PhysicsAndMathsTutor.com

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Lizbeth Gallagher
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1 1 (i) 1 A B 1 A(1 x) B(1 x) (1 x)(1 x) 1 x 1 x Cover up, substitution or equating coefficients x = 1 B = 1, B = 1/ x = ½ 1 = A/, A = / isw after correct A and B stated 1 (ii) 1 x x (1 x)(1 x) 1 1 [ ]d x (1 x) 1 kdt x ln1 x ln(1 x) kt(c) Any non-zero constant, ln(1 + x) ln(1 x) = kt (+ c) [] May be seen in separation of variables (may be implied by later working) implied by the use of factors (1 + x) and (1 x) Separating variables and substituting partial fractions. If no subsequent work integral signs needed, but allow omission of dt, but must be correctly placed if present www oe (condone absence of c) dx or When t = 0, x = 0 c = 0 1 x ln kt 1 x cao (must follow previous ) need to show (at some stage) that c = 0. s a minimum t = 0, x = 0, c = 0. Note that c = ln(-1) (usually from incorrect integration of (1 x)) or similar scores B0 Combining both their log terms correctly. Follow through their c. Allow if c = 0 clearly stated (provided that c = 0) even if B mark is not awarded, but do not allow if c omitted 1 x e kt * 1 x [7] AG www must have obtained all previous marks in this part

2 1 (iii) (1 + (0.75)) / (1 0.75)=e k substituting t = 1, x = 0.75 at any stage k = (1/)ln10 (= ( s.f.)) t = ln(.8/0.1)/k = 1.45 hours 1 (iv) 1 + x = e kt xe kt x + xe kt = e kt 1 [] * sf or better 1.45 (or better) or 1 hr 7 mins Multiplying out and collecting x terms (condone one error) x( + e kt ) = e kt 1 x = (e kt 1)/ ( + e kt ) dep* Factorising their x terms correctly OR = (1 e kt )/(1 + e kt ) * when t e kt 0 x = (1 e kt )/(1 + e kt ) 1/1 = 1 1 x 1 x e kt 1 x e kt xe kt [5] * x(1 e kt kt ) 1e dep* x = (1 e kt )/(1 + e kt ) * www (AG) as AG must be an indication of how previous line leads e kt to the required result (eg stating or showing multiplying by ) clear indication that e kt 0 so, for example, 1 0 accept as a minimum ( x ) 1or e kt 0 ( x ) 1 (NB 1 0 substitution of large values of t with no further explanation is B0) Multiplying up and expanding (condone one error) Factorising their x terms correctly www (AG) final B mark as in scheme above

3 (i) EITHER x = e t, y = te t / e t t dy dt t e Their dy / dt dx / dt in terms of t dy/dx = (te t + e t )/e t when t = 1, dy/dx = e /e = 1/e OR t = ln x, y dy / dx ln x e /ln 1 / 1 x ln x x x 9 ln / x x x 1/ 1 t = e t e t soi oe cao allow for unsimplified form even if subsequently cancelled incorrectly ie can isw cao www must be simplified to 1/e oe Any equivalent form of y in terms of x only Differentiating their y provided not eased ie need a product including p ln kx and x and subst x e t to obtain dy/dx in terms of t dy/dx = 1/e + /e =1/e 1 www cao exact only must be simplified to 1/e or e (ii) t = ln x t = (ln x)/ y = (ln x) / e y = 1 x ln x (ln x)/ [] oe cao Finding t correctly in terms of x Subst in y using their t b Required form ax ln x only NB If this work was already done in 5(i), marks can only be scored in 5(ii) if candidate specifically refers in this part to their part (i).

4 (i) Either h = (1 ½ At) dh/dt = A (1 ½ At) = A h when t = 0, h = (1 0) = 1 as required OR dh h At d h 1/ At c At c h at t =0, h = 1, 1 = (c/)² c =, h = (1 At/)² (ii) When t = 0, h = A= 0, A = 0.1 When the depth is 0.5 m, 0.5 = (1 0.05t) t = 0.5, t = (1 0.5)/0.05 = 5.86s [] Including function of a function, need to see middle step AG Separating variables correctly and integrating Including c. [Condone change of c.] Using initial conditions AG Subst and solve for A cao substitute h= 0.5 and their A and solve for t www cao accept 5.9

5 (iii) dh h B dt (1 h) (1 h) dh B dt h EITHER, LHS 1 h h h dh (h 1/ h 1/ h / )dh OR,LHS, EITHER separating variables correctly and intend to integrate both sides (may appear later) [NB reading (1+h)²as 1+h² eases the question. Do not mark as a MR] In cases where (1+h)² is MR as 1+h²or incorrectly expanded, as say 1+h+h² or 1+h², allow first for correct separation and attempt to integrate and can then score a max of M0A0A0 (for Bt+c) A0A0, max /7. expanding (1+ h) and dividing by h to form a one line function of h (indep of first ) with each term expressed as a single power of h eg must simplify say 1/ h+h/ h +h² h,condone a single error for (do not need to see integral signs) h 1/ + h 1/ + h / cao dep on second M only -do not need integral signs (1 h h )h 1/ h 1/ ( h)dh using udv uv vdu correct formula used correctly, indep of first OR h 1/ h 5/ / h 1 h h h / ( h h )dh / 5/ 1/ 4h h 5 = Bt +c h 1/ + 4h / / + h 5/ /5 = Bt + c When t = 0, h = 1 c = 56/15 h 1/ (0 + 0h + 6h ) = 56 15Bt * [7] condone a single error for if intention clear cao oe cao oe, both sides dependent on first mark cao need Bt and c for second but the constant may be on either side from correct work only (accept.7 or rounded answers here but not for final ) or c= 56/15 if constant on opposite side. NB AG must be from all correct exact work including exact c.

6 (iv) h = 0 when t = 0 B = 56/00 = When h = t = Substituting h = 0,t = 0 Accept Subst their h = 0.5, ft their B and attempt to solve t = 9.5s Accept answers that round to 9.5s www.

7 4 (i) 4 (ii) vdv/dx + 4x = 0 vdv = 4x dx ½ v = x + c When x = 1, v = 4, so c = 10 so v = 0 4x * x = cos t + sint when t = 0, x = cos 0 + sin 0 = 1* v = dx/dt = sin t + 4 cos t v = 4 cos 0 sin 0 = 4* separating variables and intending to integrate oe condone absence of c. [Not immediate v = -4x (+c)] finding c, must be convinced as AG, need to see at least the statement given here oe (condone change of c) AG following finding c convincingly Alternatively, SC v²=0-4x², by differentiation, v dv/dx= -8x vdv/dx + 4x =0 scores B if, in addition, they check the initial conditions a further is scored (ie 16=0-4). Total possible /4. AG need some justification differentiating, accept ±,±4 as coefficients but not ±1,± and not ±1/,±1 from integrating cao ww AG

8 4 (iii) cos t + sin t = Rcos(t α) = R(cos t cos α + sin t sin α ) R = 5 R cos α = 1, R sin α = tan α =, α = x = 5cos(t 1.107) v = 5sin(t 1.107) EITHER v = 0sin (t α) 0 4x = 0 0cos (t α) = 0(1 cos (t α)) = 0sin (t α) so v = 0 4x SEE APPENDIX 1 for further guidance or.4 or better (not ± unless negative rejected) correct pairs soi correct method cao radians only, 1.11 or better (or multiples of π that round to 1.11) differentiating or otherwise, ft their numerical R, α (not degrees) required form SC for v = 0 cos(t ) oe squarin their v (if of required form with same α as x), and x, and attempting to show v² = 0 4x² ft their R, α (incl. degrees) [α may not be specified]. cao www (condone the use of over-rounded α (radians) or degrees) OR multiplying out v² = (sin t + 4cos t) differentiating to find v (condone coefficient errors), squaring v = 4 sin²t16sintcost +16cos²t and x and multiplying out (need attempt at middle terms) and and 4x² = 4(cos²t + 4sintcost + 4sin²t) attempting to show v² = 0 4x² = 4cos²t +16sintcost +16sin²t (need middle term) and attempting to show that v² + 4x² = 4 (sin²t +cos²t) + 16(cos²t +sin²t) = 4+16 = 0 (or 0 4x² = v²) oe so v² = 0 4x² cao www [7]

9 4 (iv) x = 5cos(t α) or otherwise x max = 5 when cos(t α) =1, t =0, t =1.107 t = 0.55 [] ft their R oe (say by differentiation) ft their α in radians or degrees for method only cao (or answers that round to 0.554)

10 Question answer Marks Guidance 5 (i) dv/dt = kv cao condone different k (allow MR for = kv²) V = (½ kt + c) d V/dt = (½ kt + c).½ k = k(½ kt + c) = kv (ii) (½ k + c) = ½ k + c = 100 (k + c) = k + c = 00 ½ k = 100 k = 00, c = 0 V = (100t) = 10000t (1/ kt + c) constant multiple of k (or from multiplying out oe; or implicit differentiation) cao www any equivalent form (including unsimplified) Allow SCB if V=(1/ kt + c)² fully obtained by integration including convincing change of constant if used Can score M0 SCB substituting any one from t = 1, V = 10,000 or t = 0, V = 0 or t =, V = 40,000 into squared form or rooted form of equation (Allow /100 or /00) substituting any other from above Solving correct equations for both www (possible solutions are (00,0), ( 00,0), (600, 400), ( 600,400) (some from ve root)) either form www SC B for V = (100t)² oe stated without justification SCB4 if justification eg showing substitution SC those working with (k + c)² = 0,000 can score a maximum of B0 A0 (leads to k 146, c 6.8)

4754A Mark Scheme June 2014 BUT

4754A Mark Scheme June 2014 BUT 4754A Mark Scheme June 04 3x A Bx C ( x)(4 x ) x 4 x correct form of partial fractions ( condone additional coeffs eg A B BUT x 4 x ** is M0 ) Ax B Cx D x 4 x * for 3x = A(4 + x ) + (Bx + C)( x) Multiplying