FP POLAR COORDINATES: PAST QUESTIONS. The curve C hs polr eqution r = cosθ, () Sketch the curve C. () (b) Find the polr coordintes of the points where tngents to C re prllel to the initil line. (6) (c) Find the re of the region bounded by C. () (Totl mrks). The digrm bove shows the curves given by the polr equtions r =, 0, nd r =.5 + sin 0. () Find the coordintes of the points where the curves intersect. () The region S, between the curves, for which r > nd for which r < (.5 + sin θ), is shown shded in the digrm bove. City of London Acdemy
(b) Find, by integrtion, the re of the shded region S, giving your nswer in the form π + b, where nd b re simplified frctions. (7) (Totl 0 mrks). The digrm bove shows sketch of the curve with polr eqution r = + cos θ, > 0, 0 θ < π The re enclosed by the curve is 07. Find the vlue of. (Totl 8 mrks). A C R P O N I n i t i l l i n e The curve C shown in the digrm bove hs polr eqution City of London Acdemy
r ( cos ), 0.. At the point P on C, the tngent to C is prllel to the line () Show tht P hs polr coordintes,. (5) The curve C meets the line t the point A. The tngent to C t P meets the initil line t the point N. The finite region R, shown shded in the digrm bove, is bounded by the initil line, the line, the rc AP of C nd the line PN. (b) Clculte the exct re of R. (8) (Totl mrks) 5. C C O = 0 The digrm bove shows the curve C which hs polr eqution r = ( + cos θ), 0 θ < π nd the circle C with eqution r =, 0 θ < π, where is positive constnt. () Find, in terms of, the polr coordintes of the points where the curve C meets the circle C. () The regions enclosed by the curves C nd C overlp nd this common region R is shded in the figure. (b) Find, in terms of, n exct expression for the re of the shded region R. (8) City of London Acdemy
(c) In single digrm, copy the two curves in the digrm bove nd lso sketch the curve C with polr eqution r = cosθ, 0 θ < π Show clerly the coordintes of the points of intersection of C, C nd C with the initil line, θ = 0. () (Totl 5 mrks) 6. () Sketch the curve C with polr eqution r = 5 + cos θ, 0 θ π. () (b) Find the polr coordintes of the points where the tngents to C re prllel to the initil line θ = 0. Give your nswers to significnt figures where pproprite. (6) (c) Using integrtion, find the re enclosed by the curve C, giving your nswer in terms of π. (6) (Totl mrks) 7. Q C R P O I n i t i l l i n e The digrm bove shows sketch of the curve C with polr eqution r = sinθcos θ, 0 θ <. The tngent to C t the point P is perpendiculr to the initil line. City of London Acdemy
() Show tht P hs polr coordintes, 6. (6) The point Q on C hs polr coordintes,. The shded region R is bounded by OP, OQ nd C, s shown in the digrm bove. (b) Show tht the re of R is given by 6 sin cos cos d () (c) Hence, or otherwise, find the re of R, giving your nswer in the form + bπ, where nd b re rtionl numbers. (5) (Totl mrks) 8. m C O I n i t i l l i n e r cos, 0 The figure bove shows curve C with polr eqution, nd line m with polr eqution 8. The shded region, shown in the figure bove, is bounded by C nd ( ). m. Use clculus to show tht the re of the shded region is (Totl 7 mrks) 9. City of London Acdemy 5
= P l R C O = 0 r cos, 0. A curve C hs polr eqution The line l is prllel to the initil line, nd l is the tngent to C t the point P, s shown in the figure bove. () (i) Show tht, for ny point on C, r sin θ cn be expressed in terms of sin θ nd only. () (ii) Hence, using differentition, show tht the polr coordintes of P re, 6. (6) The shded region R, shown in the figure bove, is bounded by C, the line l nd the hlf-line. with eqution (b) Show tht the re of R is. 6 (8) (Totl 5 mrks) 0. City of London Acdemy 6
l P C R O = 0 Q The curve C which psses through O hs polr eqution The line l hs polr eqution r = ( + cos ), <. r = sec, < <. The line l cuts C t the points P nd Q, s shown in the digrm. () Prove tht PQ = 6. (6) The region R, shown shded in the digrm, is bounded by l nd C. (b) Use clculus to find the exct re of R. (7) (Totl mrks) City of London Acdemy 7
. The curve C hs polr eqution r = 6 cos, <, nd the line D hs polr eqution r = sec, 6 5 < 6. () Find crtesin eqution of C nd crtesin eqution of D. (5) (b) Sketch on the sme digrm the grphs of C nd D, indicting where ech cuts the initil line. () The grphs of C nd D intersect t the points P nd Q. (c) Find the polr coordintes of P nd Q. (5) (Totl mrks). C O B C I n i t i l l i n e A The digrm bove is sketch of the two curves C nd C with polr equtions C : r = ( cos ), < nd C : r = ( + cos ), <. The curves meet t the pole O, nd t the points A nd B. City of London Acdemy 8
() Find, in terms of, the polr coordintes of the points A nd B. () (b) Show tht the length of the line AB is. () The region inside C nd outside C is shown shded in the digrm bove. (c) Find, in terms of, the re of this region. (7) A bdge is designed which hs the shpe of the shded region. Given tht the length of the line AB is.5 cm, (d) clculte the re of this bdge, giving your nswer to three significnt figures. () (Totl 6 mrks). () Sketch the curve with polr eqution r = cos, < () (b) Find the re of the smller finite region enclosed between the curve nd the hlf-line = 6 (6) (c) Find the exct distnce between the two tngents which re prllel to the initil line. (8) (Totl 6 mrks) City of London Acdemy 9
. D O B A I n i t i l l i n e C A logo is designed which consists of two overlpping closed curves. The polr equtions of these curves re r = ( + cos ) nd r = (5 cos ), 0 <. The digrm bove is sketch (not to scle) of these two curves. () Write down the polr coordintes of the points A nd B where the curves meet the initil line. () (b) Find the polr coordintes of the points C nd D where the two curves meet. () (c) Show tht the re of the overlpping region, which is shded in the figure, is (9 8). (8) (Totl mrks) MARK SCHEMES. () BB City of London Acdemy 0
B shpe B Lbels (b) Tngent prllel to initil line when y = r sin θ is sttionry Consider therefore d d ( cosθ sin θ) = sin θ sin θ + cos θ( sin θ cos θ) = 0 A sin θ[cos θ cos θ - sin θ sin θ] = 0 sinθ 0 cosθ = 0 θ = 6 or 6 A Coordintes of the points,, 6 6 AA 6 (c) Are = = r d cos d sin [ ( )] A A []. ().5 sin sin 0.5 nd 5 or 8 8 5 or 6 6, A, A (b) Are = 5 8 (.5 sin ) d 8, - 9 City of London Acdemy
= (.5 sin ( cos 6 ))d - 9 5 8 8 = (.5 cos ( sin 6 )) 6 5 8 8 9 A 5 6 A 7 [0]. x A ( cos ) d 0 Applies x r 0 (d ) with correct limits. Ignore dθ. B ( + cosθ) = + 6cosθ + 9 cos θ = cos 6 cos 9 cos cos Correct underlined expression. A A x 0 9 9 6 cos cos d 9 9 6sin sin 0 9 0 (0) 0 Integrted expression with t lest out of terms of the form ± Aθ ± Bsinθ ± Cθ ± Dsinθ * Ignore the. Ignore limits. θ + 6sinθ + correct ft integrtion. Ignore the. Ignore limits. A ft 9 9 A Hence, 9 07 Integrted expression equl to 07. d * City of London Acdemy
9 07 9 As > 0, = 7 = 7 A cso Some cndidtes my chieve = 7 from incorrect working. Such cndidtes will not get full mrks [8]. () r cos = (cos cos ) or rcos = cos cos B d( r cos ) d( r cos ) ( sin cos sin ) or ( sin sin ) d d A ( sin + cos sin ) = 0 cos = stisfied by = nd r = (*) which is da 5 Alterntive for first mrks: dr sin d B dx dr r sin cos sin 8sin cos d d A Substituting r = nd 0 = f into originl eqution scores 0 mrks. (b) r d (8) ( cos cos ) d (8) sin sin A City of London Acdemy
sin 8 sin Tringle: 8 (rcos )(r sin ) = 6 7 8 A 5 6 7 ( 6) Totl re: (A)A 8 needs ttempt to expnd ( cos ) giving three terms (llow slips) Second needs integrtion of cos using cos ± Third needs correct limits my evlute two res nd subtrct needs ttempt t re of tringle nd A for co Next A is for vlue of re within curve, then finl A is co, must be exct but llow terms nd isw for incorrect collection of terms Specil cse for use of r sin gives B0A0M0A0 [] 5. () ( + cos) = Solve to obtin cos = = ± 5 nd points re (, ) nd (, ) A, A First A for r = second for both vlues in rdins. Accept.07 nd 5.59. dp or better for finl A (b) Use re = r d to give Obtin (9 + cos + cos + )d ( cos ) d City of London Acdemy
A Integrte to give + sin + sin A Use limits 5 nd nd, then double or or theirs 6 Find third re of circle = B 8 Obtin required re = A, A 8 First M for substitution, expnsion nd ttempt to use double ngles. Second M for integrting expression of the form + bcos + ccos Lose finl A only if missing in lst line (c) O 5 correct shpe B 5 nd mrked B mrked nd psses through O B First B for pproximtely symmetricl shpe bout initil line, only loop which is convex strictly within shded region [5] City of London Acdemy 5
6. () 5 5 6 5 y 5 6 7 x 5 + 5 5 6 Shpe (close curve, pprox. symmetricl bout the initil line, in ll qudrnts nd centred to the right of the pole/origin). Shpe (t lest one correct intercept r vlue... shown on sketch or perhps seen in tble). (Also llow wrt.7 or wrt 6.7). B B (b) y = r sin = 5 sin + sin cos dy d = 5 cos sin + cos (= 5 cos + cos ) A 5 cos ( cos ) + cos = 0 cos + 5 cos = 0 ( cos )(cos + ) = 0 cos =... (0.88...) Also llow rccos =.8 nd 5.0 (wrt) (Allow.8 wrt) A r = 5 + (Allow wrt 5.50) A 6 nd M: Forming qudrtic in cos. rd M: Solving term qudrtic to find vlue of cos (even if clled ). Specil cse: Working with r cos insted of r sin : st for r cos = 5 cos + cos st A for derivtive 5 sin sin cos, then no further mrks. (c) r = 5 + 0 cos + cos B 5 sin 5 0 cos cos d 0 sin Aft Aft (ft for integrtion of ( + b cos ) nd c cos respectively) 5 0 sin sin 0... 5 (50 ) or equiv. in terms of. A City of London Acdemy 6
6 st M: Attempt to integrte t lest one term. nd M: Requires use of the, correct limits (which could be 0 to, or to, or double 0 to ), nd subtrction (which could be implied). [] 7. () x = r cos = sin cos dx cos cos sin d ny correct expression A d x dx 0 0 cos (cos sin ) 0 Solving d d sin or cos or tn 6 AG A cso sin cos r = 6 6 AG Acso 6 So mny wys x my be expressing e.g. sin cos, sin ( + cos ), sin + (/) sin dx leding to mny results for d Some relevnt equtions in solving [( sin ) = 0, ( cos ) = 0, ( tn ) = 0, cos = 0] dx 0 Showing tht 6 stisfies d, llow A dx providing d correct Strting with x = r sin cn gin M0 (b) A r 6 d.6 6 sin cos d 8 sin cos = cos ( sin cos ) = cos sin = (cos + )sin cos cos sin = Answer AG A cso City of London Acdemy 7
First for use of double ngle formul for sin A Second for use of cos A = cos A Answer given: must be intermedite step, s shown, nd no incorrect work (c) Are = sin 6 sin 6 sin 8 6 sin sin 8 8 6, 6 8 6 6 6 ignore limits sin 8 (sub. limits) both co 5 A A, A sin b sin For first M, of the form (Allow if two of correct form) On epen the order of the As in nswer is s written [] 8. Use of d B r Limits re 8 B nd 6 cos = 8 (l + cos ) sin ( cos ) d A A sin 8 (0 ) 8 City of London Acdemy 8
( ) cso A [7] 9. () (i) r sin θ = cos θsin = ( sin ) sin B (= (sin θ sin θ)) (ii) d d ( (sin sin )) = (sin cos 8 sin cos ), = 0, A, = 8 sin (Proceed to sin = b) sin θ = θ = 6, r = A, A cso 6 cos d (b) sin M: Attempt d r, to get k sin θ A... 6 M: Using correct limits A =. 6 M: Full method for rectngle or tringle A R = 6 6 ( ) M: Subtrcting, either wy round d A cso 8 [5] () (ii) First A: Correct derivtive of correct expression for r sin θ or r sin θ. (b) Finl M mrk is dependent on the first nd third M s. Attempts t the tringle re by integrtion: full method is required for. Missing fctors: (or ) Mximum one mrk penlty in the question. City of London Acdemy 9
0. () O N P R = 0 Q ( + cos) = cos or r = r cos + cos = 0 or r r = 0 A (cos )(cos + ) = 0 or (r 6)(r + ) = 0 cos =, PQ = ON tn 6 cso A Note ON = = 6 (*) or A r = 6 6 or PQ = [(6) () ] = (7 ) = 6 (*) cso or ny complete equivlent r d...... 6 0... (b) = r d cos...... cos cos ( + cos) d cos d City of London Acdemy 0
sin sin = A = 6 8 (= [ + 9] 56. ) A use of their for Are of POQ = 6 or 9 B R = (8 + 9) co A 7 []. () For C: Using polr/ Crtesin reltionships to form Crtesin eqution so x + y = 6x A [Eqution in ny form: e.g. (x ) + y = 9 from sketch. 6x x y x y or ] r cos For D: nd ttempt to expnd x y = (ny form) A 5 (b) City of London Acdemy
Circle, symmetric in initil line pssing through pole B Stright line B Both pssing through (6, 0) B (c) Polrs: Meet where 6cos cos( ) = sin cos = sin sin = 0 or tn = [ = 0 or ] Points re (6, 0) nd (, ) B, A 5 [] Alterntives (only more common): () Eqution of D: Finding two points on line Using correctly in Crtesin eqution for stright line Correct Crtesin eqution City of London Acdemy
A (c) Crtesin: Eliminte x or y to form qudrtic in one vrible [x 5x + 8 = 0, y 6 y = 0] Solve to find vlues of x or y Substitute to find vlues of other vrible x or 6; y 0 or BA Points must be (6, 0) nd (, ) BA. () ( cos) = ( + cos) = cos cos = = or r = AA [Co-ordintes of points re (, ) nd (, ) ] (b) AB = rsin = A Are = r d [ ( cos ) 9 ( = cos ) ] d City of London Acdemy
[ cos cos 9( cos cos = A [ 8 0 cos 8cos )] d = = k[ 8 + 0sin B.. sin ] B Uses limits nd uses limits nd )] d correctly or uses twice smller re nd 0 correctly.(need not see 0 substituted) = [ + 0 ] or = [ + 9 ] or.0 A 7 (d) =.5 = B Are = [9 ], = 9.07 cm, A [6]. () Shpe + horiz. xis B B City of London Acdemy
(b) Are = r d = 9 cos d use of r 9 cos d = use of cos = cos 9 sin 8 = 6, A 9 = 8 6 9 = 6 or 0.0 A 6 subst. nd 6 (c) r sin = sin cos dy d = cos cos 6 sin sin (diff. r sin ), A dy d = 0 6 cos cos sin cos = 0 dy use of d = 0 6 cos cos ( cos )cos = 0 use double ngle formul 8 cos 5 cos = 0 solving cos = 0 or cos 5 = 6 or tn = 5 or sin = 6 A \ r = ( 6 5 ) = r sin = 6 6 d = A 8 use of d = r sin [6] City of London Acdemy 5
. () (5, 0), (, 0) B, B llow on digrm B only if no zeros or digrm with initil line (b) + cos = 5 cos cos = 5, = A 5 (, ), (, ) A both, ccept (c) Allow s below for method shown in either clcultion (5 cos) d ( + cos) d = (5 5cos + cos )d = (9 + cos + cos )d = (7 0cos + cos)d = ( + cos + cos)d = 7 0sin + sin, = + sin + sin A, A A = (5 cos) d + ( + cos) d dds (5 cos ) d ( 0 = cos ) d correct limits A = [7 0 + ] + [( ) 6 ] d City of London Acdemy 6
= (9 8] (*) A cso 8 [] 5. () cos cosd A = correct with limits cos cos d A = sin sin 0 A = = A 6 (b) x = cos + cos r cos dx d = sin cos sin A dx d = 0 cos = finding = or = City of London Acdemy 7
r = or r = finding r A: r =, = B: r =, = both A nd B 5 A (c) x = A WX = + = 7 (d) WXYZ = 8 B ft 7 (e) Are = 8 00 00 =. cm A [6] 6. () A O l closed loop B City of London Acdemy 8
symmetry in l B (b) Are of loop = r d = cos d = 0 cos d = 0 ( sin )cos d A sin = sin 0 A 8 = ( ) = A 7 (c) y = r sin = sin cos nd the vlue of in (0, ) for dy which d = 0 is required dy d 5 = cos sin cos 5 = cos cos cos 5 A = cos (5cos ) A cos ¹ 0 in (0, ), so cos = 5 At A, = 0.68, r =.6 ( deciml plces) A, A 7 [6]. No Report vilble for this question. City of London Acdemy 9
. The finding of the two vlues of in prt () ws usully correctly done but some cndidtes then wsted vluble time in tking their two vlues nd working in circle to find the r vlues - some of which were not equl to In prt (b) the vst mjority of the cndidtes knew how to find the re enclosed between the two rdius vectors nd the curve simply s one integrl. Others chose to split it into one re 5 from 0 to minus the re from 0 to 8 minus the re from 8 to - long wy round. Most of the cndidtes decided to use the integrtion method to find the re of the sector s well. Errors in integrtion were usully in the use of the double ngle formul lthough most used it correctly. A minority of cndidtes forgot to squre the function before integrting, but where this squring hd been done the subsequent integrtion of the trigonometric functions ws well done. Where cndidtes fell down ws in the creful ppliction of the limits to their functions writing things more netly would hve helped in number of cses. Some forgot bout the re of the sector of the circle completely. It ws, however, plesing to see mny completely correct solutions.. This question ws well nswered by cndidtes nd sttistics showed tht round 50% of the cndidture gined ll 8 mrks vilble for this question. d Most cndidtes pplied the formul r but number of them struggled to write down the correct limits to find the relevnt re for their expression. The mjority of the errors mde by cndidtes were lgebric; 6 sin θ ws sometimes missed out when ( + cos θ) ws expnded. Sometimes ws tken out from some cndidtes integrl. Most cndidtes knew 9 cos they needed to substitute for 9cos θ in order to integrte the expression but sometimes there were errors in deling with the 9 nd the. There were significnt minority of cndidtes who did not know the correct strtegy to pply in order to integrte cos θ. These cndidtes usully lost 7 of the 8 mrks vilble for this question.. Prt () required cndidtes to consider r cos θ nd to differentite to find mximum vlue. dx This ws dximplied s the tngent ws perpendiculr to the initil line nd so d = 0. A number of cndidtes chose to consider r sin θ insted nd little credit ws given for this. ( cos cos ) d In prt (b) most relised tht they needed to find. There were number of sign slips but the methods were understood, nd most tried to use double ngle nd formule. To find the shded re they needed to use the limits nd to subtrct. This gve the finite re enclosed by the rc AP nd the stright lines OP nd OA. They then needed to find the re of tringle OPN. The sum of these two res gve them the totl re required. A City of London Acdemy 0
number of cndidtes found the re of rectngle or used tringle where the sides hd incorrect lengths, prticulrly where the bse ws units insted of unit. Mny nswers for the tringle re included pi term insted of root. Full mrks in question 8 ws indictive of good grde A cndidte. 5. There were mny good solutions to this question, lthough few scored full mrks. It ws prt (b) tht cused the most difficulty nd tht ws sometimes omitted. Some cndidtes spent too long plotting points for their sketch in prt (), but most were ble to produce closed curve in pproximtely the correct position. Indiction of scle ws required to score the second mrk here. In prt (b), few cndidtes strted to differentite r cos θ insted of r sin θ, but most were ble to produce correct derivtive nd to proceed to find qudrtic in cos θ. Despite mny correct qudrtic equtions, mistkes frequently occurred from tht point onwrds. Some of these mistkes were creless, others stemmed from n pprent expecttion tht the vlues of θ would be exct. Even cndidtes who hd correct solution for one vlue of θ (.8) were often unble to produce the second vlue. The r coordinte of the required points ws often missing or wrong. In generl, cndidtes were much hppier with prt (c), where there were mny excellent solutions. There ws, however, plenty of scope for mistkes, nd these included slips in cos (cos ) squring (5 + cos ), sign or numericl errors in pplying nd integrtion slips. Sometimes the limits for integrtion were wrong or the ws omitted from the re formul, losing the lst two mrks 6. Most cndidtes strted this polr coordintes question successfully only losing n ccurcy mrk for the vlues of θ, writing them s π/ nd π/ rther thn π/ nd 5π/. r d Cndidtes were on the whole correct in the use of nd mny of them were successful in negotiting the expnsion of ( + cos θ) including the use of double ngles. The mistkes usully consisted of identifying pproprite limits nd vluble mrks were lost here. Some exmples of this included finding the whole re of C using limits of 0 nd π nd then subtrcting tht prt of C between π/ nd π/. This inevitbly resulted in errors s some cndidtes forgot bout the circle prt of the problem nd proceeded to subtrct the re rther thn dd it. Some cndidtes struggled with integrtion to estblish tht the / of the circle needed ws (6π )/. It ws plesing to note how mny cndidtes mde it correctly to the 76 8 to finl nswer lthough some found it unnecessry to simplify to 6. 7. In prt () there ws some confusion s to whether it ws r cos θ or r sin θ, or indeed r, tht needed to differentited with respect to θ. However, mny correct solutions were seen, sometimes clerly ided by the given nswers for r nd θ. Agin this ws chllenge to mrk s there were so mny different pproches with lrge number of different correct equtions tht led to the given results. Some cndidtes chose to rerrnge r sin θ cos θ, using either sin θ + cos θ =, or double ngle formul or City of London Acdemy
combintion of both, before they differentited, nd some wited until they hd differentited before mking similr move. Solutions were, therefore, often not s concise s they might hve been. Prt (b) ws not done well by lrge number of cndidtes, who hd little structure to their work. One mrk ws often wrded, perhps generously t times, for correct use of the cosine double ngle formul, but complete method ws usully only seen from the better cndidtes. Mny cndidtes spent much time on this prt, often producing very unwieldy expressions, nd cos it ws not uncommon to see incorrect double ngle formule; cos θ = ws lso seen too often. Mny cndidtes went on to gin mrks for integrtion in prt (c) lthough the downfll here cme often in not relising tht the first term could be integrted directly, or in giving the result s sin θ; tht error still llowed cndidtes to gin mrks, however, nd tht ws common score. 8. The method needed ws well understood nd most could use n pproprite formul nd identify the correct double ngle formul to crry out the indefinite integrtion. The limits, however, proved testing nd only bout 60% of the cndidtes used the correct limits of 8 nd. 9. There were some very poor nd rushed solutions to this question. In prt ()(i), surprisingly mny cndidtes were unble to link the required expression with the given polr eqution, even though they knew tht cos = sin. Most knew in prt ()(ii) tht they hd to del with rsin, but mde it more difficult for themselves by trying to differentite cos sin rther thn ( sin ) sin (s suggested by prt ()(i)). Consequentilly mny mistkes, often due to the frctionl power, were seen in differentition, nd few cndidtes proceeded legitimtely to the given polr coordintes of P. Prticulrly disppointing t this level ws the (not uncommon) mistke of squre rooting seprte terms of sum, e.g. sin sin sin sin. Completely correct solutions to prt (b) were rre. While better cndidtes did relise tht they cos sin hd to use the re of the tringle 6 6, there ws much confusion over City of London Acdemy
cos d which limits to use for the integrl, nd which re this represented. A common mistke ws to use insted of s the upper limit, even though the curve ws 0 defined only for. 0. This question ws often nswered very well. In prt () the mjority of the cndidtes were ble to find the vlues of θ t P nd Q nd usully they were then ble to prove the required result. There ws some dubious trigonometry seen here, such s OP = 6 followed by 6 tn PQ =, but mny gve convincing proof. A sizeble minority of cndidtes mde the unwrrnted ssumption in prt () tht the tngent t P ws prllel to the initil line, whilst this yielded correct vlue for θ it did not, of course, receive ny credit lthough such cndidtes were llowed to use their vlue in the following prt. In prt (b) most tried integrting 6 ( + cos θ) nd knew how to del with the cos θ by using the double ngle formul. Most chose suitble limits but common errors were to use, or 6 insted of. Some forgot to subtrct the re of the tringle nd few only clculted the re bove the initil line but, prt from creless slips, mny cndidtes gve very good solutions.. There is no doubt tht this ws the most chllenging nd lest productive question for most cndidtes. In trying to convert from polr equtions to crtesin equtions mny cndidtes often took pge of working for very little, if ny, rewrd. The eqution r 6 cos ws recognised, or plotted, s the correct circle in prt (b) but its crtesin eqution hd often not sec( r ) been found in prt (). Recognition of s stright line ws reltively rre, nd its crtesin eqution only found by the better cndidtes. Some cndidtes who hd been successful in prt () used the crtesin equtions of the grphs to find their points of intersection nd convert them to polrs coordintes, but for most cndidtes prt (c) involved solving trigonometric eqution. Most cndidtes gined very cos( ) generous first mrk, but solving the resulting eqution cos ½ ws generlly not well done; probbly sign tht confidence hd tken knock in the erlier prts. Most cos( ) cndidtes expnded to give cos sin cos ; those who progressed further to sin cos sin or sin( ), usully completed the solution, lthough cncelling sin in the former cse ws quite common. A net solution ws to use the cos cos( ). fctor formule to give City of London Acdemy
cos( ) It ws very disppointing to see cos ½ cos( ) cos = ½ or ½, even though, more disppointingly in this cse, they gve the correct nswers!. Good cndidtes were ble to score full mrks in their solutions to this question. For the mjority of cndidtes cos = ½ usully ppered in (); few ignored the rnge of nd gve A s (/, 5/) or (-/, -/). Hving n nswer to im for in (b) certinly helped mny cndidtes to work out wht they should do, but mny omitted this prt or ttempted to find n rc length. In prt (c), most used r for one or both curves, even if the ½ ws missing or limits were wrong. There were good nswers for the integrtion of trig functions cos nd cos. There ws often muddle over whether to use one or both curves in finding the required re the significnce of the vlues of θ from () ws not relised by mny. A mixture of limits 0 to for C nd 0 to / for C ws not uncommon, lthough some retrieved the sitution by then subtrcting the re / to for C. Perhps becuse time ws running out there ws evidence of crelessness over signs s cndidtes ttempted to tidy up their solutions. A very few cndidtes used r insted of r, or ttempted to integrte, thus producing nswers in or clerly not dimensionlly correct. In prt (d) the vlue of ws usully found correctly, lthough some cndidtes mistkenly used =.5. Some cndidtes left their finl nswer in the exct form, rther thn s deciml to significnt figures s requested in the question.. Prt () ws generlly correct lthough minority of students did not restrict their vlues of to those stted in the question. In prts (b) nd (c) mny cndidtes used the double ngle formule lthough some hd slight cos d ( sin ) 0 errors such s. In prt (c) mny cndidtes who solved d r ignored the instruction to give the exct distnce or used d = r not rsin.. Prts () nd (b) were generlly well done lthough mny gve n nswer for point C which ws 5,, out of rnge, giving insted of. Prt (c) is very demnding question. Mny could gin some mrks by showing tht they were ble to integrte cos 5 cos but finding the right limits nd putting together the correct res proved difficult. Correct solutions were, however, seen nd some of these were very stylish nd commendbly brief. nd 5. Most cndidtes mde good ttempt t the re, including the correct use of the double ngle City of London Acdemy
formul in the integrl. Most cndidtes ttempted to differentite r cos but severl cndidtes lost mrks by giving vlue for outside the rnge. Mny cndidtes hd difficulty in finding the length WX. Incorrect trigonometry led to the projection of OA onto the initil line being given s or nd mny cndidtes used WX = + the projection rther thn using + projection. In (d) minority of students used rther thn. 6. No Report vilble for this question. 5. W X A B O i n i t i l l i n e Z Y The digrm bove shows sketch of the crdioid C with eqution r = ( + cos ), <. Also shown re the tngents to C tht re prllel nd perpendiculr to the initil line. These tngents form rectngle WXYZ. () Find the re of the finite region, shded in the digrm bove, bounded by the curve C. (6) (b) Find the polr coordintes of the points A nd B where WZ touches the curve C. (5) (c) Hence find the length of WX. () Given tht the length of WZ is, (d) find the re of the rectngle WXYZ. () A hert-shpe is modelled by the crdioid C, where = 0 cm. The hert shpe is cut from the rectngulr crd WXYZ, shown in the digrm bove. (e) Find numericl vlue for the re of crd wsted in mking this hert shpe. () (Totl 6 mrks) City of London Acdemy 5
6. The curve C is given by r = cos,, where (r, ) re polr coordintes. () Sketch the curve C. () (b) Find the re of the region enclosed by C. (7) At the point A on C, =, where 0 < <, nd the tngent t A to C is prllel to the initil line. (c) Find, to deciml plces, the vlue of nd the corresponding vlue of r. (7) (Totl 6 mrks) City of London Acdemy 6