Thomas Whitham Sixth Form

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1 Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors

2 Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos Acos B sin Asin B A B cos Acos B sin Asin B tn A tn B tn( A B) tn Atn B tn A tn B tn( A B) tn Atn B Emple Solve the eqution sin cos cos sin for sin cos cos sin sin cos cos sin sin S T A C PV = Emple Given A B 5 nd tn A find without using clcultor the vlue of tn A B tn A tn B tn Atn B tn B t where t tn B. t

3 t t t t tn B Doule ngle formule sin sin cos cos cos sin sin cos Pge tn importnt rerrngements tn tn sin cos cos cos [see the rticle on integrtion] Emple Let Find the vlue of 5 tn without using clcultor. tn t t t t tn tn t t 8 t tn Emple Solve the eqution cos sin for cos sin sin sin sin sin sin sin sin S T A C

4 5 7 Pge R/ Methods where nd re positive constnts there re vlues of the positive R nd the cute ngle for which sin cos Rsin cos sin R cos sin cos Rsin cos sin R cos Emple Epress sin cos in the form R cos Identity method cos sin R cos cos Rcos sin Rsin R R cos R Rcos cos Rsin sin Rsin R cos cos sin Rcos cos Rsin sin sin R R tn cos sin cos sin cos in the form Rsin Emple Epress Alterntive method sin cos 5sin cos sin cos cos sin 5sin sin cos 5sin 5... tn Where 5

5 Pge Emple Solve the eqution sin cos for Answers to the nerest degree. LHS sin cos Rsin tn 8. Rsin cos R cos sin Rcos sin Rsin cos R 9 R Using the identity method you will find tht sin cos for sin 8. for sin S T A C PV=9. Geometry Prmetrics () Equtions of the form y f () or f ( y) re clled crtesin equtions. Emple y e y 5 () Equtions of the form f ( t) y g( t) where t is third vrile re clled prmetric equtions; t is the prmeter They define curve which hs points with coordintes of the form f ( t) g( t). As t vries the curve is defined.

6 Pge 5 () It is possile in some cses to otin the crtesin eqution of curve from prmetric equtions. This involves eliminting the prmeter etween the equtions. Emple A curve hs prmetric coordintes t t the crtesin eqution nd sketch the grph. The crtesin equtions re t y t From the first eqution t Sustitute into the second eqution y This is crtesin eqution of prol with verte ( -) Check point ( ) y. Find Since Emple A curve hs prmetric equtions sin sin nd to give y cos for Find the crtesin eqution of the curve nd sketch the grph. cos y use the identity sin cos y y ( -) Recognise this s the eqution of circle centre nd rdius

7 ecept tht for semi circle. Pge sin will e positive only. Hence the curve is y r = ( -) () In mny cses it isn t possile to otin coherent crtesin eqution Emple (5) Prmetric differentition Where t is prmeter t t y t t crtesin pproch. Applictions will e to tngents nd normls Emple doesn t respond well to Otin the eqution of the tngent to the curve dy d t t y t t t t y t t dy dt d dt t the point where t = d dt dy t t t dt putting t = grd t point ( ) Tngent y y dy d t t t

8 () Intersection prolems Pge 7 Emple Find the points t which the line y cuts the curve defined y prmetric coordintes t t The technique here is to sustitute prmetric coordintes y t into the eqution of the line. For intersections t t t t t t t Points re nd t Emple Refer to the emple in (5). The tngent y cuts the curve gin t point P. Find the coordintes of P For P t t t t t t t Since t = is doule root of this eqution (tngent t t = ) the cuic redily fctorises t At P 5 t P t y inspection Vector Geometry A collection of importnt results. k nd re prllel 8. ˆ where â is vector of unit mgnitude in the direction of. This cn e otherwise written s ˆ nd ˆ

9 Pge 8. P With n ssigned origin O the position of ny point P in p spce is specified uniquely y OP p nd is clled the O A position vector of P. O B AB O The point P dividing AB in the rtio m : n hs A m n m OP P m n n m B Proof: OP OA AP AB m n m m n m n m m n n m m n The midpoint of AB hs position vector m m For midpoint m n OP m

10 Pge 9. P The vector eqution of the line through A d O 5. In D spce the direction of d is given y r d A in position vector fied fied of ny point on point sclr direction the line i nd j re unit vectors in the directions nd y respectively. prmeter Any vector in this D spce cn e epressed in the form = component = y component Given v The direction rtios of v re :. v i j. In D spce i j nd k re unit vectors in the directions y nd z respectively. Any vector in this spce cn e written in the form v i j ck c r v y c j i

11 Pge The direction rtios re : : c NB A test for prllel vectors is tht their direction rtios should e the sme or reduce to the sme. 7. The sclr product of two vectors nd is written. nd is defined s:. cos where is the ngle etween nd Notice the directions of nd in the sclr product! Very importnt. nd re perpendiculr! Rememer tht c... c etc. where. my e seen ut is to e voided! Very importnt is the sclr product in component form. i.e. given i j k nd i j k.. 8. Pirs of lines in D spce r d nd r d my e (i) Prllel in which cse d kd (ii) Intersecting in which cse there re vlues of nd for which

12 (iii) Emple d d Pge Skew when neither prllel nor intersecting. Skew lines hve mutully perpendiculr line which will give closest distnce prt. Points A nd B hve position vectors nd reltive to the origin O. A further point Q hs position vector point P divides AB in the rtio : () Show on digrm the reltive positions of O A B P nd Q () Write down the position vector of P nd nother (c) Write down n epression for QP nd comment on its mgnitude nd A direction. O P Q B OP QP p q QP is prllel to OA nd is of OA Emple Find vector eqution for the line AB where A is the point ( ) nd B is the point (8 ). The perpendiculr from the origin O to AB meets it t N. Using the sclr product of AB nd ON find the coordintes of N. Deduce the coordintes of the point C (distinct from B) on AB such tht OC = OB.

13 Pge 8 8 r r or t r For N AB ON. 8. n n n 8 n n n 5 5 n n Hence N(5 -) which will e the midpoint of BC therefore for C( ) 5 8 Hence C( -8 -) Emple The lines L nd L re given y the equtions N B (8 ) P A ( ) r

14 Pge L : t r L : 8 s r (i) Write down the direction rtios of L (ii) Find unit vector in the direction of L (iii) clculte the cute ngle etween L nd L (iv) Show tht L nd L don t intersect (v) Verify tht is perpendiculr to oth lines (vi) Find points P nd Q on L nd L such tht PQ is prllel to (i) DR s of L re : - : (ii) Unit vector in direction of L = (iii) The required ngle is etween nd nd is given y the formul. cos

15 Pge cos (iv) For intersection the eqution s t t s t 8 hve to e consistent From nd eqution t = Su into st eqution s = Check in rd eqution LHS = + = RHS = 8 + = Equtions not consistent lines do not intersect. (v). is perpendiculr to L. is perpendiculr to L (vi) Tke p p p P nd q q Q 8 p q p p q p q PQ 5 nd to e prllel to k PQ

16 q p k p k 5 q p k Pge 5 q q p k p p p k p p Hence P( 5 ) nd Q( ) Alger Binomil epnsion n for n rtionl When n is not positive integer the epnsion will hve n infinite numer of terms.!! The epnsion is vlid for i.e. it is convergent for n n nn nn n nn n n... to The generl term is unlikely to e tested ut for the record it is: n! {Don t try n r! r! defined for positive integers.} Emple n n n... n r r r! s the coefficient ecuse it just won t work n! is only Otin the first four terms in the epnsions of (i) (ii)! nd write down in ech cse the rnge of vlues of for which the epnsion is convergent.

17 Pge (i) !! Epnsion convergent for i.e. (ii) y !! Epnsion convergent for i.e. Prtil Frctions For ech unrepeted liner fctor occurring in the denomintor there will rise prtil frction of the form p. The vlue of p cn e found using the cover up rule.

18 Pge 7 Emple Epress in prtil frctions For ny repeted liner fctor occurring in the denomintor there will rise two prtil frctions Q P. The vlue of Q only cn e found y using the cover up rule. The vlue of P is found y setting up denomintor free identity Emple Epress in prtil frctions P P P P P Clculus Tngents nd norml to curve t specific point Find the grdient through differentition nd use m y y Emple Find the equtions of the tngent nd norml to the grph of y ln t the point where e

19 y ln When e dy ln. d ln Pge 8 dy d y e ln( e ) e nd Tngent y e e y e Norml Grd = y e e y e Integrtion techniques Integrtion y sustitution the sustitution will e given Emple Find (i) sin d y sustituting u cos (ii) d (i) sin d sin sin (ii) cos du d y sustituting u du sin d du sin d u du u u du u A cos cos A u cos In definite integrl the limits re chnged ccording to the sustitution

20 d u u u du du u Pge 9 u du d du d u u Integrtion y prts This method is used for some products such s for emple: sin cos e ln It cn lso e used to integrte for emple: ln sin tn dv d Formul u d uv v du d d e Emple sin d sin d cos cos cos cos d cos sin C.d u du d dv sin d v cos d In words for the evlution of first second integrl Integrl derivtive First times minus Integrl of Times of second lredy found of first

21 Emple ln d Pge Whether you use the formul or words the order here hs to e chnged ln d ln Emple tn.d tn d ln d ln A ln tn d Here introduce s the second tn tn tn (Use sme method for ln d d d ln d A A sin d etc) Do them! For definite integrl Emple e d e e e e e e d. d

22 Pge e e e e e Using prtil frctions Emple Evlute (i) d (ii) d (i) C d d ln ln (ii) A A A A A d d C d d d ln ln ln ln

23 Pge d ln ln ln ln 5 ln ln 5 5 ln ln Integrtion of sin nd cos etc Here we need the rerrnged doule ngle formule for cos i.e. cos cos sin cos Emple d cos d cos sin d sin C sin C {of this ooklet see pge 7 for cos d } Using the tles of stndrd derivtives nd integrls in the formul ooklet Relevnt to Core re derivtives of inverse trig functions nd of nd cosec (Pge 5) The results cn e reversed. Emple Emple d sin d tn C C On pge the integrls of tn cot Core sec cot Emple cosec d cot C cosec nd sec re relevnt to

24 Volumes of revolution y Pge When the shded region is rotted through c out the volume of revolution will e given y V y d y.out y V dy Emple Find the volume of revolution of the grph of y cos from to out through c y V cos d cos d sin sin sin

25 Differentil equtions Pge () Equtions of the form dy d y f () or g ( ) integrte t once d d Emple dy d e y e C * Emple d y d dy sin cos A d y sin A B * * These re the generl solutions of the differentil equtions where A B C re ritrry constnts. Prticulr solutions to differentil equtions cn e found if oundry conditions re given d Emple Solve the eqution t dt d dt t t t A t A A t t given tht when {generl solution} t () Equtions which reduce to f ( ) d g( y) dy re clled vriles () seprle. Integrte oth sides ut include just one ritrry constnt. Emple dy d y dy d y ln y C

26 y e Pge 5 C y Ae e e c Ae () An importnt ppliction is to rte of growth nd rte of decy. It is importnt to recll tht rte of chnge (ROC) (with respect to time unless otherwise specified) is derivtive with respect to time). ROC positive there is growth (increse) ROC negtive there is decy (decrese) Emple The rte of decy of certin rdioctive element t ny time is proportionl to the mss of the element remining t tht instnt. After dys one third of given mss m hs disintegrted. How much is left fter further dys? Let m e the mss remining t time t. The initil mss is m (when t ) Decy implies loss of mss nd rte of decy is given y derivtive dm dt Seprte vriles m km where k dm m kdt dm kdt m ln m kt A m when t ln m A ln m kt ln m

27 ln m ln m Pge m ln m m m kt kt e kt m m e kt m when t m m m e k e k e k m m t When m m t m 9 m

28 Pge 7 Notes

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