3.6 Derivatives as Rates of Change
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1 3.6 Derivaives as Raes of Change Problem 1 John is walking along a sraigh pah. His posiion a he ime >0 is given by s = f(). He sars a =0from his house (f(0) = 0) and he graph of f is given below. (a) Describe he moion of John as precisely as you can. Soluion: Le us assume ha f() gives he posiion of John from his house a ime. Iff() < 0, hen John is wes of his house; if f() > 0, hen John is eas of his house; if f() =0, hen John is in fron of his house. For he firs hour, ha is for 0 apple apple 1 John walks eas away from his house. For he second hour, ha is, for 1 apple apple 2, John urns around and sars walking back wes. However, he does no walk all he way back o his house. 1
2 For he hird hour, ha is, for 2 apple apple 3, John realizes he needs o pick up an apple from he sore, which is way eas of his house, for Elaine. So he urns around again and walks eas o he sore. For he fourh hour, ha is, for 3, John needs o drop he apple o a Elaine s house, which is wes of his house. He urns wes and walks pas his house unil he makes i o Elaine s house. (b) When is John s velociy zero? Wha is happening o John a hose imes? Soluion: John s velociy is zero a imes =0,=1,=2, and =3(he imes where he slopes of he angen lines are zero). A hese imes, John changes direcion since he sign of he slope of he angen line o he graph of f changes hen. (c) When is John moving in he posiive direcion? When is John he furhes in he posiive direcion and he furhes in he negaive direcion? Soluion: John is moving in he posiive direcion (in his sory, eas) during imes in he region (0, 1) [ (2, 3). John is furhes in he posiive direcion a ime =3and furhes in he negaive direcion a (approximaely) ime =3.5. (d) When is John s velociy increasing? And when is i decreasing? When is John going a maximum velociy? Soluion: John s velociy is increasing on he ime inervals (0, 0.3), (1.5, 2.5). John s velociy is decreasing on he ime inervals (0.3, 1.5), (2.5, 3.5). John is moving he fases a ime =3.5, bu ha maximizes his speed, no his velociy, because he is moving in he negaive direcion a ha ime. I appears ha he maximizes his velociy a abou =2.5. Problem 2 Suppose ha a sone is hrown verically upward from a cli on Mars wih an iniial velociy of 24 f/s from a heigh of 192 f. The heigh s of he sone above he ground afer seconds is given by s() = (a) Deermine he velociy and acceleraion of he sone afer seconds. Soluion: The velociy v() is: v() =s 0 () = The acceleraion a() is: a() =v 0 () =s 00 () = 12. (b) Wha is he greaes heigh of he sone and when does i occur? Wha are he velociy and acceleraion a ha ime? 2
3 Soluion: Since he funcion s() is di ereniable everywhere (i is a polynomial), he maximum heigh mus occur a a ime when he velociy is 0. So we solve: v() = = 0 12 = 24 I is easy o check ha v() > 0 for 0 apple <2 and v() < 0 for 2 <, and so he greaes heigh of he sone really does occur a ime =2. The greaes heigh is s(2) = = 216 f. For he second quesion we have already seen ha v(2) = 0 f/sec, and we have a consan acceleraion of 12 f/sec 2.Soa(2) = 12 f/sec 2. =2 (c) When does he sone hi he ground? Wha are he velociy and acceleraion a ha ime? Soluion: The sone his he ground when s() =0. So we solve: s() = = 0 6( ) = 0 6( 8)( + 4) = 0 Since we are only considering 0, wemushave =8. A his insan, v(8) = = 72 f/sec and a(8) = 12 f/sec 2. Problem 3 (a) True or False: If he acceleraion of an objec is consan, hen is velociy is consan. Soluion: This is false in general, and is rue if and only if he acceleraion is 0. Le a denoe your non-zero consan acceleraion. If a>0henyour velociy is increasing, and if a<0henyour velociy is decreasing. See he accompanying picure for when a>0 (noe ha he unis for v() and a() are no he same!) (b) True or False: A moving objec can have negaive acceleraion and increasing speed. 3
4 Soluion: True. If your velociy is negaive (which means ha you are moving in he negaive direcion) hen a negaive acceleraion increases he magniude (or absolue value) of he velociy. Bu he magniude of he velociy is he speed. Problem 4 Suppose he average cos of producing 300 cellphones is 2 dollars per cellphone and he marginal cos a x = 300 is 1.85 dollars per cellphone. Inerpre hese coss. Soluion: The firs 300 cellphones coss, on average, 2 dollars o produce. The 301s cellphone coss 1.85 dollars o produce. Problem 5 The oal number of people, N who have conraced a common cold by a ime days afer is oubreak on an island is given by N = N() =, e 0.1 (a) Evaluae and inerpre he limi lim!1 N() Soluion: cold lim N() = lim = In he long run, people will ge he!1! e 0.1 (b) How long will i ake for he number of people who have conraced he cold o reach 40,000? Soluion: N() = = e = 40000( e 0.1 ) e 0.1 = e 0.1 =5 1 e 0.1 = =ln(.04) = ln(0.04) 0.1 = I will ake 33 days for he number of people who have conraced he cold o reach 40,000. (c) The graph of he funcion N on he inerval [0, 100] is given below. Skech (as bes you can) he graph of is derivaive, N 0 (). 4
5 Soluion: (d) Calculae N 0 (). Wha does N 0 () represen? Soluion: N (10)e e () = ( e 0.1 ) 2 = ( e 0.1 ) 2. N 0 () represens he insananeous growh rae of he number of people who have conraced he cold a he ime. (e) Evaluae and inerpre he limi lim!1 N 0 () Soluion: lim N e 0.1 () = lim!1!1 ( e 0.1 ) 2 = (0) ( (0)) 2 = 0 1 =0. In he long run, he number of people who have conraced he cold will sabilize. There will be no growh in he long run. (f) Find he average growh rae of he number of people who have conraced he disease during he ime inerval [5, 6] (or during he sixh day afer he oubreak). Soluion: AV RG = N(6) N(5) = N(6) N(5) = e 0.1(6) e 0.1(5) (g) Find he insananeous growh rae of he number of people who have conraced he disease for =5 (round o a whole number). Soluion: N 0 (5) = 0.1(5) e = 319 people per day ( e 0.1(5) ) 2 Problem 6 An oil ank is o be drained for cleaning. There are V () gallons of oil lef in he ank minues afer he draining began, where V () = 45(60 ) 2. (a) Find he average rae a which he oil drians during he firs 15 minues. 5
6 Soluion: AR = V (15) V (0) 15 0 = 45(60 15)2 45(60 0) = 45(45)2 45(60) ( ) = 15 = 4725 gallons per minue (b) Find he average rae a which he oil drains during he ime inerval [10, 15]. Soluion: AR = V (15) V (10) = 45(60 15)2 45(60 10) = 45(45)2 45(50) ( ) = 5 = 4275 gallons per minue (c) Find he rae a which he oil is flowing ou of he ank 15 minues afer he draining began. Soluion: apple d V 0 () = d 45(60 )2 ) =15 apple = 45 d d (60 )2 =15 apple d = 45 d ( ) apple = 45( ) =15 = 45( (15)) = 45( ) = 45( 90) = 4050 gallons per minue =15 (d) Find he average rae, AR, a which he oil drains during he ime inerval: (i) [15 +, 15], if 1 < <0 6
7 Soluion: (ii) [15, 15 + ], if 0 < <1 Soluion: AR = AR = V (15) V (15 + ) 15 (15 + = 45(60 15)2 45(60 (15 + )) 2 = 45(45)2 + 45(45 ) 2 = 45( ( )2 ) = 45( 90 +( )2 ) = 45( 90 + ) =( ) gallons per minue V (15 + ) V (15) = 45(60 (15 + ))2 45(60 15) 2 = 45(45 )2 45(45) 2 = 45( ( )2 ) 2025 = 45( 90 +( )2 ) = 45( 90 + ) =( ) gallons per minue (e) Use he resul in par (d) o find he limi lim!0 AR( ). Wha does his limi represen? Soluion: lim AR( ) = lim( ) = 4050 gallons per minue. This is he insananeous rae a which he oil is flowing ou of he ank 15 minues afer he draining!0!0 began. Problem 7 The graph of h() represens he heigh in fee of waer in a pool a ime minues. 7
8 (a) During wha ime inervals is he heigh of waer in he pool increasing? Decreasing? Soluion: The heigh of he waer in he pool is increasing when he funcion h is increasing or has posiive slope. This is (0, 3), (7, 10). The heigh of he waer is decreasing when he funcion h is decreasing or has negaive slope. This is (3, 7), (10, 12). (b) Assume he rae of change of he waer in he pool is 0 when =0. For wha oher values of is he rae of change of he waer in he pool 0? Soluion: =3, 7, 10. The rae of he change of waer in he pool is 0 when he slope of h is zero. This is a (c) If h() represens he heigh of he fee in waer a ime minues. Wha funcion represens he rae of change of he heigh over ime? Soluion: h 0 (), he derivaive of h(). h 0 () ells us abou he slope, or rae of change, of heigh wih respec o ime. (d) Skech a graph of he rae of change of he heigh over ime, h 0 (). Soluion: We ll learn how o skech graphs of derivaives more precisely in 4.3 bu for now we can use informaion abou he slope o help us graph h 0 (). Firs, we ll plo he poins where he slope of he graph of h() is zero. This is a =0, 3, 7, 10. 8
9 Nex, we ll use informaion abou wheher he slope is posiive or negaive o add ails o he zeros o indicae if he derivaive should be posive or negaive. Remember, he derivaive ells us abou he slope. The slope of h is posive on (0, 3), (7, 10), soh 0 () is posiive on (0, 3), (7, 10). The slope of h is negaive on (3, 7), (10, 12), soh 0 () is negaive on (3, 7), (10, 12). Finally, we ll connec our ails. If you ve already aken calculus, you may recall ha we know when our deriviive h 0 has is highes and lowes poins based on where h changes concaviy. If you haven had calculus, don worry, we ll ge here in chaper 4. For now, jus guess where you hink hese migh be. 9
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