CHAPTER : Derivatives of Eonential and Trigonometric Functions Review of Prerequisite Skills,.. a. c. d. a b a b. a. 9 Q" R 7 Q" 7R 9 9 log 6 log 6 c. log d. log 0 0 w e. log z 8 f. log a T b. a. 0 Te -intercet occurs were 0. 0 log 0 ( ) 0 0 Te -intercet is (, 0). 0 8 6 0 An eonential function is alwas ositive, so tere is no -intercet.. a. sin u r cos u r 8 6 6 c. tan u. To convert to radian measure from degree measure, multil te degree measure b a. 60 80. 80 80 c. 90 80 d. 0 80 6 e. 70 80 f. 0 80 g. 80 8 -
. 0 80 6 6. For te unit circle, sine is associated wit te -coordinate of te oint were te terminal arm of te angle meets te circle, and cosine is associated wit te -coordinate. a. sin ub c. tan u b a cos ua d. sin a ub a e. cos a ub b f. sin (u) b 7. a. Te angle is in te second quadrant, so cosine and tangent will be negative. cos u tan u Te angle is in te tird quadrant, so sine will be negative and tangent will be ositive. sin ucos u sin u 9! c. Te angle is in te fourt quadrant, so cosine will be ositive and sine will be negative. Because tan u, te oint (, ) is on te terminal arm of te angle. Te reference triangle for tis angle as a otenuse of " or!. sin u! sin u 9 sin u! tan u sin u cos u cos u! d. Te sine is onl equal to for one angle between 0 and, so. u cos 0 tan is undefined 8. a. Te eriod is or. Te amlitude is. Te eriod is or. Te amlitude is. c. Te eriod is or. Te amlitude is. d. Te eriod is or 6. Te amlitude is 7. e. Te eriod is. Te amlitude is. f. Te eriod is. Because of te absolute value being taken, te amlitude is. 9. a. Te eriod is or. Gra te function from 0 to. Te eriod is, so gra te function from 0 to. 0 0 7 0. a. tan cot sec csc LS tan cot sin cos cos sin sin cos cos sin - Cater : Derivatives of Eonential and Trigonometric Functions
cos sin RS sec csc cos? sin cos sin Terefore, tan cot sec csc. sin sin tan sec LS sin sin sin cos RS tan sec sin cos? cos sin cos sin Terefore, tan sec. sin. a. sin sin sin sin 6, 6 cos cos cos cos,. Derivatives of Eonential Functions, e,.. You can onl use te ower rule wen te term containing variables is in te base of te eonential eression. In te case of e, te eonent contains a variable.. a. e e s e t ds dt et c. e 0t dt 0e0t d. e e e. e 6 f. e ". a. e c. f() e f r() e () e d. f()!e f r() "e e a! b e. (t) e t e t r(t) te t e t f. g(t) e t gr(t) et ( e t ) e t (e t ) ( e t ) e t ( e t ). a. f r() (e e ) (6 )e6! e" ( )e 6 e d(e ) ()(e ) (e )() e e e ( ) e e f r() e e f() e f r() e a ( ) b f r(0) e () e et -
c. r(z) z( e z ) z (e z ) r() ()( e) () (e ) e e e. a. e e ( e )e e (e ) ( e ) () ()() Wen 0, te sloe of te tangent is. Te equation of te tangent is, since te -intercet was given as (0, ). c. Te answers agree ver well; te calculator does not sow a sloe of eactl 0., due to internal rounding. 6. e e Wen And wen, e. e. Te equation of te tangent is e e( ) or e 0. 0 7. Te sloe of te tangent line at an oint is given b ()(e ) (e ) e ( ). At te oint (, e ), te sloe is e (0) 0. Te equation of te tangent line at te oint A is or e 0( ) e. 8. Te sloe of te tangent line at an oint on te curve is e (e ) ( )(e ) e. Horizontal lines ave sloe equal to 0. We solve 0 ( ) e 0. Since e. 0 for all, te solutions are 0 and. Te oints on te curve at wic te tangents are orizontal are (0, 0) and (, ). e 9. If ten (e e ), r and a e e b, s a e e b c (e e )d. 0. a. e e d 9e d 7e d n n ()n ( n )e. a. d(e ) e d e d(e ) ()(e ) (e )() e e e ( ) - Cater : Derivatives of Eonential and Trigonometric Functions
d e () ( )(e ) e e c. d(e ( )) (e )() ( )(e ) e e e e e e ( ) d e () ( )(e ) e e e ( ). a. Wen t 0, N 0000 e 0 000. dn dt 000 c0 0 et 0 d 00 c. Wen t 0, dn 00 bacteria>. dt e 8 7 d. Since e t 0. 0 for all t, tere is no solution to dn dt 0. Hence, te maimum number of bacteria in te culture occurs at an endoint of te interval of domain. Wen t 0, N 0000 e 8 0 89. Te largest number of bacteria in te culture is 000 at time t 0. e. Te number of bacteria is constantl decreasing as time asses.. a. v ds dt 60a t e b 0( e t ) a dv dt 0a t e b t 0e From a., v 0( wic gives e t v t e ), 0. Tus, a 0a v 0 b 0 v. c. v T v ts` v T 0( e t ) ts` 0 lim a b ts` e t 0(), since lim 0 ts` Te terminal velocit of te skdiver is 0 m> s. e t e t 0 d. 9% of te terminal velocit is 9 (0) 8 ms. > 00 To determine wen tis velocit occurs, we solve 0( e t ) 8 e t 8 0 e t 0 e t 0 t and ln 0, wic gives t ln 0 8 s. Te skdiver s velocit is 8 m> s, s after juming. Te distance se as fallen at tis time is S 60(ln 0 e 0 ) 60 aln 0 0 b 8 7. m.. a. i. Let f() ( ). Ten, So, from te table one can see tat ii. Let f() 0.97 00.708 000.769 0 000.78 f() ( ). f() 0..8680 0.0.70 0.00.796 0.000.78?? 0.000.78 0.00.769 0.0.708 0..97 lim ( ) e. S` So, from te table one can see tat lim ( ) e. S0 Tat is, te limit aroaces te value of e.78 8 88c Te limits ave te same value because as S `, S 0. -
. a. Te given limit can be rewritten as e e 0 e 0 lim Tis eression is te limit definition of te derivative at 0 for f() e. e 0 e 0 cf r(0) d Since f r() de te value of te given limit e, is e 0. e e Again, lim is te derivative of e at. e e Tus, lim e. d 6. For Ae e, and dt Am e e. dt Amee Substituting in te differential equation gives Am e e Ame e 6Ae e 0 Ae e (m m 6) 0. Since Ae e 0, m m 6 0 (m )(m ) 0 m or m. d 7. a. sin d c (e e )d cos d cos (e e ) c. Since tan sin cos, d tan ( d sin )(cos ) (sin )( d cos ) (cos ) (e e )( )(cos ) (e e ) (cos ) (e e ) sin (e e )( )(e e ) (cos ) S(e e ) (e e )T (cos ) 8. a. Four terms: e!!.666 666! Five terms: e!!!.708! Si terms: () (cos ) (cos ) e!!!!.76 666! Seven terms: e!!!!!.78 0 6! Te eression for e in art a. is a secial case of e!!!! c. in tat it is te case wen. Ten e e e is in fact e e!!!!! c. Te value of is.. Derivatives of te General Eonential Function, b,. 0. a. d( ) ( ) ln d(. ) ln.(.) ds c. dt d(0t ) dt (0 t ) ln 0 dw d. dn ) d(06nn dn (6 n)(0 6nn )ln 0 e. d( ) ( )ln f. d(00() ) 00() ln -6 Cater : Derivatives of Eonential and Trigonometric Functions
. a. d( () ) ( )(() (ln )) (() )( ) ( ln ) d(() ) ()(() ln ) () () c. d. () ( ln ) v ( t )(t ) dv dt d((t )(t )) ( t )(t ) (t )( t ln ) t t t ln t f() ln ( fr() )( ) ( ) ln ( ) ( ) ln. f(t) 0 t? e t fr(t) (0 t )(te t ) (e t )((0) t ln 0) 0 t e t (t ln 0) Now, set f r(t) 0. So, f r(t) 0 0 t e t (t ln 0) So 0 t e t 0 and t ln 0 0. Te first equation never equals zero because solving would force one to take te natural log of bot sides, but ln 0 is undefined. So te first equation does not roduce an values for wic fr(t) 0. Te second equation does give one value. t ln 0 0 t ln 0 ln 0 t. Wen, te function f() evaluated at is f() ( ) (8). Also, d(() ) ( )ln So, at, ( )(ln ) (ln ) 8 6.6 Terefore, 6.6( ) 6.6 9.9 6.6.9 0. d(0 ) 0 ln 0 So, at, 0 ln 0 0(ln 0) 8.0 Terefore, 0.0( ) 0.0.0.0.0 0 6. a. Te alf-life of te substance is te time required for alf of te substance to deca. Tat is, it is wen 0% of te substance is left, so P(t) 0. 0 00(.) t (.)t (.) t (.) t t(ln.) ln ln. t ln t 8.80 ears Terefore, te alf-life of te substance is about.80 ears. Te roblem asks for te rate of cange wen t 8.80 ears. Pr(t) 00(.) t (ln.) Pr(.80) 00(.) (.80) (ln.) 8 9. So, te substance is decaing at a rate of about 9. ercent> ear at te time.80 ears were te alf-life is reaced. 7. P 0.(0 9 0.00 t )e dp a. dt 0.(09 0.00 t )(0.00 )e dp In 968, and 8 dt 0.(09 0.00 t )(0.00 )e 0. 0 9 dollars> annum In 978, t and dp dt 0.(0 9 0.00 )(0.00 )e 8 0.90 67 0 9 dollars> annum. In 978, te rate of increase of debt aments was $90 670 000> annum comared to $ 0 000> annum in 968. As a ratio, Rate in 978 Rate in 968 7. Te rate of increase for 978 is. 7. times larger tan tat for 968. -7
In 988, t and dp dt 0.(09 0.00 )(0.00)e 8 6.6969 0 9 dollars> annum In 998, t and dp dt 0.(09 0.00 )(0.00)e 8 9. 69 0 9 dollars> annum Rate in 998 Rate in 988 7. As a ratio, Te rate of increase. for 998 is 7. times larger tan tat for 988. c. Answers ma var. For eamle, data from te ast are not necessaril good indicators of wat will aen in te future. Interest rates cange, borrowing ma decrease, rincial ma be aid off earl. 8. Wen 0, te function f() evaluated at 0 is f(0) 0 0. Also, ) d( ( )ln So, at 0, (0)(0 )ln 0 Terefore, 0( 0) So, 0 or. From te gra, one can notice tat te values of v(t) quickl rise in te range of about 0 # t #. Te sloe for tese values is ositive and stee. Ten as te gra nears t 0 te steeness of te sloe decreases and seems to get ver close to 0. One can reason tat te car quickl accelerates for te first 0 units of time. Ten, it seems to maintain a constant acceleration for te rest of te time. To verif tis, one could differentiate and look at values were vr(t) is increasing.. Otimization Problems Involving Eonential Functions,. 7. a. Te maimum value is about 0.89. Te minimum value is 0. 8 6 0 6 8 9. 0 00 80 60 0 0 0 v(t) 0 0 0 60 80 00 0 t Te maimum value is about 0.0. Te minimum value is about 96.96.. a. f() e e on 0 # # 0 f r() e e Let f r() 0, terefore e e 0. Let e w, wen w w 0. w( w ) 0. Terefore, w 0 or w w 6 ". But w $ 0, w ". -8 Cater : Derivatives of Eonential and Trigonometric Functions
Wen w, e, " " Te maimum value is about 0 and te minimum value is about 96. Te graing aroac seems to be easier to use for te functions. It is quicker and it gives te gras of te functions in a good viewing rectangle. Te onl roblem ma come in te second function, m(), because for,. te function quickl aroaces values in te negative tousands. P(0) ln e ln ln " 0. a. P(t) e 0.0t 0 e 0.0(0) 0 e 0 ln " ln ln " 8 0.. f(0) e 0 e 0 0 f(0.) 8 0.89 f(0) e 0 e 0 8 0.0000 Absolute maimum is about 0.89 and absolute minimum is 0. m() ( )e on # # mr() e ()( )e Let mr() 0. e 0, terefore, ()( ) 0.. m() e 8 8 96 m(.) 0.e 8 0 m() 6e 8 8 0.000 0 So, te oulation at te start of te stu wen t 0 is 00 squirrels. Te question asks for lim P(t). ts` As t aroaces `, e 0.0t aroaces 0. e 0.0t So, lim P(t) ts` ts` (0) 0. Terefore, te largest oulation of squirrels tat te forest can sustain is 000 squirrels. c. A oint of inflection can onl occur wen Ps(t) 0 and concavit canges around te oint. 0 P(t) e 0.0t P(t) 0( e 0.0t ) Pr(t) 0( ( e 0.0t ) (0.06e 0.0t )) Ps (t) (.e 0.0t )((e 0.0t ) (0.06e 0.0t ) ( e 0.0t ) (0.0e 0.0t ) Ps(0) wen Solving for t after setting te second derivative equal to 0 is ver tedious. Use a graing calculator to determine te value of t for wic te second derivative is 0,.9. Evaluate P(.9). Te oint of inflection is (.9, 0). d. P(t) 0 0 0 e 0.0t 0 (.e 0.0t )( e 0.0t ) 0.e0.0t ( e 0.0t ) 0.0e0.0t ( e 0.0t ) 0.e 0.0t ( e 0.0t ) 0.0e0.0t ( e 0.0t ) 0 t 0 0 0 00 0 00 0 00 e. P grows eonentiall until te oint of inflection, ten te growt rate decreases and te curve becomes concave down.. a. P() 0 6 ( )e 0.00, 0 # # 000 Using te Algoritm for Etreme Values, we ave P(0) 0 6 0 P(000) 0 6 999e 8 7. 0 6. Now, Pr() 0 6 ()e 0.00 ( )(0.00)e 0.00 0 6 e 0.00 ( 0.00 0.00) -9
Since e 0.00.for all, Pr() 0 wen.00 0.00 0.00 0.00 00. P(00) 0 6 000e.00 8 68. 0 6 Te maimum montl rofit will be 68. 06 dollars wen 00 items are roduced and sold. Te domain for P() becomes 0 # # 00. P(00) 0 6 99e 0. 0.7 0 6 Since tere are no critical values in te domain, te maimum occurs at an endoint. Te maimum montl rofit wen 00 items are roduced and sold is 0.7 0 6 dollars.. R() 0 e 0. 0, 0 # # 8 We use te Algoritm for Etreme Values: Rr() 80e 0. 0 (0.)e 0. 0e 0. ( 0.) e 0.. 0 Since for all, Rr() 0 wen 0 or 0. 0. R(0) 0 R() 8 6. R(8) 8. Te maimum montl revenue of 6. tousand dollars is acieved wen 00 units are roduced and sold. 6. P(t) 00(e t e t ), 0 # t # Pr(t) 00(e t e t ) 00e t ( e t ) Since e t. 0 for all t, Pr(t) 0 wen e t e t t ln (0.) ln (0.) t 0.6. P(0) 0 P(0.6) 8 7. P() 8.98 Te igest ercentage of eole sreading te rumour is 7.% and occurs at te 0.6 oint. 7. a. C 0.0 0 9 e 0.07 t, 0 # t # 00 Caital investment from U.S. sources ($00 million) 6 0 8 6 0 C(t) 0 0 60 80 00 Years since 867 dc 0.0 0 9 0.07 t 0.07e dt In 97, t 80 and te growt rate was dc 0.68 0 0 9 dollars> ear. dt In 967, t 00 and te growt rate was dc. 0 9 dollars> ear. dt Te ratio of growt rates of 967 to tat of 97 is. 0 9 0.68 0 0. 9. Te growt rate of caital investment grew from 68 million dollars er ear in 97 to. billion dollars er ear in 967. c. In 967, te growt rate of investment as a ercentage of te amount invested is. 0 9 9 00 7.%. 8.00 0 d. In 977, t 0 C 9.7 0 9 dollars dc.89 0 9 dollars> ear. dt e. Statistics Canada data sows te actual amount of U.S. investment in 977 was 6. 0 9 dollars. Te error in te model is.%. f. In 007, t 0. Te eected investment and growt rates are dc C 70.90 0 9 dollars and.97 0 9 dt dollars> ear. t -0 Cater : Derivatives of Eonential and Trigonometric Functions
8. a. Te growt function is N t. Te number killed is given b K e. t After 60 minutes, N. Let T be te number of minutes after 60 minutes. Te oulation of te colon at an time, T after te first 60 minutes is P N k 60 T e T, T $ 0 dp dt T 60 a b ln et T a ln b et? T aln b et dp dt 0 wen ln T et ln T e T?. We take te natural logaritm of bot sides: ln ln a. b T ln T 7.0 T a ln b T 7.0 8. min. 0.97 At T 0, P 096. At T 8., P 78 8. For T. 8., dp is alwas negative. dt Te maimum number of bacteria in te colon occurs 8. min after te drug was introduced. At tis time te oulation numbers 78 8. wen 60 T e T P 0 60 T ln T ln Ta ln b T.7 Te colon will be obliterated.7 minutes after te drug was introduced. 9. Let t be te number of minutes assigned to stu for te first eam and 0 t minutes assigned to stu for te second eam. Te measure of stu effectiveness for te two eams is given b E(t) E (t) E (0 t), 0 # t # 0 0.(0 te 0) t 0.6(9 (0 t)e 0 t or 0 ) Er(t) 0.ae t 0 0 te 0.6ae 0 t 0 t 0 (0 t)e 0 b 0.0e t t 0 (0 t) 0.0e0 (0 0 t) t 0 b (0.0e t 0 0.0e0 t )(0 t) Er(t) 0 wen 0 t 0 t 0 (Te first factor is alwas a ositive number.) E(0). 8e. E(0) 6.6 E(0). For maimum stu effectiveness, 0 of stu sould be assigned to te firs eam and 0 of stu for te second eam. 0. Use te algoritm for finding etreme values. First, find te derivative fr(). Ten, find an critical oints b setting fr() 0 and solving for. Also, find te values of for wic fr() is undefined. Togeter tese are te critical values. Now, evaluate f() for te critical values and te endoints and. Te igest value will be te absolute maimum on te interval and te lowest value will be te absolute minimum on te interval.. a. f r() ( )(e ) (e )() e ( ) Te function is increasing wen f r(). 0 and decreasing wen f r(), 0. First, find te critical values of f r(). Solve e 0 and ( ) 0 e is never equal to zero. 0 ( ) 0. So, te critical values are 0 and. Interval e ( ),,, 0 0, So, f() is increasing on te intervals (`, ) and (0, `). Also, f() is decreasing on te interval (, 0). At 0, f r() switces from decreasing on te left of zero to increasing on te rigt of zero. So, 0 is a minimum. Since it is te onl critical oint tat is a minimum, it is te -coordinate of te -
absolute minimum value of f(). Te absolute minimum value is f(0) 0.. a. r e Setting e 0 ields no solutions for. e is a function tat is alwas increasing. So, tere is no maimum or minimum value for e. 8 6 8 6 0 6 8 r ()(e ) (e )() e ( ) Solve e 0 and ( ) 0 e is never equal to zero. 0. So tere is one critical oint:. Interval e ( ),. So is decreasing on te left of and increasing on te rigt of. So is te -coordinate of te minimum of. Te minimum value is e 8.6. Tere is no maimum value. 8 6 8 6 0 6 8 c. r ()(e ) (e )() e ( ) Solve e 0 and ( ) 0 e is never equal to zero. 0 So tere is one critical oint:. Interval e ( ),. So is decreasing on te left of and increasing on te rigt of. So is te -coordinate of te minimum of. Te minimum value is a b (e( ) ) e 8 0.7. Tere is no maimum value. 8 6 8 6 0 d. r ()(e ) (e )() e ( ) Solve e ( ) 0. Tis gives no real solutions. B looking at te gra of f(), one can see tat te function is alwas increasing. So, tere is no maimum or minimum value for e. 8 8 0 8 8. Pr() ()(e 0. ) (e 0. )() e 0. ( ) Solve e 0. 0 and ( ) 0. e 0. gives no critical oints. 0 ( )( ) 0 - Cater : Derivatives of Eonential and Trigonometric Functions
So and are te critical oints. So P() is decreasing on te left of and on Interval e 0. ( ),,,, te rigt of and it is increasing between and. So is te -coordinate of te minimum of P(). Also, is te -coordinate of te maimum of P(). Te minimum value is P() ()(e 0.() ) e 0. 8 0.6. Te maimum value is P() ()(e 0.() ) e 0. 8 0.6.. a. d(t) 0 00 80 60 0 0 t 0 0 6 8 0 Te seed is increasing wen dr(t). 0 and te seed is decreasing wen dr(t), 0. dr(t) (00t)( t )(ln ) ( t )(00) 00() t (t ln ) Solve 00() t 0 and t ln 0. 00() t gives no critical oints. t ln 0 t ln 8. So t is te critical oint. ln Interval 00() t (t ln ) t, ln t. ln So te seed of te closing door is increasing wen 0, t, and decreasing wen t. ln ln. c. Tere is a maimum at t since dr(t), 0 for t, and dr(t). 0 for Te maimum seed is d( ln ) 00 ( ln )() In 8 06. degrees> s d. Te door seems to be closed for t. 0 s.. Te solution starts in a similar wa to tat of 9. Te effectiveness function is E(t) 0.(0 te 0) t 0.6(9 ( t)e t 0 ). Te derivative simlifies to Er(t) 0.0e t t 0 0 (0 t) 0.0e ( t). Tis eression is ver difficult to solve analticall. B calculation on a graing calculator, we can determine te maimum effectiveness occurs wen t 8.6 ours. al 6. P a (L a)e klt a. We are given a 00, L 0 000, k 0.000. 0 6 P 00 9900e 0 t 99e t 0 ( 99e t ) Number of cells (tousands) We need to determine wen te derivative of te growt rate ) d (dp P is zero, i.e., wen dt dt 0. dp dt 0 (99e t ) 990000et ( 99e t ) ( 99e t ) d P dt 0 8 6 P ln ln t. ln. 0 0 6 8 0 Das 990 000et ( 99e t ) 990 000e t ( 99e t ) ()( 99et )(99e t ) ( 99e t ) 990 000et ( 99e t ) 98(990 000)e t ( 99e t ) t -
d P 0 wen dt 990000e t ( 99e t 98e t ) c. Wen t, ( 99e ) 8 0 cells> da. Wen t 8, ( 99e 8 ) 8 cells> da. Te rate of growt is slowing down as te colon is getting closer to its limiting value. Mid-Cater Review,. 8 9. a. d(e ) (e )()r (e )() e d(7e 7 ) (7e 7 )a 7 br (7e 7 )a 7 b dp dt 990 000e dp dt 990 000e 8 0 99e t e t 99 t ln 99 8.6 After.6 das, te rate of cange of te growt rate is zero. At tis time te oulation numbers 0. e7 c. ( )(e )r ( )r(e ) ( )((e )()r) ( )(e ) ( )((e ))() e e e e ( ) d. ( ) (e )r (( ) )r(e ) ( ) (e ) (( ))(e ) ( )(e ) ( )(e ) (e )( ) (e )( ) e. ( e )( e )r ( e )( (e )()r) ( e )( (e )()) ( e )( e ) ( e e e ) ( e e e ) f. (e e )(e e )r (e e ) (e e )(e e )r (e e ) (e e )(e (e )()r) (e e ) (e e )(e (e )()r) (e e ) (e e )(e (e )()) (e e ) (e e )(e (e )()) (e e ) (e e )(e e ) (e e ) (e e )(e e ) (e e ) e e 0 e 0 e (e e ) (e e 0 e 0 e ) (e e ) e e 0 e 0 e e (e e ) e0 e 0 e (e e ) (e e ) dp. a. dt 00et (t)r 00e t () 00e t Te time is needed for wen te samle of te substance is at alf of te original amount. So, find t wen P. P 00e t 00et 00 et ln 00 t ln 00 t - Cater : Derivatives of Eonential and Trigonometric Functions
Now, te question asks for t ln 00 8.06 Pra ln 00 b. dp dt Pr wen (using a calculator). ()(e )r (e )()r ()(e ) (e )() e e At te oint 0, 0e0 e 0. At te oint 0, 0e 0 So, an equation of te tangent to te curve at te oint 0 is ( 0) 0. a. r (e )r e s e r ()(e )r (e )()r ()((e ) ()r) (e )() ()((e )()) e e e s ()(e )r (e )()r e ()r ()((e )()r) (e )() (e )() ()((e )()) e e e e c. r (e )( )r ( )(e )r (e )() ( )(e ) e e e e e s (e )r ()(e )r (e )()r e e (e )() e e e e e. a. (8 )(ln 8)( )r (8 )(ln 8)() (ln 8)(8 ).((0)0. )(ln 0)(0.)r.((0) 0. )(ln 0)(0.) 0.6(ln 0)((0). ) c. fr() ( )( )r ( )( )r ( )( )(ln ) ( )() (ln )( ) ((ln )( ) ) d. Hr() 00(() )(ln )( )r 00(() )(ln )() 900(ln )() 900(ln )() e. qr() (.9)? (ln.9).9().9 (.9)? (ln.9).9() 0.9 (ln.9)(.9).9 0.9 f. fr() ( ) ( )r ( )(( ) )r ( ) ( )(ln ) ( )(( )) (ln )( )( ) ( )( ) ((ln )( ) ) 6. a. Te initial number of rabbits in te forest is given b te time t 0. R(0) 00(0 e 0) 0 00(0 ) 00() 00 Te rate of cange is te derivative,. R(t) 000 00(e t dr dt 0 00(e 0)a t r t 0 b 00(e 0)a t 0 b 0(e 0) t c. ear monts Te question asks for Rr() 0(e 8.06 d. To find te maimum number of rabbits, otimize te function. Rr(t) 0(e 0) t 0 0(e 0) t 0 e t 0 0) 0) dr dt dr dt Rr wen t. -
Wen solving, te natural log (ln) of bot sides must be taken, but (ln 0) does not eist. So tere are no solutions to te equation. Te function is terefore alwas decreasing. So, te largest number of rabbits will eist at te earliest time in te interval at time t 0. To ceck, comare R(0) and R(6). R(0) 00 and R(6) 8 0. So, te largest number of rabbits in te forest during te first ears is 00. e. 6000 000 000 0 0 0 0 0 0 Te gra is constantl decreasing. Te -intercet is at te oint (0, 00). Rabbit oulations normall grow eonentiall, but tis oulation is srinking eonentiall. Peras a large number of rabbit redators suc as snakes recentl began to aear in te forest. A large number of redators would quickl srink te rabbit oulation. 7. Te igest concentration of te drug can be found b otimizing te given function. C(t) 0e t 0e t Cr(t) (0e t )(t)r (0e t )(t)r (0e t )() (0e t )() 0e t 0e t Set te derivative of te function equal to zero and find te critical oints. 0 0e t 0e t 0e t 0e t et e t et e t (et )(e t ) ett et ln t aln b t Terefore, t (ln ) 8 0. is te critical value. Now, use te algoritm for finding etreme values. C(0) 0(e 0 e 0 ) 0 Ca aln bb 8.8 (using a calculator) C() 0.000 So, te function as a maimum wen t (ln ) 8 0.. Terefore, during te first five ours, te igest concentration occurs at about 0. ours. 8. ce k r cke k Te original function is increasing wen its derivative is ositive and decreasing wen its derivative is negative. e k. 0 for all k, PR. So, te original function reresents growt wen ck. 0, meaning tat c and k must ave te same sign. Te original function reresents deca wen c and k ave oosite signs. 9. a. A(t) 000e 0.0t 000e 0.0(0) 000 Te initial oulation is 000. at t 7 A(7) 000e 0.0(7) 7 After a week, te oulation is 7. c. at t 0 A(0) 000e 0.0(0) 9 After 0 das, te oulation is 9. 0. a. P() 760e 0.() 8 06.80 mm Hg P(7) 760e 0.(7) 8 6.8 mm Hg c. P(9) 760e 0.(9) 8 6.7 mm Hg. A 00e 0. Ar 00e 0. (0.) 0e 0. Wen 0% of te substance is gone, 0 0 00e 0. 0. e 0. ln (0.) ln e 0. ln (0.) 0. ln e -6 Cater : Derivatives of Eonential and Trigonometric Functions
ln 0. ln e 0. ln 0. 0. ln e. Ar 0e 0. Ar(.) 0e 0.(.) Ar 8 Wen 0% of te substance is gone, te rate of deca is % er ear.. f() e f r() e ()e e ( ) So e. 0. 0. Tis means tat te function is increasing wen... Wen, r () ln r ln ln ( ) ln ( ) ( ln ) ln ( ln ) ln 0. a. A P( i) t A(t) 000( 0.06) t 000(.06) t Ar(t) 000(.06) t () ln (.06) 000(.06) t ln.06 c. Ar() 000(.06) ln.06 $6.7 Ar() 000(.06) ln.06 $77.98 Ar(0) 000(.06) 0 ln.06 $0. d. No, te rate is not constant. Ar() e. ln.06 A() Ar() ln.06 A() Ar(0) ln.06 A(0) f. All te ratios are equivalent (te equal ln.06, wic is about 0.08 7), wic means tat constant.. ce r c(e ) (0)e ce r ce Ar(t) A(t). Te Derivatives of sin and cos,. 6 7 d(). a. (cos )? cos d() (sin )? 6 sin c. (cos ( ))? d( ) ( )(cos ( )) d() d. sin ()? 8 sin () d() e. cos ()? sin ()? d() cos () sin () f. (ln ) cos sin g. cos (e )? d(e ) e cos (e ) d( ). cos ( )? 9 cos ( ) is -7
i. sin 0 sin j. cos a b? d(. a. cos a b c. sin (sin )? cos d. (sin )( cos ) (sin )( ( cos )? (sin ) ( cos ) (cos ) sin ( cos ) cos cos sin cos ( cos ) ( cos ) ( cos ) sin cos cos ( cos ) cos ( cos ) cos e. (e )(sin cos ) (cos sin )(e ) e (sin cos cos sin ) e ( cos ) f. ( )(cos ) (sin )(6 ) ()(sin ) (cos )() cos 6 sin sin cos. a. Wen, fr() cos ) ( sin )(sin ) (cos )( cos ) sin cos (cos sin ) cos () ( )(cos ) ( )( sin ) (cos )( ) sin cos d(sin ) sin (sin )? f() f a b sin a b #. f ra b cos So an equation for te tangent at te oint is # a b # a #b 0 Wen 0, f() f(0) 0 sin (0) 0. f r() cos f r(0) cos (0) So an equation for te tangent at te oint 0 is 0 ( 0) 0 c. Wen f() f a b cos a?, b cos () f r() sin ()? d() sin () f ra b sin a? b sin () 0 So an equation for te tangent at te oint is () 0a b 0 d. f() sin cos, Te oint of contact is (, 0). Te sloe of te tangent line at an oint is fr() cos sin. At (, 0), te sloe of te tangent line is cos sin. An equation of te tangent line is ( ). -8 Cater : Derivatives of Eonential and Trigonometric Functions
e. f() cos a b, Te oint of tangenc is a Te sloe of te,! tangent line at an oint is f r() sin ( ). At a,! b, sin a 6 b. te sloe of te tangent line is An equation of te tangent line is " ( ). f. Wen f() f a, b sin a b cos a b ()(0) 0 f r() ( sin )(sin ) (cos )( cos ) sin cos (cos sin ) cos () f ra b cos a? b cos So an equation for te tangent wen is 0 a b 0. a. One could easil find f r() and gr() to see tat te bot equal (sin )(cos ). However, it is easier to notice a fundamental trigonometric identit. It is known tat sin cos. So, sin cos. Terefore, one can notice tat f() is in fact equal to g(). So, because f() g(), f r() gr(). f r() and gr() are negatives of eac oter. Tat is, f r() (sin )(cos ) wile gr() (sin )(cos ).. a. v(t) (sin (!t)) vr(t) sin (!t)? d(sin (!t)) dt sin (!t)? cos (!t)? d(!t) dt sin (!t)? cos (!t)? t sin (!t)? cos (!t)? sin (!t) cos (!t)!t v(t) ( cos t sin t) vr(t) ( cos t sin t)!t d( cos t (sin t) ) dt d(sin t) sin t (sin t)? dt # cos t sin t sin t (sin t)(cos t) # cos t sin t c. () sin sin sin So, treat sin sin as one function, sa f() and treat sin as anoter function, sa g(). Ten, te roduct rule ma be used wit te cain rule: r() f()gr() g()f r() (sin sin )( cos ) (sin )(sin )( cos ) (sin )(cos ) sin sin cos sin sin cos sin sin cos d. mr() ( cos )? d( (cos ) ) ( cos )? ( (cos )(sin )) ( cos )? ( sin cos ) 6. B te algoritm for finding etreme values, te maimum and minimum values occur at oints on te gra were f r() 0, or at an endoint of te interval. a. sin cos Set and solve for to find an critical oints. 0 cos sin 0 cos sin sin cos tan, Evaluate f() at te critical numbers, including te endoints of te interval. -9
So, te absolute maimum value on te interval is # wen and te absolute minimum value on te interval is # wen. 0 sin Set and solve for to find an critical oints. 0 sin 0 sin sin 6, 6 Evaluate f() at te critical numbers, including te endoints of te interval. So, te absolute maimum value on te interval is.6 wen and te absolute minimum value 6 on te interval is. wen. 8 8 0 6 f() cos 0 8. 8. f() cos sin " " 6 # 6 6 # 8.6 8. c. cos sin Set and solve for to find an critical oints. 0 cos sin 0 sin cos sin cos tan, 7 Evaluate f() at te critical numbers, including te endoints of te interval. So, te absolute maimum value on te interval is # wen and te absolute minimum value on te interval is # wen 7. 0 0 d. cos sin Set and solve for to find an critical oints. 0 cos sin 0 cos sin sin cos tan tan a b tan (tan ) Using a calculator, 8 0.6. Tis is a critical value, but tere is also one more in te interval 0 # #. Te eriod of tan is, so adding to te one solution will give anoter solution in te interval. 0.6 8.78 7 f() sin cos " " -0 Cater : Derivatives of Eonential and Trigonometric Functions
Evaluate f() at te critical numbers, including te endoints of te interval. So, te absolute maimum value on te interval is wen 8 0.6 and te absolute minimum value on te interval is wen 8.79. 8 0 8 0 0.6.79 f() sin cos 7. a. Te article will cange direction wen te velocit, sr(t), canges from ositive to negative. sr(t) 6 cos t Set sr(t) 0 and solve for t to find an critical oints. 0 6 cos t 0 cos t, t, t Also, tere is no given interval so it will be beneficial to locate all solutions. Terefore, t for some ositive k, k integer k constitutes all solutions. One can create a table and notice tat on eac side of an value of t, te function is increasing on one side and decreasing on te oter. So, eac t value is eiter a maimum or a minimum. t Te table continues in tis attern for all critical values t. So, te article canges direction at all critical values. Tat is, it canges direction for t for ositive integers k. k, k From te table or a gra, one can see tat te article s maimum velocit is 8 at te time t k. 7 s(t) 8 sin t 8 8 8 8 8. a. 0 f() Te tangent to te curve f() is orizontal at te oint(s) were f r() is zero. f r() sin cos Set f r() 0 and solve for to find an critical oints. cos sin 0 cos sin sin cos tan (Note: Te solution is not in te interval so it is not included.) Wen 0 # # f() f( ), #. So, te coordinates of te oint were te tangent to te curve of f() is orizontal is ( 9. csc, #). sin (sin ) sec (cos ) cos Now, te ower rule can be used to comute te derivates of csc and sec. ((sin ) ) r (sin ) d(sin )? (sin )? cos cos (sin ) ((sin ) ) r (sin ) d(sin )? sin? cos sin csc cot d(cos ) ((cos ) ) r (cos )? (cos )? (sin ) sin (cos ) cos? sin cos sec tan -
0. sin At te oint sin a? 6 b sin a b a! b # Terefore, at te oint te sloe of te tangent to te curve cos is #.. a. Te article will cange direction wen te velocit, sr(t) canges from ositive to negative. sr(t) 6 cos t Set sr(t) 0 and solve for t to find an critical oints. 0 6 cos t 0 cos t, t 8, 8 t Also, tere is no given interval so it will be beneficial to locate all solutions. Terefore, t for some ositive 8 k, 8 k integer k constitutes all solutions. One can create a table and notice tat on eac side of an value of t, te function is increasing on one side and decreasing on te oter. So, eac t value is eiter a maimum or a minimum. Te table continues in tis attern for all critical values t. So, te article canges direction at all critical values. Tat is, it canges direction for t for ositive integers k. k, k From te table or a gra, one can see tat te article s maimum velocit is at te time t k. t ( 6, ), 8 ( 6, ), c. At t 0, s 0, so te minimum distance from te origin is 0. Te maimum value of te sine 8 8 7 8 s(t) sin t function is, so te maimum distance from te origin is () or.. m m u u m Label te base of a triangle and te eigt. So cos u and sin u. Terefore, cos u and sin u. Te irrigation cannel forms a traezoid and te (b area of a traezoid is b ) were b and b are te bottom and to bases of te traezoid and is te eigt. b b cos u cos u cos u sin u Terefore, te area equation is given b ( cos u ) sin u A ( cos u) sin u cos u sin u sin u sin u cos usin u To maimize te cross-sectional area, differentiate: Ar (sin u)(sin u) (cos u)(cos u) cos u sin ucos ucos u Using te trig identit sin ucos u, use te fact tat sin u cos u. Ar ( cos u) cos ucos u cos ucos ucos u cos ucos u Set Ar 0 to find te critical oints. 0 cos ucos u 0 ( cos u)(cos u) Solve te two eressions for u. cos u cos u u Also, cos u u (Note: Te question onl seeks an answer around 0 #u# So, tere is no need to find all solutions. b adding k for all integer values of k.) Te area, A, wen uis 0 so tat answer is disregarded for tis roblem. - Cater : Derivatives of Eonential and Trigonometric Functions
Wen u, A sin cos sin a!? b!! Te area is maimized b te angle u.. Let O be te centre of te circle wit line segments drawn and labeled, as sown. A D!! O uu C R u In ^OCB, /COB u. Tus, sin u and cos u, R R so R sin u and R cos u. Te area A of ^ ABD is A 0 DB 00AC 0 (R ) R sin u(r R cos u) R (sin usin u cos u), were 0, u, da du R ( cos u cos u cos u sin u( sin u)). da We solve du 0: cos u sin u cos u 0 cos ucos u 0 ( cos u)(cos u ) 0 cos u or cos u u or u (not in domain). As u S 0, A S 0 and as u S, A S 0. Te maimum area of te triangle is wen u i.e., u, 6. R B! R. First find s. A cos kt B sin kt r ka sin kt kb cos kt s k A cos kt k B sin kt So, s k k A cos kt k B sin kt k (A cos kt B sin kt) k A cos kt k B sin kt k A cos kt k B sin kt 0 Terefore, s k 0.. Te Derivative of tan,. 60. a. sec a d b sec sec sec a d b sec sec c. tan ( )a d tan ( )b tan ( )a d tan ( )b tan ( ) sec ( )a d b 6 tan ( ) sec ( ) d. tan sec ( d ) tan tan sec tan ( tan sec ) tan e. sec ( )a d b tan a d b (tan ) sec ( ) tan sec f. tan ( cos )a d b sin sec a d b tan cos sin sec (tan cos sin sec ). a. Te general equation for te line tangent to te function f() at te oint (a, b) is b fr()( a). f() tan fr() sec -
f a b 0 f ra b Te equation for te line tangent to te function f() at is ( ). Te general equation for te line tangent to te function f() at te oint (a, b) is b f r()( a). f() 6 tan tan f r() 6 sec sec a d b f r() 6 sec sec f(0) 0 f r(0) Te equation for te line tangent to te function f() at 0 is.. a. sec (sin )a d sin b cos sec (sin ) tan ( ) a d tan ( )b tan ( ) sec ( ) c. a d ( )b tan ( ) sec ( ) tan (cos )a d tan (cos )b tan (cos ) sec (cos )a d cos b tan (cos ) sec (cos ) sin d. (tan cos )a d tan cos b (tan cos )(sec sin ) e. tan ( sin )a d sin b sin sec tan sin cos sin sec sin ( tan cos sin sec ) f. etan " a d tan "b e tan " (sec ")a d "b " etan " sec ". a. tan cos sin sec sin cos? cos sin? cos sin sin cos d cos cos cos ( sin ( cos ) d cos ) cos cos cos sin cos cos cos cos sin cos cos sec sin cos tan a d tan b tan sec sin cos? cos sin cos d cos 6 sin cos ( d cos ) cos 6 cos 6 sin cos cos 6 cos 6 sin cos? cos sec 6 tan sec sec ( tan ). Te sloe of f() sin tan equals zero wen te derivative equals zero. f() sin tan f r() sin (sec ) tan (cos ) sin (sec ) sin (cos ) cos sin (sec ) sin sin (sec ) sec is alwas ositive, so te derivative is 0 onl wen sin 0. So, fr() equals 0 wen 0,, and. Te solutions can be verified b eamining te gra of te derivative function sown below. - Cater : Derivatives of Eonential and Trigonometric Functions
f'() 0 6. Te local maimum oint occurs wen te derivative equals zero. sec sec 0 sec sec 6" 6 wen 6 so te local maimum 0, oint occurs wen 6 Solve for. wen. a b tan a b 0.7 Solve for wen. a b tan a b 0.7 Te local maimum occurs at te oint (, 0.7 ). 7. sec tan cos sin cos sin cos cos ( sin )(sin ) cos cos (sin sin ) cos cos sin sin cos sin cos Te denominator is never negative. sin. 0 for since sin reaces its minimum,, of at, Since te derivative of te original. function is alwas ositive in te secified interval, te function is alwas increasing in tat interval. 8. Wen tan (, ) r sec Wen, r asec b (#) So an equation for te tangent at te oint a b ( ) 0 9. Write tan sin and use te quotient rule to cos derive te derivative of te tangent function. 0. cot tan tan (0) () sec tan sec tan cos sin cos sin csc. Using te fact from question 0 tat te derivative of cot is csc, f r() csc (csc ) d(csc ) f s() 8 (csc )? 8 (csc )? (csc cot ) 8 csc cot is -
Review Eercise,. 6 6. a. r 0 e e r e c. r e? e d. r e? d( ) (6 )e e. r ()(e ) (e )() e ( ) f. sr (et )(e t ) (e t )(e t ) (e t ) et e t (e t e t ) (e t ) e t (e t ). a. 0 ln 0? ln? d( ) 6( )ln c. ()( ln ) ( )()? ( ln ) d. ( )( ln ) ( )( )? ( ln ) e. ()( ) ()( ln ) ( )()? ( ln ) ln f. ( # )( ) d( ) (# )( ) ( )a #? ln? d( #) b ( # )a b ( )a #? ln?! b # a ln! b d(). a. cos ()? sin ()? d() 6 cos () 8 sin () sec ()? d() sec () c. ( cos ) ( cos )? sin ( cos ) d. ()asec ()? d() b (tan()) () sec () tan e. (sin )ae? d() b (e )acos? d() b e sin e cos e ( sin cos ) (cos()) f. d(cos ()) (cos ())? (cos ())? sin ()? d() cos () sin (). a. f() e? f r() (e )( ) ( )(e ) e a b e a b d( cos ) Now, set f r() 0 and solve for. 0 e a b Solve and e 0 0. e is never zero. 0 0 ( ) 0 So 0 or. (Note, owever, tat cannot be zero because tis would cause division b zero in te original function.) So. Te function as a orizontal tangent at (, e). -6 Cater : Derivatives of Eonential and Trigonometric Functions
. a. e ( ) 0 Tis means tat te sloe of te tangent to f() at te oint wit -coordinate is 0. 6. a. r ()(e ) (e )() e e s ()(e ) (e )() e e e ( ) r ()(0e 0 ) (e 0 )() 0e 0 e 0 s (0)(0e 0 ) (e 0 )(0) 0e 0 00e 0 0e 0 0e 0 00e 0 0e 0 0e 0 ( ) 7. e e e (e ) (e )(e ) (e ) Now, f r() ()ae? d() b (e )() e e e ( ) f ra b e a? b e e e e (e ) e (e ) e e (e ) e e e e (e ) e ( ) 8. Te sloe of te required tangent line is. Te sloe at an oint on te curve is given b e. To find te oint(s) on te curve were te tangent as sloe, we solve: e e ln ln. Te oint of contact of te tangent is (ln,ln ). Te equation of te tangent line is ln ( ln ) or ln 0. 9. Wen, f() f a b sina b () r fr() ()(cos ) (sin )() cos sin f ra b cos sin (0) So an equation for te tangent at te oint a b 0 sin t 0. If s(t) is te function describing cos t an object s osition at time t, ten v(t) sr(t) is te function describing te object s velocit at time t. So v(t) sr(t) ( cos t)(cos t) (sin t)( sin t) ( cos t) sra b ( cos? )(cos ) ( cos? ) (sin )( sin? ) ( cos? ) ( cos )(" ) ( " )( sin ) ( cos ) ( 0)(" ) ( " )(? ) ( 0) " # 9 # # # 8? 9 is -7
So, te object s velocit at time t is # 8 8 0.98 metres er unit of time.. a. Te question asks for te time t wen Nr(t) 0. N(t) 60 000 000te t 0 Nr(t) 0 (000t)a 0 00e t 0 (t 0) 00e t 0 is never equal to zero. t 0 0 0 t Terefore, te rate of cange of te number of bacteria is equal to zero wen time t 0. dm Te question asks for Mr(t) wen t 0. dt Tat is, it asks for Mr(0). M(t) (N 000) Mr(t) (N 000) (N 000) dn From art a., dt t 0 e 00te t 0 t 000e 0 00e t 0 (t 0) Set Nr(t) 0 and solve for t.? d(n 000) dt? dn dt 0 b t (e 0 )(000) Nr(t) 00e t 0 (t 0) and N(t) 60000 000te t 0 So Mr(t) 00e 0 (t 0) (N 000) First calculate N(0). N(0) 60000 000(0)e 0 0 60000 0000e 8 7 0 So Mr(0) 00e 0 (0 0) (N(0) 000) 00e (0) (7 000) 8 606. 6. 8 0.6 So, after 0 das, about 0.6 mice are infected er da. Essentiall, almost 0 mice are infected er da wen t 0.. a. c (t) te t ; c (0) 0 cr (t) e t te t e t ( t) Since e t. 0 for all t, c r (t) 0 wen t. Since c r(t). 0 for 0 # t,, and c r(t), 0 for all t., c (t) as a maimum value of e 8 0.68 at t. c (t) t e t ; c (0) 0 c r(t) te t t e t te t ( t) c r(t) 0 wen t 0 or t. Since c r(t). 0 for 0, t, and c r(t), 0 for all t., c (t) as a maimum value of 8 0. at e t. Te larger concentration occurs for medicine c. c (0.) 0.0 c (0.) 0. In te first alf-our, te concentration of c increases from 0 to 0.0, and tat of c increases from 0 to 0.. Tus, c as te larger concentration over tis interval.. a. ( e ) r ( e ) 0 e () ( e ) (e ) 9e ( e ) e r e e c. e e r e e (e )() e e d. ( e ) r ( e ) 0 e () e ( e ). a. r ln (0.7) r (0.7) ln (0.7) c. () r () () ln () ln d. () r () ln e. e r e () ln e e f. (0) r ()0 ln 0 6(0) ln 0. a. r cos? d( ) ln cos -8 Cater : Derivatives of Eonential and Trigonometric Functions
r ( )(cos ) (sin )() cos sin c. r cos a b? d( ) d. r (cos )(cos ) (sin )(sin ) cos sin e. (cos ) d(cos ) r (cos )? cos sin f. cos (sin ) r (cos )( (sin )(cos )) (sin ) (sin ) sin cos sin 6. Comute wen to find te sloe of te line at te given oint. r sin So, at te oint r f r() is, Terefore, an equation of te line tangent to te curve at te given oint is 7. Te velocit of te object at an time t is v ds Tus, v 8 (cos (0t))(0) 80 cos (0t). Te acceleration at an time t is a dv Hence, a 80(sin (0t))(0) 800 sin (0t). d Now, s 00 s dt 800 sin (0t) 8. Since cos a b f ra b sin a b. 0 a b 0 dt. 00 (8 sin (0t)) 0. s cos (t ), v ds dt asinat bb 0 sin at b dt d s. dt and Te maimum values of te dislacement, velocit, and acceleration are, 0, and 0, resectivel. 9. Let te base angle be u, 0,u, and let te, sides of te triangle ave lengts and, as sown. Let te erimeter of te triangle be P cm. u Now, and cos u sin u so cos u and sin u. Terefore, P cos u sin u and dp sin u cos u. du For critical values, sin u cos u0 sin ucos u tan u u since 0,u,,. Wen u, #. As u S 0, cos u S, sin u S 0, and P S 0. As u S cos u S 0, sin u S and, P S 0. Terefore, te maimum value of te erimeter is! cm, and occurs wen te oter two angles are eac rad, or. 0. Let l be te lengt of te ladder, u be te angle between te foot of te ladder and te ground, and be te distance of te foot of te ladder from te fence, as sown. Tus, a dv dt 0 (cos (t )) l 0 cos at cos u and l cos u P!!!. tan u were. tan u. -9
u dl Solving ields: du 0 sin u. cos u0 Te lengt of te ladder corresonding to tis value of u is l 8. m. As u S and 0 l increases, witout bound. Terefore, te sortest ladder tat goes over te fence and reaces te wall as a lengt of. m.. Te longest ole tat can fit around te corner is determined b te minimum value of. Tus, we need to find te minimum value of l. 0.8 From te diagram, l u. 0.8 sin u and Tus, l 0 #u# cos u 0.8 sin u, : dl du sin u 0.8 cos u cos u sin u 0.8 sin ucos u. cos u sin u u wall. Relacing, tan u l cos u l. 0,u, sin u cos u, dl cos u. du sin sin u u cos u. cos usin u sin u cos. u tan u. tan u!. u 8 0.66. cos u. Solving ields: du 0 0.8 sin ucos u0 Now, l sin (0.8) 8.. Wen u0, te longest ossible ole would ave a lengt of 0.8 m. Wen u te longest ossible, ole would ave a lengt of m. Terefore, te longest ole tat can be carried orizontall around te corner is one of lengt. m.. We want to find te value of tat maimizes u. Let /ADC aand /BDC Tus, uab: tan utan (a b) tan atan b tan a tan b. A dl From te diagram, tan a9 and tan b. 9 Hence, tan u 7 6 7. We differentiate imlicitl wit resect to : sec u du 6( 7) 6() ( 7) du 6 6 sec u( 7) du tan u. tan u!. tan u 8.077 u 8 0.8. 0.8 cos (0.8) 6 B C a u b 9 7 Solving ields: 0 6 6 0 7 #. D -0 Cater : Derivatives of Eonential and Trigonometric Functions
. a. f () (sin ( )) fr() 8 sin ( ) cos ( ) f s () (8 sin ( ))(sin ( )) (cos ( ))(8 cos ( )) 8 sin ( ) 8 cos ( ) f() ( cos )(sec ) fr() ( cos )( sec? sec tan ) (sec ) ( sin ) ( cos )(sec tan ) sin (sec ) Using te roduct rule multile times, f s() ( cos )Ssec? sec tan ( sec? sec tan )T (sec tan )( sin ) ( sin )( sec? sec tan ) (sec ) ( cos ) cos sec 8 cos tan sec sin tan sec sin tan sec cos sec cos sec 8 cos tan sec 8 sin tan sec cos sec Cater Test,. 66. a. e e? ln? ( ) c. e e e e e e d. sin cos cos (sin )() cos sin e. sin ( ) sin ( )(cos( )()) 6 sin ( ) cos ( ) f. tan " sec " a " b () sec " ". Te given line is 6 or 6, so te sloe is 6. e e () 6e In order for te tangent line to be arallel to te given line, te derivative as to equal 6 at te tangent oint. 6e 6 e 0 Wen 0,. Te equation of te tangent line is 6( 0) or 6. Te tangent line is te given line.. e sin e cos Wen 0, or, so te sloe of te tangent line at (0, ) is. Te equation of te tangent line at (0, ) is ( 0) or.. v(t) 0e kt a. a(t) vr(t) 0ke kt k(0e kt ) kv(t) Tus, te acceleration is a constant multile of te velocit. As te velocit of te article decreases, te acceleration increases b a factor of k. At time t 0, v 0 cm> s. c. Wen v, we ave 0e kt e kt kt ln a b ln t ln k. ln After s ave elased, te velocit of te article k is cm> s. Te acceleration of te article is k at tis time.. a. f() (cos ) d(cos ) f r() (cos )? (cos )? (sin ) sin cos f s() ( sin )(sin ) (cos )( cos ) sin cos (sin cos ) -
f() cos cot f r() (cos )(csc ) (cot )(sin ) cos? sin cos sin? sin cos sin? sin cos cot csc cos f s () (cot )(csc cot ) (csc )(csc ) sin csc cot csc sin 6. f() (sin ) To find te absolute etreme values, first find te derivative, set it equal to zero, and solve for. d(sin ) f r() (sin )? sin cos sin Now set fr() 0 and solve for. 0 sin 0,, 0, in te interval 0 # #., Evaluate f() at te critical numbers, including te endoints of te interval. 0 f() (sin ) 0 0 So, te absolute maimum value on te interval is wen and te absolute minimum value on te interval is 0 wen 0 and. 7. f() Find te derivative, f r(), and evaluate te derivative at to find te sloe of te tangent wen. f r() ln f r() ln ln 8 0. 8. e e To find te maimum and minimum values, first find te derivative, set it equal to zero, and solve for. r ()(e ) (e )() e e e e e e e ( ) Now set r 0 and solve for. 0 e ( ) e is never equal to zero. ( ) 0 So. Terefore, te critical value is. Interval e ( ),, So f() is decreasing on te left of and increasing on te rigt of. Terefore, te function as a minimum value at a, Tere e is no maimum value. 9. f() cos sin So, f() cos sin cos. a. f r() sin ( sin )(sin ) (cos )( cos ) sin sin cos Set f r() 0 to solve for te critical values. sin sin cos 0 sin sin ( sin ) 0 sin sin sin 0 sin sin 0 ( sin )( sin ) 0 sin 0 and sin 0 So, sin. In te given interval, tis occurs wen. 6, 6 Also, sin. In te given interval, tis occurs wen. Terefore, on te given interval, te critical numbers for f() are. 6, 6, To determine te intervals were f() is increasing and were f() is decreasing, find te sloe of f() in te intervals between te endoints and te critical numbers. To do tis, it els to make a table. #, 6 6,, 6 sloe of f() So, f() is increasing on te interval 6,,, # and f() is decreasing on te 6,, 6 intervals #, and. 6 6,, - Cater : Derivatives of Eonential and Trigonometric Functions
c. From te table in art, it can be seen tat tere is a local maimum at te oint were and 6 tere is a local minimum at te oint were. 6 d. 0 Cumulative Review of Calculus. a. f() f(a ) f(a) m f( ) ( ) ( ) 8 0 6 6 6 f() f(a ) f(a) m f( ) ( ) ( ) ( ) ( ) c. f()! f(a ) f(a) m f(6 )! 9 (! 9 )(! 9 ) (! 9 ) 9 9 (! 9 ) (! 9 ) (! 9 ) d. 6 f() m lim 60 lim 60 ln. a. average velocit s(t ) s(t ) t t f(a ) f(a) ()? ( ) ( ) ( ) cange in distance cange in time () () (() () ) 6 ms > instantaneous velocit sloe of te tangent s(a ) s(a) m -
s( ) s() c ( ) ( ) (() () ) d 8 9 8 ( ) ms > f(a ) f(a). m ( ) 6 f( ) f() lim ( ) 6 f( ) f() Terefore, f().. a. Average rate of cange in distance wit resect to time is average velocit, so average velocit s(t ) s(t ) t t s() s().9().9() 9.6 ms > Instantaneous rate of cange in distance wit resect to time sloe of te tangent. f(a ) f(a) m f( ) f().9( ).9() 9.6 9.6.9 9.6 9.6.9 9.6.9 9.6 ms > c. First, we need to determine t for te given distance: 6.9.9t 9.98 t.7 t Now use te sloe of te tangent to determine te instantaneous velocit for t.7: f(.7 ) f(.7) m.9(.7 ).9(.7) 6.9.6.9 6.9.6.9.6.9.6 ms >. a. Average rate of oulation cange (t ) (t ) t t (8) (8) ((0) (0) ) 8 0 8 8 0 9 tousand fis> ear Instantaneous rate of oulation cange (t ) (t) ( ) () c ( ) ( ) (() () ) d 0 0 66 tousand fis> ear 6. a. i. f() ii. lim f() S iii. lim f() S iv. f() lim S6 No, lim f() does not eist. In order for te limit S to eist, lim f() and lim f() must eist and te S S must be equal. In tis case, lim f() `, but S lim f() `, so lim f() does not eist. S S - Cater : Derivatives of Eonential and Trigonometric Functions
7. f() is discontinuous at. lim f(), but S lim f(). S 8. a. lim S0 (0) 0 c. 9 d. lim S ( )( ) S ( )( ) S e. lim S S S S S 6 lim S S S ( ) S lim S S S! 6 ( )(! 6 ) (! 6 )(! 6 ) ( )(! 6 ) 6 9 ( )(! 6 )! 6 8 ( )( ) f.!! lim S0 S0 S0 S0 S0 9. a. f() fr() (!! )(!! ) (!! ) ( ) (!! ) (!! ) (!! ) c ( ) ( ) ( ) d c 6 6 d 6 6 6 6 6 f() f( ) f() fr() ( ) ()( ) ()( ) ( ) f( ) f() 0. a. To determine te derivative, use te ower rule: 8 -
To determine te derivative, use te cain rule: " " (6 ) " c. To determine te derivative, use te quotient rule: ( ) ( ) 6 ( ) d. To determine te derivative, use te roduct rule: ( ) ( ) ( )()( ) ( ) (0 ) ( )( ) ( ) (0 ) e. To determine te derivative, use te quotient rule: ( ) ( ) ( ) (8)( ) ( ) 6 ( ) ()( ) ( ) 6 ( ) ( ) 0( ) 9( ) ( ) 6 ( ) (0 80 6 9) ( ) ( ) (8 80 9) ( ) f. ( ) Use te cain rule ( ) 6( ). To determine te equation of te tangent line, we need to determine its sloe at te oint (, ). To do tis, determine te derivative of and evaluate for : 8 ( ) 8( ) 6( ) 6 ( ) m 6 ( ) 6 7 Since we ave a given oint and we know te sloe, use oint-sloe form to write te equation of te tangent line: ( ) 6 0 0. Te intersection oint of te two curves occurs wen 9 9 6 9 0 ( ) 0. At a oint, te sloe of te line tangent to te curve 9 9 is given b d ( 9 9) 9. At, tis sloe is () 9.. a. r(t) d dt (t 6t 00) t 6 990 is 0 ears after 980, so te rate of cange of oulation in 990 corresonds to te value r(0) (0) 6 6 eole er ear. c. Te rate of cange of te oulation will be 0 eole er ear wen t 6 0 t 6. Tis corresonds to 6 ears after 980, wic is te ear 006.. a. fr() d ( ) f s() d ( ) 0 0-6 Cater : Derivatives of Eonential and Trigonometric Functions
f() can be rewritten as f() fr() d ( ) f s() d ( ) c. f() can be rewritten as fr() d ( ) " f s() d ( ) " d. f() can be rewritten as f() fr() d ( ) f s() d ( ) 0 6 0 6. Etreme values of a function on an interval will onl occur at te endoints of te interval or at a critical oint of te function. a. fr() d ( ( ) ) f() ( ) Te onl lace were fr() 0 is at, but tat oint is outside of te interval in question. Te etreme values terefore occur at te endoints of te interval: f() ( ) f(6) (6 ) 8 Te maimum value is 8, and te minimum value is 6 f() can be rewritten as fr() d ( ) " On tis interval, $, so te fraction on te rigt is alwas less tan or equal to. Tis means tat fr(). 0 on tis interval and so te etreme values occur at te endoints. f()! f(9) 9!9 9 Te maimum value is value is. c. fr() d a e b and te minimum ( e ) Since e is never equal to zero, fr() is never zero, and so te etreme values occur at te endoints of te interval. e0 f(0) e 0 e f() e e Te maimum value is, and te minimum e value is. d. fr() d ( sin () ) 8 cos () Cosine is 0 wen its argument is a multile of or. or k k 8 k ( e )(e ) (e )(e ) ( e ) e e 9, 8 k Since P0,, 8, 8, 8, 8. Also test te function at te endoints of te interval. f(0) sin 0 f a 8 b sin f() 7-7
f a 8 b sin 7. w f a 8 b sin f a 7 8 b sin 7 f() sin () Te maimum value is, and te minimum value is. 6. a. Te velocit of te article is given b v(t) sr(t) d dt (t 0.t 6t) 9t 8t 6. Te acceleration is a(t) vr(t) d dt (9t 8t 6) 8t 8 Te object is stationar wen v(t) 0: 9t 8t 6 0 9(t 6)(t ) 0 t 6 or t Te object is advancing wen v(t). 0 and retreating wen v(t), 0. Since v(t) is te roduct of two linear factors, its sign can be determined using te signs of te factors: t-values t t 6 v(t) Object 0, t,, 0, 0. 0 Advancing, t, 6. 0, 0, 0 Retreating 6, t, 8. 0. 0. 0 Advancing c. Te velocit of te object is uncanging wen te acceleration is 0; tat is, wen a(t) 8t 8 0 t. d. Te object is decelerating wen a(t), 0, wic occurs wen 8t 8, 0 0 # t,. e. Te object is accelerating wen a(t). 0, wic occurs wen 8t 8. 0., t # 8 l Let te lengt and widt of te field be l and w, as sown. Te total amount of fencing used is ten l w. Since tere is 70 m of fencing available, tis gives l w 70 Te total area of te ens is A lw 7w w Te maimum value of tis area can be found b eressing A as a function of w and eamining its derivative to determine critical oints. A(w) 7w w, wic is defined for 0 # w and 0 # l. Since l 7 w, 0 # l gives te restriction w # 0. Te maimum area is terefore te maimum value of te function A(w) on te interval 0 # w # 0. Ar(w) d dw a7w w b 7 w Setting Ar(w) 0 sows tat w 7 is te onl critical oint of te function. Te onl values of interest are terefore: A(0) 7(0) (0) 0 A(7) 7(7) (7) 06. A(0) 7(0) (0) 0 Te maimum area is 06. m 8. r l 7 w Let te eigt and radius of te can be and r, as sown. Te total volume of te can is ten r. Te volume of te can is also give at 00 ml, so r 00 00 r -8 Cater : Derivatives of Eonential and Trigonometric Functions