72 Calculus and Structures
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1 72 Calculus and Srucures
2 CHAPTER 5 DISTANCE AND ACCUMULATED CHANGE Calculus and Srucures 73 Copyrigh
3 Chaper 5 DISTANCE AND ACCUMULATED CHANGE 5. DISTANCE a. Consan velociy Le s ake anoher look a Mary s rip o he lake on her bicycle (see Secion 4.) in Fig.. H s = s = 2 = = S s = 4 TH s = Fig. This ime she sars from her house a ime = and posiion s = 2 miles (remember Town Hall is locaed a s = ) and ravels pas her school (S) o he lake (L) a s = 5 (5 miles from Town Hall (TH)) wih a consan speed arriving here a =.25 hours (fifeen minues). We compue he speed of her bike from he familiar formula: rae x ime = disance or L s = s 2 = 5 = Speed = 2.25 miles/hr. where rae can be inerpreed as speed. Alernaively, if we know ha he speed of her bike is 2 mi/hr we can find he disance raveled in fifeen minues by he formula rae x ime = disance, or disance = 2 x.25 = 3 miles. You learned all of his in Jr. H.S. However i does no ell he whole sory. A quaniy called velociy ells us no only her speed bu also in which direcion she is going. If she rides her bike in he direcion of he lake hen her velociy is posiive and he magniude of he velociy is her speed. If she ravels away from he lake a he same speed, her velociy is negaive. If her speed is consan, hen he familiar rae x ime formula becomes, 74 Calculus and Srucures
4 Secion 5. velociy x ime = displacemen, or v x = s 2 s. The displacemen is her posiion a he end of he rip, s 2, as compared o her posiion a he beginning, s. Wih his formula velociy is posiive when raveling from lef o righ and negaive when raveling from righ o lef. Therefore if Mary reurns home from he lake a he same speed ha she raveled o he lake, he rip from he lake o Mary s house has a velociy of -2 mi/hr and (-2) x.25 = -3. She ended her rip a s 2 = 2 and began a s = 5 so he displacemen is s 2 s = 2 5 or -3 miles. Le s now draw a graph of velociy vs ime for he rip o he lake and hen back home (see Fig. 2). Noice ha he displacemen on he rip o he lake equals he signed area under he curve or 2 x.25 = 3 while he displacemen for he rip back home from =.25 o =.5 equals again he signed area under he curve or -3. Furhermore he oal signed from = o =.5 hr. equals 3 + ( 3) =. Bu Mary began her rip a s = 2 (home) and ended i a s 2 =2 (home) for a oal displacemen of s 2 s =, and you will noice ha his is value of he displacemen is also equal o he signed area under he curve in Fig. 2.. v velociy ( mi / hr ) ime ( hr ) Fig. 2 b. Variable velociy We have jus learned somehing of grea imporance o he undersanding of calculus, ha he signed area under he curve of velociy vs ime equals he oal displacemen. Alhough he velociy was consan in he example above, his holds rue even if he velociy varies in ime as he nex example illusraes. Table. Velociy of car every wo seconds Time (sec) Velociy (f/sec) Calculus and Srucures 75
5 Chaper 5 DISTANCE AND ACCUMULATED CHANGE v v Overesimae of disance (area of dark and ligh recangles ) Underesimae of disance (area of dark recangles) Shaded area esimaes disance raveled. Velociy measured every 2 seconds Overesimae of disance (area of dark and ligh recangles ) -. Underesimae of disance (area of dark recangles) Fig. 3a Shaded area esimaes disance raveled. Velociy measured every second Fig. 3b This daa is ploed in Fig. 3a. Unlike he previous rip o he lake, his ime he velociy varies over ime so we canno use he simple formula, rae x ime = disance, which only holds for consan velociies. However, if we approximae he graph of velociy vs ime by a series of consan values ha change heir values every wo seconds as in Fig. 3a, hen we can apply ha simple formula o each segmen and add he resuls. If we ake he values beneah he curve we ge an underesimae of he displacemen, Toal Displacemen = s 2 s (2) + 3(2) + 38(2) + 44(2) + 48(2) = 36 fee Noice ha his is he (5) A L approximaion o v ( ) d. We can also ge an overesimae by aking he values above he curve, This is he Toal Displacemen = (5) A R approximaion o s 3(2) + 38(2) + 44(2) + 48(2) + 5(2) = 4 fee 2 s v ( ) d. In Fig. 3b, he record of velociy vs ime for he car is recorded every second and now he signed area under he curve is a beer approximae o he acual oal displacemen. Alhough in his example he velociy was always posiive (raveling from lef o righ) i would also work if he velociy changed sign o negaive (raveling from righ o lef) where now he inegral would be inerpreed as signed area. So we have a nice applicaion of signed area under a curve: If he funcion is hough of as he velociy of a car hen he signed area under he v vs curve is he oal displacemen of he car over he duraion of he rip. Problem : The graph in Figure 4 shows he velociy record of Mary s rip o he lake on her bike. She sars ou from her house which is locaed a s = 2 mi (remember Town Hall is locaed a s = ) a = minues. If her school (S) is wo miles from her house in he direcion of he lake and he rip o he lake akes minues, 76 Calculus and Srucures
6 Secion 5.2 a) When will she pass her school? b) How far is he lake from her house? 3 c) Wha is he value of v ( ) d? Wha is he meaning of his value in erms of Mary s rip? velociy ( mi / min ) v ime ( min ) Mary s rip o he lake Fig ACCUMULATED CHANGE The velociy of a car or bike is is rae of change of disance wih respec o ime. Wha we have discovered for he signed area under he velociy vs ime curve is rue for he signed area under any rae of change vs ime curve, or, for ha maer, any (rae of change of y) vs x curve where y = f(x). The signed area under he rae of change, R(x), of y vs x, will always equal he oal change in y over he inerval of x under consideraion. To see why his is rue consider he following example. Example : Consider he funcion y = f(x) = x 2 + over he inerval, x 6 wih he y average rae of change, R avg =, compued beween successive values of f(x) a inervals of x x 2. Table 2a x f(x)= x Nex compue: 6 R avg (3) 4 2 R ( x) dx AM (2) +(2) + (2) = 36 = y Now compue he R avg =, x Table 2b beween successive values of f(x) a inervals of x. x f(x)= x R avg Calculus and Srucures 77
7 Chaper 5 DISTANCE AND ACCUMULATED CHANGE Nex compue: 6 (6) R x) dx A () +() + () +() +() +() = 36 = 37 ( M y Finally compue he R avg =, beween successive values of f(x) a inervals of x / 2. x Table 2c x ½ 3/2 2 5/2 3 7/2 4 9/2 5 /2 6 f(x)= x 2 + 5/4 2 R avg / 4 3/ 4 / 2 / 2 I ask he suden o complee his able and show again ha : Wha we observe in hese examples is ha as 6 (2) R x) dx A 36 = 37. ( M x he average rae of change of f(x), R avg = y dy, approaches he insananeous rae of change, R(x)=,, and ( ) A n M x dx R( x) dx = f(6) f() = 37 = 36 as n. This is very significan and we shall see i coming up regularly in our analysis of srucures. Laer we will discover ha i is referred o as he Fundamenal Theorem of Calculus. I works for any funcion, i.e., b a R( x) dx f ( b) f ( a) where R(x) is he insananeous rae of change of f(x) over he inerval a x b. You can see ha we have here a wonderful shorcu o finding he inegral or signed area of he rae of change of f(x) provided f(x) is given. Also if you look closely a he ables above, you will see why his works. Do you see i? b Problem 2: Compue R ( x) dx for he following funcions over he given inervals: a a) f(x) = 2 2 2x x 3x, x 3 ; b) f(x) = 3sin x, x / 2 ; c) f ( x), x x 2 Example 2: If, in hours, is he ime since he sar of a -hour period in which a baceria populaion N() increases a a rae given by, 6 R = R() = millions of baceria per hour, make an underesimae of he oal change in he number of baceria over his period using 4 hen 2. The daa is given in Table 2 and ploed in Fig. 4a and 4b. Table 3. Rae of change of R vs every wo hours. 78 Calculus and Srucures
8 Secion 5.2 ( hours) R = R() rae ( million / hour ) rae ( million / hour ) rae ( million / hour ) n = 5 4 n = 2 n = (hours ) Fig. 4a (hours ) Fig. 4b (hours ) Fig. 4c You can easily deermine ha an underesimae of he oal change in he number of baceria over (5) he hour period, N() N(), is approximaed by AL 252. million baceria, an () approximaion o R ( ) d when 4. The approximaion when 2 is A L (You should verify hese resuls). When (no shown in he Table) as shown in Fig. 4c an even more accurae approximaion o he oal change in he number of baceria is found o be, () A L 37. million baceria. Noice ha as n ges larger and ges smaller he area of he shaded recangles approaches he signed area under he curve more closely. Noe: In his example, since we do no know wha N() is, we canno use he shorcu o find he inegral. We can now make he general saemen: N( 2 ) N( ) = 2 R ( ) d Calculus and Srucures 79
9 Chaper 5 DISTANCE AND ACCUMULATED CHANGE Problems 8 Calculus and Srucures
10 Chaper 5 Problems Calculus and Srucures 8
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