Pure C4. Revision Notes


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1 Pure C4 Revision Notes Mrch 0
2
3 Contents 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 4/04/
4 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 4/04/ SDB
5 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 4/04/
6 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 4/04/ SDB
7 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 y t, t + nd y t +, t 4 intersect If they do give the point of intersection, otherwise give resons why they do not intersect Solution: If they intersect there must e vlue of t which mkes their coordintes equl, so find this vlue of t t + t 4 t 7 But for this vlue of t the ycoordintes re 47 nd 0 So we cnnot find vlue of t to give equl  nd equl ycoordintes nd therefore the curves do not intersect C4 4/04/ 5
8 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: Rerrnging we hve sin t y +, nd cost, 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: Rerrnging 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 4/04/ SDB
9 Are under curve given prmetriclly We know tht the re etween curve nd the is is given y A y y But, from the chin rule da da Integrting with respect to t A y 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 A we should write limits for t, not?? 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 4/04/ 7
10 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 rewrite ( 4 ) nd now we cn use the formul ! 4! This epnsion converges for 4 < < 4 8 C4 4/04/ SDB
11 C4 4/04/ 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 ) ( ) ( ) ( ) ( ! ) )( ( ) (! ) )( ( ) (
12 4 Differentition Reltionship etween nd So y dy 6 dy 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) sin 5 d sin 05 cos using the product rule d dy ( + 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 dy dy + y y + 0 dy y y dy 4 Eqution of the tngent is y 4( ) y 4 when nd y 0 C4 4/04/ SDB
13 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 dy t nd t + when t 9 5 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) dy t grdient 0 when t + 0 t 0 t ± points with zero grdient re (0, ) nd (, ) Eponentil functions, y ln y ln ln y dy ln dy y ln d ( ) ln C4 4/04/
14 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 π 6π 8 6 cm s Forming differentil equtions Emple: The mss of rdioctive 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 dm m km, where k is the (positive) constnt of proportionlity ln m kt + ln A C4 4/04/ SDB
15 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 cosec cot cosec tn cosec cot Integrtion using trigonometric identities Emple: Find cot Solution: cot cosec cot cosec cot + c C4 4/04/
16 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) sin cos 5 / sin 8 + sin ( ) / sin 8 sin / 6 cos 8 + ¼ cos + c 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 which is 5 times wht we wnt nd so du u to give ( sin ) 5( sin ) cos 5sin cos if we ignore the 4 C4 4/04/ SDB
17 sin 4 cos sin 5 + c 5 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 4 ln 4 d to give ( ) 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 C4 4/04/ 5
18 nd similrly cosec ln cosec + cot + c 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) Use the given sustitution involving nd u, (or find suitle sustitution) (ii) du Find either or, whichever is esier nd rerrnge to find in terms du of du, ie du (iii) 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 du u 6 6 du u (iv) du 6 u + c 6 6 C4 4/04/ SDB
19 ( 5) (v) 9 Emple: Find + + c 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 4/04/ 7
20 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 Solution: (i) u du du (ii) (ii) () Chnge limits from to u using the sustitution 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 put u + put u 5 put u + 5 or u put u or u 4 only if n is ODD put sin u only if n is EVEN (or zero) this mkes 4 4cos cos u 8 C4 4/04/ SDB
21 There re mny more possiilities use your imgintion!! Integrtion y prts The product rule for differentition is 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 Solution: (i) It does not look like product, u, ut if we tke u ln nd dv dv then u ln ln (ii) u ln du dv nd v C4 4/04/ 9
22 (iii) Are under curve ln + c ln ln We found in Core tht the re under the curve is written s the integrl 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 π y δ 0 C4 4/04/ SDB
23 nd s δ 0 we hve (using the result ove y δ y) π y δ π y 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 dy Volume of rottion out the yis 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 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 re y sec t y π π sint sec t tnt sect π [ ] 4 0 sect To find volume of revolution we need π y nd we proceed s ove writing C4 4/04/
24 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 Solution: V 8 π y 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: dy y + y y( + ) dy y + y We first chet y seprting the s nd y s onto different sides of the eqution y dy ( + ) nd then put in the integrl signs dy y + ln y + / + c See the ppendi for justifiction of this technique C4 4/04/ SDB
25 Eponentil growth nd decy Emple: A rdioctive 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 k ln m kt + ln A dm ln kt 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 4/04/
26 6 Vectors Nottion The ook nd em ppers like writing vectors in the form i 4j + 7k It is llowed, nd sensile, to rewrite 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 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 coordinte form Vectors cn lso e thought of s column vectors, 7 thus in the digrm AB Negtive vectors is the 'opposite' of nd so Adding nd sutrcting vectors 7 (i) Using digrm Geometriclly this cn e done using tringle (or prllelogrm): Adding: A 7 + B + 4 C4 4/04/ SDB
27 The sum of two vectors is clled the resultnt of those vectors 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 4/04/ 5
28 Nonprllel 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 7 then the modulus of 7 58 Or, if c 5 then the modulus of c c c 5 4 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 4/04/ SDB
29 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 4 y For two points A nd B the position vectors re A nd B To find the vector go from A O B giving + O Rtios Emple: A, B re the points (, ) nd (4, 7) M lies on AB in the rtio : Find the coordintes of M Solution : 6 B M is ( 5, 5) O A M C4 4/04/ 7
30 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 s, nd s c Then ½ ½ nd ½ ½ c O To find, go from M to O using ½ nd then from O to N using ½ c ½ + ½ c ½ c ½ B M N c Also, to find, go from B to O using nd then from O to C using c C + c c But 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, nd P, nd / / / ( ) O Q B / PQ is prllel to AB nd PQ / AB 8 C4 4/04/ SDB
31 Three dimensionl vectors Length, modulus or mgnitude of vector The length, modulus or mgnitude of the vector + +, ` 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 ( ) + ( ) + ( ) 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 coordinte form + cos θ or + + cos θ C4 4/04/ 9
32 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 λ λ 0 6 λ 0 λ 9 λ ± Emple: Find vector which is perpendiculr to, Solution: Let the vector c,, nd, p q, e perpendiculr to oth nd r p p q 0 nd q 0 r r p q + r 0 nd p + q + r 0 Adding these equtions gives 4p + r 0 Notice tht there will never e unique solution to these prolems, so hving eliminted one vrile, q, we find p in terms of r, nd then find q in terms of r r 5r p q 4 4 c is ny vector of the form r 4 5r, 4 r 0 C4 4/04/ SDB
33 nd we choose sensile vlue of r 4 to give 5 4 Angle etween vectors Emple: Find the ngle etween the vectors 4i 5j + k nd i + j k, to the nerest degree Solution: First rewrite s column vectors (if you wnt) 4 5 nd , nd cos θ cosθ 0 cos θ θ 4 o to the nerest degree Vector eqution of stright line r y is usully used s the position vector of z generl point, R R l A In the digrm the line l psses through the point A nd is prllel to the vector To go from O to R first go to A, using, nd then from A to R using some multiple of r C4 4/04/ O
34 The eqution of stright line through the point A nd prllel to the vector is r + λ Emple: Find the vector eqution of the line through the points M, (,, 4), nd N, ( 5,, 7) Solution: We re looking for the line through M (or N) which is prllel to the vector n m eqution is r λ 4 4 Emple: Show tht the point P, (, 7, 0), lies on the line r Solution: + λ 4 The coord of P is nd of the line is λ λ λ In the eqution of the line this gives y 7 nd z 0 P, (, 7, 0) does lie on the line Intersection of two lines Dimensions Emple: Find the intersection of the lines l, r λ, nd, μ + + l r Solution: We re looking for vlues of λ nd μ which give the sme nd y coordintes on ech line Equting coords λ + μ equting y coords + λ μ C4 4/04/ SDB
35 Adding 5 + λ 4 λ μ lines intersect t (, ) Dimensions This is similr to the method for dimensions with one importnt difference you cn not e certin whether the lines intersect without checking You will lwys (or nerly lwys) e le to find vlues of λ nd μ y equting coordintes nd y coordintes ut the z coordintes might or might not e equl nd must e checked Emple: Investigte whether the lines l, r + λ nd l, r nd if they do find their point of intersection 5 + μ intersect Solution: If the lines intersect we cn find vlues of λ nd μ to give the sme, y nd z coordintes in ech eqution Equting coords λ + μ, I equting y coords + λ + μ, II equting z coords + λ 5 + μ III I + II μ μ, in I λ We must now check to see if we get the sme point for the vlues of λ nd μ In l, λ gives the point (, 7, 6); in l, μ gives the point (, 7, 7) The nd y coords re equl (s epected!), ut the z coordintes re different nd so the lines do not intersect C4 4/04/
36 7 Appendi Binomil series ( + ) n for ny n proof Suppose tht f () ( + ) n + + c + + e 4 + put 0, f () n( + ) n + c + + 4e + put 0, n f () n(n )( + ) n c e + put 0, n(n ) c! f () n(n )(n )( + ) n d + 4 e + put 0, n(n )(n ) d! Continuing this process, we hve! giving f () ( + ) n nd!, etc + n +! +! +! + +! Showing tht this is convergent for <, is more difficult! + +! Derivtive of q for q rtionl Suppose tht q is ny rtionl numer,, where r nd s re integers, s 0 Then y q y s r Differentiting with respect to s r r since q q q since y s r nd y q 4 C4 4/04/ SDB
37 which follows the rule found for n, where n is n integer for negtive limits We know tht the re under ny curve, from to is pproimtely, 0 y δ If the curve is ove the is, ll the y vlues re positive, nd if < then ll vlues of δ re positive, nd so the integrl is positive ln lnln Emple: Find Solution: The integrl wnted is shown s A in the digrm By symmetry A A (A positive) nd we need to decide whether the integrl A' is +A or A From to, we re going in the direction of decresing ll δ re negtive And the grph is elow the is, the y vlues re negtive, y δ is positive 0 0 the integrl is positive nd equl to A y/ A The integrl, A A [ ln ] ln ln ln A ln Notice tht this is wht we get if we write ln in plce of ln [ ln ] ln ln ln As it will lwys e possile to use symmetry in this wy, since we cn never hve one positive nd one negtive limit (ecuse there is discontinuity t 0), it is correct to write ln for the integrl of / C4 4/04/ 5
38 Integrtion y sustitution why it works We show the generl method with n emple Choose u + But integrnd rerrnge to give leve the ecuse it ppers in this is the sme s writing the integrnd in terms of u, nd then replcing y The essentil prt of this method, writing the integrnd in terms of u, nd then replcing y, will e the sme for ll integrtions y sustitution Prmetric integrtion This is similr to integrtion y sustitution 6 C4 4/04/ SDB
39 Seprting the vriles why it works We show this with n emple If y 6y then 8y nd so Notice tht we cncel the Emple: Solve sec y Solution: sec y cos y cos cos cncelling the sin y + c C4 4/04/ 7
40 Inde Binomil epnsion convergence, 8 for ny n, 8 proof  for ny n, 4 Differentil equtions eponentil growth nd decy, forming, seprting the vriles, Differentition derivtive of q for q rtionl, 4 eponentil functions,, relted rtes of chnge, Implicit differentition, 0 Integrtion / for negtive limits, 5 y prts, 9 y sustitution why it works, 6 choosing the sustitution, 8 e, ln, prmetric, prmetric why it works, 6 reverse chin rule, 4 sec nd cosec, 5 seprting vriles why it works, 7 stndrd integrls, sustitution, definite, 8 sustitution, indefinite, 6 tn nd cot, 5 using prtil frctions, 6 using trigonometric identities, volume of revolution, 0 Prmeters re under curve, 7 eqution of circle, 6 eqution of hyperol, 6 intersection of two curves, 5 prmetric equtions of curves, 5 prmetric to Crtesin form, 6 trig functions, 6 Prmetric differentition, Prmetric integrtion, 6 res, volumes, Prtil frctions, integrtion, 6 liner repeted fctors, 4 liner unrepeted fctors, qudrtic fctors, top hevy frctions, 4 Sclr product, 9 perpendiculr vectors, 0 properties, 9 Vectors, 4 dding nd sutrcting, 4 ngle etween vectors, displcement vector, 4 distnce etween points, 9 eqution of line, intersection of two lines, length of D vector, 9 modulus, 6 nonprllel vectors, 6 prllel vectors, 5 position vector, 7 proving geometricl theorems, 8 rtios, 7 unit vector, 6 8 C4 4/04/ SDB
Math 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
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