies beween he upper and lower esimaes calculae above. In oher words, all he blackbox daa ells us is ha he car raveled beween 330 and 530 fee before sopping. As a resul, we can be compleely sure if i hi he skunk, which has 400 fee away when he braking began. 16. Roger runs a marahon. His friend Jeff rides behind him on a bicycle and clocks his speed every 15 minues. Roger sars ou srong, bu afer an hour and a half he is so exhaused ha he has o sop. Jeff s daa follow: Time since sar (min) Speed (mph) (a) Assuming ha Roger s speed is never increasing, give upper and lower esimaes for he disance Roger ran during he firs half hour. (b) Give upper and lower esimaes for he disance Roger ran in oal during he enire hour and a half. (a) Noe ha 15 minues equals 0.5 hours. Lefhand esimae = (0.5) [ ] = 5.75 miles. Righhand esimae = (0.5) [ ] = 5.5 miles. The lefhand esimae is an upper esimae, while he righhand esimae is a lower esimae. 3
4 (b) Lefhand esimae = (0.5) [ ] = 14.5 miles. Righhand esimae = (0.5) [ ] = 11.5 miles. Again, he lefhand esimae is an upper esimae, while he righhand esimae is a lower esimae. 3. A car iniially going 50 f/sec brakes a a consan rae (consan negaive acceleraion), coming o a sop in 5 seconds. (a) Graph he velociy from = 0 o = 5. (b) How far does he car ravel? (c) How far does he car ravel if is iniial velociy is doubled, bu i brakes a he same consan rae? (a) The acceleraion is consan, so he velociy graph is linear, hrough he poins ( = 0, v = 50) and ( = 5,v = 0) velociy 0 5 (b) The disance raveled is he same as he area under he graph of he velociy. The region is a riangle of base 5 and heigh 50, so he area is = 15. Thus he disance raveled is 15 fee. (c) The slope of he graph of he velociy funcion is he same, so he riangular region under i has wice he aliude and wice he base (i akes wice as much ime o sop). See he graph below. This scaling produces a riangle ha is 4 imes larger han he original, so he sopping disance is 4 imes longer, or 4 15 = 500 fee velociy Two cars sar a he same ime and ravel in he same direcion along a sraigh road. Figure 5.13 gives he velociy, v, of each car as a funcion of ime,. Which car: (a) Aains he larger maximum velociy? (b) Sops firs? 4
5 (c) Travels farher? Figure 5.13 (a) Car A has he larges maximum velociy because he peak of car A s velociy curve is higher han he peak of B s. (b) Car A sops firs because he curve represening is velociy his zero (on he axis) firs. (c) Car B ravels farher because he area under car B s velociy curve is he larger. 6. Two cars ravel in he same direcion along a sraigh road. Figure 5.14 shows he velociy, v, of each car a ime. Car B sars hours afer car A and car B reaches a maximum velociy of 50 km/hr. (a) For approximaely how long does each car ravel? (b) Esimae car A s maximum velociy. (c) Approximaely how far does each car ravel? Figure 5.14 (a) Since car B sars a =, he ick marks on he horizonal axis (which we assume are equally spaced) are hours apar. Thus car B sops a = 6 and ravels for 4 hours. (b) Car A s maximum velociy is approximaely wice ha of car B, or 100 km/hr. (c) The disance raveled is given by he area under he velociy graph. Using he formula for he area of a riangle, he disances are given approximaely by Car A disance = 1 Base Heigh = = 400 km Car B disance = 1 Base Heigh = = 100 km 5
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