MEI STRUCTURED MATHEMATICS 4758

Transcription

1 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 hour 30 minues Addiional maerials: 8 page answer bookle Graph paper MEI Examinaion Formulae and Tables (MF) TIME hour 30 minues INSTRUCTIONS TO CANDIDATES Wrie your name, cenre number and candidae number in he spaces provided on he answer bookle. Answer any hree quesions. There is an inser for use in Quesion 3. You are permied o use a graphical calculaor in his paper. Final answers should be given o a degree of accuracy appropriae o he conex. The acceleraion due o graviy is denoed by g ms. Unless oherwise insruced, when a numerical value is needed, use g = 9.8. INFORMATION FOR CANDIDATES The number of marks is given in brackes [ ] a he end of each quesion or par quesion. You are advised ha an answer may receive no marks unless you show sufficien deail of he working o indicae ha a correc mehod is being used. The oal number of marks for his paper is 7. This quesion paper consiss of 3 prined pages, blank page and an inser. HN/3 OCR 006 [R/0/66] Regisered Chariy [Turn over

2 The displacemen x a ime of an oscillaing sysem from a fixed poin is given by where l 0. (i) For wha value of l is he moion simple harmonic? Sae he general soluion in his case. [3] (ii) Find he range of values of l for which he sysem is under-damped. [3] Consider he case l. (iii) Find he general soluion of he differenial equaion. [3] When 0, x x 0 and ẋ 0, where x 0 is a posiive consan. (iv) Find he paricular soluion. [4] (v) Find he leas posiive value of for which x 0. [3] Now consider he case l 3 ẋ. lẋ 5x 0, wih he same iniial condiions. (vi) Find he paricular soluion and show ha i is never zero for 0. [8] The posiive quaniies x, y and z are relaed and vary wih ime, where 0. The value of x is described by he differenial equaion When 0, x. x. d (i) Solve he equaion o find x in erms of. [9] The quaniy y is relaed o x by he differenial equaion x dy. When 0, y 4. y (ii) Solve he equaion o find y in erms of x. Hence express y in erms of. [5] The quaniy z is relaed o x by he differenial equaion x dz z 6x. When 0, z 3. (iii) Solve his equaion for z in erms of x. Calculae he values of x, y and z when, giving your answers correc o 3 significan figures. [0] 4758 June 006

3 3 Answer pars (i) and (ii) on he inser provided. 3 Two spherical bodies, Alpha and Bea, each of radius 000 km, are in deep space. The poin A is on he surface of Alpha, and he poin B is on he surface of Bea. These poins are he closes poins on he wo bodies and he disance AB has he consan value of 8000 km. A probe is fired from A a a speed of V 0 km s in an aemp o reach B, ravelling in a sraigh line. A ime seconds afer firing, he displacemen of he probe from A is x km, and he velociy of he probe is v km s. The equaion of moion for he probe is This differenial equaion is o be invesigaed firs by means of a angen field, shown on he inser. (i) Show ha he direcion indicaors are parallel o he v-axis when v 0 (x 4000). Show also ha he direcion indicaors are parallel o he x-axis when x 4000 (v 0). Hence complee he angen field on he inser, excluding he poin (4000, 0). [6] (ii) Skech he soluion curve hrough (0, 0.05) and he soluion curve hrough (0, 0.05). Hence sae wha happens o he probe when he speed of projecion is (A) 0.05 km s, v dv (9000 x) (000 x). (B) 0.05 km s. [6] (iii) Solve he differenial equaion o find v in erms of x and V 0. [6] (iv) Given ha he probe reaches B, sae he value of x a which v is leas. Hence find from your soluion in par (iii) he range of values of V 0 for which he probe reaches B. [6] 4 The simulaneous differenial equaions d dy d x y 3 5x 4y 8 are o be solved for 0. d x (i) Show ha 3x 6. [6] d d (ii) Find he general soluion for x in erms of. Hence obain he corresponding general soluion for y. [9] (iii) Given ha x 4, y 7 when 0, find he paricular soluions for x and y and skech a graph of each soluion. [9] 4758 June 006

4 Candidae Candidae Name Cenre Number Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions INSERT Thursday 5 JUNE 006 Afernoon hours 30 minues INSTRUCTIONS TO CANDIDATES This inser should be used in Quesion 3. Wrie your name, cenre number and candidae number in he spaces provided a he op of his page and aach i o your answer bookle. This inser consiss of prined pages. HN/3 OCR 006 Regisered Chariy [Turn over

5 Inser for use wih Quesion 3 v x Inser June 006

6 Mark Scheme 4758 June 006

7 4758 Mark Scheme June 006 (i) λ = 0 B x = Acos 5+ Bsin 5 M cos 5 or sin 5 or Acosω+ Bsinω seen or GS for heir λ (ii) ( λ) 4 5 < 0 M Use of discriminan Correc inequaliy 0< λ < 5 Accep lower limi omied or 5 (iii) α + α + 5= 0 M Auxiliary equaion α = ± j x = e Ccos+ Dsin F CF for heir roos (iv) x0 = C M Condiion on x x& = e Ccos + Dsin + e Csin + Dcos M Differeniae (produc rule) 0= C+ D M Condiion on x& D = x 0 0 x = x e cos+ sin cao (v) cos + sin = 0 M an = M =.07 cao (vi) α + 6α + 5 M Auxiliary equaion α =, 5 5 x = Ee + Fe F CF for heir roos x0 = E+ F M Condiion on x 5 x& = Ee 5Fe 0= E 5F M Condiion on x& 5 E = x, F = x x 0 4 x = 5e e 4 cao x 0 x = e 5 e 4 M Aemp complee mehod 4 > 0 5> e, x > 0,e > 0 x > 0 i.e. never zero E Fully jusified (only 0 required)

8 4758 Mark Scheme June 006 (i) λ + = 0 λ = M CF x = Ae PI x = a + b B a+ ( a+ b) = + M Differeniae and subsiue a =, a+ b= M Compare a =, = b 4 e 4 0, e x = + + A F CF + PI = x = = + A M Condiion on x x = + + F Follow a non-rivial GS Alernaively: I = exp M ( d) = e Inegraing facor d e x + e x = e ( + ) d B Muliply DE by heir I e x = e ( + ) d M Aemp inegral = e + e d M Inegraion by pars 4 e 4 0, e e x = e + e + A x = + + A F Divide by heir I (mus also divide consan) = x = = + A M Condiion on x x = + + F Follow a non-rivial GS (ii) dy = y x M Separae dy = y x M Inegrae ln y = lnx+ c y = B x M Make y subjec, dealing properly wih consan = 0, x =, y = 4 y = 4 x M Condiion 3 4 e 4 4 y = + + F y = 4 (heir x in erms of ) (iii) dz + z = 6 M Divide DE by x x I = exp M Aemp inegraing facor ( x ) = x Simplified ( x z) d = 6x F Follow heir inegraing facor 3 x z = x + C z = x+ Cx F Divide by heir I (mus also divide consan) ( = 0, ) x =, z = 3 C = M Condiion on z z = x+ x cao (in erms of x) = x = 0.85 y = 3.69 B Any values (a leas 3sf) z = 3.08 B All 3 correc (and 3sf) 9 5 0

9 4758 Mark Scheme June 006 3(i) dv dv = f( x ) so ( unless f( x) = 0 ), v 0 ± Consider d v v x when v = 0, bu no if dv M dv 0 i.e. gradien parallel o v-axis (verical) E Mus conclude abou direcion dv x = 4000 v = = 0 M Consider d v when x = 4000 d x so if v 0 hen gradien parallel o x-axis (horizonal) E Mus conclude abou direcion M Add o angen field Several verical direcion indicaors on x-axis (ii) M Aemp one curve M Aemp second curve 6 V 0 = 0.05 probe reaches B B Mus be consisen wih heir curve V 0 = 0.05 probe reurns o A B Mus be consisen wih heir curve N.B. Canno score hese if curve no drawn (iii) ( = + ) vd v (9000 x) (000 x) M Separae M Inegrae v = c 9000 x x + B LHS RHS V = + + c M Condiion v 9000 x x = + + V0 450 (iv) minimum when x = 4000 B Clearly saed vmin = + + V0 M Subsiue heir x ino v or v F Their v or v when x = 4000 need v min > 0 M For v min > 0 4 v > 0 if V > M Aemp inequaliy for min V 0 > cao V

10 4758 Mark Scheme June 006 4(i) && x = x& y& M Differeniae firs equaion = x& (5x 4y+ 8) M Subsiue for &y y = x+ 3 x& M y in erms of x, x & && x = x& 5x+ 4(x+ 3 x& ) 8 M Subsiue for y && x+ x& 3x = 6 E LHS E RHS (ii) λ + λ 3= 0 M Auxiliary equaion λ = or 3 3 CF x = Ae + Be F CF for heir roos PI x = a B Consan PI 3a = 6 a = B PI correc 3 x = + Ae + Be F Their CF + PI y = x+ 3 x& M y in erms of x, x & 3 3 = 4+ Ae + Be + 3 ( 3Ae + Be ) M Differeniae x and subsiue 3 y = 7+ 5Ae + Be Consans mus correspond wih hose in x (iii) 4= + A + B M Condiion on x 7 = 7 + 5A + B M Condiion on y A=, B = 0 M Solve 3 x = + e F Follow heir GS 3 y = 7+ 0e F Follow heir GS 6 9 B Skech of x sars a 4 and decreases B Asympoe x = B Skech of y sars a 7 and decreases B Asympoe y =7 7 9