Bsics of spce nd vectors Points nd distnce One wy to describe our position in three dimensionl spce is using Crtesin coordintes x, y, z) where we hve fixed three orthogonl directions nd we move x units in the first direction, y units in the second direction, nd z units in the third direction. The x-xis consists of points of the form x, 0, 0), the y-xis consists of points of the form 0, y, 0) nd the z-xis consists of points of the form 0, 0, z). The xy-plne consists of points of the form x, y, 0), the xz-plne consists of points of the form x, 0, z) nd the yz-plne consists of points of the form 0, y, z). You should be ble to sketch picture of three dimensionl spce nd mrk ech one of these xes nd plnes. Once we cn describe position the next step is to mesure stright-line) distnce. In two dimensions we cn use the Pythgoren Theorem to get the distnce between points x 1, y 1 ) nd x 2, y 2 ) to be D = x 2 x 1 ) 2 + y 2 y 1 ) 2. On the other hnd for three dimensions we will use the Pythgoren Theorem twice) to get the distnce between points x 1, y 1, z 1 ) nd x 2, y 2, z 2 ) to be D = x 2 x 1 ) 2 + y 2 y 1 ) 2 + z 2 z 1 ) 2. A sphere is set of points which re fixed distnce, r, wy from centrl point, h, k, l). Using the distnce formul where we conveniently squre to get rid of the inconvenient squre roots) we hve tht sphere is the set of points stisfying x h) 2 + y k) 2 + z l) 2 = r 2. Sometimes we will not hve sphere given to us in this form in which cse we should rewrite it i.e., using complete the squre). If we wnt to hve ll the points in solid sphere then we hve x h) 2 + y k) 2 + z l) 2 r 2, the volume of this sphere is 4 3 πr3. The midpoint between x 1, y 1, z 1 ) nd x 2, y 2, z 2 ) is the point x1 + x 2, y 1 + y 2, z ) 1 + z 2. 2 2 2 We cn describe motion of prticle by describing position t ech time t. This gives us prmetric equtions xt), yt), zt) ). We cn tke prmetric eqution nd sk how long is the curve. This cn be found by splitting the curve into tiny little pieces, using the distnce formul on ech piece, nd dding them bck up i.e., doing integrl clculus). The limit of this process gives us length = x t) ) 2 + y t) ) 2 + z t) ) 2. Vectors Sometimes we re interested in quntities tht hve both length nd direction i.e., velocity or force). We will refer to these s vectors. Most of our discussion will be bout three dimensionl vectors but most ides generlize to ll dimensions. We note tht vectors re not tied to specific points, i.e., they cn be trnslted to ny other loction nd still be n equivlent vector. Geometriclly vector is directed line segment nd we cn dd vectors by chining them one fter nother, nd we cn scle vectors by chnging the length note tht scling by negtive number reverses the direction of the vector, this llow for subtrction of vectors). For our purposes it will be convenient to work with vector in terms of quntities, i.e., lgebriclly. This is done by writing the vector in component form. u =, b, c where is the mount of chnge in the x direction, b is the mount of chnge in the y direction nd c is the mount of chnge in the z direction. As n exmple, the vector going from the point x 1, y 1, z 1 ) to the point x 2, y 2, z 2 ) is x 2 x 1, y 2 y 1, z 2 z 1. In component form we cn esily dd nd scle vectors, working component by component, i.e.,, b, c + d, e, f = + d, b + e, c + f k, b, c = k, kb, kc The mgnitude of vector i.e., the length) cn be found in component form by trnslting the vector so tht the til is t the origin nd looking t the distnce between the tip of the vector nd the origin. In prticulr we hve, b, c = 2 + b 2 + c 2. A vector is unit vector if it hs length 1, ny vector tht is not the zero-vector 0 = 0, 0, 0 ) cn be scled to unit vector by dividing its mgnitude, i.e., u/ u. This will be used whenever we wnt to tlk bout something hppening in prticulr direction. Three importnt unit vectors re i = 1, 0, 0, j = 0, 1, 0 nd k = 0, 0, 1. These re known s the stndrd unit vectors nd we cn rewrite our vectors
s combintions of these three vectors, i.e.,, b, c =, 0, 0 + 0, b, 0 + 0, 0, c = 1, 0, 0 + b 0, 1, 0 + c 0, 0, 1 = i + bj + ck. Dot product While we cn conveniently dd nd scle vectors there is no convenient wy to multiply vectors together. There re two pproches tht ct like multipliction, we strt with the first clled the dot product. We hve vector) vector) = number. In other words the dot product tkes two vectors nd produces number. In two nd three dimensions this behves s follows this works similrly in other dimensions): 1, b 1 2, b 2 = 1 2 + b 1 b 2 1, b 1, c 1 2, b 2, c 2 = 1 2 + b 1 b 2 + c 1 c 2 With this definition it is esy to estblish some bsic fcts tht mke it look like multipliction, i.e., u v = v u, u 0 = 0, u v + w) = u v + u w. Wht mkes the dot product useful is the geometric interprettion. Let u =, b, c then we hve u u =, b, c, b, c = 2 + b 2 + c 2 = u 2 the sme result holds in other dimensions s well). Combining this with the lw of cosines we get the following: u v = u v cos θ or cos θ = u v u v This llows us to find ngles between vectors. One specil ngle between vectors is right ngle 90 or π 2 ). For such n ngle we hve tht cos θ = 0 nd this leds us to n esy test of whether two vectors re t right ngles to ech other. Nmely, two vectors re orthogonl or perpendiculr if u v = 0; this follows the convention tht the 0 vector is perpendiculr to every other vector. On side note, we sy tht two vectors re prllel if they re sclr multiples of one nother this includes the possibility of reversing direction). This cn be used to find the projection of one vector onto nother, i.e., proj v u which is the vector u projected down onto the vector v. We hve proj v u = ) u v v. v v Similrly given force vector F nd vector on which the force moves n object long D we hve tht the work is w = F D. Plnes will ply n importnt role in our clss. We think of plnes s our generliztion of lines nd when we found lines we needed point nd slope which we cn think of s point nd direction. When we find plne we will similrly need point nd direction which we cn use vector for). But for direction we don t wnt to use vector in the plne becuse there re mny possible directions. Insted we will use vector tht is perpendiculr to the plne, wht we cll the norml vector. In prticulr the norml vector n is perpendiculr to every vector in the plne. Given point x 0, y 0, z 0 ) nd n =, b, c then point x, y, z) is in the plne if nd only if the vector x x 0, y y 0, z z 0 which is vector in the plne) is orthogonl to n, i.e., or rerrnging 0 = n x x 0, y y 0, z z 0 = x x 0 ) + by y 0 ) + cz z 0 ) x + by + cz = x 0 + by 0 + cz }{{ 0 = d. } =d In this lst form it is esy to red off the norml vector to the plne. In generl when we re deling with plnes we will be working with norml vectors. So two plnes re prllel when the norml vectors re prllel, the ngle between plnes is the ngle between norml vectors, nd so on. Cross product Cross product, denoted with, gives wy to multiply vectors together nd get new vector, i.e., vector) vector)=vector). But there is big ctch, nmely it only works in three dimensions conveniently though we live in three dimensions so we cn tolerte this limittion). We hve 1, 2, 3 b 1, b 2, b 3 = 2 b 3 3 b 2, 3 b 1 1 b 3, 1 b 2 2 b 1 This cn be hrd to remember, more convenient form is to write it s determinnt of prticulr mtrix, nmely we hve 1, 2, 3 b 1, b 2, b 3 = i j k 1 2 3 b 1 b 2 b 3 i.e., the first row is the stndrd unit vectors i, j nd k; the second row is the entries of the first vector nd the third row is the entries of the second vector. There re severl wys to tke determinnts, one wy is to copy the first two columns over gin then multiply long the six digonls dd the ones where,
the digonls go down from left to right nd subtrct the ones where the digonls go up from left to right. Another lterntive is to use cofctor expnsions, i.e., we hve 1, 2, 3 b 1, b 2, b 3 = i j k 1 2 3 b 1 b 2 b 3 = i 2 3 b 2 b 3 j 1 3 b 1 b 3 + k 1 2 b 1 b 2 where to tke the determinnt of two-by-two mtrix we hve b c d = d cb. With this rule in plce it is esy to check to see tht it grees with the first definition given bove. The reson tht the cross product is useful is becuse we hve geometricl interprettion of wht is going on. A vector hs two prts, direction nd mgnitude. For the cross product we cn describe wht is going on for ech of these prts. We hve the following: Direction: Mgnitude: u v) u nd u v) v u v = u v sin θ The first property tells us tht the cross product is perpendiculr to the vectors we strted with. This is not enough to nil down the direction so we lso will insist it obeys the right hnd rule not importnt for us). This is one of the most useful properties of the cross product nd we cn use it to find norml vectors to plnes, i.e., the norml vector is perpendiculr to the plne so if we cn find two distinct vectors in the plne we cn tke the cross product nd get the norml vector. This lso gives us method to test to see if we did the cross product correctly, i.e., we must hve u u v) = 0 nd v u v) = 0. This is fst to compute nd helps us identify if we hve mde mistke in tking the cross product. The second property, the mgnitude, hs geometric interprettion. Nmely, this sys tht the mgnitude of the cross product is the re of the prllelogrm formed by the vectors u nd v. By cutting the prllelogrm in hlf we cn get the re of the tringle with two sides formed by u nd v, i.e., Are of = 1 u v. 2 We cn lso go one step further nd determine tht the re of the prllelepiped formed by the three vectors u, v nd w is w u v). We hve mong others) the following rules for cross product, notice in prticulr order mkes difference in the sign: u v = v u) u v + w) = u v + u w u 0 = 0 u u = 0 Combining the distributive property with how cross product works for the stndrd unit vectors lso gives wy to compute the cross product, i.e., i i = 0 j i = k k i = j i j = k j j = 0 k j = i i k = j j k = i k k = 0 Vector vlued functions A vector-vlued function is function which produces vector for the output. Since vector cn be broken down into its component prts this function cn be described by how it is behving in ech component, i.e., Ft) = ft), gt), ht) = ft)i + gt)j + ht)k. We cn sk clculus questions bout these functions, i.e., limits, derivtives, nti-derivtives nd definite integrls. The key is tht we will define these items in the sme wy nd since we know tht vector ddition nd scling works component-by-component nd clculus in the end boils down to fncy dditions nd scling we conclude tht it suffices to work component by component to be fir this should be crefully checked; but we will leve those detils for future courses). More prticulrly we hve the following: lim Ft) = t c lim t c ft), lim gt), lim ht) t c t c F t) = f t), g t), h t) Ft) = ft), gt), ht) Ft) = ft), gt), ht) On side note, when finding the nti-derivtive of vector vlued function we will hve seprte constnt for ech component. Of course we cn combine these constnts into single vector nd conclude tht two ntiderivtives of vector-vlued function differ by constnt vector. The vrious derivtive rules tht
we hve lerned for rel-vlued functions lso generlize s we would expect. Ft) + Gt) = F t) + G t) d )) F pt) = F pt) ) p t) pt)ft) = p t)ft) + pt)f t) Ft) Gt) = F t) Gt) + Ft) G t) Ft) Gt) = F t) Gt) + Ft) G t) Amzingly ll three of the different wys to tke products involving vectors sclr, dot nd cross) obey the sme product rule s before. Woohoo! As n ppliction, suppose we hve prmetric curve xt), yt), zt) ). Then we cn think of this s vector vlued function by putting vector from the origin to the loction of the prticle t time t, i.e., rt) = xt), yt), zt). Then we hve tht vt) = r t) represents the velocity of the prticle t time t nd t) = v t) = r t) equls the ccelertion of the prticle t time t. This cn be used to describe how the behvior of the prticle i.e., loction, direction of motion, ccelertion, nd so on). By tking ntiderivtives we cn lso find position given some informtion bout velocity nd/or ccelertion. Note tht the mgnitude of velocity is speed, nd if we wnt to find distnce trveled by the prticle from t = to t = b we cn do this by finding the integrl of speed. So we hve Length = vt) = r t). By expnding this we see we get the sme expression tht we hd before for length. We note tht rt) r t) if nd only if rt) is constnt, we will use this lter. Lines To find line we will need two items: point on the line, x 0, y 0, z 0 ); nd direction vector for the line,, b, c. Once we hve these then we cn write n eqution for the line. There re three different wys we hve discussed to write such n eqution. The first is vector formt where we find ll the points in vector form) which cn be done by strting t the given point nd dding some multiple of the direction vector, i.e., x, y, z = x 0, y 0, z 0 + t, b, c. Solving for ech component we get prmetric form, i.e., find prmetric eqution for the line. x = x 0 + t y = y 0 + bt z = z 0 + ct If we solve for t in prmetric form we cn get n expression for the line only in terms of x, y nd z known s symmetric form, i.e., x x 0 = y y 0 b = z z 0. c This ssumes tht, b, c 0, if one of them is 0 then we simply write this s combintion of equtions. For exmple if = 0 then the line would be x = x 0 nd y y 0 b = z z 0. c In ech of the forms given bove it is esy to find point on the line nd the direction vector of the line. We note tht these equtions re not unique for lines, i.e., we cn choose different point or choose prllel vector nd still hve the sme line. Given prmetric curve xt), yt), zt) ) we cn trnsform it to vector vlued function by drwing vector from the origin to the current loction on the prmetric curve, i.e., rt) = xt), yt), zt). Recll tht velocity is given by vt) = r t). Velocity encodes both the speed of the prticle nd the current direction in which the prticle is moving. If we imgine the prticle s trveling long trck nd t some time t 0 we deril from the trck, then how does the prticle move? Well it strts t the point it deriled t, rt 0 ), it will move in the direction given by velocity, r t 0 ). This movement will thus lie on the line contining the point rt 0 ) nd the direction vector r t 0 ). This is known s the tngent line. Decomposing ccelertion We cn continue to refine our converstion bout motion. If we only cre bout the direction tht the prticle is trveling in nd not the speed), then we would nturlly wnt to consider unit vector in the direction of movement. We hve the following. rt) = position vt) = r t) = velocity t) = r t) = ccelertion Tt) = r t) r = unit tngent vector t) Nt) = T t) T = unit norml vector t) The unit tngent vector is pointing in the direction of motion. Since T = 1 i.e., it is unit) we hve using
result from lst week tht T is perpendiculr to T, nd hence T is perpendiculr to N. This indictes tht N is perpendiculr to the direction vector T, tht is to sy perpendiculr to the motion of the prticle. On the other hnd since T is relted to velocity we would expect tht T should be relted to ccelertion, i.e., N should hve something to do with ccelertion. This turns out to be the cse. In prticulr we hve the following: T = r r r = T T + N N nd N = r r r Which is to sy tht we cn split ccelertion into two prts; one prt in the direction the prticle is currently moving, nd nother in direction orthogonl to how the prticle is currently moving. The first term is essentilly the projection of r onto r, i.e., we hve proj r r ) = r r r r r = r r r r r = r r r T = TT. The other term essentilly is projection onto the orthogonl direction which cn be done by using cross products. Alterntively, we cn mke the observtion tht N N = T T. From the perspective of the prticle then ll of the interesting informtion bout the motion position, velocity nd ccelertion) re ll contined in the single plne contining the point nd the vectors T nd N. The norml vector to this plne is known s the binorml vector nd is denoted B = T N. This plne is known s the osculting plne or kissing plne, i.e., it gently kisses the curve). Curvture Given curve we cn sk how bendy the curve is. Which is to sy we wnt to mesure how fst the curve is turning. We re creful here to sy tht we re not interested in our speed of rottion in regrds to how we trvel long the curve, i.e., this is not dependent on the prmeteriztion. A first pproch is to look t how quickly our direction, T, is chnging s we chnge our position long the curve, s. So curvture which is denoted with the Greek letter κ) is given by κ = dt ds. This is gret, but is hrd to compute for must curves. So using the chin rule we cn rewrite this s κ = T r. This is better but we cn even mke it more strightforwrd i.e,. T tends to hve squre roots which re best to void when tking derivtives if we cn). This is done by reclling tht T shows up s prt of N nd so is connected to N. Using this we derive κ = r r r 3. In the specil cse of prmeteriztion xt), yt)) this becomes κ = x y x y x ) 2 + y ) 2) 3/2. Coordinte systems There re three min wys to describe point in spce. We hve lredy discussed the Crtesin coordinte system lso known s rectngulr coordintes). The other two re generliztions of polr coordintes in the plne, so before jumping into them we briefly recll fcts bout polr coordintes. In the plne we cn describe our position by strting t the origin looking in the direction of the positive x-xis. We then rotte counter-clockwise) n ngle θ nd then move distnce r out. In prticulr every point cn be uniquely described by distnce r 0 nd n ngle 0 θ < 2π. In converting bck nd forth between these coordinte systems we hve the following: x = r cos θ y = r sin θ r = x 2 + y 2 tn θ = y/x It is useful to remember r 2 = x 2 + y 2. Using these reltionships it is possible though not lwys convenient!) to switch from one coordinte system to the other. On side note we do not sy θ = rctny/x) becuse the inverse tngent only hs rnge of hlf of revolution nd not full revolution. Tht mens we would need to dd some correcting term in some cses to get the correct ngle. The first generliztion to three dimensions is known s cylindricl coordintes. This cn be thought of s polr plus z. Every point cn be described by r, θ nd z we will ssume tht r 0 nd 0 θ < 2π). The wy it works is tht we use r nd θ s before to move to point in the plne. Then we move up or down ccording to z. Given wht we lredy know bout polr coordintes this mkes converting bck nd forth between Crtesin coordintes nd cylindricl coordintes very esy. We hve the following: x = r cos θ y = r sin θ z = z r = x 2 + y 2 tn θ = y/x z = z
The second generliztion to three dimensions is known s sphericl coordintes. This is more in the spirit of polr coordintes in tht we first strt t the origin nd then turn to look in the direction of the desired point. We then move sufficient distnce to get to tht point. To determine where to look we will need two ngles, one tells us how to rotte left/right in the xy-plne θ, this is the exct sme θ s in cylindricl coordintes nd we will ssume 0 θ < 2π); the second ngle then tells us how to rotte up/down, φ. The wy tht φ is mesured is s the ngle off of the positive z-xis. So we hve tht φ = 0 corresponds to the positive z-xis, φ = 1 2π corresponds to points in the xy-plne nd φ = π corresponds to the negtive z-xis, so 0 φ π. Finlly the distnce we move will be ρ 0. To be ble to convert from sphericl to Crtesin nd vice-vers we observe tht since ρ is the distnce we move tht it corresponds to the distnce from x, y, z) to 0, 0, 0) nd we hve formul for tht. We cn lso find z by drwing right tringle from the origin, to our point, to the z-xis. By using properties of right tringles we cn find z = ρ cos φ. To find x nd y we note tht the length of the other prt of the tringle is ρ sin φ nd this corresponds to the r in cylindricl/polr coordintes which cn help us solve for x nd y. Note: the bove dilogue mkes much more sense when drwing picture t the sme time.) So we hve the following: nd x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ ρ = x 2 + y 2 + z 2 tn θ = y/x ) z φ = rccos x2 + y 2 + z 2 It is useful to remember ρ 2 = x 2 + y 2 + z 2 nd tht z = ρ cos φ nd r 2 = x 2 + y 2 = ρ 2 sin 2 φ. The reson to hve multiple coordinte systems is tht some surfces re esy to describe in one coordinte system but very difficult to describe in nother. The following re some wys to describe the sme surfce in vrious coordinte systems. Sphere centered t the origin of rdius m Crtesin: x 2 + y 2 + z 2 = m 2 Cylindricl: r 2 + z 2 = m 2 Sphericl: ρ = m Cylinder with center z-xis of rdius m Crtesin: x 2 + y 2 = m 2 Cylindricl: Sphericl: r = m ρ = m csc φ Cone with ngle off the positive z-xis of α Crtesin: z = cot α x 2 + y 2 Cylindricl: Sphericl: z = cot α r φ = α In generl, given formul for surfce we cn identify the coordinte system used i.e., if we see z nd r we re in cylindricl; if we see φ or ρ we re in sphericl). Being ble to convert between coordinte systems will be useful in future chpter. Qudric surfces To understnd surfces we will often look t cross sections. These re found by looking t where plne intersects the surfce nd exmining the resulting curves). We will be prticulrly interested in trces which correspond to plnes of the form x = c, y = c or z = c. Note tht in such plnes one of the vribles is fixed so ny eqution describing surfce reduces the number of equtions involved. A specil type of surfce is cylinder. These re surfces which hve identicl trces in one of the vribles, i.e., x 2 + y 2 = 1 in three dimensions forms wht we normlly cll cylinder becuse for ech slice of the form z = c we get unit circle. These re esy to identify when written out s equtions becuse they re missing vrible. With our bckground in understnding conic sections i.e., ellipses, prbols, hyperbols nd so forth) we re redy to understnd wht is going on for cross sections of qudric surfces. These re surfces which cn be written in the form Ax 2 +By 2 +Cz 2 +Dxy+Exz+Fyz+Gx+Hy+Iz+J = 0 where A, B,..., J re constnts. But by trnsltion nd rottion we only need to consider such surfces of the form Ax 2 + By 2 + Cz 2 + J = 0 or Ax 2 + By 2 + Iz = 0. These give the following surfces: x 2 Ellipsoids: 2 + y2 b 2 + z2 c 2 = 1 Cross sections: ellipses, empty Hyperboloid of one sheet: 2 + y2 b 2 z2 c 2 = 1 Cross sections: ellipses, hyperbols Hyperboloid of two sheets: 2 y2 b 2 z2 c 2 = 1 Cross sections: ellipses, hyperbols, empty Elliptic prboloid: z = x2 2 + y2 b 2 Cross sections: ellipses, prbols, empty x 2 x 2
Hyperbolic prboloid: z = x2 2 y2 b 2 Cross sections: prbols, hyperbols Elliptic cone: z 2 = x2 2 + y2 b 2 Cross sections: ellipses, hyperbols