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1 MST09 Mthemticl methods nd models Vector lger ΛΞΠΛ± ±ΦΨΛΩffΦfifl 2 Stud guide This unit ssumes no previous knowledge of vectors. You will need to know onl sic lger nd trigonometr, nd how to use Crtesin coordintes for specifing point in plne. The recommended stud pttern is to stud one section per stud session, nd to stud the sections in the order in which the pper.
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ISBN 3 Contents Introduction 4 Descriing nd representing vectors 5. Sclrs nd vectors 5. Vector nottion 5.3 Using rrows to represent vectors 6.4 Equlit of vectors 8.5 olr representtion of two-dimensionl vectors 9 Scling nd dding vectors. Scling of vector. Addition of vectors 5.3 Algeric rules for scling nd dding vectors 6 3 Crtesin components of vector 8 3.
Vectors in two dimensions 8 3. Vectors in three dimensions 4 roducts of vectors 7 4. The dot product 7 4. The cross product 34 utcomes Solutions to the eercises Inde 4 Introduction Introduction We often need to represent phsicl quntities such s mss, force, velocit, ccelertion, time, etc., mthemticll. Most of the phsicl quntities tht we need cn e clssified into two tpes: sclrs nd vectors. Sclr quntities re quntities, like mss, temperture, energ, volume nd time, tht cn e represented single rel numer. Ther quntities, like force, velocit nd ccelertion, possess mgnitude nd direction in spce, nd cnnot e represented single rel numer; the re clled vector quntities.
The definitive vector quntit is displcement. The displcement of point specifies the position of the point in spce reltive to some reference point. We use the concept of displcement whenever we wnt to descrie sptil reltionships. Consider, for emple, the instructions written in lood on pirte s tresure mp: Tke five pces due north from the ig ok tree, then seven pces due west, nd then dig down for three metres. This is specifiction of displcement vector the displcement of the tresure from reference point (the ig ok tree). In fct, this prticulr w of specifing the displcement of the tresure is known s the Crtesin description of displcement, lthough the pirte prol didn t know tht. Alterntivel, there is the so-clled polr description of the sme displcement (equting pces with metres): Strting t the ig ok tree, dig for 9.
Pces long stright sloping line inclined t 9 elow the horizontl t ering of 54 west of north. This distnce (or mgnitude) plus direction specifiction will lso get ou to the tresure, lthough less convenientl ecuse it is more difficult to dig long sloping line. These two different specifictions re shown in Figure 0.
N 7 pces 5 pces metres 9. Pces Figure 0. Section defines vector nd discusses ws of representing vectors in two dimensions.
Section 3 discusses nother w of representing vectors, one tht esil generlizes from two to three (or more) dimensions. Sections nd 4 consider ws of operting on nd comining vectors tht is, the provide the fundmentls of vector lger. 4 5 Descriing nd representing vectors Susection. Eplins wht sclrs nd vectors re. Nd.3 then eplin how to denote vectors smolicll (i.e. Lgericll) nd how to show them in digrms.
Susection.4 eplins wht is ment sing tht two vectors re equl to one nother, which is necessr first step in the development of n lger for vectors. Susection.5 introduces method for representing vectors in two dimensions tht cn e useful in vriet of phsicl situtions. Section Descriing nd representing vectors. Sclrs nd vectors A sclr is n quntit, such s mss, time, volume nd temperture, tht cn e represented mthemticll single rel numer (nd often unit of mesurement). Rel numers themselves re emples of sclrs, nd ou cn regrd the terms sclr nd rel numer s snonmous. Emples of sclr quntities, quoted to some convenient degree of ccurc, re: the mss of the Erth, kilogrms; the temperture of melting ice, 0 degrees Celsius; m current nk ccount lnce, 53.
Pounds sterling; pi , A rel numer is defined two properties: its modulus nd its sign. Thus the mgnitude of sclr is. For emple, the mgnitude of m current ccount lnce is 53. Pounds, which sounds lot etter since it doesn t remind me tht I m in det. Note tht mgnitudes re lws non-negtive (i.e.
Positive or zero). A vector quntit is n quntit, such s force, velocit, displcement, etc., tht hs mgnitude nd direction in spce (or, in two dimensions, direction in plne). An emple is the velocit of motor cr trvelling on the M4 motorw from London to Bristol with speed of 95 km per hour in westerl direction. The mgnitude of the velocit vector is 95 (dropping units for convenience), nd the direction of the velocit vector is due west. Thus the specifiction of vector consists of: non-negtive rel numer, clled its modulus or mgnitude; direction in spce. This unit is minl concerned with just two vector quntities: displcement nd velocit.
Lter in the course ou will come cross other vector quntities such s force, torque nd momentum. Fortuntel, ll vector quntities oe ectl the sme lws of lger. Thus wht ou lern out displcements nd velocities in this unit cn e crried over to ll vector quntities. The modulus of rel numer is lso clled its mgnitude. The fmilir term speed is used to men the mgnitude of velocit. Speed is non-negtive sclr.
Vector nottion Vectors re denoted in printed tet old letters, e.g. In our written work, ou should denote vectors drwing either stright line or squiggl line under the letter, e.g. Thusifsmolisused to represent the velocit of n oject, then it must e hndwritten ou s either v or v (ut will e printed in the tet s v). It is importnt tht ou lern to write vectors using the underlining: if ou do not do so, someone reding our work m not e le to tell tht ou re referring to vector.
In prticulr, ou m lose mrks! 5 6 Section Descriing nd representing vectors The modulus or mgnitude of the vector v is denoted v or, sometimes, where there is no possiilit of miguit, v; v is non-negtive sclr.3 Using rrows to represent vectors We red v s the modulus of v or the mgnitude of v, or simpl mod v. A vector cn e convenientl represented in digrm n rrow, i.e. Stright line with n rrowhed on it. The til of the rrow m e plced t some fied origin, the direction of the rrow is chosen to represent tht of the vector, nd its length is chosen to e proportionl to the mgnitude 4 of the vector.
In Figure., which uses the origin of the Crtesin coor- dinte sstem s the fied origin, the shorter rrow represents vector of mgnitude in the positive -direction, nd the longer rrow represents Figure. Vector of mgnitude in direction t 4 rdins (45 ) to the positive -direction. (Note tht we use the convention tht positive ngles re mesured nticlockwise.) If we decide to denote these vectors letters nd, respectivel, then we cn lso put this informtion on the digrm, writing nd ner the rrowheds, s shown in Figure. Note tht in this course, nd commonl elsewhere, the rrows representing vectors re drwn using thick lines. This helps to distinguish vector rrows from other rrowed lines such s those representing the coordinte es (e.g.
Nd.) or those representing compss directions (e.g. Represent the following two vectors on digrm rrows: vector hs mgnitude 3 units nd points in the positive -direction; vector hs mgnitude 4 units nd points in the direction t 3 rdins (60 ) to the positive -direction. Vector nottion nd the use of rrows in digrms is now illustrted further specific reference to displcement vectors nd velocit vectors. Displcement is the position of point in spce reltive to some reference point or origin. For emple, the cit of Leeds is 96 km from the cit of Bristol in the direction of 5 est of north (N 5 E). The displcement of Leeds from Bristol cn e specified s the vector s = 96 km N 5 E. Here the old smol s hs een used to denote the displcement.
Note tht oth mgnitude nd direction re specified: the mgnitude of the displcement is s = 96 km, nd the direction is specified the compss ering N5 E. The displcement s = 96 km N 5 E cn e represented in digrm n rrow, s shown in Figure.3. The length of the rrow represents 96 km, which m e shown in the digrm writing s = 96. It would e wrong to write s = 96 km, ecuse the left-hnd side is vector smol nd the right-hnd side is sclr. For n two points nd Q, we cn define the displcement vector from to Q: it is the vector whose mgnitude is the distnce from to Q nd whose direction is the direction of the stright line from to Q.
A useful nottion for this vector is Q (see Figure.4). In this contet the smol Q (without n rrow) represents the length of the stright line joining nd Q, i.e. Note tht Q = Q ut Q = Q (ecuse Q Scle (km) nd Q re in opposite directions). 6 N 5 Figure.3 s s = 96 7 Section Descriing nd representing vectors Displcement vector The displcement vector Q is the vector whose mgnitude is the distnce from to Q nd whose direction is the direction of the stright line from to Q.
Figure.4 Q Q ne quer m hve occurred to ou. Wht is the displcement vector of point from itself? In other words, wht is the vector? Clerl its length is zero, ut wht is its direction? The nswer is tht it does not hve one! We define the zero vector to e the unique vector with mgnitude zero nd no direction. It is denoted 0.
Thus we cn conclude tht = 0. Zero vector The zero vector is the unique vector with mgnitude zero nd no direction. It is denoted 0. Be prticulrl creful to underline the zero vector (0 or 0 ) in our written work! A constnt velocit is lso defined mgnitude nd direction.
For instnce, in wether forecst, tpicl wind velocit might e 35 knots N from the north-west. It is not sufficient to s tht the wind velocit is 35 knots; the ovious question out such sttement would e from which direction? The vector v representing this velocit hs mgnitude 35 nd direction from the north-west nd towrds the south-est (since the 45 v = 35 ir is trvelling in the south-esterl direction). It cn e represented on digrm s shown in Figure.5. The length of the rrow represents wind v speed of 35 knots Note tht the direction of vector consists of two ttriutes: n orienttion, represented the slope of the rrow in digrms like Figures. To.5; sense, represented the rrowhed.
For instnce, the rrow representing the velocit 35 knots from the northwest in Figure.5 is line mking n ngle of 45 nticlockwise from the south direction (the orienttion) nd n rrowhed pointing towrds southest s opposed to north-west (the sense). The displcement of Birminghm from Der is 57 km in the direction S30 W.
The displcement of Leicester from Der is 3 km in the direction S45 E. Drw digrm, to suitle scle, representing these two displcements rrows. Eercise.3 A cr trvelling from London long the M with speed 70 mph heds in the direction N 60 W ner Junction 4.
Represent the velocit of the cr n rrow, drwn to suitle scle. Figure.5 Scle (knots) 7 8 Section Descriing nd representing vectors.4 Equlit of vectors Definition Two vectors re sid to e equl if the hve the sme mgnitude nd the sme direction.
You hve seen how to represent vector n rrow. This definition of equlit of vectors tells us tht the two fetures needed to define vector uniquel re its mgnitude nd direction. This mens tht n two rrows drwn t different plces on the pge ut which re equl in length, prllel nd hve the sme sense, cn e used to represent the sme vector. For instnce, the two rrows in Figure.6 re ech of length units nd point in the positive -direction. The represent two equl vectors, nd we write = d. In other words, the rrow representing vector does not hve to e drwn so tht its til is t n prticulr point. Figure.7 shows severl vectors represented rrows drwn to scle.
Find the vector equl to the vector. 3 d 3 Figure.6 3 c f e d g 3 h Solution Figure.7 We re looking for vector tht is equl in length to (i.e. One unit), prllel to nd points in the sme direction (i.e. The positive -direction). There re two rrows (nd thus vectors) other thn tht point in the positive -direction; the re c nd h. (The rrow representing d points in the negtive -direction.) The mgnitudes of c nd h re unit nd 3 units, respectivel.
Since the mgnitude of is unit, c = ut h. Note tht lthough nd c re drwn t different plces in the (, )-plne, the re equl in mgnitude nd hve the sme direction, so the re equl vectors.Eercise.4 Which vector in Figure.7 is equl to vector? 8 9.5 olr representtion of two-dimensionl vectors This susection introduces sstemtic w of specifing the mgnitude nd direction of vector in coordinte sstem. You should e fmilir with using two-dimensionl Crtesin coordinte sstem for specifing the position (, ) of point in plne, nd indeed the sme sstem is commonl used for displing vectors (s in Figures.6 nd.7). The plne polr coordinte sstem, however, is in some sense more nturl one for specifing vectors since it effectivel regrds mgnitude nd direction s two coordintes. Let r e vector on plne surfce.
Introduce Crtesin coordinte sstem, nd drw the vector s n rrow with its til t the origin, s in Figure.8. The mgnitude of r is r = r, the distnce of the tip of the rrow from.
The direction of r is specified the ngle φ mesured (usull in rdins) nticlockwise from the positive -is. Section Descriing nd representing vectors = r sin φ r = r r φ = r cos φ Figure.8 We hve not quite finished the description, ecuse vector now hs mn representtions (since rotting the line segment through n, where n is n integer, leves it unchnged). To void this miguit, we shll normll tke φ to lie in the rnge 0 sin 0 helps to sketch the Crtesin coordintes in the (, )-plne so tht ou cn cos 0 see in which qudrnt φ must lie.
The signs of sin nd cos for ngles in sin 0 these. (A simple cronm to id the memor is CAST: strting from the T C lower right, nd working nticlockwise round the qudrnts, the following qudrnt 3 qudrnt 4 re positive: Cos, All (of sin, cos, tn), Sin nd Tn.) Figure.0 9 10 Section Descriing nd representing vectors Emple. Give the polr representtion of the vectors, e nd g in Figure. E g Solution Figure.
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Ech vector is drwn s n rrow from the origin, so the polr representtion of ech vector is given the polr coordintes of its endpoint. In some cses we cn specif r nd φ simpl from inspection of Figure., ut we shll use the ove formule in order to illustrte the generl method.
The endpoint of hs Crtesin coordintes (, ), so we hve r = ( +( ) ) / =, sin φ = /r = / nd cos φ = /r = /. Thus φ = rdins, i.e. (Since 0; in the opposite direction to v if m 0 nd in the opposite direction when m 0.)? M r, φ = mr, φ,?
M r, φ = mr, mφ. Does the following proposed generl rule hold for the ddition of vectors in polr form?? R,φ + r,φ = r + r,φ + φ 4 3 Crtesin components of vector So fr we hve pproched vectors, nd the lws of vector ddition nd scling, geometricll. To dd vectors geometricll requires drwing digrms representing the vectors rrows. An lterntive, nd sometimes more convenient, lgeric pproch to representing vectors is developed in this section, first in two dimensions nd then in three.
Vectors in two dimensions We hve lred seen in Susection. How to write vectors tht re prllel to the -is or -is s sclr multiples of the Crtesin unit vectors i A= (, ) nd j, respectivel. (Recll tht i nd j re the unit vectors in the directions C of the positive - nd -es, respectivel.) f course, in generl, vectors do not lie prllel to either the -is or the -is. However, ou will see in this susection how use of the rules for vector ddition nd for sclr B multipliction llows us to write n vector in the (, )-plne in terms of the Crtesin unit vectors i nd j. Consider n ritrr vector = A in the (, )-plne, whose til is t the origin, s shown in Figure 3. The vector A is clled the position Note tht if A were in one of vector of the point A, nd its endpoint is determined the Crtesin the other three qudrnts of coordintes nd of A, i.e. The distnces B nd C, respectivel.
The plne, then one or oth of, would e negtive. 8 19 Section 3 Crtesin components of vector Furthermore, the vectors B nd C cn e written s sclings of the Crtesin unit vectors i nd j: B = i nd C = j. Hence the tringle rule (or prllelogrm rule) for the ddition of vectors llows the vector to e epressed s the sum of B nd C, i.e.s = B + C or = i + j.
The ltter is clled the component form of, nd the numers nd re clled the i- nd j-components of, respectivel. When the til of the vector is not t the origin, its components re defined in n ovious w. You m lso see these numers referred to s the - nd -components of.
Q C Q p B Figure 3. P q Referring to Figure 3., the components of re = q p nd = q p.
A shorter w of writing vector in component form is s n ordered pir of numers, (, ), where the unit vectors i nd j re not shown eplicitl. This nottion needs to e used with cre ecuse the coordintes of point in plne re lso denoted in this w, nd vectors re conceptull different from points. To void such confusion, in this course the column vector nottion = will e used insted.
In the tet, to sve spce, the column vector will often e written s = T, where the trnspose smol T here chnges the row into column. This is the w in which mn computer lger pckges displ vectors. Definition A vector = Q in the (, )-plne, where is the point (p,p )nd Q is the point (q,q ), hs component form = i + j, where = q p nd = q p,nd i nd j re the Crtesin unit vectors.
The component form m lso e written s = or = T. The numers nd re the (Crtesin) components of. 9 20 Section 3 Crtesin components of vector.Eercise 3. Write ech of the vectors in Figure.7 (pge 8) in the form = i + j nd s column vector. The mgnitude of vector given in component form is found ver esil.
For emple, the mgnitude of the vector in Figure 3. Is just the length of the line Q. This is found thgors s Theorem to e +. Mgnitude of two-dimensionl vector in component form If = Q = i + j, where nd Q hve coordintes (p,p )nd (q,q ), respectivel, then = + = (q p ) +(q p ). Vectors in component form cn lso e dded nd scled ver esil, mking use of the lgeric rules for the scling nd dding of vectors.
For emple, + = ( i + j)+( i + j) = i + j + i + j =( i + i)+( j + j) =( + )i +( + )j. So, to dd two vectors one dds their respective components. Similrl, m = m( i + j) =(m )i +(m )j, so scling vector is chieved scling its components. Adding nd scling two-dimensionl vectors in component form If = i + j, = i + j nd m is sclr, then + = ( + )i +( + )j nd m = (m )i +(m )j.
Equivlentl, using column vector nottion+ = + + nd m m =. Figure 3.3 shows four vectors in the (, )-plne. 0 21 Section 3 Crtesin components of vector 3 c d Figure 3.3 Write down the vectors in component form. Drw digrm to verif tht the scling 3.5 is the sme when otined geometricll or lgericll using components. (c) Use the tringle rule to otin the vector +.
Verif tht this vector is the sme s tht otined dding the component forms of nd. (d) Find, lgericll, the components of the vector + c. Hence find the mgnitude of the vector + c.Eercise 3.3 p + q 3 Find the numers p nd q if r =, s = nd r = s. P q 7 Find the mgnitude of the vector t if t = u + v, where u = nd v =.
Eercise 3.4 The three vectors, nd c in Figure 3.4 re specified in polr coordintes 3 =,3, = 3, 4, c =, 6. Wht re the mgnitudes of the three vectors? Write down the vectors, nd c in terms of i nd j. C (c) tin the vector + c in terms of i nd j. Figure 3.4 In Eercise 3.4 the vectors were given in polr coordinte form, which is just sstemtic w of specifing mgnitude nd direction. The process of finding the Crtesin components of vector given its mgnitude nd direction is known s resolving vector into its components.
This is j essentill wht ou did in Eercise 3.4. Thus given the mgnitude F of the vector in Figure 3.5, nd its direction φ, we cn resolve it into its i components: = cos φ nd = sin φ. Figure 3.5 Conversel, given the components nd of vector, we cn specif its You hd prctice t doing mgnitude nd direction: these clcultions in = ( + ) /, cos φ = /, sin φ = /.
22 Section 3 Crtesin components of vector You will see this ide gin in Section 4. For now, note tht if ou wish to dd two vectors in polr form it will e necessr first to resolve them into their Crtesin components (since there is no convenient formul for vector ddition in polr coordintes).
V Eercise 3.5 Resolve the vector v of mgnitude 5 shown in Figure 3.6 into its Crtesin components. Find the mgnitudes nd directions of the vectors = 3i j nd = 3i +3j. Figure Vectors in three dimensions Thus fr we hve discussed vectors in the plne, reching the component representtion of such vectors in the previous susection.
However, the world is three-dimensionl, nd few rel prolems re restricted to plne surfce. For emple, strting t point A t one corner of the cue shown in Figure 3.7, ou cn rech the opposite corner S three successive displce- ments: AQ + QB + BS. In order to work with such ddition of displcements in three dimensions, it is necessr to introduce three-dimensionl A Q coordinte sstem. Figure 3.7 S B A three-dimensionl Crtesin coordinte sstem Consider two-dimensionl Crtesin coordinte sstem. Drwthird is, the z-is, through the origin, perpendiculr to oth the - nd - es of the two-dimensionl sstem. This produces coordinte sstem with three mutull perpendiculr es, the -, - nd z-es (see Figure 3.8), intersecting t. Alterntivel, the coordinte sstem cn e chrcterized three plnes: the (, )-plne, which contins the - nd -es nd is perpendiculr to the z-is; the (, z)-plne, nlogousl defined; the (, z)-plne, gin nlogousl defined.
An point cn e represented uniquel its perpendiculr distnces from the (, )-, (, z)- nd (, z)-plnes. These distnces, clled the (Crtesin) coordintes of, re shown in Figure 3.8.
The -es shown in Figures 3.8 nd 3.9 re ment to point out of the plne of the pge. Z-is C R p 3 p p A Q (, z)-plne S B -is z 4 3 (, 3, 4) 3 -is Figure 3.8 Figure 3.9 23 Section 3 Crtesin components of vector Q, R nd S re perpendiculr to the (, )-plne, (, z)-plne nd (, z)-plne, respectivel. We denote the point the ordered triple of coordintes (p,p,p 3 ), where p = S = A, p = R = B, p 3 = Q = C.
For emple, the point (, 3, 4) is shown in Figure 3.9. When drwing Figure 3.9 it ws necessr to choose one of two possile ws for the positive z-direction to e defined; these re shown in Figure 3.0, where in oth cses the -is is ment to point into the plne of the pge, w from ou. Positive z positive z The usul convention for relting the positive directions of, nd z is given Figure 3.0 the following rule, clled the right-hnd rule.
The right hnd is held with the middle finger, first finger nd thum plced (roughl) perpendiculr to ech other, nd the other two fingers closed (see Figure 3.). If the thum nd first finger re pointing in the directions of the positive - nd -es, respectivel, then the middle finger is pointing in the direction of the positive z-is. Alterntivel, ou cn think of Figure 3.9 s showing corner of room (with the z-is pointing upwrds). If ou re stnding in the corner fcing outwrds, then the left-hnd edge of the floor is the -is, nd the righthnd edge is the -is. A coordinte sstem defined in this w is clled right-hnded sstem. Nl right-hnded sstems will e used in this course.
The sstems drwn in Figure 3.9 nd the top of Figure 3.0 re right-hnded sstems. Positive z-direction clockwise turn screw moves in positive -direction positive -direction z Figure 3. An lterntive definition of the sme positive z-direction is given the screw rule, stted s follows. Suppose tht we re turning screw into piece of wood; then clockwise rottion mkes the screw move into the wood (see Figure 3.). If we turn the screw in the sense from to s shown in Figure 3., then the direction in which the screw moves is long the positive z-direction.
For the rest of this unit the screw rule will e used to chrcterize righthnded sstem, ut ou should use whichever rule ou find esier to ppl. Turn to 3 24 Section 3 Crtesin components of vector Eercise 3.6 Decide which of the sets of perpendiculr es in Figure 3.3 define righthnded coordinte sstems. Z z z z (c) (d) Figure 3.3 (The -is points out of the plne of the pper in nd. The z-is points into nd out of the plne of the pper in (c) nd (d), respectivel.) The component form of three-dimensionl vectors The lgeric representtion of vectors cn e etended to vectors in three dimensions, such s in Figure 3.4. The vector, drwn from the origin, is the position vector of point A with three-dimensionl Crtesin coordintes (, 3 ).
A third Crtesin unit vector k is introduced to represent the positive z-direction. We now hve three Crtesin unit vectors, i, j nd k, which re perpendiculr to ech other. The vector m thus e written in component form s i z A 3 k = i + j + 3 k or = or = 3. 3 T k j Definition The position vector of point A reltive to the origin of three- Figure 3.4 dimensionl spce is the displcement of A from, i.e. The vector = A.
The i-, j- nd k-components of the position vector re the coordintes, nd 3 of the point A, respectivel. The components of vectors not sed t the origin re defined similrl, s follows. I These m sometimes e referred to s -, - nd z-components. J Definition A vector = Q in three-dimensionl spce, where is the point Note tht the component (p,p,p 3 )nd Q is the point (q,q,q 3 ), hs component form form m lso e written s = i + j + 3 k, = where = q p, = q p, 3 = q 3 p 3, nd i, j, k re the Crtesin unit vectors.
The numers, 3 re the (Crtesin) components of. Or 3 T = 3. 4 25 Section 3 Crtesin components of vector As in two dimensions, the opertions of vector lger cn e epressed in terms of components. Adding nd scling three-dimensionl vectors in component form If = i + j + 3 k, = i + j + 3 k nd m is sclr, then + = ( + )i +( + )j +( )k nd m = (m )i +(m )j +(m 3 )k.
The mgnitude of vector in terms of its components, 3 cn e found using thgors s Theorem (see Figure 3.5). The length N is +, nd A = N + NA.But A =, thus = z A 3 k i + 3 j N Figure 3.5 This cn e summrized s follows. Mgnitude of three-dimensionl vector in component form If = Q = i + j + 3 k, where the points nd Q hve coordintes (p,p,p 3 )nd(q,q,q 3 ), respectivel, then = + + = (q p ) +(q p ) +(q 3 p 3 ). 3.Eercise 3.7 Given vectors = i + j + k, = i 3j k nd c = 3i + k: epress d = 3 nd e = +4c in component form; find the mgnitudes of the vectors d nd e; (c) evlute, nd write down unit vector in the direction of; (d) find the components of vector such tht + =. Eercise 3.8 Find the mgnitude of the vector p = 26 Section 3 Crtesin components of vector Vector eqution of stright line ne useful ppliction of position vectors (in two or three dimensions) is in otining vector eqution of stright line. Find the position vector of point T ling on the stright-line segment Q (see Figure 3.6) in terms of the position vectors of nd Q. P Solution Let T e n point on Q (see Figure 3.6).
The position vector T of T reltive to the origin cn lso e written, using the tringle rule, s T = + T. Now T = sq, for some numer s, nd the point T trces out the line segment Q s s vries from 0 to. Thus the stright-line segment Q is descried the vector eqution Figure 3.6 T = + sq (0 s ).
Writing p =, q = Q, t = T, nd noting (using the tringle rule) tht Q = Q = q p, this eqution cn lso e written s t T q Q t = p + s(q p) =( s)p + sq (0 s ). Note tht if the prmeter s in Emple 3. Is llowed to rnge over ll the rel numers (.
Solve real problems by finding out how they are transformed into mathematical models and learning the methods of solution. This module covers classical mechanical models as well as some non-mechanical models such as population dynamics; and methods including vector algebra, differential equations, calculus (including several variables and vector calculus), matrices, methods for three-dimensional problems, and numerical methods. Teaching is supported and enhanced by use of a computer algebra package. This module is essential for higher level study of applied mathematics. To study this module you’ll need a sound knowledge of mathematics as developed in Essential mathematics 1 (MST124) and Essential mathematics 2 (MST125) or equivalent.
The OU strives to make all aspects of study accessible to everyone and this outlines what studying MST210 involves. You should use this information to inform your study preparations and any discussions with us about how we can meet your needs. Modules count towards OU qualifications OU qualifications are modular in structure; the credits from this undergraduate module could count towards a certificate of higher education, diploma of higher education, foundation degree or honours degree.
Browse qualifications in related subjects. What you will study This module will be of particular interest to you if you use mathematics or mathematical reasoning in your work and feel that you need a firmer grounding in it, or if you think you might find it useful to extend your application of mathematics to a wider range of problems. The module is also very suitable for those planning to teach applied mathematics. Around half of this module is about using mathematical models to represent suitable aspects of the real world; the other half is about mathematical methods that are useful in working with such models. The work on models is devoted mainly to the study of classical mechanics, although non-mechanical models – such as those used in population dynamics – are also studied.
The process of mathematical modelling, based on simplifying assumptions about the real world, is outlined. You will work in groups to create a mathematical model and to produce a mini-report. The work on methods comprises topics chosen for their usefulness in dealing with the models; the main emphasis is on solving the problems arising in the real world, rather than on axiom systems or rigorous proofs. These methods include differential equations, linear algebra, advanced calculus and numerical methods.
You’ll begin the mechanics part of the module with statics, where there are forces but no motion, and then you’ll be introduced to the fundamental laws governing the motions of bodies acted on by forces – Newton's laws of motion. These are applied to model:.
the motion of a particle moving in a straight line under the influence of known forces. undamped oscillations. the motion of a particle in space. the motions of systems of particles.
the damped and forced vibrations of a single particle. the motion (and vibrations) of several particles. In the methods part of the module you’ll cover both analytic and numerical methods.
You’ll explore the analytical (as opposed to numerical) solution of first-order and of linear, constant-coefficient, second-order ordinary differential equations, followed by systems of linear and non-linear differential equations and an introduction to methods for solving partial differential equations. The topics in algebra are vector algebra, the theory of matrices and determinants, and eigenvalues and eigenvectors. You’ll develop the elements of the calculus of functions of several variables, including vector calculus and multiple integrals, and make a start on the study of Fourier analysis.
Finally, the study of numerical techniques covers the solution of systems of linear algebraic equations, methods for finding eigenvalues and eigenvectors of matrices, and methods for approximating the solution of differential equations. You will learn Successful study of this module should improve your skills in being able to think logically, express ideas and problems in mathematical language, communicate mathematical arguments clearly, interpret mathematical results in real-world terms and find solutions to problems. Professional recognition This module may help you to gain membership of the Institute of Mathematics and its Applications (IMA). For further information, see the. Teaching and assessment Support from your tutor You will have a tutor who will help you with the study material and mark and comment on your written work, and whom you can ask for advice and guidance. We may also be able to offer group tutorials or day-schools in your locality that you are encouraged, but not obliged, to attend, and there is an online forum. Where your tutorials are held will depend on the distribution of students taking the module.
If you want to know more about study with The Open University before you register. Assessment The assessment details for this module can be found in the facts box above. You can choose whether to submit your tutor-marked assignments (TMAs) on paper or online through the eTMA system. You may want to use the eTMA system for some of your assignments but submit on paper for others.
This is entirely your choice. Entry requirements To study this module, normally you should have completed or the discontinued module MS221. There may be circumstances in which you can study this module without having first studied MST125, but you should speak to an to discuss this before registering on this module.
Knowledge of mechanics is not needed, but we do not recommend the module if you have little mathematical experience. You need a good basic working knowledge of:. algebra – you must be able to solve linear and quadratic equations with one unknown, to multiply and add polynomials, to factorise quadratic polynomials and to work with complex numbers.
geometry – you must know Pythagoras' theorem and how to use Cartesian coordinates, e.g. The equations of straight lines and circles. trigonometry – you need to know the basic properties of the three trigonometric ratios sine, cosine and tangent, and the definitions of the corresponding inverse functions. calculus – you must be able to differentiate and integrate a variety of functions, though great facility in integration is not necessary. mechanics – you should have some basic knowledge of Newtonian mechanics. You can try our to help you determine whether you are adequately prepared for this module. Open University Student Budget Account The Open University Student Budget Accounts Ltd (OUSBA) offers a convenient 'pay as you go' option to pay your OU fees, which is a secure, quick and easy way to pay.
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You then repay OUSBA interest-free and in full just before your module starts. 0% APR representative. This option could give you the extra time you may need to secure the funding to repay OUSBA. Pay by instalments – OUSBA calculates your monthly fee and number of instalments based on the cost of the module you are studying.
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Read more about. Employer sponsorship Studying with The Open University can boost your employability. OU courses are recognised and respected by employers for their excellence and the commitment they take to complete. They also value the skills that students learn and can apply in the workplace.
More than one in ten OU students are sponsored by their employer, and over 30,000 employers have used the OU to develop staff so far. If the module you’ve chosen is geared towards your job or developing your career, you could approach your employer to see if they will sponsor you by paying some or all of the fees. Your employer just needs to complete a simple form to confirm how much they will be paying and we will invoice them. You won’t need to get your employer to complete the form until after you’ve chosen your module. What's included Module books, other printed materials, algebra software, and module website.
You will need You require internet access at least once a week during the module to download module resources and assignments, and to keep up to date with module news. A calculator – you may wish to use this during the module, but you are not allowed to take a calculator into the examination. Computing requirements A computing device with a browser and broadband internet access is required for this module. Any modern browser will be suitable for most computer activities. Functionality may be limited on mobile devices. Any additional software will be provided, or is generally freely available.
However, some activities may have more specific requirements. For this reason, you will need to be able to install and run additional software on a device that meets the requirements below. A desktop or laptop computer with either:. Windows 7 or higher.
macOS 10.7 or higher The screen of the device must have a resolution of at least 1024 pixels horizontally and 768 pixels vertically. To participate in our online-discussion area you will need both a microphone and speakers/headphones.
Our website has further information including computing skills for study, computer security, acquiring a computer and Microsoft software offers for students. If you have a disability Written transcripts of any audio components and Adobe Portable Document Format (PDF) versions of printed material are available.
Some Adobe PDF components may not be available or fully accessible using a screen reader (mathematical notation may be particularly difficult to read in this way). Other alternative formats of the study materials may be available in the future.
It is important to note that use of the module software, which includes on-screen graphs and mathematical notation, will be an integral part of your study. You will need to spend considerable amounts of time using a personal computer. If you use specialist hardware or software to assist you in using a computer you are advised to about support which can be given to meet your needs. To find out more about what kind of support and adjustments might be available, or visit our website.