TCH PHN NG LOI 2

NI DUNGnh ngha tp ng loi 2Tnh cht tp ng loi 2Cch tnh tp ng loi 2nh l GreenTch phn khng ph thuc ng i.

NH NGHATrong mp Oxy, cho cung AB v 2 hm s P(x,y), Q(x,y) xc nh trn AB.Phn hoch AB bi cc im {A0, A2, .., An}, viA0 = A, An = B. Gi s Ak = (xk, yk), k = 0,,n.Gi xk = xk+1 xk , yk = yk+1 yk, k = 0,, n-1.

Trn cung AkAk+1, ly im Mk, xt tng tp

l tp ng loi 2 ca P, Q trn ABQuy c:ch tch phn trn chu tuyn (ng cong kn) C

TNH CHT TP NG LOI 2Tp ng loi 2 ph thuc vo chiu ng ii chiu ng i th tp i du.Nu C = C1 C2

CCH TNH TP NG LOI 2TH1: (C) vit dng tham s x = x(t), y = y(t), t1 :im u, t2: im cuiKhi tham s ha ng cong, lu v chiu ng i.

TH3: (C) vit dng x = x(y), y = c : im u, y = d : im cuiTH2: (C) vit dng y = y(x), x = a : im u, x = b : im cui

Nhc liKhi tham s ha cho cung trn, elippse, ngc chiu kim ng h l tham s tng dn, cng chiu kim ng h l tham s gim dn.

Cch tnh Tp ng loi 2 trong khng gian Cch tnh:(C) x = x(t), y = y(t), z = z(t), t1 :im u, t2: im cui

V D1/ Tnh:C l on ni t A(0,0) n B(1,1) theo cc ng cong sau y:on thng ABParabol: x = y2ng trn: x2+y2 = 2y, ly ngc chiu KH

a/ on thng AB: y = x, x : 0 1b/ Parabol: x = y2 , y : 0 1A(0, 0), B(1, 1)

c/ x2+y2 = 2y x2+(y 1)2 = 1, x = cost, y = 1+sint,A(0,0) ly ngc chiu KHB(1,1) t = 0

2/ Tnh:vi C l cung ellipse x2 + 3y2 = 3 i t (0, 1) n giao im u tin ca ellipse vi ng thng y = x, ly theo chiu KH. Ti giao im vi t y = x:

3/ Tnh:vi C l gt ca mt cu x2 + y2 + z2 = 6z v mp z = 3 - x ly ngc chiu KH nhn t pha dng trc Ozt : 0 22x2 + y2 = 9

CNG THC GREENnh ngha: Nu chu tuyn C(ng cong kn) l bin ca min D R2, chiu dng ca C l chiu m i trn , min D nm v bn tri.nh ngha: Min n lin l min m mi chu tuyn trong min ny c th co v 1 im trong min( khng cha l thng).

nh lD l min ng v b chn trong R2, C l bin nh hng dng ca D. Gi s P, Q v cc o hm ring lin tc trn D. Khi Lu : C c th gm nhiu chu tuyn gii hn min D.(Cng thc Green)

V D1/ Tnh:trong C l trnx2 + y2 = 1, ly ngc chiu KH.Gi D l hnh trn x2 + y2 1, khi C l bin nh hng dng ca D. p dng cng thc Green:

2/ Tnh:C = {(x, y)/ |x| + |y| = 1} , ly theo chiu KH.Gi D l hnh vung |x|+|y| 1.Khi C l bin nh hng m ca D. p dng cng thc Green :

3/ Tnh:C l na di t x2 + y2 = 2x, ngc chiu KH Nu tham s ha tnh I kh C khng kn nn khng th p dng ct Green.Gi C1 l on thng y = 0, x: 2 0D l na di hnh trn x2 + y2 2xC1

Khi C C1 l bin nh hng dng ca D.p dng ct Green: D

C1 : y = 0, x: 2 0

4/ Cho kim tra:Tnh :trong cc TH sau:C l tr x2 + y2 = R2, R > 0 ty .C l tr (x 3)2 + (y 1)2 = 2.C ={(x,y)/ max {|x|, |y|} =1}C l ng cong bao quanh gc ta ni t im (1, 0) n (, 0).Cc ng cong u ly ngc chiu KH.

C l tr x2 + y2 = R2, R > 0 ty .V P, Q v cc o hm ring khng xc nh ti (0, 0) nn khng th p dng cng thc Green trn hnh trn x2 + y2 R2 .

Tham s ha C:

Nhn xt: trn ng trn C, do x2 + y2 = R2, thay vo tp ta cLc ny : p dng ct Green cxc nh ti (0, 0).

C l tr (x 3)2 + (y 1)2 = 4.p dng ct Green trn hnh trn bin C

C ={(x,y)/ max { |x|, |y|} = 1}Khng th p dng ct Green trn min hnh vung (P, Q khng xc nh ti (0,0).Dng 1 ng trn C nh bao gc O (hoc 1 trn ln bao c ng cong C). p dng ct Green trn hnh vnh khn (HVK) gii hn bi C v C( hnh vnh khn s khng cha (0,0)).

Nu C l ng cong ty bao gc O? ly cng chiu KH(theo cu a)Nhn xt: khi tnh tp trong cu c) theo cch ny, khng s dng tham s ha ca c (C),

C l ng cong bao quanh gc O ni t (1,0) n (,0)Ni vo C bi C viC: y = 0, x : 1 Khi C C l ng cong kn bao gc O, p dng kt qu cu c).

TCH PHN KHNG PH THUC NG I4/ Tn ti hm U(x, y) tha: (Biu thc di du tp l vp ton phn ca U)vi mi chu tuyn trong Dkhng ph thuc ng ni A, BD l min m n lin. P, Q v cc o hm ring lin tc trn D. Cc iu sau tng ng:

p dng1. Thng thng ta s kim tra iu kin 1 hoc 4 (nu hm U c th on nhanh).2. Nu 1 hoc 4 tha, c 2 cch tnh tp t A n BC1: i ng ly tp thng thng i theo cc on thng // vi cc trc ta Lu min D

p dngC2: vi hm U trong k 4C1: Tm U t h :Ux = P, Uy = QC2: chn (x0, y0) ty trong DCch tm U:

V D1/ Tnh :C: on thng ni 2 im (1, -1), (2,1).Py = Qx trn R2 nn tp khng ph thuc ng i.

Cch khc: nhn thy hm U(x, y) = xy thadU = ydx + xdy trn R2 nnI = U(2, 1) U(1, -1) = 2 + 1 = 3

2/ Tnh :Theo ng khng ct ng thng x + y = 0x + y =0Py = Qx, (x,y): x + y 0

Hoc(tnh U): chn (x0, y0) = (1, 0)

3/ Tm cc hng s a, b sao cho tpkhng ph thuc ng i. Sau , vi a, b va tm c, tnh tp vi A(-1, 2), B(0,3).

4/ Tm hm s h(y) tha h(1) = 1 sao cho tpkhng ph thuc ng i. Sau , vi h va tm c, tnh tp vi A(-1,1), B(1,1) theo ng trn x2 + y2 = 2y, ly cng chiu KH.

theo na trn ng trni ng ly tp:chn ng thng ni A, B.