( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+

Size: px

Start display at page:

Download "( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+"

Eustace Fletcher
5 years ago
Views:

1 Review Eercise sin 5 cos sin an cos 5 5 an 5 9 co 0 a sinθ sin θ cos θ sin θ sin θ + an θ cos θ ( ) Since π π, < θ < anθ should be negaive. anθ ( + ) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

2 b anθ sin θ + ( ) ( ) ( + ) ( + ) 4 4 cosθ + ( ) ( ) ( + ) 4( + ) ( ) 4( + ) c π 7π θ cos π π 7π θ 6 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

3 a b + sec cos an + sec + an ( + ) ( + )( ) + π an an π + 4 an + 4 π an an 4 + sec + an θ 4 an θ cos θ sec + θ + cos cosθ a an cos 4sin Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

4 5 b ( )( ) Therefore, an, 7 an, 0 < < π 7 an ( d.p.) an π an ( ) ( d.p.) 6 a an sin + cos b 0 0 ( ) Therefore, an 0, an 0, 0 π an 0 0, π 0,π an an.075. ( d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

5 7 a Differeniaing, we have ds v 8cos4 + 8cos d Using an and subsiuing he -formulae and using a double-angle formula: cos4 cos sin ( ) ( + ) ( + ) 4 v ( ) ( ) 6 ( ) ( ) + b Solving for d s 0 we have d ± which implies ha π 5π, 6 6 To check which poin is minimum we differeniae again. d s sin 4 6sin d ( 4( ) ( )) + + ( ) ( ) ( 5 ) + ds ds We find 0 < d and 0 d > So he minimum is a 5π Therefore, an 6 0π 0π s sin + 4sin m ( d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 5

6 8 a an 8 f ( ) 5 cos + cos 5sin cos 5sin + cos 5sin b f ( ) 4 4 ( ) ( + ) ( + )( ) 4( + ) ( + )( ) 4( + ) A a saionar poin, 8 ( ) 0 an ( ) f π 6π 8 4 c From he graph of f ( ), he maimum value of f ( ) is in he range [ 60,6.5 ] If he maimum value of L is 00 lires, hen using L kf ( ), k lies in he range [ 4.8,5 ] d The poin described b 6π is he second lowes rough on he lef in he graph of f ( ). A his poin, π π f ( ) 0+ 0 sin π + sin + 0 cos 9 L kf 9.,95 Therefore, he amoun of waer pumped ( ) lies in he range [ ] Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 6

7 9 a Le f( ) cos π π f cos 0 4 f ( ) sin π π f sin 4 f ( ) 4cos π π f 4cos 0 4 f ( ) 8sin π π f 8sin 8 4 (iv) (iv) π π f ( ) 6cos f 6cos 0 4 (v) (v) π π f ( ) sin f sin 4 Talor s and Maclaurin s series need repeaed differeniaion and subsiuion. You need o displa hese in ssemaic form, boh o help ou subsiue correcl and o show our working clearl so ha he eaminer can award ou marks. f (iv) () and f (v) () are smbols which can be used for he fourh and fifh derivaives of f() respecivel. ( a) ( a) ( a) ( a) !! 4! 5! 4 5 (iv) (v) f( ) f( a) ( a)f ( a) f ( a) f ( a) f ( a) f ( a) π Subsiuing f( ) cos anda 4 π π ( ) ( ) π cos ( ) 8 ( ) π 4 π 4 π π b Le, hen Subsiuing ino he resul of par a cos (0.46 ) + (0.46 ) (0.46 ) (6 d.p.) This is he appropriae form of Talor s series for his quesion. I is given in he formula bookle. All of he even derivaives π are zero a 4 π Work ou on our calculaor and 4 hen use he ANS buon o complee he calculaion. This is a ver accurae esimae and is correc o 6 decimal places. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 7

8 0 a π Le f( ) ln(sin ) f ln ln 6 cos π π f ( ) co f co sin 6 6 π 6 π 6 f ( ) cosec f 4 f ( ) cosec co f 8 π cosec 6 π sin 6 Using he chain rule, d d ( cosec ) cosec (cosec ) d d cosec cosec co ( a) ( a) f( ) f( a) + ( a)f ( a) + f ( a) + f ( a) +!! Subsiuing π f( ) ln(sin )anda 6 This is he appropriae form of Talor s series for his quesion. I is given in he formula bookle. π π π ln(sin ) ln+ + ( 4) b π π 4 π ln π Le 0.5, hen π Work ou on our calculaor and 4 hen use he ANS buon o complee he calculaion. Subsiuing ino he resul of par a 4 ln(sin 0.5) ln+ ( ) ( ) + ( ) (6 d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 8

9 a an sec d Using he chain rule for differeniaion. d d d sec (sec ) sec secan d sec an d d d + an (sec ) sec (an ) d d d 4 4sec an + sec b Le f( ) an Using he produc rule for differeniaion d ( ) d u d v uv v + u d d d wih u sec andv an π π f an 4 4 Using he resuls in par a π π f sec ( ) 4 4 π π 4 4 π π π 4 π f sec an + sec f ( ) sec an ( ) 4 4 4( ) ( ) ( a) ( a) f( ) f( a) + ( a)f ( a) + f ( a) + f ( a)+!! π Subsiuing f( ) an and 4 π π π an π π 8 π π π sec and an 4 4 This is he firs four erms of Talor s series. You are epanding an abou he poin series. π, using Talor s 4 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 9

10 c π Le, 0 hen π π π π Subsiuing ino he resul in par b π π π 8 π an π π π + + +, as required f '''( ) f ( ) ln f ( ) + f ( )( ) + ( ) +! f ( ) f ( ) + ( ) +!! 0+ ( ) ( ) + ( ) + a f ( ) ln, f ( ), f ( ) ( ) ln + + lim b lim ( ) ( ) ( ) ( )( ) + lim ( + ) ( ) ( ) ( ) 0 0 a f ( ) sinh, f ( ) cosh ( ) ( ) ( ) ( ) sinh f f '''' '''( ) f 0 4 f 0 f ''( ) sinh, f cosh, + ( 0) + ( 0) +! 4! ''''' 5 f ( 0) ( 0) + 5! 5 sinh ( ) ( ) f ( 0)! Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 0

11 b Similarl, cosh ( ( ) ) lim cosech lim 0 sinh cosh lim lim a d d d + () ( ) 0 Differeniae () hroughou wih respec o d d d + ( ) + 0 () d d d d d Subsiuing 0, and ino() d Using he produc rule for differeniaion d du dv ( uv) v + u wih u and d d d d v d d d d, ( ) d d d d d ( ) + ( ) d d d d d ( ) d + d d d d d A 0, d Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

12 4 b Le f( ) From he daa in he quesion f( 0 ), f ( 0 ) A 0, () above becomes f (0) + 0 f (0) 4 And he resul o par a becomes f (0) f( ) f(0) + f (0) + f (0) + f (0) +!! The formula for Maclaurin s series is given in he formulae bookle. For his quesion, ou need he erms up o and including he erm in + ( ) + ( 4) a ( ) 4 d + + * Differeniae * hroughou wih respec o You need o differeniae 4 implicil wih respec o d d (4 ) (4 ) 8 d d d d + ( + ) + 8 d d d d ( + ) + 8 d d d + (4 ) () as required. d b Differeniae () hroughou wih respec o d d d d ( ) (4 ) + d d d d () When using he produc rule for d du dv differeniaion ( uv) v + u wih d d d u (4 ) and v, (4 ) mus d be differeniaed implicil wih respec o d. So (4 ) d d d 8 + (4 ) d d d d d 8 (4 ) d + d Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

13 5 c Le f( ) From he daa in he quesion f(0) A 0,,* becomes f (0) 4 A 0,,, () becomes d f (0) + 4 A d 0,,,, d d ()becomes + f (0) f (0) 8+ 6 f (0) 8 f( ) f(0) + f (0) + f (0) + f (0) +!! The formula for Maclaurin s series is given in he formulae bookle. For his quesion, ou need he erms up o and including he erm in Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

14 6 a Le f( ) From he daa in he quesion f(0) d + + () A 0,, () becomes f (0) Differeniaing () hroughou b d d d d d d A () 0,,, () becomes d has o be differeniaed implicil b. So d d ( ) ( ) d d d Using he produc rule for differeniaion d d d ( ) u v uv v + u wih d d d u and v, d d ( ) + d d d + d f (0) Differeniae () hroughou b d d d d d d d d d d d d A d 0,,, 4, d d ()becomes f (0) () Using he produc rule for differeniaion d du dv d ( uv) v + u wih u and v, d d d d d d d ( ) d d + d d d d d d + d d d d + d f( ) f(0) + f (0) + f (0) + f (0) +!! b A (0.) + (0.) ( d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

15 7 a Rearranging he differenial equaion in he quesion ( + ) + () d Le f( ) From he daa in he quesion f(0).5 A 0,.5, () becomes The righ hand side of he equaion in he quesion would be hard o repeaedl differeniae as a quoien, so mulipl boh sides b + (.5 +.5)f (0) 0+ f (0) Differeniae () hroughou b d ( + ) + ( + ) d d A () 0,.5, 0.8, () becomes d (.5 +.5) f (0) f (0) Differeniaing d b, using he chain rule d d d d d d d d d d d Differeniae () hroughou b d d d + ( + ) + ( + ) + ( + ) 0 d d d d d d d d d d ( ) ( ) d d d d () A d 0,.5, 0.8, 0.46, becomes d d () (.5.5) f (0) f (0) f (0) ɺ.75 f( ) f(0) + f (0) + f (0) + f (0) +!! ɺ ɺ ɺ b A, ( ) ( ) ( ) This is a recurring decimal. There is an eac fracion Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 5

16 d 8 a d d () Differeniae () hroughou wih respec o d d d d d d d d d d d d d d + d d d d d d d d + () d d d b Le f( ) From he daa in he quesion Using he produc rule for d d d differeniaion ( ) u v uv v + u d d d d wih u and v, d d d d d d + d d d d d d d d + d d d The wording of he quesion requires d ou o make he subjec of he d formula. There are man possible alernaive forms for he answer. f(0), f (0) A 0,,, () becomes d f (0) f (0) A d 0,,,, becomes d d () f (0) ( + ) ( 6+ ) 5 f( ) f(0) + f (0) + f (0) + f (0) +!! The formula for Maclaurin s series is given in he formulae bookle. For his quesion, ou need he erms up o and including he erm in c The series epansion up o and including he erm in can be used o esimae if is small. So i would be sensible o use i a 0. bu no a 50 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 6

17 9 a d () d d Le f( ) From he daa in he quesion f(0), f (0) 0 has o be differeniaed implicil wih respec o So d d ( ) ( ) 6 d d d A 0,, 0, () becomes d f (0) f (0) Differeniae () hroughou wih respec o d d d d d d () A d 0,, 0,, becomes d d () f (0) f (0) Differeniae () hroughou wih respec o 4 d d d d d d d d d A 0,, 0,,, d d d () becomes (iv) f (0) (iv) f (0) () Using he produc rule for differeniaion d ( ) du d v uv v + u d d d wih u 6 and v d d 6 d d d d (6 ) + 6 d d d d 6 6 d + d !! 4! (iv) f( ) f(0) f (0) f (0) f (0) f (0) The formula for Maclaurin s series is given in he formulae bookle. For his quesion, ou need he erms up o and including he erm in 4 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 7

18 9 b A ( d.p.) 0 Le u and v e du d, du 6 d, du 6 d The subsequen derivaives of u are all zero. n d v e n d n Using Leibniz s heorem, n n n d d v du d v u + n + + n n n d d d d n n n( n ) du d v n( n )( n ) du d v + n n d d 6 d d n n n( n ) n n( n )( n ) n ( e ) + n( )( e ) + (6 )( e ) + (6)( e ) 6 n e (7 + 7n + 9n n + n n n ( ) ( )( ) Le u e and v sin n du e n d dv cos d, dv sin d, dv cos d dv sin 4 d, dv cos 5 d, dv sin 6 d Using Leibniz s heorem, d dv du dv du dv u d d d d d d d d d d d d d 0 u v + 5 u v + 6 u v + u v d d d d d d d e sin+ 6e cos + 5e sin 0e cos 5e sin+ 6e cos+ e sin 8e cos 6 d d d ( ) 8e cos+ 8 e cos + e sin 8e sin 0 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 8

19 Le f ( ) ln and g( ) f ( ) ln 0 and ( ) f '( ) and g '( ) B L Hospial s rule, ln lim lim Le f ( ) ln and g( ) g 0, so we can appl L Hospial s rule. ln so ha lim( ln ) lim 0 0 f ( 0) ln0 and g ( 0), so we can appl L Hospial s rule. 0 f '( ) and g '( ) B L Hospial s rule, ln lim( ln ) lim lim ( ) lim 0 4 Le f ( ) e and g( ) sin 0 f ( 0) 0 e 0 and g ( ) f '( ) e + e and g '( ) 0 sin 0 0, so we can appl L Hospial s rule. cos ( ) ( ) e e + 0+ B L Hospial s rule, lim lim 0 sin 0 cos 5 Le f ( ) e cos and ( ) 0 f ( 0) e 0 and ( 0) 0 f ( ) e + sin and g '( ) g B L Hospial s rule, e cos e + sin lim lim 0 0 g, so we can appl L Hospial s rule. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 9

20 6 a an sin and cos + + sin + cos ( ) d d + d sin + cos + d d + an d sec d ( ) b π 4 π an 8 d d sin + cos 0 0 π an ( ) ( d.p. ) ln 0 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 0

21 7 an sin and + sin 4cos ( )( ) cos + d d + d sin 4cos d + an d sec d ( )( ) d 5 + ln an ln an π 6 π d sin 4cos ln an ln an + 5 an ln 5 an + ( ) ln ln π 6 π 5+ ln 5 ln ( ) 5 Therefore a 6 and b 5 7π 6 π Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

22 8 (, ) (,) h d h d ,.5,0.5 ( ) ( ) d + h d Therefore f (.5) 0.74 ( d.p.) 9 a (, ) ( ln,) ln e d 0 0 h d 0.6 h 0.6 h 0. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

23 9 b (, ) ( ln + 0.,.6) (ln+ 0.) e.6 d.56 + h d So.065 ( d.p.) (, ) ( ln + 0.4,.065 ) (ln + 0.4) e (.065 ) d h d So.406 ( d.p.) 0 (, v ) (,8) h dv d 0 8 dv v v0 + h d 0 v v 8.044, v 4,8.044 ( ) ( ) dv d ( ) dv v v + h d v v So v 8.06 ( d.p.) Therefore, he value of he asse is 806 (rounded off o he neares pound) five das afer i is purchased. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

24 h 0.,, ( ) ( ) 0 0 ln d h d ,.,.8 ( ) ( ) ln..8 d h d ,.,.678 ( ) ( ) ln..678 d h d Therefore, a.,.57 ( d.p.) a h 0.,, ( ) ( ) 0 0 cos d h d cos ( ) ( ) ( d.p. ) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

25 b (, ) (., ) cos d h d ,.4, ( ) ( ) ( ) cos( ) d h d (, ) (.6, ) Therefore, a.6, ( d.p.) (, P ) ( 0,000 ) 0 0 dp h cos( 0.8 0) d dp P P0 + h d 0 P P 979.5, P,979.5 ( ) ( ) dp ( ) d dp P P0 + h d P P , P, cos ( ) ( ) dp ( ) 0.5 cos( 0.8 ) d dp P P + h d P P Therefore, a das, P 0660 baceria. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 5

26 4 a (, ) ( 0, ) 0 0 d 0 h h d b (, ) ( 0., 0.8) d d d h d , 0., ( ) ( ) 0 d d d h d (, ) ( 0., ) Therefore, a 0., 0.44 (4 d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 6

27 5 (, ) (, ) d 0 h h d ,.,. ( ) ( ) d + d sin. cos d h d ( d.p.),.4,.0 ( ) ( ) d sin.4 cos.0 + d.66 + d h d ,.6,.98 0 ( ) ( ).98 ( d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 7

28 6 (, ) (,) 0 0 h 0. d 0 d + d 0 7 h d d h d Adding he wo equaions gives a f ( ) sin an h i i B Simpson s rule, sin and ( ) (.098 ) ( d.p.) b Increasing he number of inervals would give a beer esimae for he approimaion. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 8

29 8 a coshd h 0.5 i i B Simpson s rule, cosh d (4 d.p.) ( ) b Inegraing b pars wih u and v cosh : d [ ] cosh sinh sinhd [ sinh cosh] e e e + e e e + e c Percenage error e e e + e e e e + e e e + e 00 e e + e 0.% ( d.p.) Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 9

30 9 a u Differeniae hroughou wih respec o du d d + d ransforms o du + u u d du u d du u + d du d u + ln( u + ) + A ln(u + ) + B ( u ) u This is a separable equaion. You learn how o solve separable equaions in C4 Separaing he variables. Twice one arbirar consan A is anoher arbirar consan, B A e e e e C e ln(u+ ) + B B u C e e o an arbirar consan is anoher arbirar consan. Here C e B C e 4 This is he general soluion of he original differenial equaion. b a 0 C 8 C C 9 4 9e 4 This is he paricular soluion of he original differenial equaion for which a 0 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 0

31 40 a v dv + v d d dv Subsiuing v and + v ino d d Differeniaing v as a produc, d dv d ( v) + v ( ) d d d dv d + v, as ( ) d d equaion () in he quesion d v (4 + v)( + v) + v d (4 + v)( + v) (4 + v)( + v) 4+ 5v+ v dv v v v d b ( + ), as required. dv d ( + v) ln + c + v + v ln + c v ln + c This is a separable equaion and he firs sep in is soluion is o separae he variables, b collecing ogeher he erms in v and dv on one side of he equaion and he erms in and d on he oher side of he equaion. ( ) + v ( + v) dv + v c v v Subsiuing v ino he answer o par b ln + c as required ln + c Mulipl hroughou b o obain he prined answer. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

32 4 a v d + v d d Subsiue dv v and + v ino d d Differeniaing v as a produc, d ( v ) d dv d + v ( ) d d dv d + v, as ( ) d d equaion () in he quesion dv 4v ( 4 v) + v d 4 + v (4+ v) dv 4v 4v 4v v 8v v v d 4+ v 4+ v 4+ v dv v + 8v, as required d v + 4 b v + 4 6v + 8 dv dv d v + 8v v + 8v ln( v 8 v ) ln + + A ln(v 8v ) ln Hence C v + 8v + + B ln + lnc ln C This is a separable equaion and in par b ou solve i b collecing ogeher he erms in v and dv on one side of he equaion and he erms in and d on he oher side. f ( ) d ln f( ) is a sandard formula ou should f( ) know. As 6v+ 8 is he derivaive of 6v + 8 dv ln(v + 8v ) v + 8v v + 8v, An arbirar consan B can be wrien as he logarihm of anoher arbirar consan ln C. c v v Subsiuing ino he answer o par b 8 C C Mulipl each erm in he equaion b 7 a C C 00 Facorising he lef hand side of he equaion ( )( + ) 00, as required. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

33 4 a µ d µ d d d Hence d d d dµ Differeniae boh sides implicil wih respec o You ransform his equaion, making he subjec of he formula as d ou need o subsiue for d d in () Subsiuing in equaion () in he quesion dµ e d Divide b dµ + e d As µ dµ + µ e d Mulipl b ( ) dµ 4µ e, as required d Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free.

34 4 b The inegraing facor of () is e e 4 d Mulipling () hroughou b e dµ e 4µ e e e e d d ( e ) e µ d µ e e d e + C This inegraion can be carried ou b inspecion. As d (e ) 6 e, hen d e d e 6 Mulipling hroughou b e e e µ + C c As µ e + C e a 0 + C C e + e As no form of he answer has been specified in he quesion, his is an accepable answer for he paricular soluion of () Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

35 4 a v dv v + d d d dv dv dv dv dv d d d d d d Use he produc rule for differeniaion d d d d ( ) ( ) v v v v+ v+ d d d d Subsiuing for d and d d ino() d v dv dv v v d d d ( + 9 ) d dv + v d d v + d d v 5 + 9v dv d dv + d 5 + 9v 9 v, as required Divide b Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 5

36 4 b The auiliar equaion is m m 9 m ± i The complemenar funcion is given b v A cos + B sin For a paricular inegral, r dv d v p + q, p d d v p + q+ r If he righ hand side of he differenial equaion is a polnomial of degree n, hen ou can r a paricular inegral of he same degree. Here he righ hand side is a quadraic, so ou r a general quadraic p + q + r Subsiuing ino () p + 9q + 9q + 9r Equaing coefficiens of 9p q 9 Equaing coefficien of 9q 0 q 0 Equaing consan coefficiens p + 9r 0 9r p r 9 8 As p 9 A paricular inegral is 9 8! A general soluion of () is c v Acos+ Bsin A cos B sin + + Acos + Bsin+ 9 8 v v The quesion does no ask for a paricular form of he answer in par c, so his would be an acceleraion answer. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 6

37 44 a d d d d Use d d d d d d d d d d You obain an epression for d d using he chain rule. b Subsiuing, he resul of par a and he given d d d d d d 4 + d 4 + ino () d e d d d d d + d d 6 4 e d e d d d + d d 4 e, as required 6 6 Divide hroughou b 4 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 7

38 44 c The auiliar equaion is m m + 4 ( m )( m+ 4) 0 m, 4 The complemenar funcion is Ae + Be 4 For a paricular inegral r, k e If he righ hand side of he equaion is e α, ou can r k e α as a paricular inegral. This will work unless α is a soluion of he auiliar equaion. d d k e, 4 e d k Subsiuing ino d + 4 e d d 4k e + 6k e 4k e 6k k 6 e As e canno be zero, ou can divide hroughou b e A paricular inegral is e 6 The general soluion of he differenial equaion in and is Ae + Be + e 6 4 The general soluion of () is Ae + Be + e a d d In d d d d d d d d d d d d d Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 8

39 45 b d d d d d d d d d d d d d d d d + d d d d, asrequired d d + I is a common error o proceed from d d d d d o + d d d d d This is incorrec because he lef hand side has been differeniaed wih respec o and he righ hand side wih respec o. The version of he chain rule given here mus be used. c Subsiuing d d In, d d and d d + ino () d d d d + ( 6 ) sin d d d d + d d d sin d d sin, as required d d ln ln e e, using he log role n n lnf ( ) ln a lna and e f( ) Afer cancelling, divide hroughou b Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 9

40 45 d The auiliar equaion of () is m m m m ( m+ ) m+ ± i m ± i The complemenar funcion is given b + e ( Acos Bsin ) For a paricular inegral r psin + q cos pcos psin d d 4psin 4qcos d If he righ hand side of he second order differenial equaion is a k sin n or k cos n funcion, hen ou should r a paricular inegral of he form pcosn + qsinn Subsiuing ino () 4psin 4qcos + pcos qsin + 0psin + 0qcos 5sin ( 4q q + 0 p)sin + ( 4q + q + 0 q)cos 5sin (6p q)sin + (p + 6 q)cos 5sin Equaing he coefficiens of sin 6p q 5 p + 6q 0 () (4) You can solve he simulaneous equaions b an appropriae mehod. From (4) 6 p q q Subsiue ino () q q 5q 5 q Hence p q 6 The general soluion of () is e ( Acos + B sin ) + sin cos 6 ln e Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 40

41 45 d The general soluion of () is e e ( Acos(e ) + Bsin(e )) + sin(e ) cos(e ) 6 46 a Differeniaing implicil wih respec o. d d d Use d d d d Differeniaing again implicil wih respec o. 4 6 d d d d d d () Dividing he differenial equaion given in he quesion b 4, i becomes This epression is closel relaed o he lef hand side of he original differenial equaion in he quesion. This suggess o ou ha if ou divide he original equaion b 4, hen he lef hand side can jus be replaced b d d d 4 d d d Using equaion () and d + d d +, as required d b The auiliar equaion is m + 0 m ± i The complemenar funcion is given b Acos + Bsin B inspecion, a paricular inegral of () is The general soluion of () is As d d () 0, saisfies +, b d d inspecion and ou need no wrie down an working. Acos + Bsin + Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

42 46 c The general soluion of he differenial equaion in and is Acos Bsin + + () When 0, 4 A+ A Differeniaing () implicil wih respec o d Asin + Bcos d Use he chain rule d d d d ( ) ( ) d d d d When d 0, and 0 d 0 B The paricular soluion is cos + As > 0, > 0 (cos + ) As and are boh posiive, he negaive square roo need no be considered. d The maimum value of is ( + ) The maimum value of his fracion is when he denominaor has is leas value. The smalles possible value of cos is So ou can wrie down he maimum value wihou using calculus. Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

43 Challenge an and an s s + an + an s co + co s + s ( s ) + s( ) s s 4s ( )( ) 4s ( )( s ) ( )( ) s an an s Le u e cosh and v dv d, dv 6 d, dv 6 d The subsequen derivaives of v are all zero. e + e u e cosh e ( e + ) n du n e n d Using Leibniz s heorem, n n n d d v du d v u + n + + n n n d d d d n n n( n )( n ) d u dv n( n ) d u dv + n n 6 d d d d n n d u dv du + n + v n n d d d n 4 n n n n e + n n e ( )( ) ( ) + n e + e n n ( ( ) ( )( )) e + n + n n + n n n n Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 4

44 Challenge ( ) a f ( ) f ( ), u f ( ) du du d u du d u B u + u u u d u + B u f '( ) B B Inegraing boh sides wih respec o, f ( ) d A B B b f ( ) f ( ) 0 0 A B 0 A B Subracing he wo equaions gives B B B + B ( ) B + B + B + B B B B A B 0 A 9 4 Pearson Educaion Ld 08. Coping permied for purchasing insiuion onl. This maerial is no coprigh free. 44

MEI STRUCTURED MATHEMATICS 4758

OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions Thursday 5 JUNE 006 Afernoon