The Formulas of Vector Calculus John Cullinan

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1 The Fomuls of Vecto lculus John ullinn Anlytic Geomety A vecto v is n n-tuple of el numbes: v = (v 1,..., v n ). Given two vectos v, w n, ddition nd multipliction with scl t e defined by Hee is bief list of definitions v + w = (v 1,..., v n ) + (w 1,..., w n ) = (v 1 + w 1,..., v n + w n ) t v = t (v 1,..., v n ) = (tv 1,..., tv n ). (1) The dot poduct: v w = (v 1,..., v n ) (w 1,..., w n ) = n v i w i. i=1 (2) The length of vecto: v = v v = v v2 n. Vectos of length 1 e clled unit vectos. (3) The ngle θ between two vectos: v w = v w cos θ. Fo two vectos v, w with v, w 0, the enclosed ngle θ is hence given by n θ = cos 1 i=1 v iw i ( n i=1 v2 i )1/2 ( n i=1 w2 i )1/2 (4) Two vectos v, w e clled () pependicul o othogonl if v w = 0. We denote this by v w. (b) pllel if v w = v w. We denote this by v w. Note tht two vectos v, w e pllel if nd only if thee is scl t such tht v = t w. The zeo vecto 0 = (0,..., 0) is by definition pllel nd pependicul to evey vecto. Lines nd Plnes Let u, v be two vectos with v 0. Then u + t v = (u 1 + tv 1,..., u n + tv n ); t yields line in n. This line psses though the point u in the diection of v. Let u, v, w n be vectos whee v nd w e not pllel. Then u + s v + t w = (u 1 + sv 1 + tw 1,..., u n + sv n + tw n ) defines plne though the point u, spnned by v nd w. The notion of pependicul vecto yields nothe fom of epesenting plne in 3. Let 0 = (x 0, y 0, z 0 ) be fixed point in the plne nd n vecto pependicul to the plne. An bity point = (x, y, z) on the plne stisfies n ( 0 ) = 0. Moeove, the eqution n ( 0 ) = 0 defines plne in 3 fo ny fixed n, v 3, povided n 0. In 3, the epesenttion of line l(t) = (x, y, z) = u + t v cn be solved fo t in ech coodinte if none of the vlues v 1, v 2, v 3 e zeo: t = x u 1 v 1 t = y u 2 v 2 t = z u 3 v 3 This yields nothe epesenttion of line in 3, which is clled symmetic eqution: x u 1 = y u 2 = z u 3. v 1 v 2 v 3 The pmetic eqution epesenting plne cn be tnsfomed similly giving the stndd eqution of plne: x + by + cz = d, whee the noml vecto n is (, b, c). 1

2 oss Poduct At point p of ny plne in 3 thee is exctly one line pependicul to the plne. If v nd w spn the plne, vecto spnning this pependicul line is given by the coss poduct v w. Fo two vectos v, w 3 the coss poduct v w is defined by v w = (v 1, v 2, v 3 ) (w 1, w 2, w 3 ) = (v 2 w 3 v 3 w 2, v 3 w 1 v 1 w 3, v 1 w 2 v 2 w 1 ) Altentively, the coss poduct is given by the 3 3 deteminnt v w = v i j k 1 v 2 v. 3 w 1 w 2 w 3 v w is pependicul to both v nd w v, w nd v w e oiented ccoding to the ight hnd ule if θ is the ngle between v nd w, then v w = v w sin θ. The length of v w is equl to the e of the pllelogm spnned by v nd w. The coss poduct is (1) line: t (v w) = (t v) w = v (t w) nd u (v + w) = u v + u w. (2) nti-commuttive: v w = w v Anothe emkble popety of the coss poduct is the following: u (v w) = (u v) w. Geometiclly, u (v w) is the volume of the pllelepiped given by u, v, nd w. Sphees nd ylindes The vecto nottion llows fo n elegnt desciption of othe geometic objects such s sphees nd cylindes. Fo scl 0 nd vecto c 3, the eqution x c 2 = 2 yields sphee of dius centeed t c. Fo scl 0 nd vectos c, n 3 with n = 1, the eqution n (x c) 2 = 2 yields cylinde of dius centeed ound the line given by c + t n. Pmeteized uves in 3 Velocity, Speed, nd Acceletion Given thee diffeentible functions x, y, z : [, b], the vecto (t) = (x(t), y(t), z(t)) cn be undestood s descibing pticle moving though spce; the pticle s position depends on the (time) pmete t. This peception gives ise to the following definitions. We define the velocity of t time t to be (t) = (x (t), y (t), z (t)) speed of t time t to be (t) = [x (t)] 2 + [y (t)] 2 + [z (t)] 2 cceletion of t time t to be (t) = (x (t), y (t), z (t)). Pticles moving t constnt speed Suppose tht (t) = c fo ll t fo cuve (t). Diffeentiting (t) = c, we obseve tht (t) is othogonl to (t) fo ll t: (t) = 1 (t) (t) = 0 fo ll t. Smooth uves A cuve : [, b] 3 is clled smooth if its speed vnishes t most t the endpoints: (t) 0 fo ll t (, b). Aclength The clength l(, b) of cuve : [, b] 3 is given by l(, b) = b (t) dt A cuve : [, b] is clled pmeteized with espect to clength if (t) = 1 fo ll t [, b]. uvtue Let : [, b] 3 be pmeteized with espect to clength. Then κ(t) = (t) is clled the cuvtue of t t. Diffeentiting the eqution (t) = 1 shows tht (t) nd (t) e othogonl fo ll t. If κ(t) 0, the vecto n(t) = (t) (t) is well-defined unit vecto. 2

3 Noml nd Binoml Vectos If κ(t) 0, we cn define the Noml vecto of the cuve t time t to be n(t) Binoml vecto of the cuve t time t to be b(t) = (t) n(t) The binoml vecto is obviously unit vecto, so we cn pply the sme esoning s befoe to see tht b(t) nd b (t) e othogonl. On the othe hnd, diffeentiting b(t) = (t) n(t) we get: Hence b (t) is pllel to n(t). b (t) = (t) n(t) + (t) n (t) = (t) n (t). Tosion The eqution b (t) = τ(t) n(t) defines the tosion τ of the cuve t time t. Fenet Fme Wheneve κ(t) 0, the vectos (t), n(t), b(t) e mutully othogonl unit vectos. They spn the Fenet Fme. uvtue of uves NOT pmeteized with espect to Aclength Let : [, b] 3 be my smooth cuve. Its cuvtue κ(t) t time t is given by κ(t) = (t) (t) (t) 3. el-vlued Functions Domin nd nge A function f : n ssigns unique el vlue f(x 1,..., x n ) to ech point (x 1,..., x n ) of set D in n. The set D is clled the domin of f. The set = {f(x 1,..., x n ) (x 1,..., x n ) D} is clled the nge of f. Limits A el numbe L is sid to be the limit of f t (, b,... ) if fo ll sequences ( m, b m,... ) with lim m m =, lim m b m = b,..., the following holds: lim f( m, b m,... ) = L. m We denote this by lim f(x 1, x 2,... ) = L. (x 1,x 2,... ) (,b,... ) Equivlently, if fo evey el numbe ɛ > 0 thee is nothe el numbe δ > 0 such tht (x 1, x 2,... ) (, b,... ) < δ f(x 1, x 2,... ) L < ɛ. ontinuity A function f : n with domin D is sid to be continuous t (, b,... ) D if lim f(x 1, x 2,... ) = f(, b,... ) (x 1,x 2,... ) (,b,... ) Ptil Deivtives Let f : n be function with domin D nd (x 1,..., x n ) D. The ptil deivtive of f t (x 1,..., x n ) with espect to x i is given by the limit f f(x 1,..., x i + h,..., x n ) f(x 1,..., x i,..., x n ) = lim. x i h 0 h Diffeentibility A function f : n with domin D is clled diffeentible t (x 1,..., x n ) D if ll ptil deivtives exist nd e continuous t (x 1,..., x n ). Diectionl Deivitve Let f : n be function with domin D nd D. Suppose tht u is unit vecto in n. The diectionl deivtive of f t in the diection of u is given by d f( + tu) dt. t=0 Gdient Vecto Let f : n be function with domin D tht is diffeentible t D. The gdient vecto f of f t is given by f ( f = x 1,..., f ) x n 3

4 Diectionl Deivtives nd the Gdient Vecto Let f : n be function with domin D tht is diffeentible t D. Using the chin ule, the diectionl deivtive is given by d f( + tu) dt = f u t=0 Highe Deivtives nd liut s Theoem Let f : n be function with domin D nd suppose the ptil deivtives of f e themselves diffeentible. Then diffeentiting f x i with espect to x j is the sme s diffeentiting f x j with espect to x i : 2 f = 2 f. x i x j x j x i Globl nd Locl Extem A function f : n with domin D hs globl mximum t D if f(x) f() fo ll x D globl minimum t D if f(x) f() fo ll x D locl mximum t D if thee is disc centeed t such tht f(x) f() fo ll x locl minimum t D if thee is disc centeed t such tht f(x) f() fo ll x iticl Points Let f : D n be diffeentible. We cll point D citicl point if f = 0. If f hs n extemum t, then is citicl, but the convese is not necessily tue. Functions of Two Vibles Second Deivtive Test Let f : D 2 nd its deivtives be diffeentible nd let (x 0, y 0 ) D be citicl point of f. Let ( 2 f 2 ( f 2 ) 2) D(x 0, y 0 ) = x 2 y 2 f x y (x0,y 0) if D(x 0, y 0 ) > 0 nd 2 f x 2 < 0, then f hs mximum t (x 0, y 0 ). (x0,y 0) if D(x 0, y 0 ) > 0 nd 2 f x 2 > 0, then f hs minimum t (x 0, y 0 ). (x0,y 0) if D(x 0, y 0 ) < 0, then f hs sddle t (x 0, y 0 ). if D(x 0, y 0 ) = 0, then the second deivtive test gives no infomtion bout the ntue of the citicl point. The Double Integl Let = [, b] [c, d] nd let f : be continuous. The double integl of f ove is defined to be f(x, y) da = lim (x i x i 1 )(y j y j 1 )f(x i, yj ) P 0 whee P = P [,b] P [c,d] is ptition of, P = P [,b] P [c,d] is its nom, nd (x i, y j ) [x i 1, x i ] [y j 1, y j ]. i,j Fubini s Theoem Let f : 2 be continuous nd = [, b] [c, d]. The double integl is given by b d d b f(x, y) da = f(x, y) dydx = f(x, y) dxdy Level Sufces Let f(x, y, z) : 3 be function of thee vibles. The level set {(x, y, z) f(x, y, z) = c} fo constnt c genelly yields sufce in 3. c Tngent Plnes of Level Sufces onside the level set {(x, y, z) f(x, y, z) = c} of function f : 3. (1) The gdient vecto f(x 0, y 0, z 0 ) t point (x 0, y 0, z 0 ) is pependicul to the plne tngent to the level sufce t (x 0, y 0, z 0 ). 4 c

5 (2) The tngent plne is given by the eqution f(x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0. Using Lgnge Multiplies to Find Extem with onstints Let f : 3 be diffeentible. To find the mximum nd minimum vlue of f subject to the constint g(x, y, z) = c, the gdients f nd g must be pllel. An lgoithm to find the mximum nd minimum vlues is hence given by: (1) Find ll points (x, y, z) such tht f = λ g, fo some λ, nd g(x, y, z) = c. (2) Evlute f t these points. The lgest (smllest) vlue is the mximum (minimum) of f subject to the constint g(x, y, z) = c. The Tiple Integl Let g : 3 be continuous nd = [, b] [c, d] [e, f]. The tiple integl of g ove is defined to be g(x, y, z) dv = lim (x i x i 1 )(y j y j 1 )(z k z k 1 )g(x i, yj, zk) P 0 i,j,k hin ule Let (t) be smooth cuve in n nd let f : n be function of sevel vibles. Then df dt = f (t). Moe genelly, suppose ech of the vibles x i is function of the vibles t 1,..., t m, then f t j = n i=1 f x i x i t j hnge of Vibles Fomul in Two Dimensions Let (x(u, v), y(u, v)) be pmete tnsfomtion with Jcobin mtix ( ) x/ u x/ v J = y/ u y/ v which mps S 2 into 2. Then f(x, y) dxdy = S f(u, v) det J dudv Pol oodintes The pmete tnsfomtion (x(, θ), y(, θ)) = ( cos θ, sin θ) expesses (x, y) in pol coodintes. Then f(x, y) dxdy = f(, θ) ddθ Sufce Ae Let f : 2 be diffeentible. The sufce e of f ove D is given by [ ] 2 [ ] 2 f f dxdy x y Pth Integls Let f : 2 be diffeentible nd let (t) pmeteize smooth cuve in 2. The pth integl of f long is given by f((t)) (t) dt S 5

6 Vecto Fields A vecto field F ssigns to ech point in domin n vecto in n. Line Integls Let F be vecto field on domin n nd let be smooth cuve in pmeteized by (t). The line integl of F long is given by F((t)) (t) dt onsevtive Vecto Fields nd the Potentil Function If F = f fo some function f : 2, then F is clled consevtive vecto field with potentil function f. Line Integls of onsevtive Vecto Fields Let F = f be consevtive on domin n. Fo ny continuous cuve in fom u to b which is pmeteized by (t), we hve F (t) dt = f(v) f(u). In pticul (1) If is closed then F dt = 0. (2) if is nothe continuous cuve in fom u to v pmeteized by s(t), then F s dt = F dt. The integl is sid to be pth-independent. Line Integls of Vecto Fields in 2 Let F = (P, Q) be vecto field on domin 2 nd let be smooth cuve in given by (x(t), y(t)). Then b F(P (x(t), y(t)), Q(x(t), y(t))) (x (t), y (t)) dt = P (x, y) dx + Q(x, y) dy onsevtive Vecto Fields in 2 If F = (P, Q) = f, then by liut s theoem, P y = Q x. Geen s Theoem Let be simply connected egion with positively-oiented boundy. Then Q P dx + Q dy = x P y dxdy Stokes Theoem If S is sufce in 3 with boundy S pmeteized by nd F is vecto field in 3, then F d = ( F) ds S Divegence (Guss ) Theoem If V is compct volume in 3 with boundy V = S nd F is vecto field in 3, then ( F) dv = F ds V S V 6

13.5. Torsion of a curve Tangential and Normal Components of Acceleration

13.5. Torsion of a curve Tangential and Normal Components of Acceleration 13.5 osion of cuve ngentil nd oml Components of Acceletion Recll: Length of cuve '( t) Ac length function s( t) b t u du '( t) Ac length pmetiztion ( s) with '( s) 1 '( t) Unit tngent vecto '( t) Cuvtue: