Bsics of spce nd vectors Points nd distnce One wy to describe our position in three dimensionl spce is using rtesin 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 ometimes we will not hve sphere given to us in this form in which cse we should rewrite it (ie, 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 ( x(t), y(t), z(t) ) 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 (ie, doing integrl clculus) The limit of this process gives us length = b (x (t) ) 2 + ( y (t) ) 2 + ( z (t) ) 2 dt Vectors ometimes we re interested in quntities tht hve both length nd direction (ie, 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, ie, 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, ie, 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, ie,, b, c + d, e, f = + d, b + e, c + f k, b, c = k, kb, kc The mgnitude of vector (ie, 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, ie, 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, ie,, 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, ie, 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) ombining 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, ie, proj v u which is the vector u projected down onto the vector v We hve proj v u = ( ) u v v v v imilrly 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, ie, or rerrnging 0 = n x x 0, y y 0, z z 0 = (x x 0 ) + b(y y 0 ) + c(z 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 o two plnes re prllel when the norml vectors re prllel, the ngle between plnes is the ngle between norml vectors, nd so on ross product ross product, denoted with, gives wy to multiply vectors together nd get new vector, ie, (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 ie, 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, ie, 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, ie, 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, ie, 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, ie, 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 ombining the distributive property with how cross product works for the stndrd unit vectors lso gives wy to compute the cross product, ie, 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 ince vector cn be broken down into its component prts this function cn be described by how it is behving in ech component, ie, F(t) = f(t), g(t), h(t) = f(t)i + g(t)j + h(t)k We cn sk clculus questions bout these functions, ie, 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 F(t) = t c lim t c f(t), lim g(t), lim h(t) t c t c F (t) = f (t), g (t), h (t) F(t) dt = f(t) dt, g(t) dt, h(t) dt b b F(t) dt = f(t) dt, b g(t) dt, b h(t) dt 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 d ( ) F(t) + G(t) = F (t) + G (t) dt d ( ( )) ( F p(t) = F p(t) ) p (t) dt d ( ) p(t)f(t) = p (t)f(t) + p(t)f (t) dt d ( ) F(t) G(t) = F (t) G(t) + F(t) G (t) dt d ( ) F(t) G(t) = F (t) G(t) + F(t) G (t) dt 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 ( x(t), y(t), z(t) ) 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, ie, r(t) = x(t), y(t), z(t) Then we hve tht v(t) = 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 (ie, 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 o we hve Length = b v(t) dt = b r (t) dt By expnding this we see we get the sme expression tht we hd before for length We note tht r(t) r (t) if nd only if r(t) 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, ie, x, y, z = x 0, y 0, z 0 + t, b, c olving for ech component we get prmetric form, ie, 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, ie, 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, ie, we cn choose different point or choose prllel vector nd still hve the sme line Given prmetric curve ( x(t), y(t), z(t) ) we cn trnsform it to vector vlued function by drwing vector from the origin to the current loction on the prmetric curve, ie, r(t) = x(t), y(t), z(t) ecll tht velocity is given by v(t) = 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, r(t 0 ), it will move in the direction given by velocity, r (t 0 ) This movement will thus lie on the line contining the point r(t 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 r(t) = position v(t) = r (t) = velocity (t) = r (t) = ccelertion T(t) = r (t) r = unit tngent vector (t) N(t) = T (t) T = unit norml vector (t) The unit tngent vector is pointing in the direction of motion ince T = 1 (ie, 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, ie, 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, ie, 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, ie, it gently kisses the curve) urvture 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, ie, 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 o 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 o using the chin rule we cn rewrite this s κ = T r This is better but we cn even mke it more strightforwrd (ie, 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 (x(t), y(t)) this becomes κ = x y x y ( (x ) 2 + (y ) 2) 3/2 oordinte systems There re three min wys to describe point in spce We hve lredy discussed the rtesin 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 = rctn(y/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 rtesin 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 o 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 rtesin 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) o 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 phere centered t the origin of rdius m rtesin: x 2 + y 2 + z 2 = m 2 ylindricl: r 2 + z 2 = m 2 phericl: ρ = m ylinder with center z-xis of rdius m rtesin: x 2 + y 2 = m 2 ylindricl: phericl: r = m ρ = m csc φ one with ngle off the positive z-xis of α rtesin: z = cot α x 2 + y 2 ylindricl: phericl: z = cot α r φ = α In generl, given formul for surfce we cn identify the coordinte system used (ie, 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 curve(s) 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, ie, 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 (ie, 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 +z 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 + z 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 ross sections: ellipses, empty Hyperboloid of one sheet: 2 + y2 b 2 z2 c 2 = 1 ross sections: ellipses, hyperbols Hyperboloid of two sheets: 2 y2 b 2 z2 c 2 = 1 ross sections: ellipses, hyperbols, empty Elliptic prboloid: z = x2 2 + y2 b 2 ross sections: ellipses, prbols, empty x 2 x 2
Hyperbolic prboloid: z = x2 2 y2 b 2 ross sections: prbols, hyperbols Elliptic cone: z 2 = x2 2 + y2 b 2 ross sections: ellipses, hyperbols
Multivrible differentition Functions of severl vribles Up to this point we hve looked t functions of single vrible, for exmple prmetric curve hs single prmeter, usully thought of s time We re now redy to look t functions of severl vribles, ie, multi-vrible functions These re functions which tke severl inputs nd produce n output (usully the output will be single number; t the end of the semester we will look t the cse when the output is vector), ie, f(x, y) or g(x, y, z) The domin of function is the set of ll inputs for which we get vlid output The domin will either be given to us (ie, we re interested in restriction of the function) or we will determine the domin bsed on the function given to us There re three problems tht will possibly occur, nmely: Division by zero: 0 qure root of negtive: (< 0) Log of non-positive number: ln( 0) Generlly speking, s long s n input voids these three problems then the point will be in our domin For function f(x, y) the domin will be some subset of the plne The rnge of function is the set of ll possible outputs Generlly speking, the domin is usully very esy to find while the rnge tkes lot of work! A function f(x, y) cn be represented grphiclly by plotting ll points of the form ( x, y, f(x, y) ) The result will be surfce in three dimensions One wy to understnd this surfce is to look t level curves, these correspond to wht we previously clled trces with respect to z Nmely they re the curves in the plne tht correspond to solutions f(x, y) = k The collection of these curves produces the contour mp By exmining the contour mp we cn understnd how our function behves The contour mp is similr to topogrphicl mp used by hikers o for exmple where lines re bunched together re plces where the function rpidly chnges When we move long line we do not chnge the output Being ble to connect surfces with contour plots is useful bility which comes with lot of prctice For functions g(x, y, z) we could similrly try to plot ll points of the form ( x, y, z, g(x, y, z) ) ; but this is four dimensionl object which is very hrd to drw! However, we cn still get useful informtion by looking t level surfces which correspond to the set of points stisfying g(x, y, z) = c By understnding level surfces we cn get some intuition bout wht is hppening with our function Prtil derivtives One wy to understnd function is to look t slices Previously we sw tht level curves cme from looking t slices of the form z = k We could lso look t slices of the from x = x 0 or y = y 0 The surfce corresponding to our function when intersected with such plne gives curve in tht plne And we love curves! In prticulr, we cn find the slope of the tngent lines to these curves This leds to the ide of prtil derivtives Given z = f(x, y) we hve f z (x, y) = x x = f f(x + h, y) f(x, y) x(x, y) = lim, h 0 h f z (x, y) = y y = f f(x, y + h) f(x, y) y(x, y) = lim h 0 h Tking prtil derivtives is just like tking derivtives of single vrible functions s long s we remember the following rule: When tking prtil derivtive with respect to vrible, tret ll the other vribles s constnts Nottionlly we use the symbol (pronounced prtil ) This cts similrly to d, ie, previously we hd dy z dx nd now we hve x But both nottions re sking the sme thing, how does one vrible chnge s we perturb the other vrible We will lwys use the d for functions of single vrible nd lwys use the for functions of two or more vribles The other nottion, f x, is similr to wht ws previously f, we need the subscript to help specify which vrible we re tking derivtive with respect to We cn lso tke higher order prtil derivtives, including mixed prtil derivtives The nottion helps us to keep trck of which derivtives we tke nd in which order we tke them, for exmple there re four second order prtil derivtives, 2 f x 2 = f xx, 2 f y x = f xy, 2 f x y = f yx, 2 f y 2 = f yy imilr nottion works for higher order prtil derivtives When the function is nice (which will essentilly lwys be the cse in our clss) then f xy = f yx In other words the order of tking prtil derivtives does not mtter This is true whenever f xy nd f yx re continuous in neighborhood Finlly we note tht the sme nottion nd ides work for functions of three or more vribles Limits Often we hve to del with expressions which re mbiguous, the clssic exmple of this is 0/0 There is no vlue for this becuse ny vlue could work
o if we cnnot know wht it is, the we cn sk the question wht should it be This is where limits come in, nmely we look t wht is hppening to the expression t points nerby nd bsed on wht is hppening we cn indicte wht the vlue should be or indicte tht there is no vlue tht it should be (ie, if the points nerby re giving mbiguous possibilities) Nottionlly we hve lim f(x, y) = L (x,y) (,b) which is red s the limit s (x, y) pproches (, b) of f(x, y) is L Intuitively wht this mens is tht s (x, y) gets close to (, b) then f(x, y) gets close to L When we moved to limits of two vribles we opened up whole new cn of wesome Previously we could only tlk bout pproching vlue from either the left or the right Now we cn pproch from the left, from the top, from different ngle, long spirl, long prbol, long nything we wnt If the limit exists then we will lwys get the sme nswer However, if we ever get two different nswers when we pproch in two different wys then the limit does not exist (DNE for short) As n exmple x 2 y 2 lim (x,y) (0,0) x 2 + y 2 = DNE becuse if we pproch long the x-xis it ppers to pproch 1, but if we pproch long the y-xis it ppers to pproch 1 which re totlly not the sme A function is continuous when wht hppens is wht we expected to hppen, ie, ontinuous lim f(x, y) = f(, b) (x,y) (,b) Polynomils re continuous, s re sin, cos, rctn, e nd mny others We cn dd/subtrct/multiply continuous functions nd the result will be continuous function; we cn lso divide nd the result is continuous wherever the denomintor is not 0 We cn lso compose continuous functions (put function in function) nd get continuous functions When deling with limits we first check to see if it is continuous nd plug in the point to see wht we get; if we get vlue then we re done nd if we get 0 0 then the limit does not exist If we get 0/0 then we strt looking t different wys to pproch the limiting point; if we get two different vlues we re done If we re still not done then we need to try to rewrite the function or perhps bound one prt; this is nontrivil nd there is whole clss dedicted to teching this (we will mostly void this sitution!) For future reference we will need some nottion Given set, n interior point x is point where we cn put smll bll centered t x completely inside of ; boundry point y is point where every smll bll centered round y contins points both inside nd outside of An open set is set where every point is n interior point; closed set is point which includes ll the boundry points A bounded set is set which cn be plced inside of single lrge bll Differentibility A function is differentible if loclly the function is liner (ie, flt) Put nother wy, if we zoom in close enough the function becomes more nd more like plne More mthemticlly we hve f(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) }{{} = tngent plne / liner pproximtion + g 1 (x, y)(x x 0 ) + g 2 (x, y)(y y 0 ) }{{} = error where g 1, g 2 0 s (x, y) (x 0, y 0 ), which is to sy tht for points (x, y) ner (x 0, y 0 ) the error is tiny The choice of the prtil derivtives is driven by the need to mtch up with the cross sections with the plnes x = x 0 nd y = y 0 o we need to hve prtil derivtives t the point we re interested in If the prtil derivtives exist nd re continuous in neighborhood round (x 0, y 0 ) then the function is differentible t tht point ince the prtil derivtives will ply n importnt rule in wht is to follow we hve specil vector which consists of the prtil derivtives, this is the grdient vector Given f we denote the grdient of f by f f If z = f(x, y) then f = If w = g(x, y, z) then g = x, f y g x, g y, g z The grdient will tke over mny of the roles tht ws previously done by the derivtive For exmple the bove formultion of differentibility becomes f(x, y) f(x 0, y 0 ) + f(x 0, y 0 ) x x 0, y y 0 In lter chpter we will see it is useful to think of s vector of prtil derivtives, ie, = x, y,, x so tht the grdient corresponds to scling this vector by f We note tht the grdient vector stisfies mny bsic properties we would expect of derivtive including
(f + g) = f + g, showing is liner (cf) = c f, for some constnt c (fg) = g f + f g, the product rule Finlly, we note tht if function is differentible t point then it must lso be continuous t tht point Properties of the grdient If we look t the individul terms in the grdient we see tht f x is indicting how the function is behving in the x-direction Tht is to sy tht if we strt from n initil point nd move in the (positive) x-direction then the initil rte of chnge of the function is given by this prtil derivtive imilrly we hve tht f y indictes how the function is behving in the y-direction But wht if we wnt to know how the function is behving in some other direction? To nswer this kind of question we will use directionl derivtives, f(p + h u) f(p) D u f(p) = lim h 0 h Where D u f(p) is the directionl derivtive of f in the direction u from the point p = (x 0, y 0 ), or lterntively the rte of chnge of f s we strt out from p in the direction u To indicte direction we will lwys use unit vectors If we think of f s elevtion nd p s indicting our ltitude/longitude position nd u indicting which direction we wnt to go, then D u f(p) indictes how steep it is from our current loction if we move in the direction indicted by u We hve lredy mstered doing directionl derivtives in the x nd y directions (these re our old friends, the prtil derivtives) The nice thing is tht when our function is differentible finding directionl derivtives in ny direction is just s esy In prticulr we hve the following: D u f(p) = f(p) u (Usully the hrdest thing bout these problems is to remember to mke sure tht u is unit vector) ecll tht u v = u v cos θ, nd since u is unit vector we hve D u f(p) = f(p) cos θ ince 1 cos θ 1 this gives bounds on how lrge the directionl derivtive cn be, nd in which direction we chieve mximum nd minimum vlues In prticulr we hve the following: f(p) is mximum rte of increse t p f(p) points in the direction which gives the mximum rte of increse t p f(p) is mximum rte of decrese t p f(p) points in the direction which gives the mximum rte of decrese t p In prticulr the grdient encodes informtion bout how to move to chieve mximum rtes of increse nd/or decrese On the other hnd D u f(p) = 0 long level curves or level surfces This is becuse on level curve or level surfce the function is constnt (ie, unchnging) o if u is tngent to level curve we cn conclude tht f(p) is perpendiculr to u Or put more succinctly: f(p) is perpendiculr to level curves/surfces This gives n esy wy to find the norml for tngent plnes to surfce, nmely given surfce described by F(p) = k we use F(p) s the norml vector hin rule The chin rule pplies when we hve function inside of function For exmple z = f(x, y), x = x(t) nd y = y(t) In this cse, z is ultimtely function of t nd so it is nturl to sk how does z vry s we vry t, or in other words wht is dz dt From differentibility we cn rerrnge the terms to get z = z z x + y + EO x y where z = f(x, y) f(x 0, y 0 ), x = x x 0 nd y = y y 0 Also we note tht the error term is very smll, much smller thn our other terms nd becomes insignificnt ner (x 0, y 0 ) If we now divide both sides by t nd tke limit s t 0 we hve dz dt = z dx x dt + z dy y dt In the bove expression we hve used both dz nd z We use the d when we re treting the function s depending on single vrible nd the when we re treting the function s depending on two or more vribles; however they both re sking to do the sme thing, nmely tke derivtive The sme ide works in more complicted situtions For instnce we could hve tht z = f(x, y) nd tht x = x(s, t) nd y = y(s, t) so tht z depends on both s nd t In this cse we hve z s = z x x s + z y y s z t = z x x t + z y y t In generl whenever we del with functions within functions (possibly within functions themselves, n inception of functions if you will) then we cn find
how the vribles chnge with respect to one nother One esy wy to keep trck is to form tree showing the dependencies mong the vribles Then to find the derivtive of the top vrible with respect to one of the leves we simply dd up the product of the prtil derivtives of the pths (s discussed in clss) One specil cse tht we hve is when we hve implicit reltionships mong the vribles For exmple, F(x, y) = 0 defines y s function of x o F depends on x nd y but the reltionship lso shows tht y is function of x Tking the derivtive of both sides with respect to x using the chin rule we get F x + F y dy dx = 0 or dy dx = F x F y imilrly we get the following: F(x, y, z) = k = z x = F x F z nd z y = F y F z These types of formul re useful when we wnt to pproximte the chnge in output given tht we know the pproximte chnges of our input In prticulr this cn be used for error tolernce but we cn lso use this to give pproximte vlues for the function ner point tht we cn esily evlute the function Tngent plnes re trying to mimic the function so tht it grees loclly with the function in both the vlue of the function nd the first order prtil derivtives of the function We cn lso try to find function tht mtches the vlue, the first order prtil derivtives nd the second order prtil derivtives These re done by using the Tylor polynomils The second order Tylor polynomils re shown below (where the function nd derivtives re ll evluted t the point (x 0, y 0 )): z = f + f x (x x 0 ) + f y (y y 0 ) + 1 2 f xx(x x 0 ) 2 + f xy (x x 0 )(y y 0 ) + 1 2 f yy(y y 0 ) 2 Tngent plnes nd other miscellny We hve seen tngent plnes done in two different wys When we did differentibility for function z = f(x, y) we sid tht function loclly looks like plne long with some possible error, the plne we got ws s follows: z = f(x 0, y 0 )+ f x (x 0, y 0 )(x x 0 )+ f y (x 0, y 0 )(y y 0 ) The second wy we hve seen these tngent plnes is when deling with F(x, y, z) = k, which we cn think of s level surfce In this cse for point p = (x 0, y 0, z 0 ) we cn use the properties of grdient to note tht F(p) will be our norml vector so tht our tngent plne is F(p) x x 0, y y 0, z z 0 = 0 or if we expnd out the bove we get the following: F x (p)(x x 0) + F y (p)(y y 0) + F z (p)(z z 0) = 0 It is importnt to note tht these two definitions re comptible, ie, if we wnted the tngent plne for z = f(x, y) we would get the sme plne s if we worked with the function F(x, y, z) = f(x, y) z = 0 We cn rerrnge our terms bove to get the following: z f x x + f y y or in differentil form dz = f x dx + f y dy Optimiztion The gol of optimiztion is to mximize or minimize function There re two types of mximums A globl mximum is point where the function evluted t tht point is t lest s lrge or lrger thn the function evluted nywhere else A locl mximum is point where the function evluted t tht point is t lest s lrge or lrger thn the function evluted t points nerby imilr definitions pply for minimums The nice fct is tht optimiztion works similrly to single vrible clculus, ie, we generlly look for criticl points nd then pply test of some sort One nice fct tht still holds is tht if function is continuous on closed nd bounded set then the function must chieve mximum nd minimum vlue on tht set At mximum (similrly minimum) we cnnot get bigger, so the grdient should not be nonzero (otherwise moving in the direction of the grdient llows us to increse nd moving in the direction opposite the grdient llows us to decrese) o we hve tht mximums nd minimum will occur t criticl points, these include: Where f = 0 (ie, criticl points) Where f is undefined At boundry For function we find criticl points by looking t f nd where it is 0, equivlently where the prtil derivtives re 0 Once we hve the criticl points the next step is to determine wht type of criticl point
it is In single vrible clculus we hd the first nd second derivtive tests; in multi-vrible clculus we hve the econd Prtils Test This is done by noting tht t criticl point the prtil derivtives re equl to zero nd so nerby using the second order Tylor polynomil we hve f(x, y) f(x 0, y 0 ) + 1 2 [ x y] [ fxx f xy f yx f yy ][ ] x y We cn use properties of 2 2 mtrices (ie, eigenvlues) to link the locl behvior to the determinnt of this mtrix, ie, D = f xx f yy (f xy ) 2 We hve the following possibilities If D > 0 nd f xx < 0 (or f yy < 0) then it is locl mximum o if mximizing the function f(x, y) given tht g(x, y) = k then the method of Lgrnge multipliers reduces down to solving the following equtions: f(x, y) = λ g(x, y) nd g(x, y) = k This leds to lrge system of nonliner equtions (ie, one for ech prtil derivtive nd one for the constrint) When in doubt on how to solve such system the following technique tends to work: olve for λ nd set the vrious terms equl to lmbd equl to one nother This gives nother reltionship between x nd y tht cn be used with g(x, y) = k to solve for the possible points yielding mximum nd/or minimum Once we hve these points we plug into the function, the lrgest vlue is the mximum nd the smllest vlue is the minimum If D > 0 nd f xx > 0 (or f yy > 0) then it is locl minimum If D < 0 then it is sddle, neither mx or min If D = 0 the test is inconclusive When finding the mximum nd/or minimum on closed nd bounded set we know tht it must exist, so we find ll the plces where it could exist nd then test ech point In short we do the following: 1 Find ll criticl points in the interior using f 2 Find ll criticl points on boundry (reduce dimensions down) 3 Plug in list of criticl points into function 4 Lrgest number on list is mx; smllest is min Lgrnge multipliers The technique of Lgrnge multipliers is used to solve optimiztion problems with constrint These re usully esy to identify, ie, there will be two functions one tht is being mximized nd the other tht is constrint on the vribles (ie, given tht or such tht ) Essentilly wht it will boil down to is tht if the grdients of the function we re optimizing nd the function tht is our constrint re not prllel then by slightly perturbing long our constrint we cn increse or decrese our vlue o we cn conclude tht if we re t point where we re mximizing or minimizing the two grdient vectors must be prllel (this includes the possibility tht one of them is 0)
Multivrible integrtion Multivrible integrtion Integrtion is ment to nswer the question how much, depending on the problem nd how we set up the integrl we cn be finding how much volume, how much surfce re, how much mss, etc The philosophy of integrtion boils down to breking up the quntity tht we re interested in finding into smll mngeble prts, ech prt of which is esy to find, nd then dding them up to get the totl We wnt to do this for multivrible functions o we strt by considering function f(x, y) nd rectngulr region tht we wnt to integrte over In this cse we will interpret f(x, y) s height nd we wnt to find volume Our region tht we will integrte over (ie, the bse ), which we will denote by, will consist of the points x b nd c y d We subdivide up into smll pieces so tht for these pieces the volume becomes essentilly tht of the volume of tll nd skinny box, nmely f(x k, y k ) A k The point (x k, y k ) is point inside of the smll subdivision nd A k is the re of the subdivision o to find n pproximtion for the totl we now dd these ltogether to get Volume f(x k, y k ) A k This method gives wy to pproximte integrls when we cnnot directly integrte using tools of clculus Also this is essentilly wy tht computers do numericl integrtion, computers just love dding lots of numbers together To get better pproximtion we tke the limit, where here the limiting process is refining our subdivision, nottionlly this is P 0 (this nottion is not importnt!) nd we hve Volume = lim f(xk, y k ) A k = P 0 f(x, y) da This ssumes tht the limit exists, fortuntely for us if we know tht f is continuous on then the limit exists, or we sy f is integrble on In prticulr for the functions tht we re interested in the integrl will lwys exist ince integrtion boils down to dding, nd dding behves nicely we get the following properties: kf(x, y) da = k f(x, y) da ( ) f(x, y) + g(x, y) da = f(x, y) da + g(x, y) da If the region tht we re working on cn be split into two prts which only overlp on their borders, ie, cn be broken into two pieces 1 nd 2, then f(x, y) da = f(x, y) da + f(x, y) da 1 2 If f g on then f(x, y) da Iterted integrtion g(x, y) da Let us continue with trying to integrte f(x, y) over the region with x b nd c y d Insted of breking the rectngle down into ever more refined smller rectngles (s we did lst week), we will find it more convenient to find the volume by slicing the function For exmple by tking the volume tht we re trying to find nd looking t thin strips where we hold y constnt Then the volume of one smll strip is A(y) y where A(y) is the re of the cross section Tking limits we cn conclude Volume = d c A(y) dy On the other hnd we hve tht our cross section will look like function of x s x rnges between nd b, in prticulr it will be the function f(x, y) (remember tht in our cross section tht we re holding y fixed) o we hve A(y) = b Putting these together we hve f(x, y) da = f(x, y) dx d b c f(x, y) dx dy This is nested integrl or n iterted integrl When evluting this integrl we lwys work from the inside out, tht is we perform the inside integrl nd then evlute the bounds nd then we go to the next integrl We could hve strted this whole converstion by slicing in different wy, ie, holding x constnt The sme ides crry through nd we cn conclude tht f(x, y) da = b d c f(x, y) dy dx Notice tht we chnged both the order on the bounds nd the order of the d terms Nottion is importnt The inside integrl goes with the inside d term nd then we work our wy out step by step Also, it is useful to keep trck of bounds while doing these integrls, for exmple we could be more specific bout
the bounds (so tht we re less likely to mke mistke) As n exmple we hve b d c f(x, y) dy dx = x=b y=d x= y=c f(x, y) dy dx When f(x, y) = g(x)h(y) we cn use properties of constnts with respect to integrtion to conclude d b c ( b g(x)h(y) dx dy = )( d ) g(x) dx h(y) dy c We note tht for some functions (for exmple ones involving bsolute vlue) it is sometimes more convenient to brek the integrl up into pieces On similr note we cn use symmetry in some cses to simplify n integrl Integrtion beyond rectngles While we love our flt things, we will hve to del with things which re not flt, this includes hving regions tht re not rectngles We will del with two generl cses A region is y-simple when it cn be described by x b nd φ 1 (x) y φ 2 (x) For such region it is good for us to hold x constnt nd tke thin verticl strips Doing this we get f(x, y) da = b φ2 (x) φ 1 (x) f(x, y) dy dx A region is x-simple when it cn be described by c y d nd ψ 1 (y) x ψ 2 (y) For such region it is good for us to hold y constnt nd tke thin horizontl strips Doing this we get f(x, y) da = d ψ2 (y) c ψ 1 (y) f(x, y) dx dy Unlike rectngles, chnging the order of integrtion is not s simple s swpping few symbols round To chnge the order of integrtion we need to chnge the wy tht we describe our region We hve the following generl procedure: 1 Write down the current bounds 2 Drw picture, clerly indicting the region tht we re integrting nd (idelly) how we re currently integrting 3 elbel ny bounding curves s needed, ie, chnge y = f(x) to x = f 1 (y) 4 Use the picture to determine how to write down the new bounds Work from the outside in 5 Woohoo! Bounds chnged hnging the bounds cn tke some previously impossible function to integrte nd help us to integrte etting up nd chnging the bounds re some of the most importnt ides from this chpter nd you cn expect to be tested on them Note tht when we chnge bounds it might require us to brek our integrl up into severl pieces (ie, whenever bounding curve chnges) onversely it might lso llow us to consolidte severl integrls Integrtion in polr coordintes We cn lso integrte in polr coordintes The importnt prt bout this process is how we chop our region up In rtesin coordintes the ide is to subdivide the region into smll rectngles so tht the re of ech rectngle is dx dy or dy dx In polr coordintes when we subdivide we brek things into smll pieces of θ, ie, dθ, nd smll pieces of r, ie, dr o tht insted of smll rectngles we re looking t pieces of circulr wedges We know how to find the re of wedges (ie, given circle of rdius r nd centrl ngle of α, the re of the wedge is 1 2 αr2 ) nd so we cn determine tht the re of our little piece is r dr dθ In summry we hve: dy dx in rtesin coordintes da = r dr dθ in polr coordintes We lso need to rewrite the function tht we re trying to integrte in terms of r nd θ which cn be done using x = r cos θ nd y = r sin θ (nother often used fct is x 2 + y 2 = r 2 ), so we hve f(x, y) da = f(r cos θ, r sin θ) r dr dθ Where we lso need to describe our region in terms of r nd θ While it is possible to hve to integrte r dθ dr this will hppen rrely in prctice o in generl to describe region we do the following: 1 Find the bounds for θ, these will either be given or re found by looking for the intersection of curves 2 Once the bounds for θ re known pick typicl θ between the bounds nd look for how r vries, ie, from the closest curve in the direction of θ to the frthest curve in the direction of θ (If there ever is trnsition between curves we simple brek the integrl into pieces) The best time to use polr coordintes re in the following situtions: We re told to do problem in polr coordintes, or re given curves describing our region in polr coordintes
Our region cn esily be described using polr coordintes, prticulrly true when we hve circles centered t the origin or on the x or y xis If we hve x 2 + y 2 on the inside of some function nd we cn t integrte If we cnnot mke progress in rtesin coordintes Applictions of integrtion There re severl questions tht we cn nswer with integrtion, for exmple, Volume = (height) da where the height is usully mesured s the distnce from the surfce z = f(x, y) to the xy-plne But remembering in this formt hs the dvntge of hndling more generl situtions, ie, finding the volume between two surfces In this cse we figure out the region we integrte over nd for the height we do top bottom We cn lso use double integrtion to nswer questions bout regions in the plne We cn think of region s corresponding to lmin, ie, thin plte which hs vrying thickness or density which we denote using δ(x, y) The first problem to consider is the mss If the density is constnt the mss is simply found by multiplying the density times the re When the density vries we cn pproximte the mss by the following: subdivide the lmin into pieces (smll prts of size A); pproximte the mss of ech piece (ie, δ(x, y) A); dd the msses up (ie, δ(x, y) A) Tking the limits of finer subdivision this sum becomes n integrl nd we hve Mss = (density) da = δ(x, y) da If we were to spin this region round line we could look t vrious quntities ssocited with this ction For exmple the (first) moment mesures torsionl effects, specificlly the turning effect provided by this force To find the moment of prticle we tke force times distnce to where we rotte round To find the moment of the lmin we repet the sme ide s bove by finding the moment of ech smll piece of subdivision nd dding Tking the limits of finer subdivisions this becomes n integrl nd we hve Moment = (distnce)(density) da We let M x denote the moment with respect to spinning round the x-xis nd M y denote the moment with respect to spinning round the y-xis ince the distnce from (x, y) to the x-xis is y nd the distnce to the y-xis is x we immeditely hve M x = yδ(x, y) da nd M y = xδ(x, y) da Given moment we cn find the center of mss (x, y) (the point t which the lmin blnces) by x = M y M = xδ(x, y) da, δ(x, y) da y = M x M = yδ(x, y) da δ(x, y) da Alterntively we cn think of x s the weighted verge of x where we hve weighted ech x vlue ccording to the mss t tht point When finding the center of mss it is convenient to use symmetry We need symmetry of both the region nd the density function We cn lso find the second moment, or the moment of inerti This works similr to the moment, the only difference being insted of using (distnce) we use (distnce) 2 o we hve Inerti = (distnce) 2 (density) da We let I x denote the inerti with respect to spinning round the x-xis, I y denote the inerti with respect to spinning round the y-xis, nd I z the inerti with respect to spinning round the z-xis (ie, spinning the plne round the origin) ince the distnce from (x, y) to the x-xis is y, the distnce to the y-xis is x, nd the distnce to the origin is x 2 + y 2 we immeditely hve I x = y 2 δ(x, y) da, I y = x 2 δ(x, y) da, I z = (x 2 + y 2 )δ(x, y) da = I y + I x hnging gers, let us go bck to the cse when we hve z = f(x, y) s surfce We cn then sk the question bout how much surfce re lies bove prticulr region We pproch this s lwys by subdividing into little pieces, pproximting ech piece, nd dding bck up For exmple if we subdivide the region into little rectngles of size x by y we cn look t wht is hppening t the surfce bove this smll rectngle Becuse we re looking t smll rectngle the piece bove it will be lmost flt nd so cn be well pproximted by using prllelogrm In prticulr the prllelogrm whose two sides re formed by the vectors x, 0, x fx nd 0, y, y f y