Geometry and Calculus of Variations Lecture notes
|
|
- Warren Mosley
- 5 years ago
- Views:
Transcription
1 Sum /3/12 23:06 pge 1 #1 Geometry n Clculus of Vritions Lecture notes Mrch 12, 2012 Abstrct This is the type version of Dr A.Gilbert s lecture notes. c 1 Lecture Spce Curves Definition: A spce curve is smooth mp x : I R 3, where I is given intervl in R. We llow I = R. Definition: The following is efine nlogously to plnr curves: velocity, spee, regulr curve, orienttion, reprmetriztion, rc-length, unit tngent, frme. Exmple: Let x(t) = (t, t 2, t 3 ) T (cubic curve). It is esy to see tht x (t) = (1, 2t, 3t 2 ) T, v(t) = 1 + 4t2 + 9t 4 n for the rclength we hve s(t) = t u2 + 9u 4 u Exmple: x(t) = (cos t cos 4t, cos t sin 4t, sin t). Notice tht x(t) = 1 i.e. lies on the unit sphere. x(t) encircles the x 3 xis while climbing from equtor to north pole. By irect computtion v(t) = cos 2 t x = ( sin t cos 4t, sin t sin 4t, cos t) T + cos t( 4 sin t, 4 cos t, 0) T. 1.2 Curvture Unit tngent T = x /v is efine s before. Defn: Let x be regulr spce curve. Curvture k is efine by k(t) = 1 v T. Notice tht for plne curves k = ± 1 v T wheres for spce curves k 0. This is preliminry version of the notes. It will be upte weekly. 1
2 Sum /3/12 23:06 pge 2 #2 () The cubic curve (b) The spheric curve Figure 1: Exmples of spce curves 1.3 Biregulr curves We wnt to efine Frenet frme long x, i.e. group of three mutully orthogonl unit vectors in R 3. We cn o so if x is biregulr. Definition: Regulr curve, i.e. the unit tngent T exists, is biregulr if T 0. Next we efine the Unit norml: unlike plne curves there is whole plne, clle norml plne contining vectors perpeniculr to T. Among the norml vectors we give specil nmes to two prticulr ones: norml n binorml. For biregulr x the unit norml is the vector in the irection T N = T T. Clerly N exists, for T 0. From the efinition it follows tht N n T re orthogonl (to see this ifferentite T (t) 2 = 1 by t). At ech point x(t) we now hve 2 orthogonl unit vectors T n N. To complete the Frenet frme we introuce the binorml B. B = T N it is the vector prouct of T n N. The Frenet Frme is the orthogonl set (3 3 orthogonl mtrix) ( T, N, B) of tngent vectors to R 3 t x(t). Recll tht tngent vectors to R 3 t x(t) nee not point long curve, but unit tngent vector T oes. 2
3 Sum /3/12 23:06 pge 3 #3 Figure 2: Helix with k = = Formule for T n N Let x(t), t I be prmetristion of the curve x then ifferentiting x n using efinitions of T, N n k we conclue x = v T x = v T + v T = v T + v 2 k N. In prticulr if s = t, i.e. the curve is rc length prmetrise, then x = T x = T = k N, So N is unit vector orthogonl T in plne forme by x n x = v T. Exmples: 1) Helix: x(t) = ( cos t, sin t, kt). Clerly x (t) = ( cos t, sin t, 0) points long N. 2) Cubic curve: x(t) = (t, t 2, t 3 ) T we hve t t = 0, x (0) = (1, 0, 0), x (0) = (0, 1, 0). 2 Lecture Structurl equtions We hve efine the Frenet frme for biregulr curve x s orthogonl 3 3 mtrix ( T, N, B) where 3
4 Sum /3/12 23:06 pge 4 #4 T = x v N = T T B = T N We wnt to erive the structurl equtions tht is to relte ( T, N, B ) to orthogonl 3 3 mtrix ( T, N, B). Step 1: Eqution for T From the efinition of curvture n the unit norml we hve combining these two equtions we get tht T = T N k = T v T = kv N Step 2: Eqution for B. It is quite esy to see tht for (t) n b(t) t ( b) = b + b reclling the efinition of B n utilizing this formul we get B = t ( T N) = T N + T N = T N since T n N re prllel. Thus T N is prllel to N which yiels B = τvn for some sclr function τ. Step 3: Eqution for N Notice tht N = B T. Thus N = B T + B T = τvn T + B (kvn) = kvt + τvb Summrizing we get the structurl equtions ( T, N, B ) = ( T, N, B)v 0 k 0 k 0 τ 0 τ 0 (1) Defn: τ : I R is the torsion. Defn: The plne etermine by the unit tngent vector T n the unit norml vector N is clle the osculting plne. Theorem: Curvture n torsion re unchnge by reprmetriztion. 4
5 Sum /3/12 23:06 pge 5 #5 Proof: Step 1: Velocity n spee Let us tke t = g(u), u is the new prmeter n g 0 then x 1 (u) = x(g(u)) n x 1 = x v t g = 1 v 1 g Step 2: Tngent By efinition thus T 1 = x 1 v 1 = T 1 = x 1 1 x T = v 1 v t x t g v 1 = T vg v 1 = T vg v g = ε T where ε = g g = ±1: +1 if g is orienttion preserving or 1 orienttion reversing. Step 3: Curvture The the u-erivtive of the tngents equlity n utilize the structurl equtions v 1 k 1N1 = T 1 u = ε T t g = εvkg N Tknig the mgnitue of both sies n noting tht both N 1 n N re unit vectors we conclue v 1 k 1 = ε vk g k 1 = v v 1 k g = k Moreover N 1 = εvkg v 1 k 1 N = ε v v 1 g N = ε g g = ε2 N = N. Step 4: Torsion Finlly we wnt to show tht torsion is unchnge. By efinition B 1 = T 1 N 1 = ε T N = ε B. Differentiting with respect u then B 1 = ε B t g n using the structurl equtions tking the mgnitue n the proof of the Theorem is now complete. B 1 = v 1 τ 1 N1 = ε( vτ N)g = εb g τ 1 = ε v v 1 g τ = ε g g τ = ε2 τ = τ 2.2 Interprettion of curvture n Torsion The trjectory of prticle moving in spce prouces curve. The osculting plne is forme by unit vectors T n N. For generl spce curves the prticle tries to escpe the osculting plne, for otherwise it is plne curve. When k is constntly zero, T never chnges n the curve is stright line. As the nme curvture suggests, k mesures the rte t which ny nonstright curve tens to eprt from its tngent. When τ is constntly zero, the osculting plne never chnges, n we hve plne curve, with constnt binorml B. Thus the torsion mesures the rte t which twiste curve tens to eprt from its osculting plne. Without loss of generlity we my tke the rc length prmetristion of x n ssume tht T, N, B t x(s 0 ) coincie with the irections of x, y n z-xis. Let X be the projection of the curve onto the 5
6 Sum /3/12 23:06 pge 6 #6 osculting plne t x(s 0 ). Then Then X(s) = x(s) x(s 0 ) B[(x(s) x(s 0 )) B]. X = x B( x B), X = x B( x B) In prticulr t s = s 0, using the fct tht B is orthogonl to T n N, we get X = x = T, X = x = k N. Thus the curvture of X t s = s 0 is k = et[ X, X ]. Thus the projection of the curve onto the osculting plne t x(t 0 ) hs curvture k. Now consier the projection onto the norml plne pssing through x(t 0 ), i.e. forme by N(t 0 ) n B(t 0 ). There the projecte curve hs Frenet frme (t, n), with t = N(t 0 ), n = B(t 0 ) t t = t 0 stisfies t = τn, n = τt. thus τ mesures the rottion in norml plne bout T -xis s we pove long the curve. 2.3 Sphericl imge n Drboux vector Let us efine the vector then ω = v k 2 + τ 2 n ω = vτ T + vk B, T = ω T N = ω N B = ω B which cn be reily verifie. ω is clle Drboux vector. It is the ngulr velocity of the prticle when the point is moving long the curve. 2.4 Formule for k n r Proposition 1 Let x be biregulr curve then k = 1 v 3 x x τ = ( x x ) x v 6 k 2 Proof: Let s ifferentite x three times n use structurl equtions x = v T x = v T + v T = v T + v 2 k N, x = v T + v T + 2vv k N + v 2 k N + v 2 k N, = v T + v vk N + 2vv k N + v 2 k N + v 2 k( kv T + τv B). Hence x x = v T (v T + v 2 k N) = v 3 k T N = v 3 k B. Now tking the mgnitue of this vector we conclue k = 1 v 3 x x. Finlly using the fct tht B is orthogonl to N n T n tking the sclr prouct of x x with x we conclue [ x x ] x = v 3 kv 3 kτ = v 6 k 2 τ n the proof follows. 6
7 Sum /3/12 23:06 pge 7 #7 3 Lecture Locl behviour Theorem 2 If we choose the Frenet frme (the triheron) such tht T (0) = e 1, N(0) = e 2, B(0) = e 3 n tke the rc-length prmetriztion then in this coorintes x(s) s e 1 + k 2 s2 e 2 + kr 6 s3 e 3 here e 1, e 2, e 3 is the stnr orthogonl bsis in R 3. To see this we use Tylor s expnsion x(s) = x(0) + x (0)s + x (0) s 2 + x (0) s ! 3! n use the structurl equtions (remember v = 1!) x = T x = T = kn x = k N + kn = k N + k( kt + rb). Thus ssuming tht x(0) = 0 we get x(s) = T s + k N 2! s2 + k N + k( kt + rb) s ! = T (s k2 3! s3 ) + N( k 2! s2 + k 3! s3 ) + rk 3! s Assuming tht s is very smll we cn ignore the higher powers of s within the coefficients of T, N n B thus x(s) = s e 1 + k 2 s2 e 2 + kr 6 s3 e 3 + = An the result follows. Thus loclly ny biregulr curve is cubic! 3.2 Implicitly efine curves ss k 2 s2 kτ 6 s3 Let f 1, f 2 be functions of three vribles, f 1 : (x, y, z) R, f 2 : (x, y, z) R. Consier C the intersection of the zero level sets of f 1 n f 2, then it is n implicitly efine spce curve, provie tht it is not empty. The tngent plne to the grph of f 1 hs norml f 1 thus the tngent plne, Π 1, t (x 0, y 0, z 0 ) is f 1 (x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0. Similrly Π 2, given by f 2 (x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0 is the tngent plne to the grph of f 2 t (x 0, y 0, z 0 ). If (x 0, y 0, z 0 ) C then the both Π 1 n Π 2 re tngent tp C t (x 0, y 0, z 0 ). Clerly the tngent vector is colliner to the vector proucts of the normls of Π 1 n Π 2, thus T = f 1(x 0, y 0, z 0 ) f 2 (x 0, y 0, z 0 ) f 1 (x 0, y 0, z 0 ) f 2 (x 0, y 0, z 0 ). Defn: An implicitly efine spce curve is si to be regulr if f 1 f 2 0. Clerly the tngent vector is efine for regulr implicitly efine spce curves. 7
8 Sum /3/12 23:06 pge 8 #8 () (b) (c) Figure 3: Exmples of implicit spce curves 3.3 Exmples ) Let f 1 = x x 2 2 x 1 n f 2 = x 3 x 1 b) Let f 1 = x x x n f 2 = x 1 + x 2 + x 3 1. Thus f 1 = 0 is the unit sphere n f 2 = 0 is plne so intersection is circle in R 3. c) Another exmple is the intersection of S 1 : x 2 = x 2 1 n S 2 : x 3 = x Torsion Exmples Exmple 1: Biregulr plne curve hs τ = 0. This is simple exercise. The converse sttement is lso true. A biregulr curve with τ = 0 is plnr! (Hint: Use structurl equtions). Exmple 2: A helix hs constnt curvture n torsion. It is stightforwr to verify tht for helix x(t) = ( cos t, sin t, bt) T k = 2 + b 2, τ = b 2 + b 2. Conversely let us show tht if k n τ re constnts then x is helix. If τ = 0 then x is plne curve with constnt curvture thus it is circle, for k = θ s, θ is n rgument function (Lecture 6) n hence θ = ks + C where C is n rbitrry constnt. Hence x = T = (cos(kt + C), sin(kt + C)) T implying tht x = 1 k (sin(kt + C) + A, cos(kt + C)) + B)T which is circle of rius R = 1/k centere t (A, B). Thus we ssume tht τ 0. Let s tke the rc length prmetristion of the curve. Then Drboux vector ω = τ T + k B is constnt, for ω = τ T + k B = τ(k N) + k( τ N) = 0. Without loss of generlity we my ssume tht ω is colliner to e 3 so ω = e 3 k2 + τ 2. Next ω T = (τt + kb) T = τ, thus the unit tngent vector mkes constnt ngle α with the e 3 τ xis n cos α =. k 2 +τ 2 Furthermore x 3 ( x ω) = = x ω = T s s ω = cos α thus x 3 (s) = s cos α + C, where C is n rbitrry constnt. Next let us consier the projection of the curve onto the plne orthogonl to ω = e 3, i.e. X = x ( x ω)ω. For this plne curve we hve X = x ( x ω)ω, X = x 8
9 Sum /3/12 23:06 pge 9 #9 In prticulr the spee of X is V = ( x ) 2 2( x ω) 2 + ( x ω) 2 = 1 ( x ω) 2 = 1 τ 2 thus V is constnt. If (t, n) is the Frenet frme of X the projecte curve, then from Frenet equtions we hve k N = x = X = s ( X ) = s (V t) = V t = V k X n, where k X is the curvture of the projecte curve X. Equting the right n left hn sies n tking the mgnitues gives k X = k/v = const. Thus X is circle. 4 Lecture Surfces Defn: A locl surfce in R 3 is smooth, injective mp x : D R 3 with continuous inverse x 1 : x(d) D where D is omin (open, connecte set) in R 2. x : (u, v) x(u, v) Remrks: 1) The ssumption of smoothness implies tht x is somewht istorte version of D 2) x is injective points of x(d) re lbelle (coorinte) by corresponing points of D 3) the inverse mpping x 1 is continuous prevents ner self-intersection Uner these conitions x is homeomorphism, i.e. bijection with continuous inverse x Surfces in R 3 If locl surfce x lies in set Σ R 3 then x is locl surfce in Σ. A surfce in R n is subset Σ R 3 such tht for ech point p of Σ there exists locl surfce in Σ whose imge contins neighborhoo N of p in Σ. 4.3 Exmples Sphere: Let us consier x(u, v) = sin u cos v sin u sin v cos u, (u, v) D = (0, π) (0, 2π). The trce of x is unit sphere excluing semi-circle connecting North n South poles. Line u = const is line of ltitue n line v = const is line of longitue. Ellipsoi: Let us consier x(u, v) = sin u cos v b sin u sin v c cos u, (u, v) D = (0, π) (0, 2π). Here, b, c re non-zero constnts. The trce is n ellipsoi with n rc omitte. 9
10 Sum /3/12 23:06 pge 10 #10 Grph of function: Now let f : D R be smooth function. Consier x(u, v) = u v f(u, v). Note tht not ll surfces cn be represente s grphs. See the exmple of unit sphere. p(t) Surfce of revolution: Let t q(t) 0 be curve with trce in the (x 1, x 2 ) plne n rotte it bout the x 1 xis to get p(u) x(u, v) = q(u) cos v. q(u) sin v 4.4 Regulrity For surfce x, the prtil velocities t x(u, v) re tngent vectors to R 3 t x(u, v) given by x u = x 1 u x 2 u x 3 u, x v = x 1 v x 2 v x 3 v. Interprettion: The spce curves (lines) u x(u, v 0 ) n v x(u 0, v), for fixe u 0 n v 0 hve velocities x u n x v. Defn: The locl surfce is regulr if x u n x v re linerly inepenent t ech point of D. 4.5 Displcement n re [ ] δu Consier smll isplcement t ech the point p δv 0 = (u 0, v 0 ) D. Then it is mppe by the [ ] δu mtrix [x u, x v ] to [x u, x v ]. The regulrity implies tht x δv u n x v hve istinct irections t ech p 0 D. Hence smll rectngle (u 0 + δu) (v 0 + δv) in (u, v) spce is pproximtely mppe, by x to the prllelogrm with sies x u δu, x v δv. If x u, x v re linerly inepenent then the mtrix [x u, x v ], which pproximtes x, efines n invertible liner trnsformtion between bove rectngle n prllelogrm. The Inverse Function Theorem sys tht the sme hols for the exct mp x: Aroun ech point p 0 D, there exists neighborhoo u on which x is smooth bijection between N n x(n), where N is neighborhoo of x(p 0 ). 4.6 Regulrity of previous exmples The sphere is regulr locl surfce since x u n x v re linerly inepenent. Similrly it follows tht ellipsoi is regulr too. 10
11 Sum /3/12 23:06 pge 11 #11 5 Lecture Regulr implicitly efine surfces Defn: Let f : R 3 R be smooth function n c R be such tht the level set Σ = {x R 3 : f(x) = c} is non-empty. If f 0, x Σ then Σ is regulr implicitly efine surfce. Theorem: A regulr implicitly efine curve surfce is surfce in R 3. Geometric inpterprettion: f is the norml to Σ, if x3 x 3 surfce contins no verticl irections surfce loclly grph of x 3 (x 1, x 2 ). Exmples: f(x) = x 2 1 f(x) = x 2 3 x 2 1 x 2 2 Quric surfce k 1 x k 2 x k 3 x 2 3 = 1 6 Lecture Curve in surfce The curves lying in surfce provie importnt informtion bout the surfce. Defn: Let Σ be surfce. The curve α : I R 3 is curve in Σ if α(i) Σ. When Σ = x(d) is locl surfce we cn pull bck α to D, since it is much esier to work out in D thn in x(d) Lemm Let α be curve in Σ, Then there exists unique smooth curve w : I D so tht α(t) = x(w(t)). Proof: Define w = x 1 α : I D, so α = x (x 1 α). Exmple: D = (0, 2π) (0, ), x(u, v) = (cos u, sin u, v) T. Tke w 1 (t) = (2t, π/t) T, x(w 1 (t)) = (cos 2t, sin 2t, π/4) T w 2 (t) = (π/2, t) T, x(w 2 (t)) = (0, 1, t) T w 3 (t) = (2t, t) T, x(w 3 (t)) = (cos 2t, sin 2t, t) T 6.2 Tngent spce to surfce Σ Recpitulting: T p R 3, the tngent spce to R 3 t p, is spce of ll tngent vectors (p, v) to R 3 t p. Defn: Let p Σ. A tngent vector v to R 3 t p is tngent to Σ if v is the velocity t p of some curve in Σ. The set of ll such tngent vectors to Σ t p is the tngent spce (plne) to Σ t p, written T p Σ. Next we chrcterise of tngent spce: Lemm Let x(d) be locl surfce in Σ n let p = x(w 0 ) x(d). Then T p Σ is the subspce of T p R 3 spnne by x u (w 0 ) n x v (w 0 ). Proof: Step1: T p Σ Spn(x u, x v ). Let t w(t) be curve in D, such tht w(0) = w 0. Then t x(w(t)) is curve in x(d), pssing through p when t = 0, nt its velocity t p is By efinition the velocity is in T p Σ. t [x(w(t))] t=0 = x u (w 0 )u (0) + x v (w 0 )v (0). 11
12 Sum /3/12 23:06 pge 12 #12 Step2: Spn(x u, x v ) T p Σ. For ny λ, µ R : is velocity of curve in x(d). Summriging Spn(x u, x v ) = T p Σ. λx u (w 0 ) + µx v (w 0 ) = t [x(w 0 + t(λ, µ) T )] t=0 7 Lecture Vector fiels on surfce Defn: A vector fiel Z on surfce Σ is n ssignment to ech p Σ of tngent vector Z(p) to R 3 t p. A unit norml vector fiel N is vector fiel on Σ such tht t ech p Σ : N(p) = 1, N(p) T p Σ On regulr locl surfce x, smooth unit norml fiel is given by N = x u x v x u x v where s for regulr implicitly efine surfce N = f f. Exmple: The norml vector fiel of the sphere x(u, v) = N n for the grph of function z = g(x, y) it is N = f. f = ( gx, gy,1) 1+g 2 x +gy First Funmentl form of Surfce Definition: Let p : x( w 0 ) be point in regulr surfce Σ. The first funmentl form is symmetric biliner form on T p Σ efine by I p (X, Y) = X Y, X, Y T p Σ where The mtrix tht efines this biliner form is [ E F F G 7.3 Exmples Sphere Cyliner Grph Surfce of revolution E = x u x u, F = x u x v, G = x v x v. ] 12
13 Sum /3/12 23:06 pge 13 #13 8 Lecture Arclength Let z(t) = x( w), w(t) = [ u(t) v(t) Hence the rclength of z from z() to z(b) is ] be regulr curve lying in Σ. Its velocity is [ ] z u (t) = (x u x v ) (t) v (t) z (t) t = [ ] E F Exmple 1: For the sphere = F G w (t) = ( 1, 2) hence s = π 0 [1 + 4 [ sin2 t] 1/2 t. ] E F Exmple 2: For the cyliner = F G (1, 1) T hence s = 2π t = 2 2π. 8.2 Reprmetristion (E(u ) 2 + 2F u v + G(v ) 2 ) 1/2 t. [ sin 2 u [ ]. Tke w(t) = (π t, 2t), t (o, π). ]. For w(t) = (t, t) T, t (0, 2π) w (t) = We cn reprmetrise s usul: if s z(s), lying in Σ, is rclength prmetrise then its unit spee n I(z, z ) = 1 Exmple: Cone hs prmetristion x(u, v) = u cos v [ ] [ ] u sin v E F 2 0 n = F G 0 u u 2. Let w(t) = (t, 2 ln t), t (0, ). Then w = (1, 2 t ) n hence s(t) = t 2 = 2t so rc length prmetristion is z 1 (s) = x(w 1 (s)), where w 1 (s) = ( s 2, 2 ln( s 2 ))T Isometry Defn: If S 1 n S 2 re surfces, smooth mp φ : S 1 S 2 is clle locl isometry if it tkes ny curve in S 1 to curve of the sme length in S 2. If locl isometry φ exists we sy S 1 n S 2 re loclly isomorphic. Theorem: A smooth φ : S 1 S 2 is locl isometry if n only if the mtrices of first funmentl forms re equl. [ ] E F Exmple: Plne (u, v, 0) n cyliner (cos u, sin u, v) re isomorphic sonce = F G 8.4 Geoesics [ Motivtion Look for nlogue in surfce Σ of stright lines in plne. Stright lines re chrcterise by: zero ccelertion n istnce minimising between two points. Definition: A curve z in Σ R 3 is geoesic of Σ if its ccelertion z is lwys norml to Σ. ]. Since z is norml to Σ we get z z (s z is tngent to Σ) thus z 2 z = const. Hence we conclue tht geoesics hve constnt spee. t = 2 z z = 0 implying tht 13
14 Sum /3/12 23:06 pge 14 #14 9 Lecture 19 Let us fin the geoesics of the sphere n cyliner. Sphere: First consier the sphere S of rius r, x x = r 2. If z is geoesic then z (t) is colliner to the norml of S t p = z(t) i.e. to z(t) since z is colliner to the norml. Using this observtion we compute t ( z z) = z z + z z. Clerly z z = 0. But z is colliner to the norml, by the efinition of geoesic, n z is colliner to the norml hence z z =. Combining we get t ( z z) = 0. This ientity, in prticulr, implies tht the plne Π pssing through 0 n forme by the vectors z n z hs constnt norml N (iepenent of t). Since z Π it implies tht Thus the geoesics re the gret circles. N z(t) = 0. Cyliner: It is simple exercise. See lso Lecture Distnce minimistion Theorem: Let Σ = x(d) be regulr, locl surfce n z = x w regulr curve connecting A = z(), B = z(b). Then the rclength of z J[ z] = is sttionry implies tht z is geoesic. ( z (s) z (s)) 1/2 s We wnt to show tht the ccelertion z is colliner to n. Let z be curve for which the minimum is relise. Without loss of generlity we my ssume tht z is of spee one, i.e. z z = 1 or equivlently E(u ) 2 + 2F u v + G(v ) 2 = 1 where z(t) = x w(t) n w(t) = (u(t), v(t)), t (, b) is curve in omin D. Introuce z(t, ε) = x(w + εh) where ε > 0 is smll n h() = h(b) = 0 thus z(t, ε) leves the points A, B on the surfce unchnge. Using Tylor s expnsion in ε t ε = 0, it is esy to see tht z(t, ε) = z(t, 0) + ε z ε (t, 0) + ε2 (... ) By efinition z ε (t, ε) = x x(w + εh) = ε u h 1 + x v h 2 where h = (h 1, h 2 ). Hence t ε = 0 we hve z ε (t, 0) = x u(w(t))h 1 (t) + x v (w(t))h 2 (t)- liner combintion of prtil velocities hence z ε (t, 0) T pσ. Denote this tngent vector by T = z ε (t, 0) = x u(w(t))h 1 (t) + x v (w(t))h 2 (t). Thus z(t, ε) = z + εt + ε 2 (... ) n z (t, ε) = z (t) + εt + ε 2 (... ). It follows tht z (t, ε) z (t, ε) = [ z (t) + εt + ε 2 (... )] [ z (t) + εt + ε 2 (... )] = z (t) z (t) + 2 z(t) T + ε 2 ( ) therefore 14
15 Sum /3/12 23:06 pge 15 #15 ε z (τ, ε) z (τ, ε)τ = z (τ, ε) z ε 2T z (τ, ε) + ε(... ) (τ, ε)τ = 2 z (τ, ε) z (τ, ε) t ε = 0 we get T z z (τ) z (τ) = 0 Since z is of spee one, i.e. z (τ) z (τ) = 1, we conclue fter integrtion by prts 0 = T z τ = [T(b) z (b) T() z()] T z τ = T z τ for T(b) = T() = 0 (recll tht T = x u (w(t))h 1 (t) + x v (w(t))h 2 (t) n h() = h(b) = 0). Decompose vector z = z n + z T where z n is colliner to norml n n z T T pσ-the tngentil component. Then 0 = T z = T [ z n + z T ] = T z T Since T T p Σ is n rbitrry tngent vector, for h is rbitrry, we my choose T = z [ z T ] 2 τ = 0 thus the tngentil component of z vnishes n thus z is colliner to the norml n. 10 Lecture Energy The energy functionl of curves in Σ is E[ z] = where we set g( z (τ)) = z (τ) z (τ). g( z (τ))τ = z (τ) z (τ)τ T so tht The erivtion of this sttement is similr to the one for the rc length function J (see lecture 19). As bove we tke z(t, ε) = x(w(t) + εh(t)). Then ε thus t ε = 0 we hve tht g( z (τ, ε))τ = z g( z (t, ε)) z g( z (τ)) T τ = 0. z(τ, ε)τ ε Noting tht z g( z ) = z we conclue, fter integrtion by prts tht 0 = z T = z T hence the tngentil component of the ccelertion z vnishes n z is colliner to the norml n. In other wors if E[ z] is sttionry on curves then τ ( g) is norml to Σ. Now g = 2 z hence E sttionry lso implies tht z is norml to Σ i.e. extremls of E re lso geoesics of Σ. Remrk: For the rclength we hve to impose the constrint z = 1 otherwise we get the trces of geoesics. Plus the wkwr term. For E the constnt spee conition comes utomticlly n no wkwr terms re present. 15
16 Sum /3/12 23:06 pge 16 # Surfces of revolution The question of fining the geoesics of given surfce my be very complicte. However if some itionl properties of surfce is known it cn be reuce to system of ODE s which my be solve explicitly. Such exmple is surfce of revolution. For convenience we tke (θ, z) s inepenent vribles x(θ, z) = R(z) cos θ R(z) sin θ z Here z is the height of the point n R(z) > 0 is given function. The prtil velocities n the first funmentl form re x θ (θ, z) = R(z) sin θ R(z) cos θ 0, x z (θ, z) = R (z) cos θ R (z) sin θ 1 E = R 2, F = 0, G = (R ) hence the energy functionl is E = R 2 (θ ) 2 + (R 2 + 1) z 2 Consier the Euler-Lgrnge equtions (remember tht the unknown functions re θ(t), z(t)) t (2R2 θ ) = 0 t (2(R 2 + 1)z ) z (R2 (θ ) 2 + (R 2 + 1) z 2 ) = 0 Unlike the first eqution the secon one is very complicte. Inste of simplifying it we concentrte on the first one n R 2 (θ ) 2 + (R 2 + 1) z 2 = 1 (since the spee of geoesic is constnt). Thus t (2R2 θ ) = 0 R 2 (θ ) 2 + (R 2 + 1) z 2 = 1 The first eqution mens tht R 2 θ = C, C is constnt. If C = 0 then θ = k-constnt giving the meriins s geoesics. Notice tht z = 0 is not necessrily geoesic. If C 0 then θ = C R 2, then fter iviing the secon eqution by θ 2 we get ( ) z R 2 + (R ) θ = 1 (θ ) 2 = R4 C 2 or equivlently ( ) z 2 [ ] R 4 θ = C 2 1 R2 (R 2 + 1). 16
17 Sum /3/12 23:06 pge 17 #17 Now ssume tht z is function of θ, i.e. in the omin D of (θ, z)-coorintes, z cn be represente s grph over θ-xis, then by chin rule we hve or equivlently z t = z θ θ t θ = z θ fter substituting into the eqution bove results z ( ) 2 [ ] z R 4 = θ C 2 1 R2 (R 2 + 1). This is n ODE with unknown function z = z(θ) n ny solution of this eqution represents geoesic in (θ, z)-plne (the omin D) Cyliner An interesting cse is when R(z) = ρ-constnt which correspons to the cyliner. Then R = 0 n the ODE is ( ) 2 z = ρ2 [ ρ 2 θ C 2 C 2] = A Using the first Euler-Lgrnge eqution we euce θ = C ρ -constnt thus θ(t) = C 2 ρ t + D where D is n 2 rbitrry constnt. Thereby z = Aθ + B = AC ρ t + AD + B where A, B re constnts. Thus 2 ρ cos( C ρ t + D) 2 z(t) = ρ sin( C ρ t + D) 2 t + AD + B AC ρ 2 is the generl form of the geoesic for cyliner. It is helix if C 0, A 0, verticl line if C = 0, A 0 n circle if C 0, A = 0. 17
Section 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationEULER-LAGRANGE EQUATIONS. Contents. 2. Variational formulation 2 3. Constrained systems and d Alembert principle Legendre transform 6
EULER-LAGRANGE EQUATIONS EUGENE LERMAN Contents 1. Clssicl system of N prticles in R 3 1 2. Vritionl formultion 2 3. Constrine systems n Alembert principle. 4 4. Legenre trnsform 6 1. Clssicl system of
More information1. Review. t 2. t 1. v w = vw cos( ) where is the angle between v and w. The above leads to the Schwarz inequality: v w vw.
1. Review 1.1. The Geometry of Curves. AprmetriccurveinR 3 is mp R R 3 t (t) = (x(t),y(t),z(t)). We sy tht is di erentile if x, y, z re di erentile. We sy tht it is C 1 if, in ddition, the derivtives re
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationMath 32B Discussion Session Session 7 Notes August 28, 2018
Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls
More informationMath 211A Homework. Edward Burkard. = tan (2x + z)
Mth A Homework Ewr Burkr Eercises 5-C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z =
More informationCurves. Differential Geometry Lia Vas
Differentil Geometry Li Vs Curves Differentil Geometry Introduction. Differentil geometry is mthemticl discipline tht uses methods of multivrible clculus nd liner lgebr to study problems in geometry. In
More informationSturm-Liouville Theory
LECTURE 1 Sturm-Liouville Theory In the two preceing lectures I emonstrte the utility of Fourier series in solving PDE/BVPs. As we ll now see, Fourier series re just the tip of the iceerg of the theory
More informationLECTURE 3. Orthogonal Functions. n X. It should be noted, however, that the vectors f i need not be orthogonal nor need they have unit length for
ECTURE 3 Orthogonl Functions 1. Orthogonl Bses The pproprite setting for our iscussion of orthogonl functions is tht of liner lgebr. So let me recll some relevnt fcts bout nite imensionl vector spces.
More information3.4 Conic sections. In polar coordinates (r, θ) conics are parameterized as. Next we consider the objects resulting from
3.4 Conic sections Net we consier the objects resulting from + by + cy + + ey + f 0. Such type of cures re clle conics, becuse they rise from ifferent slices through cone In polr coorintes r, θ) conics
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationConservation Law. Chapter Goal. 6.2 Theory
Chpter 6 Conservtion Lw 6.1 Gol Our long term gol is to unerstn how mthemticl moels re erive. Here, we will stuy how certin quntity chnges with time in given region (sptil omin). We then first erive the
More informationOverview of Calculus
Overview of Clculus June 6, 2016 1 Limits Clculus begins with the notion of limit. In symbols, lim f(x) = L x c In wors, however close you emn tht the function f evlute t x, f(x), to be to the limit L
More information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
More informationMath 6455 Oct 10, Differential Geometry I Fall 2006, Georgia Tech
Mth 6455 Oct 10, 2006 1 Differentil Geometry I Fll 2006, Georgi Tech Lecture Notes 12 Riemnnin Metrics 0.1 Definition If M is smooth mnifold then by Riemnnin metric g on M we men smooth ssignment of n
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More information1.3 The Lemma of DuBois-Reymond
28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBois-Reymond We needed extr regulrity to integrte by prts nd obtin the Euler- Lgrnge eqution. The following result shows tht, t lest sometimes, the extr
More informationIntroduction to Complex Variables Class Notes Instructor: Louis Block
Introuction to omplex Vribles lss Notes Instructor: Louis Block Definition 1. (n remrk) We consier the complex plne consisting of ll z = (x, y) = x + iy, where x n y re rel. We write x = Rez (the rel prt
More informationPHYS 4390: GENERAL RELATIVITY LECTURE 6: TENSOR CALCULUS
PHYS 4390: GENERAL RELATIVITY LECTURE 6: TENSOR CALCULUS To strt on tensor clculus, we need to define differentition on mnifold.a good question to sk is if the prtil derivtive of tensor tensor on mnifold?
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More information4. Calculus of Variations
4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the
More informationWhen e = 0 we obtain the case of a circle.
3.4 Conic sections Circles belong to specil clss of cures clle conic sections. Other such cures re the ellipse, prbol, n hyperbol. We will briefly escribe the stnr conics. These re chosen to he simple
More informationIf we have a function f(x) which is well-defined for some a x b, its integral over those two values is defined as
Y. D. Chong (26) MH28: Complex Methos for the Sciences 2. Integrls If we hve function f(x) which is well-efine for some x, its integrl over those two vlues is efine s N ( ) f(x) = lim x f(x n ) where x
More informationLecture 3: Curves in Calculus. Table of contents
Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up
More informationdf dt f () b f () a dt
Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem
More informationdt. However, we might also be curious about dy
Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationHomework Assignment 5 Solution Set
Homework Assignment 5 Solution Set PHYCS 44 3 Februry, 4 Problem Griffiths 3.8 The first imge chrge gurntees potentil of zero on the surfce. The secon imge chrge won t chnge the contribution to the potentil
More informationSection 6.3 The Fundamental Theorem, Part I
Section 6.3 The Funmentl Theorem, Prt I (3//8) Overview: The Funmentl Theorem of Clculus shows tht ifferentition n integrtion re, in sense, inverse opertions. It is presente in two prts. We previewe Prt
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationNavigation Mathematics: Angular and Linear Velocity EE 570: Location and Navigation
Lecture Nvigtion Mthemtics: Angulr n Liner Velocity EE 57: Loction n Nvigtion Lecture Notes Upte on Februry, 26 Kevin Weewr n Aly El-Osery, Electricl Engineering Dept., New Mexico Tech In collbortion with
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationTHREE-DIMENSIONAL KINEMATICS OF RIGID BODIES
THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES 1. TRANSLATION Figure shows rigid body trnslting in three-dimensionl spce. Any two points in the body, such s A nd B, will move long prllel stright lines if
More information1. Gauss-Jacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. Guss-Jcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationChapter 3. Vector Spaces
3.4 Liner Trnsformtions 1 Chpter 3. Vector Spces 3.4 Liner Trnsformtions Note. We hve lredy studied liner trnsformtions from R n into R m. Now we look t liner trnsformtions from one generl vector spce
More informationBest Approximation in the 2-norm
Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationMath Advanced Calculus II
Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationChapter Five - Eigenvalues, Eigenfunctions, and All That
Chpter Five - Eigenvlues, Eigenfunctions, n All Tht The prtil ifferentil eqution methos escrie in the previous chpter is specil cse of more generl setting in which we hve n eqution of the form L 1 xux,tl
More informationElementary Differential Geometry: Lecture Notes. Gilbert Weinstein
Elementry Differentil Geometry: Lecture Notes Gilbert Weinstein Contents Prefce 5 Chpter 1. Curves 7 1. Preliminries 7 2. Locl Theory for Curves in R 3 8 3. Plne Curves 10 4. Fenchel s Theorem 15 Exercises
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More informationNotes on the Eigenfunction Method for solving differential equations
Notes on the Eigenfunction Metho for solving ifferentil equtions Reminer: Wereconsieringtheinfinite-imensionlHilbertspceL 2 ([, b] of ll squre-integrble functions over the intervl [, b] (ie, b f(x 2
More informationUS01CMTH02 UNIT Curvature
Stu mteril of BSc(Semester - I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT- 1 Curvture Let f : I R be sufficiently
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More information440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationNote 16. Stokes theorem Differential Geometry, 2005
Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationdx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.
Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd
More informationCalculus of Variations
Clculus of Vritions Com S 477/577 Notes) Yn-Bin Ji Dec 4, 2017 1 Introduction A functionl ssigns rel number to ech function or curve) in some clss. One might sy tht functionl is function of nother function
More informationBasics of space and vectors. Points and distance. Vectors
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
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationMUST-KNOW MATERIAL FOR CALCULUS
MUST-KNOW MATERIAL FOR CALCULUS MISCELLANEOUS: intervl nottion: (, b), [, b], (, b], (, ), etc. Rewrite ricls s frctionl exponents: 3 x = x 1/3, x3 = x 3/2 etc. An impliction If A then B is equivlent to
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationSurface Integrals of Vector Fields
Mth 32B iscussion ession Week 7 Notes Februry 21 nd 23, 2017 In lst week s notes we introduced surfce integrls, integrting sclr-vlued functions over prmetrized surfces. As with our previous integrls, we
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationAn instructive toy model: two paradoxes
Tel Aviv University, 2006 Gussin rndom vectors 27 3 Level crossings... the fmous ice formul, undoubtedly one of the most importnt results in the ppliction of smooth stochstic processes..j. Adler nd J.E.
More informationIntroduction to the Calculus of Variations
Introduction to the Clculus of Vritions Jim Fischer Mrch 20, 1999 Abstrct This is self-contined pper which introduces fundmentl problem in the clculus of vritions, the problem of finding extreme vlues
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm-6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More informationTHE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem
More informationRecitation 3: Applications of the Derivative. 1 Higher-Order Derivatives and their Applications
Mth 1c TA: Pdric Brtlett Recittion 3: Applictions of the Derivtive Week 3 Cltech 013 1 Higher-Order Derivtives nd their Applictions Another thing we could wnt to do with the derivtive, motivted by wht
More informationMath 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008
Mth 520 Finl Exm Topic Outline Sections 1 3 (Xio/Dums/Liw) Spring 2008 The finl exm will be held on Tuesdy, My 13, 2-5pm in 117 McMilln Wht will be covered The finl exm will cover the mteril from ll of
More informationLecture XVII. Vector functions, vector and scalar fields Definition 1 A vector-valued function is a map associating vectors to real numbers, that is
Lecture XVII Abstrct We introduce the concepts of vector functions, sclr nd vector fields nd stress their relevnce in pplied sciences. We study curves in three-dimensionl Eucliden spce nd introduce the
More informationFINAL REVIEW. 1. Vector Fields, Work, and Flux Suggested Problems:
FINAL EVIEW 1. Vector Fields, Work, nd Flux uggested Problems: { 14.1 7, 13, 16 14.2 17, 25, 27, 29, 36, 45 We dene vector eld F (x, y) to be vector vlued function tht mps ech point in the plne to two
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationMatrix & Vector Basic Linear Algebra & Calculus
Mtrix & Vector Bsic Liner lgebr & lculus Wht is mtrix? rectngulr rry of numbers (we will concentrte on rel numbers). nxm mtrix hs n rows n m columns M x4 M M M M M M M M M M M M 4 4 4 First row Secon row
More informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationVector Functions. Exercises Chapter 13 Vector Functions
33 Chpter 13 Vector Functions 13 Vector Functions shown for t between n 2π Both strt n en t the sme point, but the first helix tkes two full turns to get there, becuse its z coorinte grows more slowly
More informationSolution to HW 4, Ma 1c Prac 2016
Solution to HW 4 M c Prc 6 Remrk: every function ppering in this homework set is sufficiently nice t lest C following the jrgon from the textbook we cn pply ll kinds of theorems from the textbook without
More informationLecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature
Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics - A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the
More informationMath 142: Final Exam Formulas to Know
Mth 4: Finl Exm Formuls to Know This ocument tells you every formul/strtegy tht you shoul know in orer to o well on your finl. Stuy it well! The helpful rules/formuls from the vrious review sheets my be
More informationNumerical Linear Algebra Assignment 008
Numericl Liner Algebr Assignment 008 Nguyen Qun B Hong Students t Fculty of Mth nd Computer Science, Ho Chi Minh University of Science, Vietnm emil. nguyenqunbhong@gmil.com blog. http://hongnguyenqunb.wordpress.com
More information1.1 Functions. 0.1 Lines. 1.2 Linear Functions. 1.3 Rates of change. 0.2 Fractions. 0.3 Rules of exponents. 1.4 Applications of Functions to Economics
0.1 Lines Definition. Here re two forms of the eqution of line: y = mx + b y = m(x x 0 ) + y 0 ( m = slope, b = y-intercept, (x 0, y 0 ) = some given point ) slope-intercept point-slope There re two importnt
More informationProblem set 1: Solutions Math 207B, Winter 2016
Problem set 1: Solutions Mth 27B, Winter 216 1. Define f : R 2 R by f(,) = nd f(x,y) = xy3 x 2 +y 6 if (x,y) (,). ()Show tht thedirectionl derivtives of f t (,)exist inevery direction. Wht is its Gâteux
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationCHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS LEARNING OBJECTIVES After stuying this chpter, you will be ble to: Unerstn the bsics
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationPDE Notes. Paul Carnig. January ODE s vs PDE s 1
PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationCalculus of Variations: The Direct Approach
Clculus of Vritions: The Direct Approch Lecture by Andrejs Treibergs, Notes by Bryn Wilson June 7, 2010 The originl lecture slides re vilble online t: http://www.mth.uth.edu/~treiberg/directmethodslides.pdf
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationThings to Memorize: A Partial List. January 27, 2017
Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More information