# Pure Core 4. Revision Notes

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Pure Core 4 Revision Notes Mrch 06 Pure Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
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Pure Core 4 Revision Notes Mrch 06 Pure Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd series 8 Binomil series ( + ) n for ny n 8 4 Differentition 0 Reltionship etween nd 0 Implicit differentition 0 Prmetric differentition Eponentil functions, Relted rtes of chnge Forming differentil equtions 5 Integrtion Integrls of e nd Stndrd integrls Integrtion using trigonometric identities Integrtion y reverse chin rule 4 Integrls of tn nd cot 5 Integrls of sec nd cosec 5 Integrtion using prtil frctions 6 Integrtion y sustitution, indefinite 6 Integrtion y sustitution, definite 8 Choosing the sustitution 8 Integrtion y prts 9 Are under curve 0 Volume of revolution 0 Volume of revolution out the is Volume of revolution out the y is Prmetric integrtion Differentil equtions Seprting the vriles Eponentil growth nd decy C4 MAR 06 SDB 6 Vectors 4 Nottion 4 Definitions, dding nd sutrcting, etceter 4 Prllel nd non - prllel vectors 5 Modulus of vector nd unit vectors 6 Position vectors 7 Rtios 7 Proving geometricl theorems 8 Three dimensionl vectors 9 Length, modulus or mgnitude of vector 9 Distnce etween two points 9 Sclr product 9 Perpendiculr vectors 0 Angle etween vectors Vector eqution of stright line Intersection of two lines Dimensions Dimensions 7 Appendi 4 Binomil series ( + ) n for ny n proof 4 Derivtive of q for q rtionl 4 for negtive limits 5 Integrtion y sustitution why it works 6 Prmetric integrtion 6 Seprting the vriles why it works 7 Inde 8 C4 MAR 06 SDB Alger Prtil frctions ) You must strt with proper frction: ie the degree of the numertor must e less thn the degree of the denomintor If this is not the cse you must first do long division to find quotient nd reminder ) () Liner fctors (not repeted) ( )( ) A + ( ) () Liner repeted fctors (squred) ( ) ( ) A ( ) B ( ) + + (c) Qudrtic fctors) ( A + B + )( ) ( + ) + A + B or + ( + + c)( ) ( + + c) Emple: Epress 5 + ( )( + ) in prtil frctions Solution: The degree of the numertor,, is less thn the degree of the denomintor,, so we do not need long division nd cn write 5 + ( )( + ) A B + C + multiply oth sides y ( )(+ ) A( + ) + (B + C)( ) 5 + A A clever vlue!, put 5 A + C C esy vlue, put 0 A B B equte coefficients of 5 + ( )( + ) NB You cn put in ny vlue for, so you cn lwys find s mny equtions s you need to solve for A, B, C, D C4 MAR 06 SDB 7 + Emple: Epress ( )( ) in prtil frctions Solution: The degree of the numertor,, is less thn the degree of the denomintor,, so we do not need long division nd cn write 7 + ( )( ) A + B ( ) + C 7 + A( ) + B( ) + C( )( ) multiply y denomintor 9 + 5B B clever vlue, put ¼ 7 / + 5 A A clever vlue, put ½ 9A B + C C esy vlue, put ( )( ) + ( ) Emple: Epress in prtil frctions Solution: Firstly the degree of the numertor is not less thn the degree of the denomintor so we must divide top y ottom 9 ) + 9 ( Fctorise to give 9 ( )( + ) nd write 6 6 A B + 9 ( )( + ) + multiplying y denomintor 6 A( + ) + B( ) 6 6A A clever vlue, put 6 6B B clever vlue, put C4 MAR 06 SDB Coordinte Geometry Prmetric equtions If we define nd y in terms of single vrile (the letters t or θ re often used) then this vrile is clled prmeter: we then hve the prmetric eqution of curve Emple: y t, + t is the prmetric eqution of curve Find (i) the points where the curve meets the -is, (ii) the points of intersection of the curve with the line y + Solution: (i) The curve meets the -is when y 0 t t ± curve meets the -is t (, 0) nd ( +, 0) (ii) Sustitute for nd y in the eqution of the line t ( + t) + t t 8 0 (t 4)(t + ) 0 t or 4 the points of intersection re (0, ) nd (6, ) Emple: Find whether the curves t +, y t nd s, y s intersect If they do give the point of intersection, otherwise give resons why they do not intersect Solution: If they intersect there must e vlues of t nd s (not necessrily the sme), which mke their -coordintes equl, so find these vlues of t nd s t + s s t + 4 The y-coordintes must lso e equl for the sme vlues of t nd s t s (t + 4) t + t t 0 (t )(t + ) 0 t, s 0 or t, s Curves intersect t t giving (9, 7) Check s 0 gives (9, 7) Or curves intersect t t giving (, ) Check s giving (, ) C4 MAR 06 SDB 5 Conversion from prmetric to Crtesin form Eliminte the prmeter (t or θ or ) to form n eqution etween nd y only Emple: Find the Crtesin eqution of the curve given y y t, t + Solution: t + t, nd y t y ( ), which is the Crtesin eqution of prol with verte t (, ) With trigonometric prmetric equtions the formule sin t + cos t nd sec t tn t will often e useful Emple: Find the Crtesin eqution of the curve given y y sin t +, cos t Solution: Re-rrnging we hve sin t y +, nd cos t, which together with sin t + cos t y + + ( + ) + (y ) 9 which is the Crtesin eqution of circle with centre (, ) nd rdius Emple: Find the Crtesin eqution of the curve given y y tn t, 4sec t Hence sketch the curve y Solution: Re-rrnging we hve tn t, nd sect, 4 which together with sec t tn t y 4 4 y which is the stndrd eqution of hyperol with centre (0, 0) nd -intercepts (4, 0), ( 4, 0) C4 MAR 06 SDB Are under curve given prmetriclly We know tht the re etween curve nd the -is is given y A y da y But, from the chin rule Integrting with respect to t A y da da da y Emple: Find the re etween the curve y t, t + t, the -is nd the lines 0 nd Solution: The re is A y A y we should write limits for t, not?? 0 Firstly we need to find y nd in terms of t y t nd t + Secondly we re integrting with respect to t nd so the limits of integrtion must e for vlues of t 0 t 0, nd t + t t + t 0 (t )(t + t + ) 0 t so the limits for t re from 0 to A y 0 0 (t ) (t + ) 0 t 4 t t 5 5 t t 0 5 Note tht in simple prolems you my e le to eliminte t nd find y in the usul mnner However there will e some prolems where this is difficult nd the ove technique will e etter C4 MAR 06 SDB 7 Sequences nd series Binomil series ( + ) n for ny n ( + ) n n( n ) n( n )( n ) + n + + +!! This converges provided tht Emple: Epnd ( + ), giving the first four terms, nd stte the vlues of for which the series is convergent Solution: ( + ) + ( ) + ( ) ( ) ( ) ( ) ( 4) ( ) + ( ) +!! This series is convergent when / Emple: Use the previous emple to find n pproimtion for Solution: Notice tht ( + ) when So writing in the epnsion The correct nswer to deciml plces is not d eh? Emple: Epnd ( 4 ), giving ll terms up to nd including the term in, nd stte for wht vlues of the series is convergent Solution: As the formul holds for ( + ) n we first re-write ( 4 ) nd now we cn use the formul ! 4! This epnsion converges for 4 4 8 C4 MAR 06 SDB C4 MAR 06 SDB 9 Emple: Find the epnsion of 6 in scending powers of up to Solution: First write in prtil frctions 6 + +, which must now e written s ) ( ) ( ) ( ) ( ! ) )( ( ) (! ) )( ( ) ( 4 Differentition Reltionship etween nd So y 6 6 Implicit differentition This is just the chin rule when we do not know eplicitly wht y is s function of Emples: The following emples use the chin rule (or implicit differentition) d(y ) d(sin y) d(5 y) d + d(y ) d(sin y) y 0y + 5 cos y using the product rule d ( y) ( + y) ( + y) ( + y) + Emple: Find the grdient of, nd the eqution of, the tngent to the curve + y y t the point (, ) Solution: Differentiting + y y with respect to gives + y y + 0 y y 4 Eqution of the tngent is y 4( ) y 4 when nd y 0 C4 MAR 06 SDB Prmetric differentition Emple: A curve hs prmetric equtions t + t, y t t (i) Find the eqution of the norml t the point where t (ii) Find the points with zero grdient Solution: (i) When t, 6 nd y t nd t + when t t t Thus the grdient of the norml t the point (6, ) is 5 / 9 nd its eqution is y 5 / 9 ( 6) 5 + 9y 48 (ii) t grdient 0 when t + 0 t 0 t ± points with zero grdient re (0, ) nd (, ) Eponentil functions, y ln y ln ln y ln y ln d ( ) ln C4 MAR 06 SDB Relted rtes of chnge We cn use the chin rule to relte one rte of chnge to nother Emple: A sphericl snowll is melting t rte of 96 cm s when its rdius is cm Find the rte t which its surfce re is decresing t tht moment Solution: We know tht V 4 / πr nd tht A 4π r Using the chin rule we hve dv dv dr dr dv 4π r, since 4π r dr dr dv 4π r dr 96 4 π dr dr 6π Using the chin rule gin da da dr dr da 8 π r, since 8π r dr dr da 8 π 6 cm s 6π Forming differentil equtions Emple: The mss of rdio-ctive sustnce t time t is decying t rte which is proportionl to the mss present t time t Find differentil eqution connecting the mss m nd the time t dm Solution: Rememer tht mens the rte t which the mss is incresing so in this cse we must consider the rte of decy s negtive increse dm m dm km, where k is the (positive) constnt of proportionlity m dm kt ln m kt + ln A m Ae kt C4 MAR 06 SDB 5 Integrtion Integrls of e nd e e + c ln + c for further tretment of this result, see the ppendi Emple: Find + Solution: + + ½ + ln + c Stndrd integrls must e in RADIANS when integrting trigonometric functions f () n n + f ( ) f () + n sin f ( ) cos ln cos sin e e sec tn sec sec tn cosec cot cosec cosec cot Integrtion using trigonometric identities Emple: Find cot Solution: cot cosec cot cosec cot + c C4 MAR 06 SDB Emple: Find sin Solution: sin ½ ( cos ) sin ( cos ) ½ ¼ sin + c You cnnot chnge to in the ove result to find sin see net emple Emple: Find sin Solution: sin ½ ( cos ) ½ ( cos 6) sin ( cos 6) ½ / sin 6 + c Emple: Find sin cos 5 Solution: Using the formul sin A cos B sin(a + B) + sin(a B) This formul is NOT in the formul ooklet you cn use the formule for sin(a ± B) nd dd them sin cos 5 / sin 8 + sin ( ) / 6 cos 8 + ¼ cos + c / sin 8 sin Integrtion y reverse chin rule Some integrls which re not stndrd functions cn e integrted y thinking of the chin rule for differentition Emple: Find sin 4 cos Solution: sin 4 cos If we think of u sin, then the integrnd looks like constnts, which would integrte to give / 5 u 5 so we differentite u 5 sin 5 d 5 4 sin 5 sin cos which is 5 times wht we wnt nd so u 4 du 4 to give ( ) ( ) 5sin cos sin 4 cos sin 5 + c 5 if we ignore the 4 C4 MAR 06 SDB Emple: Find ( Solution: ( ) ) If we think of u ( ), then the integrnd looks like constnts, which would integrte to ln u so we differentite ln u ln d to give ( ) 4 ln 4 which is 4 times wht we wnt nd so ( ) ¼ ln + c ( ) f '( ) In generl ln f ( ) + c f ( ) u du if we ignore the Integrls of tn nd cot sin tn cos sin cos, f '( ) nd we now hve ln f ( ) f ( ) + c tn ln cos + c tn ln sec + c cot cn e integrted y similr method to give cot ln sin + c Integrls of sec nd cosec sec (sec + tn ) sec + sec tn sec sec + tn sec + tn The top is now the derivtive of the ottom f '( ) nd we hve ln f ( ) + c f ( ) sec ln sec + tn + c nd similrly cosec ln cosec + cot + c C4 MAR 06 SDB 5 Integrtion using prtil frctions For use with lgeric frctions where the denomintor fctorises 6 Emple: Find + 6 Solution: First epress in prtil frctions A B + + ( )( + ) + 6 A( + ) + B( ) put A, put B ln + 4 ln + + c 4 Integrtion y sustitution, indefinite (i) (ii) (iii) Use the given sustitution involving nd u, (or find suitle sustitution) du Find either or, whichever is esier nd re-rrnge to find in terms du of du, ie du Use the sustitution in (i) to mke the integrnd function of u, nd use your nswer to (ii) to replce y du (iv) Simplify nd integrte the function of u (v) Use the sustitution in (i) to write your nswer in terms of Emple: Find 5 using the sustitution u 5 Solution: (i) u 5 (ii) du du 6 6 (iii) We cn see tht there n will cncel, nd 5 u u du u 6 6 du u (iv) du 6 u + c 6 (v) ( 5) 9 + c 6 C4 MAR 06 SDB Emple: Find + using the sustitution tn u Solution: (i) tn u (ii) sec u sec u du du (iii) + sec + tn u u du sec u (iv) du since + tn u sec u sec u du u + c (v) tn + c Emple: Find 4 using the sustitution u 4 Solution: (i) u 4 d (ii) We know tht ( u ) du u so differentiting gives du u u du (iii) We cn see tht n will cncel nd 4 u so 4 u du u (iv) du u + c (v) 4 + c A justifiction of this technique is given in the ppendi C4 MAR 06 SDB 7 Integrtion y sustitution, definite If the integrl hs limits then proceed s efore ut rememer to chnge the limits from vlues of to the corresponding vlues of u Add (ii) () Chnge limits from to u, nd new (v) Put in limits for u Emple: Find 6 using the sustitution u Solution: (i) u du du (ii) (ii) () Chnge limits from to u u 4, nd 6 u u + du (iii) u 4 6 (iv) u + u du u u 4 4 (v) [ + 64] [ + 8] to SF Choosing the sustitution In generl put u equl to the wkwrd it ut there re some specil cses where this will not help ( + ) 5 put u + () put u + 5 put u + 5 or u + 5 n 4 put u or u 4 only if n is ODD put sin u only if n is EVEN (or zero) this mkes 4 4 cos u cos u There re mny more possiilities use your imgintion!! 8 C4 MAR 06 SDB Integrtion y prts The product rule for differentition is d(uv) u dv u dv + v du uv v du u dv d(uv) v du To integrte y prts dv (i) choose u nd (ii) (iii) du find v nd sustitute in formul nd integrte Emple: Find sin Solution: (i) Choose u, ecuse it disppers when differentited dv nd choose sin du (ii) u nd dv sin v sin cos dv du (iii) u uv v sin cos ( cos ) cos + sin + c Emple: Find ln dv u, ut if we tke u ln nd then dv u ln ln du dv nd v ln ln + c Solution: (i) It does not look like product, (ii) (iii) dv u ln ln C4 MAR 06 SDB 9 Are under curve We found in Core tht the re under the curve is written s the integrl y We cn consider the re s pproimtely the sum of the rectngles shown If ech rectngle hs wih δ nd if the heights of the rectngles re y, y,, y n then the re of the rectngles is pproimtely the re under the curve yδ y nd s δ 0 we hve yδ y This lst result is true for ny function y Volume of revolution y y f () y + δ If the curve of y f () is rotted out the is then the volume of the shpe formed cn e found y considering mny slices ech of wih δ: one slice is shown The volume of this slice ( disc) is pproimtely π y δ Sum of volumes of ll slices from to nd s δ 0 we hve (using the result ove π y δ π y πy δ y δ y ) 0 C4 MAR 06 SDB Volume of revolution out the is Volume when y f (), etween nd, is rotted out the is is V π y Volume of revolution out the y is Volume when y f (), etween y c nd y d, is rotted out the y is is V d c π Volume of rottion out the y-is is not in the syllus ut is included for completeness Prmetric integrtion When nd y re given in prmetric form we cn find integrls using the techniques in integrtion y sustitution y y think of cncelling the s See the ppendi for justifiction of this result Emple: If tn t nd y sin t, find the re under the curve from t 0 to t π / 4 Solution: The re y for some limits on y We know tht y sin t, nd lso tht tn t sec t re y y π π 4 4 sin t sec t tn t sect [ ] π sec t To find volume of revolution we need π y nd we proceed s ove writing πy πy C4 MAR 06 SDB Emple: The curve shown hs prmetric equtions t, y t The section etween 0 nd 8 ove the -is is rotted out the -is through π rdins Find the volume generted 0 0 y Solution: V 8 π y t ± nd 8 t, ut the curve is ove the -is y t 0 t 0, t +, or lso y t, t t 6 7 V π y π t t π t 8 t π 8 π ( 8 ) 4 Differentil equtions Seprting the vriles Emple: Solve the differentil eqution Solution: y + y y( + ) y + y We first chet y seprting the s nd y s onto different sides of the eqution ( + ) nd then put in the integrl signs y y + ln y + / + c See the ppendi for justifiction of this technique C4 MAR 06 SDB Eponentil growth nd decy Emple: A rdio-ctive sustnce decys t rte which is proportionl to the mss of the sustnce present Initilly 5 grms re present nd fter 8 hours the mss hs decresed to 0 grms Find the mss fter dy Solution: Let m grms e the mss of the sustnce t time t dm is the rte of increse of m so, since the mss is decresing, dm dm km k m m dm k ln m kt + ln A ln m kt A m Ae kt When t 0, m 5 A 5 m 5e kt When t 8, m 0 0 5e 8k e 8k 0 8 8k ln 0 8 k So when t 4, m 5e Answer 8 grms fter dy C4 MAR 06 SDB 6 Vectors Nottion The ook nd em ppers like writing vectors in the form i 4j + 7k It is llowed, nd sensile, to re-write vectors in column form ie i 4j + 7k 4 7 Definitions, dding nd sutrcting, etceter A vector hs oth mgnitude (length) nd direction If you lwys think of vector s trnsltion you will not go fr wrong Directed line segments The vector AB is the vector from A to B, (or the trnsltion which tkes A to B) This is sometimes clled the displcement vector from A to B A B Vectors in co-ordinte form Vectors cn lso e thought of s column vectors, thus in the digrm Negtive vectors 7 AB 7 AB is the 'opposite' of BA nd so BA AB Adding nd sutrcting vectors (i) Using digrm Geometriclly this cn e done using tringle (or prllelogrm): Adding: A 7 + B + The sum of two vectors is clled the resultnt of those vectors 4 C4 MAR 06 SDB Sutrcting: (ii) Using coordintes c + d + c + d nd c d c d Prllel nd non - prllel vectors Prllel vectors Two vectors re prllel if they hve the sme direction one is multiple of the other Emple: Which two of the following vectors re prllel? 6, 4, 6 4 Solution: Notice tht ut is not multiple of 4 nd so 6 is prllel to 4 nd so cnnot e prllel to the other two vectors 4 Emple: Find vector of length 5 in the direction of 4 Solution: hs length nd so the required vector of length 5 5 is 4 9 C4 MAR 06 SDB 5 Non-prllel vectors If nd re not prllel nd if α + β γ + δ, then α γ δ β (α γ) (δ β) ut nd re not prllel nd one cnnot e multiple of the other (α γ) 0 (δ β) α γ nd δ β Emple: If nd re not prllel nd if + + β α + 5, find the vlues of α nd β Solution: Since nd re not prllel, the coefficients of nd must lnce out α 5 α 7 nd + β β Modulus of vector nd unit vectors Modulus The modulus of vector is its mgnitude or length If AB 7 then the modulus of AB is AB AB Or, if c 5 then the modulus of c is c c () Unit vectors A unit vector is one with length Emple: Find unit vector in the direction of 5 Solution: hs length + 5, 5 nd so the required unit vector is C4 MAR 06 SDB Position vectors If A is the point (, 4) then the position vector of A is the vector from the origin to A, usully written s OA 4 For two points A nd B the position vectors re OA nd OB To find the vector AB go from A O B giving AB + O y A B Rtios Emple: A, B re the points (, ) nd (4, 7) M lies on AB in the rtio : Find the coordintes of M Solution : AB 6 B AM AB OM OA + AM OM 5 5 M is ( 5, 5) O M A C4 MAR 06 SDB 7 Proving geometricl theorems Emple: In tringle OBC let M nd N e the midpoints of OB nd OC Prove tht BC MN nd tht BC is prllel to MN Solution: Write the vectors OB s, nd OC s c Then OM ½ OB ½ nd ON ½ OC ½ c O To find MN, go from M to O using ½ nd then from O to N using ½ c MN ½ + ½ c MN ½ c ½ B M N c Also, to find BC, go from B to O using nd then from O to C using c C BC + c c But MN ½ + ½ c ½ (c ) ½ BC BC is prllel to MN nd BC is twice s long s MN Emple: P lies on OA in the rtio :, nd Q lies on OB in the rtio : Prove tht PQ is prllel to AB nd tht PQ / AB A Solution: Let OA, nd OB P AB OA + OB OP OA, OQ OB nd PQ OP + OQ / / / ( ) O Q B / AB PQ is prllel to AB nd PQ / AB 8 C4 MAR 06 SDB Three dimensionl vectors Length, modulus or mgnitude of vector The length, modulus or mgnitude of the vector OA OA + +, ` sort of three dimensionl Pythgors is Distnce etween two points To find the distnce etween A, (,, ) nd B, (,, ) we need to find the length of the vector AB AB AB AB ( ) + ( ) + ( ) Sclr product cos θ where nd re the lengths of nd nd θ is the ngle mesured from to θ Note tht (i) cos 0 o (ii) ( + c) + c (iii) since cosθ cos ( θ) In co-ordinte form + cos θ or + + cos θ C4 MAR 06 SDB 9 Perpendiculr vectors If nd re perpendiculr then θ 90 o nd cos θ 0 thus perpendiculr to 0 nd 0 either is perpendiculr to or or 0 Emple: Find the vlues of λ so tht i λj + k nd i + λj + 6k re perpendiculr Solution: Since nd re perpendiculr 0 λ + 6 λ 0 6 λ 0 λ 9 λ ± Emple: Find vector which is perpendiculr to, Solution: Let the vector c, p q r 0 nd, nd, p q, e perpendiculr to oth nd r p q r p q + r 0 nd p + q + r 0 0 Adding these equtions gives 4p + r 0 Notice tht there will never e

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