Giáo trình vi phân tích hàm nhiều biến nguyễn cao đạt

Ldi teaHIEU TRUdNG Trifcrng Dai Hoc Dan Lap Guru Long Tritdng Dai Hoc Dan Lap Cilu Long trong nd Inc thnc hien nhiem vu mot trung tam dao tao cii nhan, ky sU chink quy cd chat luqng coo,

(Litu hanh noi bo)

2002

BO GIAO DUO VA DAO TAO

TRUONG DAI HOC DAN LAP CUU LONG

— oOo—

TKb'vi£N I NGUYEN CAO DAT - TRAN NGOC HAN

VO PHUfdC HAU - DINH NGOC THANH

Ldi tea

HIEU TRUdNG Trifcrng Dai Hoc Dan Lap Guru Long

Tritdng Dai Hoc Dan Lap Cilu Long trong nd Inc thnc hien nhiem vu mot trung tam dao tao cii nhan, ky sU chink quy cd chat luqng coo, mot trung tam gdn dao tao vdi nghien cilu ling dung khoa hoc ky thuqt tien tien, mot trung tam truyen ba khoa hoc trong viing Dong Bang Song Cdu Long.

Mot trong nhilng you to trong yeu la doi ngu gidng vien cua TrUdng vdi trinh do, kinh nghiem gidng day, uy tin khoa hoc cua ho cung vdi de citong, gi-do trinh dnoc soati thdo theo chitang trinh quy dinh cua Bq Gido Due vd Dao Tao.

An phdm nay la gido trinh ndm hoc 2001-2002 cua

Bq mdn Loan - Tin hoc, Ban Khoa Hoc Co Ban, Tritdng Dai Hoc Dan Lap Cilu Long.

Dai Hoc Dan Lap Cilu Long in gido trinh nay de sinh vien nhd tritdng doc, nghien cilu.

GS NGUYfiN CONG BINH.

Trong chifOng nay, chung ta khao sat khong gian lR n, vdi

n > 1, nhtflamot tap hop cac diem va cung lakhong gian cac vectcf,

lam co sb d£ khao sat phep tinh vi tich phan ham nhieu bien Trtrdchet, khong gian IRn difOc gidi thieu tong quat cimg cac tap con dftcbiet cua nd Sau do, mat phAng IR2 va khong gian IR3 se dtfqc khaosatchi tiet voi cac khai niem can ban ve die d’ng va mdt.

Chtfcmg 1 Khong gian IR7'

AtAB = {{ta 1 + (l-t)b 1 ,ta2 + (l-t)b2, ,tan + ('i t)bn ) | t 6 IR}

vamotdoan thing, ky hieu AB, vdi

AB = {(taid- (1 — t)6i,ta2 + (1 —t)&25 tan + (1 — t)^n) | 0 < t < 1}

y2(a 2- - 6i)21=1

con ddc biet cua IRA

Diem Mt (tai d- (1 — t)6i, ta2 d~ (1 — t)&2i •••» t^n d- (1 — t)&n), voi

0 < t < 1, nim trong doan AB va di chuyen tif B (ling vdi t = 0) d£n A (ting vdi t = 1) Ngiroc lai, diim Af'(s6i + (1 - s)ai,s62 +

(1 - s)a2,sbn d- (1 - s)an) vdi 0 < s < 1 cung nim trong doan AB

nhifng di chuyen tir A (ting vdi s = 0) den B (tfrtg vdi s = 1)

(Hi) Qua cdu - mat cau Ung vdi mot diem A(ai, ,an) va mot so thifc r > 0 Ta goi qua cdu dong tam A ban kind r, ky hieu

B'(A, r) la tap hop nhu’ng diim M(zi,xn) sao cho khoang each tif

M den A nhd hon hay bing ban kinh r, nghia la,

{M^xi,x n) | MA2 < r2}

/ M(zi, ,zn) ^2(xi - aj2 < r2 >

(ii) 0 Mot o trong IRA la tich cua n khoang (bi chan) trong

IR Neu tit ca cac khoang la khoang dong, 6 tirong dug difOc goi la

o dong. Neu tat cicac khoang la md, 6 tirong dng du’oc goi la 6 mA.

B(A,r)

S(A,r)

0 va qua cau trong 7R3

trong do A(ai,an), B(6i,&n), P(pi,Pn) va Q(gi,Q

n)-Trong 7R2, mat cau chinh la vong iron va qua cau chinh la

dia iron (mien trong mat phaug gidi han bdi vong tron, chtra ca vong tron hay khong tiiy qua cau nay la dong hay md)

Ttrong tif, ta cd qua cau md B(A,r) va mat cau S(A, r) tam

A ban kinh r,

1.2 Khong gian vectcf

Mot dogn thdng dinb hudng AB, voigoc A va ngon B ch&nghan, dirqc goila mot vector biiqc, ky hieu AB c\tAB du’pc goi laj/m cua vecto AB vado dai doan thAng AB du’Oc goi la modun cua AB,

ky hieu | AB | Nhu’ vay, ting vdi mdi cap di£m A, B, ta cd hai vecto

AB va BA.

Hai vecto AB va PQ diroc goi la bang nhau neu

Chifcfng 1 Khong gian lRn

bi - ai = q, - Pi Vi = 1, 2,n

= r2 ai) 2

n M^} , ,xn) ^Xi-ai)2<r 2

i=l n M{x{ , ,x n ) ^(xi

Chifdng 1 Khong gian IRn

8

dt AB

PQ

a 4- b (ai 4- bi, az + b2,an 4- 6n)

Vd’i haivectcra = (ctj,an), b = (6b 6n) va vdi k e IR, tadinh nghia cac phep toan co ng, nhan vd hitdng va tick vo hit dng cua

chung nhtrsau :

a = (6j — <z j, 62 — 0,2, ,b n

bi = kcii Vi = 1,2, ,n

va ta viet a|| b Hem nfta, ne'11 k > 0, ta ndia, b cung htfrfng, ky hieu

a b Trtfcmg hap k < 0, ta noi a, b ngitefe hitting va viet a b Ngoai ra, ta quy irdcrang vecta 0 cung hirdng vdi moi vecto

Nhir vay, hai vecto se b&ng nhau khi chiing cd ciing phircmg, cung hu’d’ng va cd modun bang nhau Do do, cac vecto’ ttf do cd the bi6u diSn bang vecto’ buoc cd gdc tai mot diem cho trirdc nen ta cd the dong nhat tap hefp cac vecto tn’ do vtii tap hop IR n bang eachdong nhat mdi vectoa vdi diem A saocho a =OA.

~ Gn)

Khi do, (&i — <11,62 ~ ^2, — ^n) dtrqc goi la toa do cua

cac vecto AB, PQ ma ta goi chung la mot vecto (tit do) a va ta viet

Dac biet mot vecto c<) gdc va ngon trung nhau, nghia la vecto

cd modun bang 0, difOc goi la vecto khong, ky hieu la 0

Hai vectoa = (ai, , un),b = (61, ,6n) du’oc goi la song song(hay cung phitong) neu cd mot sothtfc k sao cho

a

Tif do, ta dinh nghiti goc 0 < 0 <

Tap hop cac vecto vdi cac phep cong va nhan vo hu’dng nhif tren cd cau true dai so cua mot khong gian vector nghia la

tv tao bdi hai vecto a va b

(ii) a.b

(iii) a.a = |a|

Ngoai ra do bat ding thifc Cauchy-Schwarz, ta cd

(vi) (h 4- &).a = h.a 4- k.a

(vii) A.(a 4- b) = h.a + A.b

(viii) l.a = a

vdi moi vecto’ a, b, c va cac so thu’e A, k.

Ti'ch vo htfdng cd cac tinh chat (vdi moi vecto a, b, c)

(i) a.(b 4- c) = a.b 4- a.c

b.a

(A.di,k.<7.2, h.an)

52 a-ibi

1=1

(i) a l| b 3k E JR,a = A?.b.

HAM NHIEU BIEN

1.3 Ham so theo nhieu bien so

Mot anh xa

Cuoi ciing, xet n vecttf e^, i — 1, 2, , n, troiig do tpa dp thii

i cua ei bang 1, cac tpa dp khac bang 0, nghia la

(1,0, ,0)(0,1,

11(1,0, ,0) + ®2(0,l, ,0) + + a:„(0,0,1)

n

52

i=l

va ta gpi {ei,e2, , en} la ccf sd chink tdc cua JR n

bang cong thu’c

a.bHJb| =

Ta cd cac tinh chat can ban ve vecto tren IR n nhu’san :

(i) ||x|| > 0,Vx 6 vd ||x|| = 0 x = O (nhde lai rang O

la didm cd tat cd cac to a do bdng 0 ma ta con goi Id gdc to a do)

/(x) = ||x|| =

\i=l

Theo thoi quen, ngudi ta thudng it quan tain den mien anh I

va trong mot so tru’bng hop ngu'di ta cung khdng de y den mien xac dinh D ma chi quan tarn den bieu thu’c /(aq,x n ) Khi do, mien

anh duqc ngam hieu la ca IR va mien xac dinh cua f du’qc ngam hieu

la tap con D Ion nhat cua lR n sao cho bieu thhc /(zi,cd nghia.

Cu the vdi n = 2, 3, ta cd

\^-y-n = 3 :■ f(x,y,z) = ||(z, y, z)|| = x2 y 2 + z2

X y

{y C- 7Rn|</(x,y) < r}

{y G^K(x,y) <r}

{y e!Rn\d^,y)= r}

(b) Ham phep chieu : lacac anh xa

(Hi) ||x + y|| < ||xjj + ||y||, Vx, y 6 IR"

Tinh chat (iii) du’o’c goi la ”Bat ddng thidc tam gidc” va dtftfc chiing minh clu’a vao Bat dang thitc Cauchy-Schwarz

Ngoai ra, vdi hai diem x,y £ thi <Z(x, y) = ||x - y|| chinh

la khodng each giifa x va y Qua cau cung nhu’ mat cau tam x G JR n

ban ki'nh r > 0 dittfc viet lai theo ky hieu khoang each nhu’ sau

Chu’ cfng I Khong gian lR n

f ; IR n

xy

x 2 4- y2du’o’c ngam hieu la anh xa tif D = IR 2\ {(0, 0)} vao IR

Ngoai ra, ta cd the xac dinh them gia tri cua f tai diem (0, 0)

de nhan du’Oc mot ham so xacdinh tren toan khong gian.ZR2 Ching

trong do x — (zi,zn), 1 < i < n f du’Oc goi la phep chieu xuong

thanh phan thd i va thu’bng du’Oc ky hieu la

Cu thi vdi n = 2, ta cd cac phep chieu

IR

13han, ta cd ham so

1.4 Ham vecto’ theo nhieu bi£n so.

Ta cd hai tru’dng hop dac biet quan trong sau :

0 '

Mot anh xa f : D C IRn -> IRm dtfqc goi la mot hdrn vecto1

theo n bien so thu'e hay van tat la ham vecto Bay gib, tfng vdi mdi

x 6 D, f(x) dtfqc xac dinh bang m thanh phan cua nd ma ta lan Itfqt

ky hieu la fa (x), AW, ■•-, fm (x) Mdi mot fa, 1 < i < m, la mot ham

so thtfc theo n bien so thtfc va dtfqc goi la ham thanh phan thd i cua

f va ta viet

1.5 Phiftfng trinh tham so cua difc/ng trong JRn.

Nhac lai rang mot du’d’ng thing d du’Oc xac dinh bang hai diem

phan biet A, B Cu thi trong IR2 va 7R3, ta cd

Trong IR2 : Vdi A(xA,yA),B(xB,yB) thi difbng thing d qua

A va B cd the viet lai bang khai niem vecto la

x y

xA + t(xB - xA)

yA + t(yB - yA)

Chifcfng 1 Khong gian IRn

Tru’bng vecto’ du’qc xii dung nhieu trong vat ly va co hoc Khi

do, (^(x), , /m(x)) thu’bng dupe coi nhu la mot vecto cd diem gdc tai x Cac trtfbng vecto quan trong se du’Oc gidi thieu trong cudi chu’Ong 6.

M(x,y)Ed <=> AM||AB

x - xA

y -yA 3t G IR,

Zvj 4- Vfi.

4- Vx

z fi X

fi x

z fi x

fi x

quuq §u£>jiqd ipq quip oux ((Sp'Zp'Tu) = e Xuq)

= e £>paA oom ipA Suos Suos ba (gy7 Suoi} (Vz 1 Vfi‘Fa^y Abij)

SU014 (Vfi lVx)y uiaip 40m Bnb ip p Sui’qo Supjip ‘)Bnb Suo^

Sp-j 4- Vz

ar 3 7 z»’? +

^D‘7 4- Vx

1?P BA yi Suoio (uBqo iq Suoqq XBq UBqo iq) SuBoqq

;oui B| / op Suojo ‘ar <- I : v ‘^7 1|UB 3B3 IPA ‘^Bub Suoj,

■p Bno Guwyd up upoo bj b iou bi ‘l = |B| n£u

‘■jaiq ovq p bud Bwotnyd up jopaa bj 16S b f>439A P ^upjip Bin

ubiS Suoyyj -j

x y

X-XA y-VA X y

X

y z

Ta ndiC la mot dieting cong trong IR 2 (hay JR3) xac dinh boi phu’Ong

trinh tham so va viet

I i

J Chu'cfng 1 Khong gian IR n

- ■ •

^+rcosi

yA + rsint

t 6 I■ hay C : <

a2® - ary = a2xA - aryA

Bat a = a2,/3 = — ai va 7 = aryA — a2xA, d&ng thifc tren trcf thanh

va ta cd

y) G d <=> ax 4- fly + 7 = 0

Nhan hai ve cua cac dAng thtfc trong (1) Ian liro’t cho a 2 va -ai rbi cong ve theo ve, ta cd

1.6 Phifcmg trinh chmh t4c cua dtfcmg trong IR2.

Xet du’bng thAng d trong IR2 cho bdi phu’Ong trinh tham so

cost

sin t

t

x y z Chircfng 1 Khong gian IRn

ax 4- fly 4- 7 = 0

IM&NG BAI HQC kN UPCUl ] THU View ISfdiBg.k'o

thi C = /r(i)) I I G IR } xac dinh mot du’bng cong trong

fl?3 va ta viet

-—M

A/tt\

C : y - /(x) = 0, x E I

2 + y2+ ax + by + c = 0

1.7 Tiep tuyen vo'i difo’ng cong.

Xet dtfong cong C trong 7R2 cho boi phifo’ng trinh tham sb

Chinh xac ve mat thuat ngtf, M dirpc goi la tien ve Mq khi

khoang each MMq tien ve 0 Bay gib vdi

/(«)

9(1)

Gia sir cac ham /, g deu kha vi tai to- Goi M' la giao di6m cua

mi’a difbng thing Mq M va vbng trbn tarn Mq ban kinh 1 thi M q M' la

cosin chi phirong va cimg htrd’ng vdi ntia dirbng thing Mq M Bay gib

C cd tiep tuyen tai M q ndu M' tien v& mot trong hai diem 7,+ ,T“

khi M tien ve Mq tren C.

x y

Chifcrng 1 Khong gian IR n

va mot diem M0(/(t0),p(t0)) G C Dirbng thing M q M ndi M o vdi

diem Mg(t)) bat ky tren C duoc goi la mot cat tuyen qua M q cua C Bay gib, khi cho M chay tren C ve M q , neu cat tuyen Mq M

quay ve mot vi tri gidi han M qu thi ta ndi M qu latiep tuyen tai M q

vdi C

M

M o

nghia la, M —> Mq tren (C') va

Ttfctag til, ta cung cd

1)2 _(_ ^ g(^)-g(^o

a = (/'W^'tio))

va phu’Ong trinh tham so cua tiep tuyen Mqu la

va lan lifOt cho t tp

Vd'i i = (1,0, 0),j = (0,1,0), k = (0,0,1) la ccf sd chinh tSc cua jR3 va bang ky hieu cua dinh thtfc, ta cd

ai

x y z

a=(//(io)^/W,A'(io))

x y z

cong C trong R3

= /(i)

= g(t) t g c IR h^t)

V(5i f, g, h lacachamkha vitai £ I. Taco tiep tuyen tai Mo(/(Zo), vdi C cd vectochi phirongla

Vdi cac vecto trong IR?, ngoai phep cpng, nhSn vdi vd htrdng

va tich vo hifdng, ngirdi ta con xac dinh tich cd hifdng hai vecto

a = (d!, ao, 03), b = (&i, bz, 63), ky hieu a x b, bang bieu thifc

Chu'cfng 1 Khong gian lRn

x b

Chon diem A (hay A') sao cho

ta nhan difoc MA— a x b.

1.9 Phu’o'ng trinh mat

Tich co hitdngco cac tinh chat sau

(i) a.(a x b) = b.(a x b) = 0, nghia la a, b 1

(ii) |a x b| = |a|.|b|sin0 vdi 0 la gdc tao bdi hai vecto a,b.Hay ndi khacdi, moduncua vecto tich a x b chinh la dien tich hinh binh hanh cd hai canh la a, b

bang tri so dien tich hinh binh hanh P

Vdi mat phang P xac dinh bdi ba diem khong thang hang

d Su^qd reui ■eno ps uieip quii^ Su.o.nqd oo ecj eA '^yf 3 (Vs) !^A

(vn - Dfiyi + (vfi _ aftys

(Pa: — 3x)"i -f- (Pa: — #a:)’S

ta noi S la mot mattrong 7R3 cho bcri phuvng trinh tham so

Tirong til vol anh xa F : D C iR3 —> 1R, dat

(i) Xet mat can S tarn , y A , z A)ban kinh r > 0 trong 7R3 vdi co so chinh tac i = (1,0, 0), j = (0,1,0), k = (0,0,1) Ung vdi mdi

va ta nhan dtfpc phifOng trinh chinh tac cua P :

Tong quat, vdi cac anh xa f,g,h : D —> 7R, trong do D la mot

cua 7R2 va dat

y - va

ys - yA

yc - yA

nen phtfcmgtrinhchinh taccua S la

xaY + (y Va)2 + (z - za) 2 - r

o Mat try : S : F(x,y) =0

vdi F(x, y) laphu’etag tr'inh chinh tAc cua mot conic trong IP2

x 2 u 2 z2

o Mat non : S :— +f

-a 2 b 2 c 2

1.10 Difcfng cong trong mat

Duong cong C trong IP3 duoc goi la nam trong mat S neu

C Q S Ngudi ta thuemgdung cac duefng cong nam trong S d£ bi6u dien rung nhu khao sat mat S Ta co cac du’d’ng cong thu'b'ng diingsau :

y z

X

y z

X

y z

y(s,to) h{s,t0 )

Chtfcmg 1 Khong gian IRn

vdi I = {t e 1R | (50, i) e D} va J = {s E 1R | (s,i0) E D}, la cac

difcmg cong nam trong S.

thi vdi mdi (soJo) E D, cac dn’Ong cong

/(5o3)

g(s0,t) tel h(s 0,t)

la hai du’bng cong nftm trong S di qua M C'i va C'2 Ian lu’pt cd cacvecto tiep tuyen tai M la a va b D£ dang nhan thay rang OM-L a, b, hay ndikhac di, n = a x b ciing phu’Ong vdi OM.

y2 C

2zo

la cac dito'ng cao

la mot du’bng cong nam trong S Dirbng cong nay la phdng vi nd hoan toan nam trong mot mat phing (cu the trong trirdng hop nay

la mat phang P : z — zq = 0). TifOng ttf vd’i cac dtrong cong C : F(x,yo,z) — 0 va C : F(xn,y,z) = 0 Trtfbng hop dac biet khi

F(x, y,z) = z - f(x, y), ta ndi CZo vdi

Chu’ o'ng 1 Khong gian ]R n

lamot hyperbol nhan phtftfngj lam phirong cua true thifc khi z0 > 0

va nhan phnong i lam phtfemg cua true thifc khi z0 < 0. Khi z0 = 0,

ta nhan dtfOc hai difbng thangcat nhau

1.11 Mat phAng ti£p xue

Xet mot mat S va mot di£m Mq( xq , yo, zq) € S Neu nididircmgcong C trong mat S di qua Mq (nghiala Mq E C) deu cd mot vecto tiep tuyen vc va neu tat ca cac vectc?nay dong phdng (nghia

lachungluon song song vdi mot mat phing, chang han P) thi ta ndi

S nhan mat phAng tt qua Mo va song song vdi P lam mat phdng tiep xue va baygib phapveetd cuatt (vacung la cua F) difpc goi la phap vectcf cua S tai M q

Khi S nhan mot mat phing tiep xue tt tai Mo(xo, y0, z0) thi duoc hoan toan xac dinh bang hai vectotiep tuyen a = (di,a2, as)

C : y2 - x

o

30 Chuang 1 Khong gian lR n

( tt ): <

Vi du Vdi paraboloit hyperbolic trong vi du trtfdc

n = a x b = k

De tim mat phang tiep xuc, neu co, cua S tai Mo(0, 0,0), ta

xet hai dirbng cong

va b = (61,62,63) khong ciing phu’Ong cua hai du’bng cong trong S di qua M q Dodo, tt cd phu’Ong trinh tham so

C :

C :

thi C va C la hai dtfbng cong trong S di qua Mq Day la cac parabol trong cac mat phing xOz va yOz nen cd cac vecto tiep tuyen lan Itftft la

x y z

a=(l,0,0) = i

b = (O,l,O)=j

- x0) + n2(?/ - t/ ) + n^z - z0) = 07T : 721(2!

Do do, mat phing tiep xuc cua 5, neu cd, chinh la mat phAng

xOy Phap vecto cua S tai Mo la

va dong thdi nhan 11 = (rii, 722,723) = a x b lam mot phap vecto Ta suy ra phu’o'ng trinh chinh tac

Bai tap.

1 Chifng minh rang cac vectc? sau la khong song song

2 Cho hai vecto (cii, 6i), (a-2- b?) song song Chtfng minh rang

4 Chifng minh rKng tarn giac vdi ba dinh (1, 2, 3), (4, 0,5) va (3, 6,4)

cd mot goc vuong Tim so do hai goc con lai.

(d)(0,0, »0) va (1,1, ,1) trong IRn

5 Trong cac bai tap sau, hay tinh a +b; a - 2b; |a|, |b|; a.b; goc giifa

a, b; cac vecto don vi (vecto cd modun = 1) cd phuong a, b,

(a)(0,0,0) va (2,-1,-3) (6)(1,2) va (tt, 3)

(c)(3,5,1) va (2,3,4)

Chifcmg 1 Khong gian IRn

3 Tim khoang each giifa cac diem sau

1

Chu’ cfng 1 Khong gian IR ‘

32

e2, b = e2 + 2.e3

a)a = 2ei + e2 - 3e3 6)a = (3,-6,2)

10 Tim a.b va cosin cua goc gitfa a va b vdi a, b cho bed

(a) a = ei

(b) a = (1, -3, 0), b = ei 4- e2 + e3

7 Viet bat ding thifc trong dinh nghia qua cau dong, qua cau mo

va mat cau vdi tarn va ban kinh nhtf sau :

u)3ei + 2e2, 5e2 + e3 6)(1, 0,0), 5ei — 6e2 + 10e3

|a±b|2 = |a|2 + |b|2

6 Cho vecto a trong 7R3 hop vdi 3 true Ox,Oy,Oz ba gdc a,(3,y

Chung minh ring vecto n = (cosa)ei + (cos/3)e2 + (cos7)e3 ciing

phuong vdi vecto a va |n| = 1

9 Vdi moi vecto a 0, chung minh ring vecto u = — la vecto ciing

lal

phuong vdi a va cd modun bing 1 Tim u vdi a da cho

11 a) Su dung cong thuc [a ± b|2 = (a ± b).(a ± b), chung minh

|a ± b|2 = |a|2 ± 2a.b + |b|2

8 Tim tarn va ban kinh ciia cac mat cau thoa phuong trinh : a)2x2 + 2y 2 + 2z2 + + = 9 b)x24-y 2 + z2 — 2az = 0 c)x 2 4- y 2 4- z 2 — 2x— 4y 4- 6z = —10

a)(2,l,-7),4 d)(-l, 0,3), 5/2

b) Suy ra (Dinh ly Pythagore)

b)x 2 4- y 2 4- z

khi va chi khi a ± b

c) Sir dung cau a), chu’ng minh

c) ^(f) = (t,t2,t3,<),io-0

12 Viet phu’Ong trinh tham so cua difdng thAng

a) qua P(2, -4, -1) va song song vdi vecto (1,1,1)

b) qua P(l,2,-2),Q(-l,0,l)

c) qua (4, -1, 0) va song song vdi difdng thAng x = 2 2t, y — 2-t,z =4t

d) qua (0, —7, 0) va vuong gdc vdi cac vecto (1,2, 3), (3,4, 5)

e) qua (1, 2,1) va song song vdi Ox

|a + b|2 + |a - b|2 = 2|a|2 + 2|b|2

Chu'cfng 1 Khong gian lRn

d)^o =

la6)aei + be-2, cej + de?

17 Tinh dien ti'ch hinh binh hanh co ba dinh lien tiep la

d) g(t) = (In | sin11, In | cos

16 Ti'nh ti'ch u x v vdi u, v

a)(3)-6,l),(l,0,2)

c)(2, —3,2), ei T 2e2 + 03

19 Tim phirong trinh tham so va phtrong trinh chmh tac cua cac mat phAng Ve cac mat phing do

a) qua (0, 0, 0) va cd phap vecto (1, 0, 0)

b) song song mat xOy va qua (0, 0,1)

Chu’o'ng 2 Ham lien tuc 35

Chieo’ng 2

DAY HOI TU TRONG lRn

Nhac lai rang, mot anh xa

du’Oc g'.'i la hlot day trong III”.

Mucdich chinh cua chu’o’ng nay la nham khao sat str lien tuc

cua ham nhieu bien ciing cac tinh chat can ban cua nd Tu’Ong til

nhu’ ddi vdi ham mot bien, ta se bi£u dien khai niem gidi han va lien

tuc cua ham nhieu bien bang day hoi tu (trong IR” ).

2.1 Dinh nghia Day (x(m)) dtftfcgoila hoi tuneu cd motx

sao cho

G IRn

x : IN

||x(m) — x|| —> 0 khi m —» oo

Goi Xi(m) la toa do thu i cua x(m), vdi i = l,2, ,n Ta

cd n day sb (xi(m))mG1jv, ma ta goi la cac day thank phan cua day(x(m))mGjv- Nhu vay mot day trong 2Rn*duoc hoan toan xac dinh bang n day sothuc

-> Ui n

x(m)

Chwcfng 2 Ham lien tuc

ta nhan duoc su lien he can ban giifa day hoi tu trong IRn va day so

hoi tu nhu sau

x du’Oc goi la giRz han ciia day (x(m)) vata viet

2.2 M6nh d£ Day (x(7n))m6jv trong IRn hoi tu neu va chi neu t&'t cd cac day thanh phan (x^m))™^ deu hoi tu Khi do, x =

(®1, X2,Xn) Id gitfi han cua day (x(m)) neu vd chi neu

lim xArrt} = x^ Vi = 1, 2, , n

m—>oo

Chu'ong 2 Hani lien tuc 37

phan ky trong JR2 vi cd mot day thanh

HAM LltN TUC.

Tru’dc het, ta khao sat cac ham f : D C IR n —> IR.

2.3 Dinh nghia.

(i) f duvc goi la lien tuc tai die'm xq 6 D neu

Vdi (zq, j/o) G JR2 bat ky, tif bat dAng thiic tarn giac, ta cd

Chii y De don gian ky hieu, ta cd the viet day (x,

cho (x(m)) khi khong gay nham lAn

Ve > 0, > 0, Vx G D, ||x - x || < 6 => |/(x) - /(x0)| < e

\\{x,y) - (xq,7/0)l| < => \f(x,y) - f(xo,yo)\ < c,

(ii) f diroc goi la lien tuc tren D neu f lien tuc tai moi diem

Chitcfng 2 Ham lien tuc

38

ta cd ling vdi e > 0, chon = e th'i vdi moi (x,y) E 7R2,

||(«,z/) - (zo,vo)|| < 5 => |pri(s,7/) - pn(«o,2/o)| <

e-Vay pry la ham lien tuc tren IR2 Tu’crng tit cho ham pr2- □

2.4 Dinh ly Cho f : D C IRn —> IR va xq G D Ta cd

Chufng minh

Tom lai,

nghia la, f lien tuc tai (ajo,tvo) va do do, lien tuc tren IR2

(ii) Xet cac phep chieu pr1,7)r2 : IR2 —> IR va sii dung cac bat ding thu'c

Chiicfng 2 Ham lien tuc 39

®o?/oZo

Rd rang x

ra f lien tuc tai

(xo,yo)-3c > 0, V5 > 0,3x5 e D, ||x5 - x0|| < 6 va [/(x^) - /(x0)| > e

nghia la, /(xm) —> /(x0) khi m—> oo

Chieu ddo : Diing phanchu’ng, gia str f khong lien tuc tai x0,nghia la,