Pure C4. Revision Notes
|
|
- Zoe Williams
- 8 years ago
- Views:
Transcription
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 y-coordintes re 47 nd 0 So we cnnot find vlue of t to give equl - nd equl y-coordintes 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: Re-rrnging 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: 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 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 re-write ( 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 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 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 re-rrnge 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 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 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 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 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 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 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, 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 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 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 co-ordinte 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 re-write 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 co-ord 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 co-ordintes on ech line Equting co-ords λ + μ equting y co-ords + 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 co-ords re equl (s epected!), ut the z co-ordintes 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 non-prllel 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.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationSection 7-4 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationPROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -
More informationVectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.
Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued
More informationMultiplication and Division - Left to Right. Addition and Subtraction - Left to Right.
Order of Opertions r of Opertions Alger P lese Prenthesis - Do ll grouped opertions first. E cuse Eponents - Second M D er Multipliction nd Division - Left to Right. A unt S hniqu Addition nd Sutrction
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationHarvard College. Math 21a: Multivariable Calculus Formula and Theorem Review
Hrvrd College Mth 21: Multivrible Clculus Formul nd Theorem Review Tommy McWillim, 13 tmcwillim@college.hrvrd.edu December 15, 2009 1 Contents Tble of Contents 4 9 Vectors nd the Geometry of Spce 5 9.1
More informationSection 5-4 Trigonometric Functions
5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationGeometry 7-1 Geometric Mean and the Pythagorean Theorem
Geometry 7-1 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationScalar and Vector Quantities. A scalar is a quantity having only magnitude (and possibly phase). LECTURE 2a: VECTOR ANALYSIS Vector Algebra
Sclr nd Vector Quntities : VECTO NLYSIS Vector lgebr sclr is quntit hving onl mgnitude (nd possibl phse). Emples: voltge, current, chrge, energ, temperture vector is quntit hving direction in ddition to
More informationModule Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials
MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic
More informationWarm-up for Differential Calculus
Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationLECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.
LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 64-83.
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More information2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of
More informationaddition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.
APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The
More informationVector differentiation. Chapters 6, 7
Chpter 2 Vectors Courtesy NASA/JPL-Cltech Summry (see exmples in Hw 1, 2, 3) Circ 1900 A.D., J. Willird Gis invented useful comintion of mgnitude nd direction clled vectors nd their higher-dimensionl counterprts
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationThe Velocity Factor of an Insulated Two-Wire Transmission Line
The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More informationRIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS
RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationRadius of the Earth - Radii Used in Geodesy James R. Clynch February 2006
dius of the Erth - dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.
More informationReview Problems for the Final of Math 121, Fall 2014
Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationSummary: Vectors. This theorem is used to find any points (or position vectors) on a given line (direction vector). Two ways RT can be applied:
Summ: Vectos ) Rtio Theoem (RT) This theoem is used to find n points (o position vectos) on given line (diection vecto). Two ws RT cn e pplied: Cse : If the point lies BETWEEN two known position vectos
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationAngles 2.1. Exercise 2.1... Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example
2.1 Angles Reognise lternte n orresponing ngles Key wors prllel lternte orresponing vertilly opposite Rememer, prllel lines re stright lines whih never meet or ross. The rrows show tht the lines re prllel
More informationPHY 140A: Solid State Physics. Solution to Homework #2
PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationSection 1: Crystal Structure
Phsics 927 Section 1: Crstl Structure A solid is sid to be crstl if toms re rrnged in such w tht their positions re ectl periodic. This concept is illustrted in Fig.1 using two-dimensionl (2D) structure.
More informationChapter. Contents: A Constructing decimal numbers
Chpter 9 Deimls Contents: A Construting deiml numers B Representing deiml numers C Deiml urreny D Using numer line E Ordering deimls F Rounding deiml numers G Converting deimls to frtions H Converting
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More information2012 Mathematics. Higher. Finalised Marking Instructions
0 Mthemts Higher Finlised Mrking Instructions Scottish Quliftions Authority 0 The informtion in this publtion my be reproduced to support SQA quliftions only on non-commercil bsis. If it is to be used
More informationVectors and dyadics. Chapter 2. Summary. 2.1 Examples of scalars, vectors, and dyadics
Chpter 2 Vectors nd dydics Summry Circ 1900 A.D., J. Willird Gis proposed the ide of vectors nd their higher-dimensionl counterprts dydics, tridics, ndpolydics. Vectors descrie three-dimensionl spce nd
More informationAll pay auctions with certain and uncertain prizes a comment
CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 1-2015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin
More informationPentominoes. Pentominoes. Bruce Baguley Cascade Math Systems, LLC. The pentominoes are a simple-looking set of objects through which some powerful
Pentominoes Bruce Bguley Cscde Mth Systems, LLC Astrct. Pentominoes nd their reltives the polyominoes, polycues, nd polyhypercues will e used to explore nd pply vrious importnt mthemticl concepts. In this
More informationMODULE 3. 0, y = 0 for all y
Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)
More informationVectors and dyadics. Chapter 2. Summary. 2.1 Examples of scalars, vectors, and dyadics
Chpter 2 Vectors nd dydics Summry Circ 1900 A.D., J. Willird Gis proposed the ide of vectors nd their higher-dimensionl counterprts dydics, tridics, ndpolydics. Vectors descrie three-dimensionl spce nd
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationDistributions. (corresponding to the cumulative distribution function for the discrete case).
Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More informationFAULT TREES AND RELIABILITY BLOCK DIAGRAMS. Harry G. Kwatny. Department of Mechanical Engineering & Mechanics Drexel University
SYSTEM FAULT AND Hrry G. Kwtny Deprtment of Mechnicl Engineering & Mechnics Drexel University OUTLINE SYSTEM RBD Definition RBDs nd Fult Trees System Structure Structure Functions Pths nd Cutsets Reliility
More information5.6 POSITIVE INTEGRAL EXPONENTS
54 (5 ) Chpter 5 Polynoils nd Eponents 5.6 POSITIVE INTEGRAL EXPONENTS In this section The product rule for positive integrl eponents ws presented in Section 5., nd the quotient rule ws presented in Section
More informationIntroduction to Integration Part 2: The Definite Integral
Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the
More informationExponential and Logarithmic Functions
Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationFUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation
FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationRoots of Polynomials. Ch. 7. Roots of Polynomials. Roots of Polynomials. dy dt. a dt. y = General form:
Roots o Polynomils C. 7 Generl orm: Roots o Polynomils ( ) n n order o te polynomil i constnt coeicients n Roots Rel or Comple. For n n t order polynomil n rel or comple roots. I n is odd At lest rel root
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More informationEuler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems
Euler Euler Everywhere Using the Euler-Lgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch
More information6 Energy Methods And The Energy of Waves MATH 22C
6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this
More informationDIFFERENTIATING UNDER THE INTEGRAL SIGN
DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I hd lerned to do integrls by vrious methods shown in book tht my high school physics techer Mr. Bder hd given me. [It] showed how to differentite prmeters
More information10.6 Applications of Quadratic Equations
10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationExam 1 Study Guide. Differentiation and Anti-differentiation Rules from Calculus I
Exm Stuy Guie Mth 2020 - Clculus II, Winter 204 The following is list of importnt concepts from ech section tht will be teste on exm. This is not complete list of the mteril tht you shoul know for the
More informationNumerical Methods of Approximating Definite Integrals
6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils,
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationCHAPTER 11 Numerical Differentiation and Integration
CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods
More informationBrillouin Zones. Physics 3P41 Chris Wiebe
Brillouin Zones Physics 3P41 Chris Wiebe Direct spce to reciprocl spce * = 2 i j πδ ij Rel (direct) spce Reciprocl spce Note: The rel spce nd reciprocl spce vectors re not necessrily in the sme direction
More informationCHAPTER 9: Moments of Inertia
HPTER 9: Moments of nerti! Moment of nerti of res! Second Moment, or Moment of nerti, of n re! Prllel-is Theorem! Rdius of Grtion of n re! Determintion of the Moment of nerti of n re ntegrtion! Moments
More information