Chaper 5 Eponenial and Logarihmic Funcions Chaper 5 Prerequisie Skills Chaper 5 Prerequisie Skills Quesion 1 Page 50 a) b) c) Answers may vary. For eample: The equaion of he inverse is y = log since log =. Chaper 5 Prerequisie Skills Quesion Page 50 a) y = : Domain: (, ) Range: (0, ) y = log : Domain: (0, ) Range: (, ) b) y = : -inercep: none y-inercep: 1 y = log : -inercep: 1 y-inercep: none c) y = : The funcion is increasing on he inerval:! (, ) y = log : The funcion is increasing on he inerval: y! (, ) d) y = : y = 0 y = log : = 0 MHR Calculus and Vecors 1 Soluions 477
Chaper 5 Prerequisie Skills Quesion 3 Page 50 a) b) c) 3 = 8 3.5 =! 11.3 1.5 =!.8 d) log10 =! 3.3 e) log 7 =!.8 f) log 4.5 =!. Chaper 5 Prerequisie Skills Quesion 4 Page 50 a) b) c) 3 = 8 3.5 =! 11.3 1.5 =!.8 d) log10 =! 3.3 e) log 7 =!.8 f) log 4.5 =!. Chaper 5 Prerequisie Skills Quesion 5 Page 50 a) b) c) d) y = 3 ( ) = 3 y = ( ) = 4 4 y = ( ) = y = ( ) =!! 4 MHR Calculus and Vecors 1 Soluions 478
Chaper 5 Prerequisie Skills Quesion 6 Page 50 a) b) c) d) e) log10 5 log 5 = log 10 =!.3 log10 66 log4 66 = log 4 10 =! 3.0 log1010 log310 = log 3 10 =!.096 log10 7 log 7 = log 10 =!.807 log10 75 log3 75 = log 3 10 =! 3.930! 1 log 0.1 log = % 10 log 5 =! 1.431 10 f) 5 $ 10 log! 1 log 0.5 = 4 log 0.5 10 g) 1 $ % 10 = h) log10 5 log0.5 5 = log 0.5 10 =!.3 Chaper 5 Prerequisie Skills Quesion 7 Page 50 a) b) ( )( ) 3! 3 h k hk = h k 3 3 3 6 ( a )( ab ) = ( a )( a b ) = a b 5 6 MHR Calculus and Vecors 1 Soluions 479
c) d) e) f) g) 3!! 6 ( )( ) ( )( ) y y = ( y ) y 3 3 4 1 1 8u v u 4uv v 1 = y 3! =! 1 11 18 3!!! 6 ( g )( gh ) = ( g )( g h ) 1 = 6 h + ( ) = + 4 3 + 4! 3 = 4 ( ) =!! 4 ( ) = 5 6 h) a b ab = a b! 1 Chaper 5 Prerequisie Skills Quesion 8 Page 50 a) log 5 + log = log(5! ) = log10 = 1! 4 log 4 log 3 = log $ % 3 ' = log 8 b) = 3! 50 log 50 log 0.08 = log $ % 0.08 ' = log 65 c) 5 5 5 = 4 5 d) 3 log(0.01) = 3(log 0.01) =! 6 MHR Calculus and Vecors 1 Soluions 480
e) 3 1 1 log 1000 + log 100 = (log1000) + (log100) 3 3 = + 3 13 = 6 f) log + log 5 = log(! 5) = log10 = Chaper 5 Prerequisie Skills Quesion 9 Page 50! a a) log a log a = log$ % a '! 1 = log$ % ' b) c) d) e)! aba ab '! a = log$ % b ' log ab + log a log ab = log$ % 8! a 4log a 4log a = log$ 4 % a = log a 4 = 4log a 6 3 3 6 3log a b + 3log ab = log[( a b )( a b )] = ' 9 9 log( a b ) = 9log ab log a b + log b = log[( a b)( b )] = 3 log(4 a b ) MHR Calculus and Vecors 1 Soluions 481
Chaper 5 Prerequisie Skills Quesion 10 Page 51 a) b) c) + 1 = 4 = ( ) = + =! + 1 + 1 4 = 64 + 1 3 4 = (4 ) + 1 = 3 = 1! 5 3 = 7 1! 5 3 3 = (3 ) 3! 5 = 13 = 4 d) log! log = log 5 log = log(5 ) = 10 e) log 5 + log = 3 log 5 + log = log1000 log 5 = log1000 5 = 1000 = 00 f)! 3log 5 = 3log = 3log( 5) = 3log10 = 3 Chaper 5 Prerequisie Skills Quesion 11 Page 51 a) = 1.06 log = log1.06 log = log1.06 =! 11.9 MHR Calculus and Vecors 1 Soluions 48
b) c) 50 = 5 log50 = log5 log50 = log5 =! 1.! 1 10 = $ %! 1 log10 = log $ % log10 =! 1 log $ % =! ' 3.3! 4 d) 75 = 5() 75 log$ % = log() 5 ' 1 log $ % 3! = ' 4 log =! 6.3! 4 Chaper 5 Prerequisie Skills Quesion 1 Page 51 a) i) 100 baceria b) C ii) 00 baceria iii) 400 baceria c) Answers may vary. For eample: The formula for an eponenial growh funcion is P = P 0 a, where P is he baceria populaion, P 0 is he iniial baceria populaion, a is he eponenial base or growh rae, and is he ime for he populaion o grow, in his case, doubling ime. The iniial populaion of baceria is 50, so P 0 = 50. The populaion doubles eponenially, so a =. The populaion doubles afer 3 days, so = number of days 3. 3 Therefore, he correc equaion is P = 50(). MHR Calculus and Vecors 1 Soluions 483
Chaper 5 Prerequisie Skills Quesion 13 Page 51 a) Time (min) Amoun Remaining (g) 0 100 5 50 10 5 15 1.5 0 6.5! b) A() = 100 1 $ % 5 c) i)! 1 A( 30) = 100 $ % 30 5 6! 1 = 100 $ % = 1.565 Afer half an hour, he amoun remaining is 1.565 g. ii) Half a day is 1 60 min = 70 min.! 1 A( 70) = 100 $ % 70 5 4 =! 4.484' 10 Afer half a day, he amoun remaining is 4.484! 10 4 g. MHR Calculus and Vecors 1 Soluions 484
Chaper 5 Secion 1 Raes of Change and he Number e Chaper 5 Secion 1 Quesion 1 Page 56 a) b) c) d) MHR Calculus and Vecors 1 Soluions 485
Chaper 5 Secion 1 Quesion Page 56 a) b) c) d) Chaper 5 Secion 1 Quesion 3 Page 56 a) f ( ) = : {!!} b) No c) No f ( ) = e : {!!} MHR Calculus and Vecors 1 Soluions 486
Chaper 5 Secion 1 Quesion 4 Page 57 a) B. The graph of he derivaive of a quadraic funcion is a sraigh line. b) C. The graph of he derivaive of a line is of he form y = a, where a is a consan. c) D. The graph of he derivaive of an eponenial funcion is also an eponenial funcion. d) A. The graph of he derivaive of a cubic funcion is a quadraic funcion. Chaper 5 Secion 1 Quesion 5 Page 57 a) b > e b) 0 b < e Chaper 5 Secion 1 Quesion 6 Page 58 a) b) Answers may vary. For eample: The graph of he rae of change of y in he -ais. c)! 1 y = $ % will be a compression and a reflecion of he graph of MHR Calculus and Vecors 1 Soluions 487
Chaper 5 Secion 1 Quesion 7 Page 58 Answers may vary. For eample: If 0 < b < 1, he graph of y = b will be above he -ais and he graph of he rae of change of his funcion will be below he -ais. If b > 1, he graph of y = b and he graph of he rae of change of his funcion will boh be above he -ais. Chaper 5 Secion 1 Quesion 8 Page 58 a) b) c) Answers may vary. For eample: The graph of he combined funcion g( ) will be a horizonal sraigh line. MHR Calculus and Vecors 1 Soluions 488
d) Answers may vary. For eample: The graph of g( ) = ln 4 is a consan funcion. Therefore he graph is a horizonal sraigh line. Chaper 5 Secion 1 Quesion 9 Page 58 a) Answers may vary. For eample: No. The shape of he graph of g will no change. The shape of he graph of g will be a horizonal sraigh line. If he base is oher han 4, he graph will be parallel o he graph of g and shifed up or down depending on he numerical value of he base. If he value of he base is greaer han 4, he graph will be shifed up. If he value of he base is greaer han 1 and less han 4, he graph will be shifed down, bu will sill be above he -ais. If he value of he base is greaer han 0 and less han 1, he graph will be shifed down and will be below he -ais. b) The graph is he line g( ) = ln e, which is he horizonal sraigh line f!( ) g( ) = where g( ) = 1. f ( ) Chaper 5 Secion 1 Quesion 10 Page 58 Soluions o he Achievemen Checks are shown in he Teacher s Resource. Chaper 5 Secion 1 Quesion 11 Page 58 a) Answers may vary. For eample: The graph of he funcion g! ( ) will be he horizonal sraigh line, y = 0. b) Answers may vary. For eample: If f ( ) = ln, hen f!( ) = ln, and g( ) =. g( ) = ln, which is jus a consan so g! ( ) = 0. This is applicable for any base. Therefore, he graph of g! ( ) will be he graph of y = 0. The graph of g( ) is a consan funcion. The derivaive of a consan funcion is 0. MHR Calculus and Vecors 1 Soluions 489
c) Answers may vary. For eample: The funcion g( ) will be a horizonal sraigh line for any value of b, b > 0 and he derivaive funcion, g! ( ), will be g! ( ) = 0, for any value of b, b > 0, since he derivaive of any consan funcion is he horizonal sraigh line y = 0. Chaper 5 Secion 1 Quesion 1 Page 58 a) {!!} b) {y < 0 < 1, y!!} c) As he value of c increases, he graph of he funcion is shifed o he righ. Chaper 5 Secion 1 Quesion 13 Page 58 Answers may vary. For eample: Some answers include: Leonhard Euler; 177; e is used in probabiliy Chaper 5 Secion 1 Quesion 14 Page 58 B Chaper 5 Secion 1 Quesion 15 Page 58 D MHR Calculus and Vecors 1 Soluions 490
Chaper 5 Secion The Naural Logarihm Chaper 5 Secion Quesion 1 Page 65 a) b) i) Domain: {!!} ii) Range: {y < 0, y!!} iii) -inerceps: none; y-inercep: 1 iv) Horizonal asympoe: y = 0 v) Decreasing on he inerval (, ) vi) Maimum or minimum poins: none vii) Poins of inflecion: none Chaper 5 Secion Quesion Page 65 a) b) i) Domain: { > 0,!!} ii) Range: {y!!} iii) -inercep: 1; y-inerceps: none iv) Verical asympoe: = 0 v) Decreasing on he inerval (0, ) MHR Calculus and Vecors 1 Soluions 491
vi) Maimum or minimum poins: none vii) Poins of inflecion: none Chaper 5 Secion Quesion 3 Page 65 Answers may vary. For eample: No. The funcion f ( ) and he funcion g( ) are no inverse funcions. They are no reflecions of each oher in he line y =. Chaper 5 Secion Quesion 4 Page 65 Answers may vary. For eample: a) e 4!= 55 b) e 5!= 150 c) e!= 7.5 d) e!!= 0.1 Chaper 5 Secion Quesion 5 Page 65 a) b) c) d) 4 e =! 54.598 5 e =! 148.413 e =! 7.389 e! =! 0.135 Chaper 5 Secion Quesion 6 Page 65 a) ln 7 =! 1.946 b) ln 00 =! 5.98 c)! 1 ln $ = % 4! 1.386 d) ln(! 4) is undefined Chaper 5 Secion Quesion 7 Page 65 Answers may vary. For eample: The value of ln 0 is, which is undefined. Also, he domain of he funcion (0, ), so when = 0 he funcion is undefined. y = ln is he inerval MHR Calculus and Vecors 1 Soluions 49
Chaper 5 Secion Quesion 8 Page 65 a) ln( ) = ln e e = b) ln( e ) + ln( e ) = ln( e ) = ln e = c) d) e ln( + 1) = + 1 ln(3 ) e e = e (ln( )) 3 ( ln ) = 6 Chaper 5 Secion Quesion 9 Page 65 a) e = 5 ln e = ln5 =! 1.609 4 b) 1000 = 0e e 4 = 50 4 ln e = ln 50 = 4ln50 =! 15.648 c) ln( e ) = 0.44 = 0.44 d) 7.316 = e ln( ) = 7.316 = 3.658 Chaper 5 Secion Quesion 10 Page 65 a) 3 = 15 ln 3 = ln15 ln15 = ln 3 =!.465 MHR Calculus and Vecors 1 Soluions 493
b) 3 = 15 log3 = log15 log15 = log3 =!.465 c) Answers may vary. For eample: The value of can be found by aking naural logarihms of boh sides of he equaion or by aking common logarihms of boh sides of he equaion. Chaper 5 Secion Quesion 11 Page 65 4 a) V ( ) = V e ma V! ma 4 = Vmae 1 $ % $ % ' ' 1 =! 4ln $ % ' =!.8 I will ake.8 s.! 4 ln e = ln! 4 b) V ( ) = V e ma! V! ma 4 = Vmae 10! 1 4 ln $ e % = ln $ % ' 10 ' 1 =! 4ln $ % 10 ' =! 9. I will ake 9. s. MHR Calculus and Vecors 1 Soluions 494
Chaper 5 Secion Quesion 1 Page 66 a) Use he EpReg funcion: The equaion of he funcion is: T ( ) = 00e! k = 00( 0.867) Taking he logarihm of boh sides, ln e! k = ln(0.867)! = ln(0.867) k 1 k =! ln(0.867) k = 7 b) T (10) = 00e!10 7 T (10) = 00(0.867) 10!= 48!= 48 A 10 min, he emperaure is 48ºC. c) T (15) = 00e!15 7!= 3 A 15 min, he emperaure is 3ºC. Answers may vary. For eample: The pizza will reach room emperaure (1ºC) afer a long ime. MHR Calculus and Vecors 1 Soluions 495
Chaper 5 Secion Quesion 13 Page 66 a) ln + ln 3 =! 1.7918 b) ln 6 =! 1.7918 c) Answers may vary. For eample: The resuls seem o verify he Law of Logarihms for Muliplicaion. In erms of naural logarihms, he Law of Logarihms for Muliplicaion of naural logarihms is ln( a! b) = ln a + ln b, a > 0, b > 0. Chaper 5 Secion Quesion 14 Page 66 a) i) ii) N 10 (ln )! 5700 = (ln )! 0 5700 = N0e ln $ e % ln 0.1 ' 5700ln 0.1 =! ln =! 18 935 The age is 18 935 years. N 100 (ln )! 5700 = (ln )! 0 5700 = N0e ln $ e % ln 0.01 ' 5700ln 0.01 =! ln =! 37 870 The age is 37 870 years. iii) N (ln )! 5700 = (ln )! 0 5700 = N0e ln $ e % ln 0.5 ' 5700ln 0.5 =! ln = 5700 The age is 5700 years. b) Answers may vary. For eample: No. The half-life of C-14 is approimaely 5700 years. I will ake 5700 years for he sample o have a C-14 o C-1 raio of half of oday s level and i will ake 11 400 years for he sample o have a C-14 o C-1 raio of one quarer of oday s level. MHR Calculus and Vecors 1 Soluions 496
c) ( )! N 5700 ln $ N0 = ' % ln Chaper 5 Secion Quesion 15 Page 66 a) b) Answers may vary. For eample: The shape of he graph is similar o he shape of he normal disribuion curve. c) The maimum value is y = 1 and his occurs when = 0. d) Answers may vary. For eample: From he graph, use a rapezoid o esimae he area wih a base of 3 unis, op 0.5 unis, and heigh 1 uni.! A = h a + b $ %! = 1 3+ 0.5 $ %!= 1.75 This gives an esimae of 1.75 unis. Noe: The oal area is given by he inegral of he error funcion and is! = 1.77 square unis. MHR Calculus and Vecors 1 Soluions 497
e) From he graph i can be seen ha an esimae of he area beween = 1 and = +1 can be made by using he sum of he area of a recangle and rapezoid. area of recangle =! 0.367 = 0.734 area of rapezoid = (1! 0.376) $ + 0.5 % ' = 0.78 Therefore, he esimaed area beween = 1 and = 1 is 0.734 unis + 0.78 unis!= 1.5 unis. Noe: The Empirical Rule for normal disribuions saes ha his area should be 68% of he oal area under he curve. Chaper 5 Secion Quesion 16 Page 66 D Chaper 5 Secion Quesion 17 Page 66 C MHR Calculus and Vecors 1 Soluions 498
Chaper 5 Secion 3 Derivaives of Eponenial Funcions Chaper 5 Secion 3 Quesion 1 Page 74 a) g!() = 4 ln 4 b) f!() = 11 ln11 c) dy d =! 1 $ % ln 1 d) N!() = 3e e) h! ( ) = e f) dy d =! ln! Chaper 5 Secion 3 Quesion Page 74 a) f!( ) = e ; f!!() = e ; f!!! () = e b) f (n) () = e Chaper 5 Secion 3 Quesion 3 Page 74 dy d = 5 ln5 dy = 5 ln5 d =!= 40. The insananeous rae of change is 40.. Chaper 5 Secion 3 Quesion 4 Page 74 dy d = 1 e dy d =4 = 1 e4!= 7.3 The slope is 7.3. MHR Calculus and Vecors 1 Soluions 499
Chaper 5 Secion 3 Quesion 5 Page 74 dy dy d = 8 ln8 d = 1 1 = 8 ln8 = ( )3ln = 6 ln When = 1, y = Subsiue 1 = 1, y 1 =, and m = 6 ln in y! y 1 = m(! 1 ). Therefore, y! = 6 ln! 1 % $ ' ( ) + (! 3ln ) y = 6 ln Chaper 5 Secion 3 Quesion 6 Page 74 a) N() = 10( ) ; is he ime in days; N() is he number of frui flies b) N ( ) = 10( 7 7 ) = 180 Afer 7 days, here will be 180 frui flies. c) Rae of increase = N!() N!(7) = 10( 7 )ln = 10( )ln!= 887 A 7 days, he rae is 887 frui flies per day. d) 500 = 10( ) ln50 = ln = ln50 ln!= 5.64 I will ake 5.64 days for he populaion o reach 500 flies. MHR Calculus and Vecors 1 Soluions 500
e) N!(5.64) = 10( 5.64 )ln!= 346 A 5.64 days, he rae is 346 frui flies per day. Chaper 5 Secion 3 Quesion 7 Page 75 a) i) 0 = 10( )ln = ln! $ ln ln % = ln!= 1.53 The ime is 1.53 days. ii) 000 = 10( )ln = 00 ln! ln 00 $ ln % = ln!= 8.17 The ime is 8.17 days. b) Answers may vary. For eample: Since he growh rae increases eponenially, i is mos desirable o begin an eerminaion program very soon afer days. A 8 days, he populaion becomes ou of conrol. MHR Calculus and Vecors 1 Soluions 501
Chaper 5 Secion 3 Quesion 8 Page 75 f!() = 1 e f!(ln3) = 1 eln3 = 3 Therefore, he slope of he perpendicular line is 3. When = ln 3, y = 3. 3 Subsiue 1 = ln 3, y1 =, and m = in y! y1 = m( 1 ). 3 3 y! = (! ln 3) 3 3 y = + ln 3 + 3 3 Chaper 5 Secion 3 Quesion 9 Page 75 y =! 3 + 3 ln3+ 3 y!=! 3 +.341 Chaper 5 Secion 3 Quesion 10 Page 75 a) Answers may vary. For eample: The shape of he graph of g( ) is a horizonal sraigh line. b) f!() = kb ln b ; g() = kb ln b kb = ln b c) The simplified form of he funcion is he graph of a horizonal sraigh line: g( ) = ln b. MHR Calculus and Vecors 1 Soluions 50
Chaper 5 Secion 3 Quesion 11 Page 75 Answers may vary. For eample: Use a graphing calculaor. Le k = 5 and b = 3. Then g() = ln3!= 1.0986 Chaper 5 Secion 3 Quesion 1 Page 75 a) g( ) = 1; Answers may vary. For eample: The derivaive of he eponenial funcion of he form f ( ) = ke is f!( ) = ke. The simplified form of he funcion g( ) is he funcion f!( ) g( ) = f ( ) ke = ke = 1 b) g( ) = 1 Chaper 5 Secion 3 Quesion 13 Page 75 a) f (n) () = b (ln b) n b) Answers may vary. For eample: f!() = b ln b, f!!() = b (ln b), f!!! () = b (ln b) 3,..., f (n) () = b (ln b) n MHR Calculus and Vecors 1 Soluions 503
Chaper 5 Secion 3 Quesion 14 Page 75 a) Answers may vary. For eample: Boh funcions have he same y-value of 16, when he -value is 4. Boh of he funcions are increasing funcions ha do no have a local maimum or minimum poin, or a poin of inflecion. The funcion g() = is increasing more rapidly han he funcion f ( ) = over he given inerval, 4 16. f ( ) =, 4 16 ( ) g =, 4 16 b) Answers may vary. For eample: No. The derivaives of he wo funcions will no be similar. The derivaive of he quadraic funcion f ( ) = is f!( ) =. When graphed, he derivaive funcion is a linear funcion wih a slope of. The derivaive of he eponenial funcion g( ) = is g!() = ln. When graphed, he derivaive funcion is also an eponenial funcion. c) f!( ) =, 4 16 g!() = ln, 4 16 d) Answers may vary. For eample: Yes. There are wo -values for which he slope of f ( ) will be approimaely he same as he slope of g( ) when rounded o five decimal places. When he -value is 0.485 09, he slope of f ( ) and he slope of g( ) is 0.970 18. When he -value is 3.1 43, he slope of f ( ) and he slope of g( ) is 6.44 87. The -values can be found using a graphing calculaor. MHR Calculus and Vecors 1 Soluions 504
Chaper 5 Secion 3 Quesion 15 Page 75 a) P(h) = 101.3e! kh 95.6 = 101.3e!1000k ln 95.6 % $ 101.3 ' = (!1000k)ln e k =!0.001ln 95.6 % $ 101.3 ' k!= 0.000 057 9!(0.000 057 9)(000) b) P(000) = 101.3e!= 90. The pressure is 90. kpa. 0.000 057 9h c) P!(h) = 101.3(0.000 057 9)e 0.000 057 9h!= 0.00587e 0.000 057 9(1500) d) P!(1500) = 0.005 87e!= 0.005 38 The rae is 0.005 38 kpa/m. Chaper 5 Secion 3 Quesion 16 Page 76 a) N( ) = 50( ) where N is he number of visiors and is he ime in weeks. b) i) N = 4 ( 4) 50( ) = 800 Afer 4 weeks, here will be 800 visiors. 1 ii) N ( 1) = 50( ) Afer 1 weeks, here will be 04 800 visiors. c) N!() = 50( )ln i) N!(4) = 50( 4 )ln!= 555 The rae is 555 visiors per week. ii) N!(1) = 50( 1 )ln!= 141957 The rae is 141 957 visiors per week. d) Answers may vary. For eample: No. This rend will no coninue indefiniely. The number of people visiing he sie will evenually level off. MHR Calculus and Vecors 1 Soluions 505
Chaper 5 Secion 3 Quesion 17 Page 76 a) Answers may vary. For eample: One Inerne sie claims he curren populaion is growing a a rae of 05 000 per day, or 8500 per hour, or 140 per minue, or.3 people per second. b) Answers may vary. For eample: i) Equaion form: P() = 4e 0.019, where P is he populaion, in billions, and is he ime, in years, since 1975. ii) Graphical form: c) i) P(50) = 4e 0.019(50) = 10.34 838 64 The populaion would be 10.34 838 64 billion. ii) P(55) = 4e 0.019(55)!= 85 930.51 67 The populaion would be 85 930.51 67 billion. iii) P(105) = 4e 0.019(105)!= 1 148 008 96 The populaion would be 1 148 008 96 billion. d) Answers may vary. For eample: No. This model is no susainable over he long erm. Oher facors ha could affec his rend are he amoun of resources available o susain he populaion and he available areas on he earh ha could susain his number of people. e) Answers may vary. For eample: If he resources, such as food, sar o diminish hen he populaion increase would slow down, since he deah rae would increase relaive o birh. Poor nuriion is one conribuing facor o low birh raes. The facor 0.019 would be reduced. MHR Calculus and Vecors 1 Soluions 506
Chaper 5 Secion 3 Quesion 18 Page 76 a) Answers may vary. For eample: i) 10!6 = 4e 0.019 ln(.510!7 ) = 0.019!=!800 Since = 0 in 1975, he year when he populaion would have been 1000 is prediced as 1975 800 = 1175. ii) 10!7 = 4e 0.019 ln(.510!8 ) = 0.019!=!91 Since = 0 in 1975, he year when he populaion would have been 100 is prediced as 1975 91 = 1054. iii)!10 9 = 4e 0.019 ln(5!10 10 ) = 0.019!= 117 Since = 0 in 1975, he year when he populaion would have been is prediced as 1975 117 = 848. b) Answers may vary. For eample: No. The answers in par a) do no seem reasonable. Chaper 5 Secion 3 Quesion 19 Page 76 Answers may vary. For eample: Sudens may use a graphing calculaor o graph of. = 1: angen is y =!1 where he slope is 1 = : angen is y = 0.5! 0.31 where he slope is 0.5 = 0.5: angen is y =!1.69 where he slope is y = ln and use he angen funcion for some values 1 Graph he derivaive funcion of y = ln (i.e., f ( ) =! ) and compare he ordered pairs wih he (, slope) values for y = ln. They are he same: MHR Calculus and Vecors 1 Soluions 507
Chaper 5 Secion 3 Quesion 0 Page 76 D Chaper 5 Secion 3 Quesion 1 Page 76 E MHR Calculus and Vecors 1 Soluions 508
Chaper 5 Secion 4 Differeniaion Rules for Eponenial Funcions Chaper 5 Secion 4 Quesion 1 Page 8 a) y = e ln b b) dy d = e lnb ln b Chaper 5 Secion 4 Quesion Page 8 a) b) c) dy =! 3e d! 3 f ( ) = 4e! 4 5 dy d = e! (!)e! = e + e! d) dy d = ln + 3 ln3 e) f () = 3e! ( ) 3 f () = ()3e! 3( ) ln = 6e! 3( 3 )ln f) g) h) dy d = 4e + 4e = 4e ( +1) dy d = 5 e! ln5! 5 (e! ) =!5 e! (1! ln5) f ( ) = e + e! 6e! 3 MHR Calculus and Vecors 1 Soluions 509
Chaper 5 Secion 4 Quesion 3 Page 8 a) dy d =!e! sin + e! cos = e! (cos! sin ) b) dy d =! sin (ecos ) c) f!() = e ( 3 + ) + e ( 3) = e ( 4 +1) d) g!() = 4e cos + e cos (sin ) = 4e cos ( sin 1) Chaper 5 Secion 4 Quesion 4 Page 8 f!( ) = e e! = = If f ( ) 0 hen e e 0. Therefore, e (1! e ) = 0 e = 1 since e > 0 = ln0.5 f (ln0.5) = e ln0.5! e ln0.5 = 0.5! 0.5 = 0.5 Therefore, by using derivaive ess, here is a local maimum of y = 0.5 when = ln(0.5). Chaper 5 Secion 4 Quesion 5 Page 8 Adding wo eponenial funcions gives an eponenial funcion. f!() = e + e Se f!() = 0. e (1+ e ) = 0 e mus equal 1 or 0 bu since e > 0, he funcion has no local erema. MHR Calculus and Vecors 1 Soluions 510
Chaper 5 Secion 4 Quesion 6 Page 8 Graph of y = e! e Graph of y = e + e Chaper 5 Secion 4 Quesion 7 Page 83 a) P(3) = 50e 0.5(3)!= 4 Afer 3 days, here will be 4 baceria. b) Iniial populaion = 50 (i.e., = 0) 10(50) = 50e 0.5 10 = e 0.5 ln10 = 0.5 = ln10 0.5!= 4.6 The ime is 4.6 days. c) e 0.5 = 10 0.5 = ln10 = ln10!= 4.6 P() = 50e 0.5!= 50(10) 4.6 d) P(5) = 50(10) 5 4.6!= 611 Afer 5 days, here will be 611 baceria. MHR Calculus and Vecors 1 Soluions 511
e) P(5) = 50e 0.5(5)!= 609 Afer 5 days, here will be 609 baceria. Answers may vary. For eample: The funcion from par c) approimaes he relaionship beween e and from he iniial funcion. Chaper 5 Secion 4 Quesion 8 Page 83 0.065() a) i) A( ) = 3000e = 3416. 49 Afer years, he amoun will be $3416.49. 0.065(5) ii) A( 5) = 3000e = 415. 09 Afer 5 years, he amoun will be $415.09. 0.065(5) iii) A( 5) = 3000e = 15 35. 6 Afer 5 years, he amoun will be $15 35.6. b) 6000 = 3000e 0.065 e 0.065 = 0.065 = ln = ln 0.065!= 10.7 I will ake 10.7 years. c) A!() = 3000(0.065)e 0.065 A!() = 195e 0.065 A!(10.7) = 195e 0.065(10.7) = 390.9 The invesmen is growing a a rae of $390.9 per year. MHR Calculus and Vecors 1 Soluions 51
Chaper 5 Secion 4 Quesion 9 Page 83 a) f!( ) = e (cos + sin ) f!!() = e (cos ) f!!! () = e (cos sin ) f (4) () =!4e (sin ) f (5) () =!4e (cos + sin ) f (6) () =!8e (cos ) b) Answers may vary. For eample: The firs and fifh derivaives have he epression cos + sin. The hird derivaive has he epression cos sin. The second and hird derivaives have he same coefficien e. The fourh and fifh derivaives have he same coefficien 4e The second and sih derivaives have he epression cos. The fourh derivaive has he epression sin. The derivaives all have he epression e in hem. c) i) f (7) () =!8e (cos! sin ) ii) f (8) () = 16e (sin ) d) Answers may vary. For eample: n!1 f (n) 4 () = (!4) e (cos + sin ), for n!{1,5,9,13,...} n! f (n) 4 () = (!4) e cos, for n!{,6,10,14,...} n!3 f (n) 4 () = (!4) e (cos! sin ), for n!{3,7,11,15,...} n f (n) 4 () = (!4) e (sin ), for n!{4,8,1,16,...} Chaper 5 Secion 4 Quesion 10 Page 83 Answers may vary. For eample: Laura s moorcycle depreciaes in value he fases when she firs drives i off he lo. The rae of depreciaion a ime = 0 is calculaed as $500 per year. Therefore, her moorcycle is depreciaing a he rae of $500 per year when = 0. Graph of V!() MHR Calculus and Vecors 1 Soluions 513
Chaper 5 Secion 4 Quesion 11 Page 83 a) P 0 = 000 1! $ 4 4000 = 000 a % 1 a 4 = a = 16 b) P 1 1! $ 6% = 000! 16 $ 6 %!= 3175 Afer 10 min, here will be approimaely 3175 algae. c) i) P!() = P 0 (a )ln a P!(1) = 000(16 1 )ln16!= 88 73 Afer 1 h, he rae of change of he populaion will be approimaely 88 73 algae per hour. ii) P!(3) = 000(16 3 )ln16!= 713 047 Afer 3 h, he rae of change of he populaion will be approimaely 713 047 algae per hour. Chaper 5 Secion 4 Quesion 1 Page 83 a) Answers may vary. For eample: Cheryl has ried o differeniae he eponenial funcion using he power rule. The power rule canno be used o differeniae an eponenial funcion, since he eponen is a variable. b) Answers may vary. For eample: Cheryl saw a erm ha was in eponen form and hough ha she could use he power rule. c) The derivaive of an eponenial funcion y = a is dy d = a ln a, so he correc answer is dy d = 10 ln10. MHR Calculus and Vecors 1 Soluions 514
Chaper 5 Secion 4 Quesion 13 Page 83 dy d =!e! d y d = 4 e!! e! Poins of inflecion occur for -values ha saisfy d y d = 0. 0 = 4 e!! e! 4 = since e! > 0 = ± 1 If = 1, y = 1 e ; If =! 1, y = 1 e. Therefore, he poins of inflecion are! 1, 1 % $ e ' and 1, 1 % $ e '. Chaper 5 Secion 4 Quesion 14 Page 84 a) dy d = e cos! e sin Le dy = 0 o find he -values of he local erema. d 0 = e (cos! sin ) cos = sin since e > 0 =! 4 and = 5! 4 in he inerval 0!!!. MHR Calculus and Vecors 1 Soluions 515
d y d = e (! sin ) + e cos! e cos! e sin =!e sin A =! 4, d y is negaive, so here is a local maimum. d A = 5! 4, d y is posiive, so here is a local minimum. d Using a graphing calculaor: Two local erema occur on he inerval: a local maimum a (0.785, 1.551) or 5! local minimum a (3.97, 35.889) or $ 4,!35.889 % ' over he inerval [0, π].! $ 4, 1.551 % ' ; and a b) The local maimum for f () occurs! 4 rad o he righ and 7! 4 rad o he lef of where he local maimums (0, 1) and (π, 1) occur for he funcion y = cos over he inerval [0, π]. The local minimum for he funcion y = e cos occurs! 4 rad o he righ of where he local minimum (π, 1) occurs for he funcion y = cos over he inerval [0, π] MHR Calculus and Vecors 1 Soluions 516
Chaper 5 Secion 4 Quesion 15 Page 84 a) 0.75(V ma ) = V ma 1! e! % 8 $ ' e! 8 = 0.5! 8 = ln(0.5) =!8ln(0.5)!= 11.1 The ime required is 11.1 h. 1 8 b) V ( ) Vmae! = 8 Chaper 5 Secion 4 Quesion 16 Page 84 a) Answers may vary. For eample: The funcion is an increasing funcion on he inerval (!, ). The funcion is concave down on he inerval (!, 0) and concave up on he inerval (0,!). b) Answers may vary. For eample: The shape of he derivaive of he funcion will be concave up on he inerval (!, ) wih a local minimum value when = 0. MHR Calculus and Vecors 1 Soluions 517
Chaper 5 Secion 4 Quesion 17 Page 84 a) b) Answers may vary. For eample: The shape of he funcion will be concave up on he inerval (!, ) wih a local minimum value when = 0. Answers may vary. For eample: The derivaive of he funcion is an increasing funcion on he inerval (!, ). The derivaive funcion is concave down on he inerval (!, 0) and concave up on he inerval (0,!). Chaper 5 Secion 4 Quesion 18 Page 84 a) i) ii) d ( d sinh ) = e! (!1)e! = e + e! = cosh d ( d cosh ) = e + (!1)e! = e! e! = sinh b) Answers may vary. For eample: The predicions in par a) were correc. MHR Calculus and Vecors 1 Soluions 518
Chaper 5 Secion 4 Quesion 19 Page 84 a) Take he derivaive wih respec o of boh sides of he equaion = e y. b) Since = e y by aking he derivaives of boh sides, y! dy 1 = e $ % d dy 1 = y d e dy 1 = d Chaper 5 Secion 4 Quesion 0 Page 84 A Chaper 5 Secion 4 Quesion 1 Page 84 E MHR Calculus and Vecors 1 Soluions 519
Chaper 5 Secion 5 Making Connecions: Eponenial Models Chaper 5 Secion 5 Quesion 1 Page 89 a) N(10) = 100e!(10) b) 73 = 100e!(10) ln(0.73) =!10 = ln(0.73) 10!= 0.031 The disinegraion consan is 0.031/min. 100 = 100e!0.031 ln 0.5 =!0.031 =! ln0.5 0.031!= The half-life is min.!! 1$ c) N() = 100 %!! 1 = 100 $ %!!= 100 1 $ % d) N ( ) =! N0e! log 1 e $ % loge log0.5 $ % '0.031 '0.031 N!(5) = (0.031)(100)e (0.031)(5)!=.65 The sample is decaying a.65 mg/min afer 5 min. MHR Calculus and Vecors 1 Soluions 50
Chaper 5 Secion 5 Quesion Page 90! a) i) M Rn (1) = 100 1 $ % 1 3.8!= 83.3 Afer 1 day, here will be 83.3 mg of radon. ii) M (7) = 100! 1 $ Rn % 7 3.8!= 7.9 Afer 1 week, here will be 7.9 mg of radon.! b) 0.5(100) = 100 1 $ %! 1$ 3.8 1 % = 4! 1$ 3.8! 1 $ % = % 3.8 = = 7.6 I will ake 7.6 days.!! 1$ $ c) M Rn () = 100 % % 3.8 M Rn ' () = 100!! 1$ $ 3.8 % % 1 3.8 (.8 3.8 M Rn ' () = 100! 1$ 3.8 1 3.8 % ln! 1$ % ln 1 i) M Rn! (1) = 100 1% 3.8 1 3.8 $ ' ln!= (15. The rae of decay is 15. mg/day. 1 MHR Calculus and Vecors 1 Soluions 51
ii) M Rn! (7) = 100 1% 3.8 1 3.8 $ ' ln!= (5.1 The rae of decay is 5.1 mg/day. 7 7.6 M Rn! (7.6) = 100 1% 3.8 1 3.8 $ ' ln!= (4.6 The rae of decay is 4.6 mg/day. Chaper 5 Secion 5 Quesion 3 Page 90 a) i) Since iniially here is 100 mg of radon, here is 0 mg of polonium. 1 ( ii) M Po (1) = 100 1! 1 + % 3.8 * - * $ ' - )*,-!= 16.7 There will be 16.7 mg of polonium. 1 ( 1% % 3.8 * b) M Po () = 100 * 1! $ ' * $ ' )* ( M Po. () = 100! 1 * 1% % * 3.8 $ ' * $ ' )* M Po. () =! 100 1% 3.8 1 3.8 $ ' ln + - - -,-!.8 3.8 + 1% $ ' ln 1 - - -,- Answers may vary. For eample: The firs derivaive of he funcion is he rae of change of he amoun of polonium in milligrams per day. MHR Calculus and Vecors 1 Soluions 5
Chaper 5 Secion 5 Quesion 4 Page 90 a) b) Answers may vary. For eample No. The wo funcions are no inverses of each oher. They are no a reflecion of each oher in he line y =. The coordinaes are (3.8, 50). The poin of inersecion is he half-life of radon. c) Answers may vary. For eample: A he poin of inersecion, which is he half-life of radon, he derivaives of each funcion are equal in value, bu opposie in sign ( ± 9.1 mg/day). The rae of change of radon is negaive, since he amoun of radon is decreasing, and he rae of change of polonium is posiive, since he amoun of polonium is increasing. This makes sense from a physical perspecive since he radon is being convered ino polonium, so he rae of decay of radon mus equal he rae of growh of polonium. MHR Calculus and Vecors 1 Soluions 53
d) Answers may vary. For eample: The shape of his graph is he horizonal sraigh line y = 100. This makes sense from a physical perspecive since he sum of he amoun of radon and he amoun of polonium will always be equal o 100 as he radon decays. Chaper 5 Secion 5 Quesion 5 Page 90 a) Answers may vary. For eample: Yes. The funcion in he graph is an eample of damped harmonic moion. The curve is sinusoidal wih diminishing ampliude as he ime is increasing. b) i).7 ms ii) 0.00 7 s c) f = 1 0.00 7!= 440 The frequency is 440 Hz. d) I() = 4cos(!(440))e k Chaper 5 Secion 5 Quesion 6 Page 91 a) k = 101./s I() = 4cos[!(440)]e 101. b) Answers may vary. For eample: I found he value by subsiuing he I() value of ha occurs when = 0.006 804 5 ms ino he equaion in quesion 5 par d). c) MHR Calculus and Vecors 1 Soluions 54
Chaper 5 Secion 5 Quesion 7 Page 91 a) The frequency of he sound is no a funcion of ime. Therefore, i does no diminish over ime. b) Pich decay could look like he following graph. As he frequency diminishes, he period will increase leading o a sreched sinusoidal curve. Chaper 5 Secion 5 Quesion 8 Page 91 a) v() = h!() = 0.5e 0.5 sin + e 0.5 cos Graph his funcion o find he maimum velociy. The maimum value is a = 0 s and is v ma = 1 m/s. MHR Calculus and Vecors 1 Soluions 55
b) v() =!0.5e!0.5 sin + e!0.5 cos a() = v () ( ) = d e!0.5 (!0.5sin + cos) d = e!0.5 (!0.5cos! sin)! 0.5e!0.5 (!0.5sin + cos) = e!0.5 [sin(!1+ 0.5) + cos(!0.5! 0.5)] = e!0.5 [!0.75sin! cos] Graph a() = e!0.5 [!0.75sin! cos] and find he maimum. F = ma!= 60(0.1 4)!= 1.7 The greaes force is 1.7 N. Chaper 5 Secion 5 Quesion 9 Page 91 a) b) Answers may vary. For eample: As he shock absorbers wear ou wih ime, he verical displacemen of he shock absorber will increase since he ampliude of he funcion is larger. c) Answers will vary. For eample: As he verical displacemen of he shock absorber increases wih wear, he modelling equaion will change o reflec he increases. MHR Calculus and Vecors 1 Soluions 56
Chaper 5 Secion 5 Quesion 10 Page 91 a) Answers may vary. For eample: Yes. Rocco s moion is an eample of damped harmonic moion. The curve is sinusoidal wih diminishing ampliude as he ime is increasing. b) Answers may vary. For eample: No. Biff will no be able o rescue Rocco. Rocco will no swing back o wihin 1 m from where he sared falling.! Rocco s iniial horizonal posiion, in meres: (0) = 5cos! $ % = 5 e'0.1(0) Rocco will swing back owards his horizonal posiion when! =! so when = 4 s.! Rocco s horizonal posiion a = 4 s, in meres: (4) = 5cos! $ % e'0.1(4)!= 3.35 Rocco will be 5 m 3.35 m = 1.65 m away from his iniial posiion so no wihin 1 m. MHR Calculus and Vecors 1 Soluions 57
c) Find he horizonal velociy and find when he speed is less han m/s.!() = 5! sin!! $ % ' ( e0.1 + 5( 0.1)cos $ % ' ( e0.1 ) 5! = e 0.1 sin! $ % ' ( + 0.5cos!, + $ % ' (. * - Rocco is a he boom of he swing a = 1 s, 3 s, 5 s, 5!!(1) = e 0.1(1) $ % ' (!(3) = e 0.1(3) 5! $ % ' ( = 7.11 = 5.8 5!!(5) = e 0.1(5) $ % ' (!(7) = e 0.1(7) 5! $ % ' ( = 4.76 = 3.90 5!!(9) = e 0.1(9) $ % ' (!(11) = e 0.1(11) 5! $ % ' ( = 3.19 =.61 5!!(13) = e 0.1(13) $ % ' (!(15) = e 0.1(15) 5! $ % ' ( =.14 = 1.75 Therefore, Rocco mus swing back and forh 3.75 imes before he can safely drop o he ground. The graph below shows ha he slope of he angen a = 15 is 1.755 m/s. i.e.,! ( ), he horizonal velociy a he boom of he swing is less han m/s afer 15 s. d) MHR Calculus and Vecors 1 Soluions 58
Chaper 5 Secion 5 Quesion 11 Page 9 a) The period of Rocco s vine is 4 s. Therefore, use T =! l g wih T = 4 s and g = 9.8 m/s o find l. 4 =! l 9.8! l = 9.8 $!%!= 4.0 The vine is abou 4.0 m long. b) i) Answers may vary. For eample: If he vine were shorer, Rocco s posiion graph would have a shorer period and a smaller ampliude. ii) Answers may vary. For eample: If he vine were longer, Rocco s posiion graph would have a longer period and a larger ampliude. c) Answers may vary. For eample: If he vine were longer, Rocco could swing back o wihin 1 m of Biff. A shorer vine would slow o m/s in fewer swings. Chaper 5 Secion 5 Quesion 1 Page 9 ( I() = I pk 1! e! 1000 $ 00 * ) * I() = I pk ( ) 1! e!5 +, % ' + -, - a) i) 0.50I pk = I pk 1! e!5 ln(0.5) =!5 = ln(0.5)!5!= 0.14 I will ake 0.14 s. $ % MHR Calculus and Vecors 1 Soluions 59
ii) 0.90I pk = I pk 1! e!5 $ %!0.10 =!e!5 ln(0.1) =!5 = ln(0.1)!5!= 0.46 I will ake 0.46 s. b) I!() = I pk e R ) $ % L ' ( R, + $ % L ' (. * + -. = I pk [e 5 (5)] = 5I pk e 5 i) I!(0.14) = 5I pk e 5(0.14)!=.5I pk The rae is.5i pk A/s. ii) I!(0.46) = 5I pk e 5(0.46)!= 0.5I pk The rae is 0.5I pk A/s. Chaper 5 Secion 5 Quesion 13 Page 9 Soluions o he Achievemen Checks are shown in he Teacher s Resource. Chaper 5 Secion 5 Quesion 14 Page 93 a) The graph has he shape of a logisic funcion. MHR Calculus and Vecors 1 Soluions 530
b) 755.6 P()!= 1+1.9e!0.5 c) Answers may vary. For eample: The curve appears o fi he daa very well as shown in he graph. d) There is a horizonal asympoe a y = 756. e) Answers may vary. For eample: The rabbi populaion will no reach 756. MHR Calculus and Vecors 1 Soluions 531
Chaper 5 Secion 5 Quesion 15 Page 93 a), b) 800 755.6 f( ) = 1+1.9! e -0.5! -1! 9747.4! e f' ( ) = -1! +51.6! e + 33.8 e A: (5.10, 94.40 ) A 5 Answers may vary. For eample: The graph has a maimum value a (5.1, 94.4). The growh rae of he rabbis will increase o a maimum and hen decrease o zero. c) The rabbi populaion was growing he fases a he fifh year. d) The rabbi populaion was growing a he rae of 94.4 rabbis per year. Chaper 5 Secion 5 Quesion 16 Page 93 755.6 P() = 1+1.9e!0.5 = 755.6(1+1.9e!0.5 )!1 P () =!755.6(1+1.9e!0.5 )! (1.9)(!0.5)e!0.5 = 4873.6e!0.5 (1+1.9e!0.5 ) 4873.6 = e 0.5 + 5.8 +166.41e!0.5 = 4873.6 e + 5.8 + 166.41 e MHR Calculus and Vecors 1 Soluions 53
Chaper 5 Secion 5 Quesion 17 Page 93 Answers may vary. For eample: a) Assume ha in he pack of 5 wolves, only he dominan male and female breed. Each year, he female gives birh o a lier of 5 pups. According o quesion 14, he maimum populaion for he rabbis is 755. Assume ha 50 rabbis are consumed per year by 10 wolves and he rae of consumpion is proporional wih he wolf populaion. Year Rabbi Populaion Wolf Populaion 15 705 10 16 630 15 17 530 0 18 455 5 19 305 30 0 30 35 Rabbi Populaion Wolf Populaion b) The rabbi populaion is seadily declining as he wolf populaion is seadily increasing. c) Answers will vary. Rabbi: y = 97.9 + 70.5 Wolf: y = 5 + 10 Boh models are linear. The wolf populaion in increasing as he rabbi populaion declines. The rae of decline is greaer in he rabbi populaion han he rae of increase in he wolf populaion. MHR Calculus and Vecors 1 Soluions 533
Chaper 5 Secion 5 Quesion 18 Page 93 B MHR Calculus and Vecors 1 Soluions 534
Chaper 5 Review Chaper 5 Review Quesion 1 Page 94 a) b) 3 y = ln 3 y = 3 c) y = 3 d) y = Chaper 5 Review Quesion Page 94 Answers may vary. For eample: a) Choose he values of n o be n!!. Evaluae he limi for values of n ha are larger and larger. As he values of n become larger, he value of he limi will approach he value e. b) Choose n = 100 000. lim 1+ 1 % n!100 000 $ n ' n 1 % = 1+ $ 100 000 ' e =.7 100 000 Chaper 5 Review Quesion 3 Page 94 a) b) MHR Calculus and Vecors 1 Soluions 535
c) y =! ln Answers may vary. For eample:!! ln( e ) = = e!(! ln ) Chaper 5 Review Quesion 4 Page 94 a) 3 e! = 0.050 b) ln(6.) = 1.85 c) 3! ln e 4 $ % = 0.75 d) ln(0.61) e = 0.61 Chaper 5 Review Quesion 5 Page 94 a) = 1.10 b) = 0.01 c) =.3 d) = 9.1 Chaper 5 Review Quesion 6 Page 94 a) 50 baceria b) P(4) = 50e 0.1(4)!= 81 Afer 4 days, here will be 81 baceria. c) 100 = 50e 0.1 e 0.1 = 0.1 = ln = ln 0.1!= 5.78 I will ake abou 6 days for he populaion o double. MHR Calculus and Vecors 1 Soluions 536
d) P() = 50 log ( ) e 0.1! = 50 = (50) loge(0.1 ) log 5.8 $ % Chaper 5 Review Quesion 7 Page 95 a) i) f!() = 1 % $ ' ln 1 b) ii) g!() = e Chaper 5 Review Quesion 8 Page 95 dy d = ln3(3 ) slope = dy d =1 = ln3(3 1 ) = 6ln3 A = 1, m = 6ln3, and y = 6. Subsiue hese values ino y = m + b. b = 6 6ln3 Therefore, he equaion of he angen is y = 6 ln3+ 6! 6ln3. MHR Calculus and Vecors 1 Soluions 537
Chaper 5 Review Quesion 9 Page 95 dy d =!3e slope = dy d =ln =!3e ln =!6 A = ln, m = 6, and y = 6. Subsiue hese values ino y = m + b. b = 6 + 6 ln Therefore, he equaion of he angen is y = 6 6 + 6 ln. Chaper 5 Review Quesion 10 Page 95 5 9 a) A(5) = 1000() = 1469.73 The value afer 5 years is $1469.73. 9 b) i) 000 = 1000() = 1 9 = 9 I will ake 9 years o double in value. 9 ii) 3000 = 1000() 9 3 = () ln3 = 9 ln = 9ln3 ln!= 14.6 I will ake 14.6 years o riple in value. MHR Calculus and Vecors 1 Soluions 538
1 c) A() = 1000( 9 ) A!() = 1000 9 ( ) 8 9 ( ln) = 1000 9 ( 9 )ln i) A!(9) = 1000 9 9 ( 9 )ln = 154.03 The rae is $154.03 per year. ii) A!(14.6) = 1000 14.6 9 ( 9 )ln = 30.97 The rae is $30.97 per year. Chaper 5 Review Quesion 11 Page 95 a) b) dy = (6!!+1 )e3 d f! = e ( ) ( 1) dy c) = 3! e d! d) dy d = e (cos()! sin())!! 1$ $ e) g() = 3% %!! 1$ $ g (() = 4 3% %! g (() = 4 1 $ 3% 4 4 ' e sin 3! 1$ 3%! ln 1 $ 3% ' esin (cos )! ln 1 $ 3% ' esin (cos ) MHR Calculus and Vecors 1 Soluions 539
Chaper 5 Review Quesion 1 Page 95 Local erema occur when dy d = 0. dy d = e 0 = e = 0 since e > 0 When = 0, y = 1. d y d = e + 4 e which is posiive for = 0. Therefore, here is a local minimum a (0, 1). Chaper 5 Review Quesion 13 Page 95 Local erema occur when dy d = 0. dy d = e 0 = e Bu e > 0 for all values of. Therefore, here are no local erema. Chaper 5 Review Quesion 14 Page 95 a) $900 b) V (1) = 900e! 1 3 = 644.88 The value afer one year is $644.88. c) 450 = 900e! 3 e! 3 = 0.5! 3 = ln(0.5) =!3ln(0.5)!=.1 I will ake.1 years for he compuer o be worh half is original value. MHR Calculus and Vecors 1 Soluions 540
d) V!() = 1 $ % 3' ( 900e = 300e 3 V!(.1) = 300e.1 3!= 149.98 The rae of depreciaion is $149.98 per year. 3 Chaper 5 Review Quesion 15 Page 95! a) 70 = 80 1 $ % 5! 1$ h 7 % = 8 5 h ln! 1 $ % = ln! 7 $ 8 % 5 h h = 5ln(0.5) ln(0.875) h!= 6 The half-life is 6 days. b)! 1 N( ) = 80 $ % 6!! 1$ $ c) N() = 80 % % N '() = 80!! 1$ $ 6 % % 1 6 ( 5 6 N '() = 80! 1$ 6! 1 $ 6 % ln % N '(5) = 80! 1$ 6! 1 $ 6 % ln % 5! 1$! % ln 1 $ %!= (1.9 The rae of decay is 1.9 mg/day. MHR Calculus and Vecors 1 Soluions 541
Chaper 5 Review Quesion 16 Page 95 a) (0) = 3cos(0)e!0.05(0) = 3 The horizonal disance is 3 m. b)!() = 3(0.05)cos()e 0.05 3sin()e 0.05 = 0.15cos()e 0.05 3sin()e 0.05 The maimum value occurs a =!,where!!% $ ' The greaes speed is.8 m/s.!= (.8. c) Kara's maimum horizonal disance occurs a = 0,!,!,..., n! for n!!. A = 0,!,!, 3!, 4!, 5!, () > 1. (6!) = 3cos(6!)e!0.05(6!)!= 1.169 (7!) = 3cos(7!)e!0.05(7!)!=!0.999 Therefore Kara will be wihin a disance of 1 m in her maimum displacemen a!= s. I will ake her 3.5 swings. Graph y = (), y = 1, and y =!1. d) MHR Calculus and Vecors 1 Soluions 54
Chaper 5 Pracice Tes Chaper 5 Pracice Tes Quesion 1 Page 96 A Chaper 5 Pracice Tes Quesion Page 96 C Chaper 5 Pracice Tes Quesion 3 Page 96 A Chaper 5 Pracice Tes Quesion 4 Page 96 D Chaper 5 Pracice Tes Quesion 5 Page 96 a) b) dy = e d 1! f ( ) = e + 3 e + e! e 3!! Chaper 5 Pracice Tes Quesion 6 Page 96 Local erema occur when dy d = 0. dy d = e!! e! 0 = e! (1! ) = 0, = 1 If = 0, hen y = 0. If = 1, hen y = e. d y d = e!! 4e!! 4e! + 4 e! A = 0, d y is posiive. d A = 1, d y is negaive. d There is a local minimum a (0, 0). There is a local maimum a (1, e ) or (1, 0.135). MHR Calculus and Vecors 1 Soluions 543
Chaper 5 Pracice Tes Quesion 7 Page 96 a) P ( 0) = 50 Iniially, 50 people had he virus. 7 b) P(7) = 50()!= 566 Afer 1 week, 566 people will be infeced. c) P!() = 50 1 % $ ' () ln = 5() ln 7 P!(7) = 5() ln!= 196 Afer 1 week, he virus will be spreading o 196 people/day. d) 1000 = 50() () = 0 ln = ln 0! ln 0$ = ln %!= 8.64 I will ake 8.64 days for 1000 people o be infeced. Chaper 5 Pracice Tes Quesion 8 Page 96 a) b) Answers may vary. For eample: Boh graphs are decreasing funcions. The y-inercep of he funcion inercep of he funcion! 1 y = ln $ % is (, 0). ' y =! e is (0, ). The - MHR Calculus and Vecors 1 Soluions 544
Chaper 5 Pracice Tes Quesion 9 Page 96 f!() = e slope = f!(ln ) = e ln = 4 A = ln, m =!4, and y =!4. Subsiue hese values in y = m + b. b =!4 + 4ln Therefore, he equaion of he angen is y =!4! 4 + 4 ln. Chaper 5 Pracice Tes Quesion 10 Page 96 a) N() = N 0 e! 64 = 100e!10!10 = ln(0.64) = ln(0.64) 10!= 0.045 The disinegraion consan is 0.045/min. b) N 0 = N 0 e! = ln0.5 =! ln0.5 0.045!= 15.4 The half-life is 15.4 min.! 1$ c) N() = N 0 % 15.4 d) N!() = N 0 e N!(15) = (0.045)(5)e (0.045)(15)!= 0.57 Afer 15 min, he sample is decaying a 0.57 mg/min. MHR Calculus and Vecors 1 Soluions 545
Chaper 5 Pracice Tes Quesion 11 Page 96 a) V (10) = 1000(1.05) 10 = 168.89 The value is $168.89 afer 10 years. b) 000 = 1000(1.05) = ln ln1.05!= 14. I will ake 14. years o double in value. c) V () = 1000(1.05) V!() = 1000(1.05) ln1.05 d) V!(10) = 1000(1.05) 10 ln1.05 = 79.47 The rae is $79.47 per year. Chaper 5 Pracice Tes Quesion 1 Page 97 a) b) c) dy e d = + (sin cos sin ) dy = e! +! d! ( 1) dy sin e ( cos ) d = + Chaper 5 Pracice Tes Quesion 13 Page 97 a) 15 = 0e! 16! 16 = ln(0.75) =!16ln(0.75)!= 4.6 I will ake 4.6 h. b) V!() = 0 1 $ % 16' ( e 16 = 5 4 e 16 MHR Calculus and Vecors 1 Soluions 546
c) V!(1) = 5 4 e 1 16!= 1.174 The rae of change of he volage is 1.174 V/h afer 1 h. d) Answers may vary. For eample: 16 The volage is dropping a a slower rae since he slope of he funcion y = e! becomes less negaive wih ime. e) Answers may vary. For eample: V!() = 5 4 e 16!= 1.103 The rae of change of he volage is 1.103 V/h afer h. Chaper 5 Pracice Tes Quesion 14 Page 97 a) N 0 = 1000 N() = N 0 e k 1500 = 1000e k (1) k = ln1.5 b) 000 = 1000e (ln1.5) = 1.5 = ln ln1.5!= 1.7 I will ake 1.7 days. c) N! = (ln1.5) ( ) 1000 e (ln1.5) d) N!(5) = 1000e (ln1.5)5 (ln1.5)!= 3079 The growh rae is 3079 baceria per day. MHR Calculus and Vecors 1 Soluions 547
Chaper 5 Pracice Tes Quesion 15 Page 97 a) d() = 5e! b) c) A = 1, A =, d(1) = 5e!(1) d() = 5e!()!= 1.84!= 0.68 A 1 s, he displacemen is 1.84 m. A s, he displacemen is 0.68 m. d) d!() = 5e e) d!(1) = 5e (1)!= 1.84 The rae is 1.84 m/s. Chaper 5 Pracice Tes Quesion 16 Page 97 A!() = 30e A!(1) = 30e 1!= 11.04 The ampliude is changing a 11.04 cm/cm. MHR Calculus and Vecors 1 Soluions 548
Chapers 4 and 5 Review Chapers 4 and 5 Review Quesion 1 Page 98 Answers may vary. For eample: a) y =!cos() b) The insananeous rae of change is zero a he poin This poin is a local maimum poin on he graph.! $, % '. 5! c) The insananeous rae of change is a maimum a he poin $ 4, 0 % '. This poin is a zero of he second derivaive of f on a secion of he graph where he funcion is increasing. 3! d) The insananeous rae of change is a minimum a he poin $ 4, 0 % '. This poin is a zero of he second derivaive of f on a secion of he graph where he funcion is decreasing. 3! e) A poin on he graph ha is a poin of inflecion is he poin $ 4, 0 % '. The funcion is decreasing a his poin and he curve changes from concave down o concave up a his poin. Chapers 4 and 5 Review Quesion Page 98 dy a) 1 cos d = + b) Since he slope m is given by dy d =! dy = 1+ cos(! ) d =! = 0 f (! ) =! Subsiue hese values in he equaion y = m + b. b =! So he equaion of he angen is y =!. c) Answers may vary. For eample: The angen o f ( ) a = 3! will no have he same equaion. The angen line will be parallel o he angen equaion in par b), and he equaion will be y = 3!. MHR Calculus and Vecors 1 Soluions 549
Chapers 4 and 5 Review Quesion 3 Page 98 a) f!( ) = cos b) g! = ( ) (cos sin ) c) Use he double angle ideniy cos = cos! sin. f!() = cos = (cos sin ) = g!() d) Answers may vary. For eample: The original funcions are equal. The epansion for he double angle ideniy for f () is sin = sin cos. Chapers 4 and 5 Review Quesion 4 Page 98 Answers may vary. For eample: A() = sin + cos and A!() = cos sin. When he funcions are graphed in he same viewing screen of a graphing calculaor he rae of change of he ampliude never eceeds he maimum value of he ampliude iself. A() = sin + cos A!() = cos sin Chapers 4 and 5 Review Quesion 5 Page 98 a) s() = 8sin! $ 30 % ' = 8sin! $ 15 % ' is he ime in s, s() is he norh-souh posiion of he rider in m b) s!() = 8 15 cos $ % 15 ' ( MHR Calculus and Vecors 1 Soluions 550
c) The maimum speed occurs when s!!() = 0. s!!() = 8! 15 sin! $ % 15 ' ( 0 = sin! $ % 15 ' (! = k!, k )! 15 = 15k Choose any value of k and subsiue ino s!() o find he maimum value. Take he absolue value since speed is always posiive. s!(15) = 8 15 cos $ % 15 (15) ' ( = 8 15!= 1.68 The maimum speed is 1.68 m/s. d) From par c), he maimum speed occurs for = 15k, k!!. The posiions a which he maimum speed occurs are: s(0) = 0 s(15) = 0 s(30) = 0 Therefore, he posiion of he rider is 0 m each ime he maimum speed is reached. Chapers 4 and 5 Review Quesion 6 Page 98! a) D() = cos! 6 $ % + 9 assuming = 0 a 9:00 A.M., is in hours, and D is he deph in fee. b) D!() =! 3 sin! $ % 6 ' (! c) The maimum speed occurs when sin! 6 $ % = '1.! 6 = 3! + k!, k!! = 9 +1k The deph is rising fases a = 9 h, = 1 h, = 33 h, MHR Calculus and Vecors 1 Soluions 551
d) A = 9, D!(9) =! (9)! sin 3 $ % 6 ' ( =! 3!= 1.047 The waer is rising a 1.047 f/h. Chapers 4 and 5 Review Quesion 7 Page 99 % % a) A!() = 0() $ sin $ 1000! ' ' cos % 1 $ 1000! ' 1000! = 1 5! sin % $ 1000! ' cos % $ 1000! ' b) A!() = 1 5 sin $ % 1000! ' ( cos $ % 1000! ' ( = 0! $ sin 1000! % = 0 1000! =! = 1000!!= 9869.60! $ cos 1000! % = 0 1000! = ' = 500!!= 4934.8 The firs value of is 4934.8 s. c) MHR Calculus and Vecors 1 Soluions 55
Chapers 4 and 5 Review Quesion 8 Page 99 Answers may vary. For eample: a) f!() = lim h0 f ( + h) f () h = lim h0 ( +h) h ( h ) = lim h0 h = lim h0 $ = lim h0 % = ln f!(0) = ln!= 0.693 ( h 1) h ( h 1) h ' ) ( b) The resul will be ln 3 = 1.099, which is greaer han he previous resul. c) f!() = lim h0 = lim h0 = lim h0 = lim h0 $ = 3 lim h0 % = 3 ln3 f!( 0) = ln3 = 1.0986 f ( + h) f () h 3 ( +h) 3 h 3 (3 h ) 3 h 3 (3 h 1) h (3 h 1) h ' ) ( MHR Calculus and Vecors 1 Soluions 553
Chapers 4 and 5 Review Quesion 9 Page 99 a) $.00 b) c) A = 1(1 + 0.5) =.5 Afer 1 year, i will be worh $.5. n! 1 A = 1 1+ $ where n is he number of compounding periods. % n If n = 3, A = $.37. If n = 4, A = $.44. If n = 5, A = $.49. If n = 100, A = $.70. d) Since lim 1+ 1 n! $ % n' ( n = e, which is abou.718, he amoun A is approaching $.7. Chapers 4 and 5 Review Quesion 10 Page 99 a) b) sin cos ln( e ) + ln( e ) = sin + cos = 1! 1! ln 5 $ %! 1 ln( e ) e = 5 $ $ % % = 5 Chapers 4 and 5 Review Quesion 11 Page 99 a) Answers may vary. For eample: No. The smoke deecor is no likely o fail while I own i. The half-life of americium-41 is 43. years. b) Afer 50 years, m = 0.(0.5) 50 43.!= 0.1846 There will be 0.1846 mg. MHR Calculus and Vecors 1 Soluions 554
c) 0.05 = 0.(0.5) 0.5 = (0.5) 43. 43. 43. = ln0.5 ln0.5! = 43. ln0.5 $ ln0.5 %!= 864.4 I will ake 864.4 years. Chapers 4 and 5 Review Quesion 1 Page 99 a) f!() = 1 ln1 b) g!() = 3 % $ 4 ' c) h! ( ) = 5e d) i!() = ln ln 3 % $ 4 ' MHR Calculus and Vecors 1 Soluions 555