Collection of formulas Matematikk 3 (IRF30017)

Transcription

1 ollection of formuls Mtemtikk 3 (IF30017) onic sections onic sections on stndrd form with foci on the x-xis: Ellipse: Hyperbol: Prbol: x y2 b 2 = 1, > b, foci: (±c, 0), c = 2 b 2. x 2 2 y2 b 2 = 1, foci: (±c, 0), c = 2 + b 2, symptotes : y = ±(b/)x. y = x2, focus: (0, p), directrix (styrelinje): y = p. 4p In the cse of the ellipse, is clled the semimjor xis (store hlvkse) nd b the semiminor xis (lille hlvkse). The method of Lgrnge multipliers Assume tht f(x 1,..., x n ) nd g(x 1,..., x n ) re dierentible functions nd tht g 0 when g = 0. The sttionry points of f subject to the constrint g = 0 re found by solving the n + 1 sclr equtions f = λ g, g = 0 for the n + 1 unknowns λ, x 1,..., x n. The sttionry points re cndies for locl mxim nd minim of f subject to g = 0. ouble nd triple integrls rtesin (x, y, z), cylindricl (r, θ, z) nd sphericl (ρ, φ, θ) coordintes of point P : From cylindricl to rtesin: x = r cos θ, y = r sin θ, z = z. From sphericl to cylindricl: r = ρ sin φ, θ = θ, z = ρ cos φ. From sphericl to rtesin: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. r = x 2 + y 2 is the distnce to the z xis nd ρ = x 2 + y 2 + z 2 is the distnce to the origin ( OP ). θ [0, 2π] is the polr ngulr coordinte of the projection of P on the xy-plne nd φ [0, π] is the ngle between the z-xis nd OP. Are nd volume elements: da = dx dy = r dr dθ = J(u, v) du dv, dv = dx dy dz = r dz dr dθ = ρ 2 sin φ dρ dφ dθ = J(u, v, w) du dv dw, (x, y) J(u, v) = (u, v) = x x u v (x, y, z) y y, J(u, v, w) = (u, v, w) Applictions of double nd triple integrls: Are of : A = Averge of f over : f = 1 A u v da, Volume of : V = f(x, y) da, Averge of f over : f = 1 V Object with mss density δ(x, y, z) occupying region in spce: Mss: M = M yz = δ(x, y, z) dv, enter of mss: x = M yz M, x δ(x, y, z) dv, M xz = y δ(x, y, z) dv, M xy = y = M xz M, dv f(x, y, z) dv z = M xy M, z δ(x, y, z) dv

2 Prmetric curves nd line integrls Below the following prmetriztion of curve in spce is ssumed: Tngent vector: Arc length: L = : r(t) = g(t)i + h(t)j + k(t)k, t b v(t) = dr = g (t)i + h (t)j + k (t)k, Unit tngent vector: T = v v, v = v v, b eltions between dierentils: Line integrl of sclr f(x, y, z) long : v, Arc length prmeter: s(t) = dr = dxi + dyj + dzk, dr = Tds, ds = v f(x, y, z) ds = b f(r(t)) v(t), t f(r(t)) = f(g(t), h(t), k(t)) Line integrl of vector eld F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P (x, y, z)k long : denitions {}}{ F T ds = F dr = The line integrl of the x-component of F long : M(x, y, z) dx = b Mdx + Ndy + P dz = b how to clculte {}}{ F(r(t)) v M(r(t)) dx b = M(g(t), h(t), k(t)) g (t) Nmes on line integrls: work, ow, circultion nd ux Let F be vector eld in n nd prmetrized curve in the sme spce. The line integrl is clled the F dr work done on n object moving long the curve if F is force eld ow long if F is velocity eld circultion long if F is velocity eld nd is closed curve (for closed curve the line integrl is often written ) v(t ) Flux integrl in two dimensions: Let F = Mi + Nj be vector eld nd simple closed curve in the plne ( 2 ) with unit norml n oriented outwrds. The following line integrl is the ux of F cross the : ux = F n ds = Flux integrl in three dimensions: see surfce integrls below. del, divergence nd curl el opertor: 3 : M dy N dx = i x + j y + k z, n : = The following denitions ssume tht F = Mi + Nj + P k is vector eld in spce ( 3 ), but the divergence generlizes nturlly to spce of rbitrry dimensions ( n ): ivergence of F : div F = F = M x + N y + P i j k, url of F : curl F = F = z x y z M N P Identities: ( f) = 0, ( F) = 0 n i=1 e i x i

3 onservtive elds nd pth independence The following sttements re equivlent if F is vector eld in spce whose components re continuous in connected nd simply connected domin nd is curve in the sme domin: 1. F is conservtive (this is nother wy to sy tht the integrl F dr is pth independent) 2. F is curl-free, F = 0 (this provides component test for conservtive elds, in the plne write F = Mi + Nj + 0k) 3. F is grdient eld: F = f (the function f(x, y, z) is clled potentil function for F) B 4. F dr = f(b) f(a) for ll curves from A to B A 5. F dr = 0 for ll closed curves Green's theorem Let be region in the plne bounded by the piecewise smooth, simple closed curve nd let F = M(x, y)i + N(x, y)j + 0k be vector eld with M nd N hving continuous prtil derivtives. ircultion-curl form: or Flux-divergence form: or urfce integrls F dr = Mdx + Ndy = F n ds = Mdy Ndx = F k da ( N x M y F da ( M x + N y ) dx dy ) dx dy Let be smooth surfce in spce ( 3 ). The re element dσ depends on the description of : 1) dσ = r u r v du dv if is given prmetriclly s r(u, v) = f 1 (u, v)i + f 2 (u, v)j + f 3 (u, v)k 2) dσ = G G k dx dy if is given implicitly by the eqution G(x, y, z) = 0 3) dσ = gx 2 + gy dx dy if is given explicitly s the the grph z = g(x, y) Below the cse 3) of n explicitly dened surfce is ssumed. Let be the shddow of on the xy-plne. The re of is: A = dσ = gx 2 + gy dx dy The integrl of sclr f(x, y, z) over : A surfce hs two unit norml elds: f(x, y, z)dσ = f(x, y, g(x, y)) gx 2 + gy dx dy n = ± G G = ± g xi g y j + k gx 2 + gy For given choice of n the ux of F = Mi + Nj + P k cross is: Flux = F n dσ = sgn(n k) Mg x Ng y + P dx dy

4 toke's theorem nd the divergence theorem Let be n oriented piecewise smoooth surfce in spce hving piecewise smooth boundrry curve tht is right-hnded reltive to. tokes theorem: F dr = F n dσ Let be region in spce with piecewise smooth boundrry surfce hving n outwrd unit norml eld n. ivergence theorem: F n dσ = F dv In both thorems the components of F = Mi + Nj + P k hve continuous prtil derivtives. Modeling in physics Numericl methods onsider the rst-order dierentil eqution: du = f(u, t) Let u n be numericl pproximtion to u(t n ), where t n = t 0 + n t. Euler method: Or more compctly: Use the tngent t the previous point to estimte the next: ( ) du u n+1 = u n + t = u n + f(u n, t n ) t n u n+1 = u n + f n t First order method (locl error: t 2, globl error: t). Midpoint method: Use Euler's method with hlf time step to estimte the slope t the midpoint (tril step), then pply this to estimte the next point: ( ) du u n+ 1 = u t n + 2 n 2 = u n f(u n, t n ) t, (tril step), ( ) du u n+1 = u n + t = u n + f(u n+ 1, t n + t 2 2 ) t Or more compctly: n+ 1 2 u n+ 1 2 = u n f n t, (tril step), u n+1 = u n + f n+ 1 2 t econd order method (locl error: t 3, globl error: t 2 ). Higher order dierentil equtions A second order dierentil eqution cn be rewritten s system of two coupled rst order equtions: d 2 u 2 = f ( u, du ), t I. II. du = v, dv = f (u, v, t) The numericl schemes bove cn then be pplied to nd u n+1 nd v n+1 from u n nd v n.

5 imensionless vribles An ordinry dierentil eqution for x(t) cn be written on dimensionless form by introducing length scle L nd time scle τ: d x = L x, t = τ t 2 x 2 = L d 2 x τ 2 d t, 2 where in I units im(x) = im(l) = m, im(t) = im(τ) = s nd im( x) = im( t) = 1. I bse units: m, s, kg. I derived units: N=kg ms 2 (Newton's 2nd lw), J=N m (work-energy theorem). ome solutions of selected dierentil equtions Hrmonic oscilltor eqution (ordinry, liner, homogeneous): d 2 x 2 + w2 x = 0 x(t) = A cos (wt + φ) Amplitude: A [m], ngulr frequency: w [rd/s], frequency: f = w 2π [Hz], period: T = 1 f = 2π w, phse: φ [rd]. One-dimensionl wve eqution Mechnicl wves on string: (prtil, liner, homogeneous): 1 2 y v 2 t 2 = 2 y x 2 Hrmonic wve trvelling to the right: y(x, t) = A cos (kx wt + φ), w = v k. Wve number: k [m 1 ], wve length: λ = 2π k [m]. tnding wves with boundrry conditions y(0, t) = y(l, t) = 0: y(x, t) = A sin(kx) cos(wt), w = v k, k = nπ L, n = 1, 2, 3,... One-dimensionl het eqution / diusion eqution (prtil, liner, homogeneous): u t = c2 2 u x 2 u(x, t) = A sin(kx) e (ck) 2t, k = nπ, n = 1, 2, 3,... L The solutions bove stisfy the boundrry conditions u(0, t) = u(l, t) = 0.

6 From previous courses clr product nd vector product When = 1 i + 2 j + 3 k nd b = b 1 i + b 2 j + b 3 k: i j k b = b cos α = 1 b b b 3, b = b sin α = b 1 b 2 b 3 tright line in spce Prmetriztion of line through the point P 0 (x 0, y 0, z 0 ) prllell to v = [, b, c]: r(t) = r 0 + tv = (x 0 + t)i + (y 0 + bt)j + (z 0 + ct)k, t A possible prmetriztion of stright line from r 1 to r 2 : Plne in spce r(t) = r 1 + (r 2 r 1 )t, 0 t 1 Eqution for plne through the point P 0 (x 0, y 0, z 0 ) norml to n = [, b, c]: ircle in the plne P 0 P n = 0 (x x 0 ) + (y y 0 )b + (z z 0 )c = 0 Eqution for circle with rdius nd center in (x 0, y 0 ): (x x 0 ) 2 + (y y 0 ) 2 = 2 Tylor expnsion Tylor series of function f(x) bout the point x = : f(x) = k=0 Tylor polynom of degree n: P n (x) = n k=0 f (k) () k! f (k) () (x ) k = f() + f ()(x ) + 1 k! 2! f ()(x ) (x ) k = f() + f ()(x ) + 1 2! f ()(x ) f (n) () (x ) n n! Liner pproximtion to f(x) round x = : ome trigonometric identities f(x) f() + f ()(x ) if 1 2 f ()(x ) 2 f ()(x ) sin 2 u + cos 2 u = 1, sin(u + v) = sin u cos v + cos u sin v, cos(u + v) = cos u cos v sin u sin v, sin(2u) = 2 sin u cos u, cos(2u) = cos 2 u sin 2 u, cos 2 u = (1 + cos(2u))/2, sin 2 u = (1 cos(2u))/2