Description

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

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.

Related Documents

Share

Transcript

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

We Need Your Support

Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks