n Ng Minh o hm ring

MC LC CHNG 2. O HM RING ....................................................................................... 1

2.1. Hm nhiu bin ...................................................................................................... 1 2.1.1. Hm hai bin .................................................................................................... 1 2.1.2. th ............................................................................................................... 3 2.1.3. ng mc ....................................................................................................... 5 2.1.4. Hm ba hoc nhiu bin ................................................................................... 8

2.2. Gii hn v s lin tc ........................................................................................... 9 2.2.1. Gii hn ........................................................................................................... 9 2.2.2. S lin tc ....................................................................................................... 13 2.2.3. Hm ba hoc nhiu bin ................................................................................. 14

2.3. o hm ring ...................................................................................................... 15 2.3.1. Khi nim v o hm ring ........................................................................... 15 2.3.2. ngha ca cc o hm ring ....................................................................... 17 2.3.3. Hm nhiu hn hai bin .................................................................................. 18 2.3.4. o hm cp cao ............................................................................................ 19

2.4. Mt phng tip din v xp x tuyn tnh ........................................................... 20 2.4.1. Mt phng tip din ........................................................................................ 20 2.4.2. Xp x tuyn tnh ............................................................................................. 22 2.4.3. Vi phn ........................................................................................................... 23 2.4.4. Cc hm ba hoc nhiu bin ........................................................................... 25

2.5. Quy tc dy chuyn .............................................................................................. 25 2.5.1. o hm hm n ............................................................................................ 28

2.6. o hm theo hng v vc t gradient ............................................................. 29 2.6.1. o hm theo hng ...................................................................................... 29 2.6.2. Vc t gradient ............................................................................................... 31 2.6.3. Hm ba bin ................................................................................................... 32 2.6.4. Cc i ca o hm theo hng .................................................................... 33 2.6.5. Mt phng tip din ca mt mc ................................................................... 34 2.6.6. Tm quan trng ca vc t gradient ................................................................ 35

2.7. Cc gi tr ln nht v nh nht .......................................................................... 36 2.7.1. Cc i a phng v cc tiu a phng .................................................... 36 2.7.2. Cc tr tuyt i .............................................................................................. 41

2.8. Nhn t Lagrange ................................................................................................ 42 2.8.1. Phng php nhn t Lagrange ..................................................................... 42 2.8.2. Hai rng buc ................................................................................................. 46

n Ng Minh o hm ring

Trang 1

CHNG 2. O HM RING

2.1. Hm nhiu bin

Trong phn ny chng ta xem xt cc hm hai hoc nhiu bin theo bn cch nhn

Li ni (Theo m t bng li)

S (Theo bng cc gi tr)

i s (Theo biu thc tng minh)

Trc quan (Theo th)

2.1.1. Hm hai bin

Nhit ca mt im trn b mt ca tri t ti bt k thi gian no ph thuc vo kinh

x v v y ca im . Chng ta c xem l hm ca hai bin x v y, hoc nh l hm

ca mt cp (x, y). Chng ta biu th s ph thuc ny bng cch vit T = f(x, y).

Th tch V ca hnh tr trn ph thuc vo bn knh r v chiu cao h ca n, v V = pr2h.

Chng ta ni rng V l hm ca r v h, v vit V(r, h) = pr2h.

nh ngha Hm hai bin l mt quy lut cho tng ng mi cp c th t cc s thc (x, y)

trong tp D vi duy nht mt s thc f(x, y). Tp D l min xc nh ca f v min gi tr ca

n l tp tt c cc gi tr c th ca f, tc l tp {(, ) | (, ) }.

Chng ta thng vit = (, ) ch ra s tng minh gi tr ca f ti im tng qut

(, ). Cc bin x v y l cc bin c lp v z l bin ph thuc.

Hm hai bin l hm m min xc nh l tp con ca

R2 v min gi tr l tp con ca R. Mt cch trc quan, hm

hai bin ging nh cc mi tn (Hnh 1), trong min D

c biu th nh l mt tp con ca mt phng xy v min

gi tr l tp cc s trn ng thng thc, c biu th bi

trc z. V d, nu (, ) biu th nhit ti mi im (, )

trong mt tm kim loi phng vi hnh dng ca D, chng ta

c th xem trc z l nhit k hin th nhit o c.

Nu hm f c cho bi cng thc v cha ch ra min xc nh th min xc nh ca f

c hiu l tp tt c cc cp (, ) sao cho biu thc cho nhn gi tr thc.

V d 1 Vi mi hm sau y, nh gi (3,2) v phc tho min xc nh.

(a) (, ) =

(b) (, ) = ( )

Li gii (a) (3, 2) =

=

Biu thc f c ngha nu mu s khc 0 v gi tr trong

cn bc hai khng m. V vy min xc nh ca f l

= {(, )| + + 1 0, 1}

Bt phng trnh x + y + 1 0, hay y x 1, m t

nhng im nm ti hoc pha trn ng thng y = x 1.

n Ng Minh o hm ring

Trang 2

Rng buc x 1 ngha l phi b i nhng im trn ng thng x = 1. (Xem Hnh 2).

(b) (3,2) = 3 ln(2 3) = 31 = 0

Bi v ( ) xc nh khi > 0, tc l

< , vy min xc nh ca f l = {(, ) | < }.

y l tp cc im thuc v bn tri ca parabol = .

(Xem Hnh 3.) Khng phi tt c cc hm u c th c

biu th bi cng thc tng minh. Hm trong v d sau y

c m t bng li v bng c lng s theo cc gi tr

ca n.

V d 2 nhng vng c thi tit ma ng khc nghit, ch s gi lnh (wind-chill

index) thng c s dng m t mc nghim trng ca ci lnh. Ch s W ny l nhit

cm nhn ph thuc vo nhit thc t T v tc gi v. V vy, W l hm ca T v v v

chng ta c th vit = (, ). Bng 1 ghi gi tr ca W c bin son bi Dch v Thi

tit Quc gia ca Hoa K (National Weather Service) v Dch v Kh tng ca Canada

(Meteorological Service).

V d, bng cho thy nu nhit l -5oC v tc gi l 50 km/h, th s cm thy lnh

nh nhit khong -15oC khi khng c gi. V vy ( 5, 50) = 15.

V d 3 Nm 1928, Charles Cobb v Paul Douglas cng b mt nghin cu, trong h

m hnh ha s pht trin ca nn kinh t M trong giai on 1899 -1922. Chng c coi

l mt cch nhn n gin ha ca nn kinh t trong sn lng c xc nh bi s lng

lao ng tham gia v s vn u t. Trong khi c nhiu yu t khc nh hng n hiu qu

kinh t, m hnh ca h c chng minh l kh chnh xc. Hm m h s dng m hnh

sn xut c dng

[1] P(L, K) = bLK1-

trong P l tng sn lng (gi tr tin t ca tt c cc hng ha sn xut trong mt nm), L

l s lng lao ng (tng s ngi-gi lm vic trong mt nm), v K l s vn u t (gi tr

n Ng Minh o hm ring

Trang 3

tin t ca tt c cc my mc, thit b, v cc ta nh). Trong phn 2.3 chng ta s ch ra rng,

lm th no dng thc ca phng trnh [1] dn ti cc gi nh kinh t nht nh.

Cobb v Douglas s dng d liu kinh t c cng b

ca chnh ph c c Bng 2. H ly nm 1899 lm c

s v P, L v K cho nm 1899 l mi gn gi tr 100. Cc gi tr

trong nhiu nm khc c th hin theo t l phn trm ca cc

con s ca 1899.

Cobb v Douglas s dng phng php bnh phng ti

thiu Table 2 xy dng hm

[2] P(L, K) = 1.01L0.75K0.25

Nu chng ta s dng m hnh a ra bi cc hm trong

phng trnh [2] tnh ton sn lng trong nhng nm 1910

v 1920, chng ta nhn c gi tr

P(147, 208) = 1.01(147)0.75(208)0.25 161.9

P(194, 407) = 1.01(194)0.75(407)0.25 235.8

chng l kh gn vi gi tr thc t l 159 v 231.

Hm sn lng [1] sau c s dng trong nhiu

phm vi khc nhau, t cc cng ty t nhn n kinh t ton cu.

Min xc nh ca n l {(L, K) | L 0, K 0} bi v L v K

biu th lao ng v vn nn khng bao gi m.

V d 4 Tm min xc nh v min gi tr ca g(, ) = 9 .

Li gii Min xc nh ca g l

= {(, )| 9 0} = {(, )| + 9}

l a trn tm (0, 0) bn knh bng 3. (Xem Hnh 4.)

Min gi tr ca g l | = 9 , (, )

Bi v 9 9 nn 9 3.

Do min gi tr ca g l { | 0 3} = [0, 3].

2.1.2. th

Mt cch khc hnh dung c trng ca hai bin l xem xt th ca n.

nh ngha Nu f l hm hai bin c min xc nh l D th th ca n l tp tt c cc

im (x, y, z) R3 sao cho z = f(x, y) v (x, y) D.

Nh vy, th ca hm mt bin l ng cong vi phng

trnh y = f(x) th th ca hm hai bin l mt cong vi phng trnh

z = f(x, y).

Chng ta c th hnh dung rng hnh chiu ln mt phng xy ca

th S ca hm f chnh l min D (Hnh 5).

n Ng Minh o hm ring

Trang 4

V d 5 Phc ha th hm f(x, y) = 6 3x 2y.

Li gii th ca f c phng trnh z = 6 3x 2y hay 3x + 2y + z = 6, l mt

phng. v mt phng, ta tm cc im chn (intercepts). Cho y = z = 0,

ta nhn c x = 2, l x-chn. Tng t, y-chn bng 3 v z-chn bng 6.

iu ny gip chng ta phc ha phn ca th nm trong phn tm

u tin ca khng gian (first octant) nh trong Hnh 6.

Hm trong V d 5 l trng hp c bit ca hm

f(x, y) = ax + by + c,

n c gi l hm tuyn tnh. th ca cc hm c phng trnh

z = ax + by + c hay ax + by z + c = 0

l cc mt phng. Tng t nh hm tuyn tnh mt bin, hm tuyn tnh hai bin ng vai tr

rt quan trng trong cc php ton vi phn v tch phn.

V d 6 Phc ha th ca hm g(, ) = 9 .

Li gii th c phng trnh = 9 . Bnh phng hai v ta nhn c

= 9 hay + + = 9, chnh l phng trnh ca mt cu

tm ti gc ta v bn knh bng 3. Nhng v z 0 nn th ca

hm g ch l na pha trn ca mt cu.

Ch Ton b mt cu khng th biu th bi mt hm hai bin

x v y. Nh trong V d 6, bn cu (hemisphere) trn c biu th bi

phng trnh (, ) = 9 , cn bn cu di c biu th

bi phng trnh (, ) = 9 .

V d 7 S dng my tnh v th ca hm Cobb-Douglas (, ) = 1.01. .

Li gii Hnh 8 biu th th ca P theo cc gi tr ca nhn cng L v vn K trong

phm vi t 0 n 300. My tnh v mt cong bng cch v ra

cc vt dc. Chng ta thy rng gi tr ca hm P tng theo c

hai s tng ca L v K, nh l d on. Trong MATLAB, chng

ta s dng cc cu lnh sau:

x = 0:10:300; y = x;

[X,Y] = m eshgrid(x,y);

Z = 1.01.*X.^ 0.75.*Y.^ 0.25;

surf(X,Y,Z)

V d 8 Tm min xc nh, min gi tr v v th hm s (, ) = 4 + .

Li gii Min xc nh ca h l ton b mt phng R2. Min gi tr l [0, +). th ca

n c phng trnh = 4 + , y chnh l mt paraboloid elliptic.

Cc vt ct ngang l cc ellipse, cc vt ct dc l cc parabola (Hnh 9).

Cc chng trnh my tnh cho php v th ca hm hai bin. Trong

hu ht cc chng trnh nh vy, cc vt dc trong cc mt phng x = k v

n Ng Minh o hm ring

Trang 5

y = k c v vi cc gi tr cch u nhau ca k v mt phn ca th c loi b bng cch

s dng loi b dng n.

Hnh 10 biu th cc th ca mt s hm c v bi my tnh. Ch rng chng ta

c th nhn c nhng hnh nh tt hn khi chng ta s dng vic quay hnh v chn im

quan st thch hp.

Trong cc hnh (a) v (b), th rt phng v bm st vo mt phng xy, ngoi tr gn

ln cn ca gc ta , bi v l rt nh khi x hoc y l ln.

2.1.3. ng mc

Cho n nay chng ta c hai phng php hnh dung hm: s mi tn v th.

Mt phng php th ba, mn tng t nhng ngi lm bn , l mt bn cc chu

tuyn trn cc gi tr biu th cao c gn kt vi cc ng mc (level curves).

nh ngha ng mc ca hm hai bin f l nhng ng cong c phng trnh f(x, y) = k,

y k l hng s (thuc min gi tr ca f).

Mi ng mc f(x, y) = k l tp tt c cc im trn min xc nh ca f m ti f nhn

gi tr k. Ni khc i, n biu th nhng ch m th ca f c chiu cao l k.

T Hnh 11, chng ta c th thy mi quan h gia ng mc v cc vt ngang.

ng mc f(x, y) = k nh l giao tuyn ca th ca f vi mt phng ngang z = k c

chiu xung mt phng xy.

n Ng Minh o hm ring

Trang 6

Mt v d quen thuc ca ng mc l chng xut hin trong bn a hnh ca mt

khu vc min ni, nh

bn trong Hnh 12.

ng mc l mc

cao so vi mt nc bin.

Nu bn i b dc theo

mt trong nhng ng

cong, bn khng ln cng

khng xung.

Mt v d quen

thuc na l hm nhit

c gii thiu trong on m u ca phn ny. y cc ng cong c gi l ng

nhit (isothermals) v chng kt ni cc min c cng mt nhit . Hnh 13 cho thy mt bn

thi tit ca th gii cho thy nhit trung bnh trong thng Ging. Cc ng ng nhit

l nhng ng cong phn cch cc di mu.

V d 9 Hnh 14 biu th bn ng mc ca hm f. S dng n c lng cc

gi tr f(1, 3) v f(4, 5).

Li gii im (1, 3) thuc phn gia hai ng mc vi cc gi tr 70 v 80, v vy ta

c lng f(1, 3) 73. Tng t f(4, 5) 56.

n Ng Minh o hm ring

Trang 7

V d 10 Phc ha ng mc ca hm f(x, y) = 6 3x 2y vi cc gi tr k = -6, 0, 6, 12.

Li gii Cc ng mc l 6 3x 2y = k hay 3x + 2y + (k 6) = 0

y l h cc ng thng vi dc

. Bn ng

mc ring ng vi k = -6, 0, 6 v 12 l 3x + 2y 12 = 0, 3x

+ 2y 6 = 0, 3x + 2y = 0 v 3x + 2y + 6 = 0.

Chng c phc ha trn Hnh 15. Cc ng mc

l song song v cch u nhau bi th ca f l mt phng.

V d 11 Phc tho cc ng mc ca hm

(, ) = 9 vi k = 0, 1, 2, 3.

Li gii ng mc l 9 = hay + = 9 . y l h cc ng

trn ng tm vi tm (0, 0) bn knh 9 .

Cc trng hp k = 0, 1, 2, 3 c biu th trn Hnh 16.

Hy th hnh dung nhng ng cong ny c nng ln to

thnh mt mt cong v so snh vi th ca mt bn cu trong

Hnh 7.

V d 12 Phc tho cc ng mc ca hm (, ) = 4 + + 1

Li gii ng mc l 4 + + 1 = hay

( )+

= 1

y vi k > 1, biu th mt h

cc ellipse vi cc bn trc

(semiaxes) l 1 v 1.

Hnh 17(a) cho thy mt bn

ng mc ca h c v bi my

tnh. Hnh 17(b) cho thy nhng

ng mc c nng ti th

ca h (mt paraboloid elliptic),

chng tr thnh cc vt ngang.

V d 13 V ng mc ca hm Cobb-Douglas trong V d 3.

Li gii Trong Hnh 18, cc ng ng mc ca hm Cobb-Douglas

P(L, K) = 1.01L0.75K0.25 c v bi my tnh.

n Ng Minh o hm ring

Trang 8

Cc ng mc c gn nhn theo cc gi tr ca sn phm P. V d, ng mc c

nhn 140 biu th tt c cc gi tr ca nhn cng L v u t K c sn phm P = 140. Chng

ta thy rng, i vi mt gi tr c nh ca P, th L tng K gim, v ngc li.

Ty theo mc ch, mt bn ng mc hu ch hn mt th. l chc chn ng

trong V d 13. (So snh Hnh 18 vi Hnh 8.) N cng ng trong vic c tnh gi tr ca

hm, nh trong V d 9.

Hnh 19 cho thy mt s ng mc c my tnh to ra cng vi cc th tng ng.

Ch rng cc ng mc trong phn (c) t li vi nhau gn ngun gc ta . Tng ng vi

thc t l cc th trong phn (d) l rt dc khi gn gc ta .

2.1.4. Hm ba hoc nhiu bin

Mt hm ba bin, f, l quy lut gn mi b ba c th t (x, y, z) trn min D R3 vi

duy nht mt gi tr thc (, , ). V d, nhit T ti mi im trn b mt Tri t ph

thuc vo kinh x, v y v thi im t, v vy c th vit = (, , ).

V d 14 Tm min xc nh ca (, , ) = ln( ) + .

Li gii Biu thc f(x, y, z) c xc nh khi z y > 0, v vy min xc nh ca f l

= {(, , ) | > }

y l na khng gian bao gm tt c cc im nm v pha trn mt phng z = y.

Rt kh cm nhn th ca hm ba bin, v n nm trong khng gian bn chiu. Tuy

nhin, chng ta c c mt s ci nhn su sc vo f bng cch kim tra cc mt mc (level

surfaces) ca n, l nhng mt cong c phng trnh f(x, y, z) = k, vi k l mt hng s. Nu

im (x, y, z) di chuyn dc theo mt mt mc, gi tr ca f(x, y, z) vn khng i.

n Ng Minh o hm ring

Trang 9

V d 15 Tm mt mc ca hm f(x, y, z) = x2 + y2 + z2.

Li gii Cc mt mc l x2 + y2 + z2 = k, vi k 0. l h cc mt cu ng tm vi

bn knh (Xem Hnh 20). V vy, khi (x, y, z) thay i trn bt k mt cu tm O, gi tr ca

f(x, y, z) l khng i.

Hm n bin l quy lut gn mi b n-s thc (x1, x2, ..., xn)

vi mt s thc z = f(x1, x2, ..., xn).

Ta k hiu Rn l tp tt c cc b n-s thc. V d, nu mt

cng ty s dng n loi nguyn liu lm ra mt sn phm, ci l gi

ca nguyn liu th i, xi l s n v nguyn liu th i, khi gi

thnh C ca mi sn phm l hm ca n bin x1, x2, ..., xn.

[3] C = f(x1, x2, ..., xn) = c1x1 + c2x2 + ... + cnxn.

Hm f c gi tr thc vi min xc nh l tp con ca R3. i khi ta s dng k hiu vc

t biu th hm dng gn hn:

Nu = , , , , ta vit () thay cho (, , , ). Vi k hiu nh vy,

chng ta c th nh ngha hm trong phng trnh [3] nh sau: () = .

y = , , , v c.x l k hiu tch v hng ca cc vc t c v x trong Vn.

Xem s tng ng mt mt gia im (, , , ) trong R3 vi vc t v tr

, , , trong Vn, chng ta c ba cch quan nim v hm f c xc nh trong tp con

ca Rn:

1. Nh l hm ca n bin , , ,

2. Nh l hm ca mt bin im (, , , )

3. Nh l hm ca mt bin vc t = , , ,

2.2. Gii hn v s lin tc

2.2.1. Gii hn

Chng ta xem xt hai hm

(, ) =

v (, ) =

khi c x v y ng thi dn v 0, tc l im (x, y) dn v gc ta .

Bng 1 v Bng 2 lit k cc gi tr ca f(x, y) v g(x, y), chnh xc ti ba ch s thp

phn, i vi cc im (x, y) gn gc ta . Ch rng hm khng xc nh ti gc ta .

n Ng Minh o hm ring

Trang 10

N biu th rng, khi (x, y) dn n (0, 0) th cc gi tr ca f(x, y) dn n 1, trong khi

cc gi tr ca g(x, y) khng dn ti gi tr no. N ch ra rng, cc bng chng s l chnh

xc v ta vit

lim(,) (,)

= 1 v lim

(,) (,)

khng tn ti.

Tng qut, chng ta k hiu

lim(,) (,)

(, ) =

biu th rng gi tr ca f(x, y) dn n L khi im (x, y) dn ti im (a, b) dc theo bt k

ng no nm trn trong min xc nh ca f. Ni khc i, chng ta c th lm cho gi tr ca

f(x, y) gn vi L bng cch chn im (x, y) gn im (a, b). nh ngha chnh xc c

pht biu nh sau:

[1] nh ngha Gi s f l hm hai bin vi min xc nh D cha im (a, b). Chng ta

ni rng "gii hn ca f(x, y) bng L khi (x, y) dn ti (a, b)" v ta vit lim(,) (,)

(, ) = ,

nu vi mi e > 0 bt k, tm c s > 0 sao cho nu (x, y) D v ( ) + ( ) <

th |(, ) | < .

Ngoi ra, ngi ta cn dng k hiu lim

(, ) = v (, ) khi (, ) (, ).

Ch rng |f(x, y) L| l khong cch gia cc s f(x, y) v L, v ( ) + ( )

l khong cch gia im (x, y) v im (a, b). V vy nh ngha 1 ni rng khong cch gia

cc s f(x, y) v L c th nh ty bng cch cho khong cch gia im (x, y) v im (a, b)

nh (nhng khc 0). Hnh 1 minh ha nh ngha 1 theo ngha ca biu mi tn. Vi mi

khong nh (L - e, L + e) cha L, chng ta c th tm c min hnh trn D [c th tr i

im (a, b)] vi tm (a, b) v bn knh > 0 sao cho f nh x tt c cc im trong D [c th

tr i im (a, b)] vo trong khong (L - e, L + e).

Mt minh ha khc ca nh ngha 1 c cho trong Hnh 2, mt cong S l th

ca f. Vi e > 0 cho trc, ta c th tm c > 0 sao cho nu (x, y) thuc min D v (x, y)

(a, b) th phn tng ng ca S nm gia cc mt phng z = L e v L + e.

Vi hm mt bin, x ch c th dn n a theo hai pha t bn tri hoc bn phi. Nh li

rng nu lim

() lim

() th khng tn ti lim

().

n Ng Minh o hm ring

Trang 11

Vi cc hm hai bin th vic khng n gin bi v chng ta c th cho (x, y) dn n

(a, b) t mun vn hng khc nhau (Hnh 3), min l (x, y) vn

thuc min xc nh ca f.

nh ngha 1 ch cp ti khong cch gia (x, y) v (a,

b) m khng quan tm n hng ca s dn n. Do , nu

gii hn tn ti th f(x, y) phi dn ti cng mt gii hn, khng

ph thuc (x, y) dn ti (a, b) nh th no. V th nu chng ta

tm thy hai ng dn n (a, b) ca (x, y) c hai gii hn khc

nhau th lim(,) (,)

(, ) khng tn ti.

Nu f(x, y) L1 khi (x, y) (a, b) dc theo C1 v f(x, y) L2 khi (x, y) (a, b) dc

theo C2 m L1 L2 th lim(,) (,)

(, ) khng tn ti.

V d 1 Chng t rng lim(,) (,)

khng tn ti.

Li gii Gi s (, ) = ( )/( + ). Trc ht ta xt s dn n (0, 0) dc

theo trc x. Sau cho y = 0 ta c f(x, 0) = x2/x2 = 1 vi mi x 0,

v vy f(x, y) 1 khi (x, y) (0, 0) dc theo trc x.

Gi chng ta dn dn dc theo trc y bng cch t x = 0. V

f(0, y) = -y2/y2 = -1 vi mi y 0 nn f(x, y) -1 khi (x, y) (0, 0)

dc theo trc y (Hnh 4). Bi v f c hai gii hn khc nhau dc theo

hai ng khc nhau nn gii hn trn khng tn ti.

V d 2 Nu f(x, y) = xy/(x2 + y2), tn ti hay khng gii hn lim(,) (,)

(, )?

Li gii Nu y = 0 th f(x, 0) = 0/x2 = 0, vy f(x, y) 0 khi (x, y) (0, 0) dc theo trc x.

Nu x = 0 th f(0, y) = 0/y2 = 0, vy f(x, y) 0 khi (x, y) (0, 0) dc theo trc y.

Mc d chng ta nhn c cng mt gii hn, nhng iu

khng chng t gii hn cho l bng 0. Gi chng ta xt s dn

n (0, 0) dc theo ng y = x.

Vi x 0 th (, ) =

=

,

v vy (, )

khi (x, y) (0, 0) dc theo y = x (Hnh 5).

V vy gii hn cho khng tn ti.

Hnh 6 lm r cho V d 2. Sn cong xut hin trn ng

y = x tng ng vi thc t l f(x, y) = 1/2 i vi mi im (x, y)

trn ng , ngoi tr gc ta .

V d 3 Cho (, ) =

, c hay khng gii hn lim

(,) (,)(, )?

n Ng Minh o hm ring

Trang 12

Li gii Nh li li gii trong V d 2, chng ta tit kim thi gian bng cch cho (x, y)

dn ti (0, 0) dc theo mi ng nghing i qua gc ta y = mx, y m l dc:

(, ) = (, ) =( )

( )=

=

0

khi (x, y) (0, 0) dc theo y = mx. V th f c cng mt gii hn dc theo mi ng nghing

i qua gc ta . Nhng iu khng chng t gii hn

cho bng 0. Gi chng ta cho (x, y) dn ti (0, 0) dc theo

parabola x = y2, ta c

(, ) = (, ) =

()=

=

vy (, )

khi (x, y) (0, 0) dc theo x = y2.

Vy gii hn cho khng tn ti.

Hnh 7 l th ca hm trong V d 3. Ch rng sn dc nm trn parabola x = y2.

By gi chng ta hy xem xt cc gii hn m tn ti. Cng nh i vi hm mt bin,

vic tm gii hn cho cc hm hai bin c th c n gin ha bng cch s dng cc tnh

cht ca gii hn. Cc quy tc tm gii hn ca hm mt bin c th c m rng n cc hm

hai bin: Gii hn ca mt tng bng tng ca cc gii hn, gii hn ca mt tch bng tch ca

cc gii hn. c bit, cc cng thc sau y l ng khi (x, y) (a, b):

[2] lim = lim = lim =

nh l Squeeze vn cn ng.

Nu (, ) (, ) (, )

v lim (, ) = lim (, ) =

th lim (, ) =

V d 4 Tm lim(,) (,)

nu n tn ti.

Li gii Nh trong V d 3, chng ta c th rng gii hn dc theo bt k ng thng

no i qua gc ta u bng 0. iu khng chng minh c gii hn cho bng 0,

nhng cc gii hn dc theo cc parabola y = x2 v x = y2 cng bng 0, v vy chng ta bt u

nghi ng rng gii hn l tn ti v bng 0.

Cho e > 0. Chng ta cn tm > 0 sao cho nu 0 < + < th

0< ,

tc l, nu 0 < + < th ||

< .

Mc d + v 0, nn /( + ) 1, v vy

[3] ||

3|| = 3 3 +

V th nu ta chn = e/3 v gi s 0 < + < th

0 3 + < 3 = 3

=

T theo nh ngha 1, lim(,) (,)

= 0.

Chng ta cng c th s dng nh l Squeeze chng minh.

n Ng Minh o hm ring

Trang 13

Tht vy, t [3], ch n [2], ta c iu cn chng minh.

2.2.2. S lin tc

S lin tc ca hm hai bin c nh ngha tng t nh i vi hm mt bin.

[4] nh ngha Hm hai bin f c gi l lin tc ti (a, b) nu lim(,) (,)

(, ) = (, ).

Chng ta ni f lin tc trn D nu n lin tc ti mi im (a, b) thuc D.

ngha trc quan ca s lin tc l nu im (x, y) thay i mt lng nh th gi tr ca

f(x, y) cng thay i mt s lng nh. iu ny c ngha rng mt cong l th ca mt hm

lin tc khng c l hoc b rch.

S dng cc thuc tnh ca gii hn, bn c th thy tng, hiu, tch v thng cc hm

lin tc l lin tc trn min xc nh ca chng. Hy s dng tnh cht ny a ra mt s

v d v hm lin tc.

Hm a thc (polynomial) hai bin l tng ca cc hng thc dng cxmyn, trong c l

hng s, cn m v n l cc s nguyn. Hm phn thc (rational) l t s ca cc a thc. V d,

f(x, y) = x4 + 5x3y2 + 6xy4 7y + 6 l hm a thc, cn (, ) =

l hm phn thc.

Cc gii hn trong [2] chng t rng cc hm f(x, y) = x, g(x, y) = y v h(x, y) = c l cc

hm lin tc. Bi v mi a thc u c xy dng t cc hm n gin f, g v h bng cc

php nhn v cng, nn mi hm a thc hai bin u lin tc trn R2. Tng t, mi hm phn

thc u lin tc trn min xc nh ca n bi n l thng ca hai hm lin tc.

V d 5 nh gi lim(,) (,)

( + 3 + 2)

Li gii Bi v hm (, ) = + 3 + 2 l a thc nn n t khp ni, v

vy ta c th tm gii hn bng cch thay trc tip:

lim(,) (,)

( + 3 + 2) = 12 12 + 3.1 + 2.2 = 11

V d 6 Hm (, ) =

lin tc ti nhng u?

Li gii Hm f khng lin tc ti (0, 0) bi v n khng xc nh ti . Do f l hm phn

thc nn min lin tc ca n l tp D = {(x, y) | (x, y) (0, 0)}.

V d 7 Gi s

(, ) =

+ (, ) (0,0)

0 (, ) = (0,0)

y g c xc nh ti (0, 0) nhng g vn khng lin tc bi v lim(,) (,)

(, ) khng

tn ti (xem V d 1).

V d 8 Gi s

(, ) =

3

+ (, ) (0,0)

0 (, ) = (0,0)

n Ng Minh o hm ring

Trang 14

Chng ta bit rng f lin tc vi (x, y) (0, 0) v n l hm phn thc. T V d 4 ta c

lim(,) (,)

(, ) = lim(,) (,)

3

+ = 0 = (0,0)

V vy f lin tc ti (0, 0), v do n lin tc trn ton R2.

Ging nh hm mt bin, php ly hm hp ca hai hm l mt cch nhn c hm

th ba. Thc t, c th ch ra rng nu f l hm hai bin lin tc v g l hm mt bin lin tc

xc nh trn min gi tr ca f, th hm hp (composite) h = gof c xc nh bi

(, ) = ((, ))

cng l hm lin tc.

V d 9 Tm min lin tc ca hm h(x, y) = arctan(y/x).

Li gii Hm f(x, y) = y/x l hm phn thc nn n lin tc ngoi tr trn ng thng

x = 0. Hm () = arctan l lin tc khp ni. V vy hm

hp g(f(x, y)) = arctan(y/x) = h(x, y) lin tc ngoi tr trn

ng thng x = 0. Hnh 9 ch ra s t gy trn th ca

hm h trn trc y.

2.2.3. Hm ba hoc nhiu bin

Mi kt qu bit u c th m rng cho hm ba hoc nhiu bin.

K hiu lim(,,) (,,)

(, , ) = ngha l cc gi tr ca f(x, y, z) dn ti s L khi im

(x, y, z) dn ti im (a, b, c) dc theo bt k ng no trong min xc nh ca f.

V khong cch gia hai im (x, y, z) v (a, b, c) trong R3 bng

( ) + ( ) + ( ) ,

nn ta c th nh ngha chnh xc nh sau:

Vi mi s e > 0, tn ti s > 0 sao cho, nu (x, y, z) thuc min xc nh ca f v

0 < ( ) + ( ) + ( ) <

th |(, , ) | < e.

Hm f lin tc ti (a, b, c) nu lim(,,) (,,)

(, , ) = (, , ),

V d, hm

(, , ) =1

+ + 1

l hm phn thc ca ba bin v v vy n lin tc ti mi im trong R3, ngoi tr x2 + y2 + z2 = 1.

Ni khc i, n khng lin tc trn mt cu tm ti gc ta , bn knh bng 1.

Nu s dng k hiu vc t trnh by trong phn 2.1, chng ta c th nh ngha gii

hn ca hm hai hoc ba bin dng n gin nh sau.

n Ng Minh o hm ring

Trang 15

[5] nh ngha Nu f xc nh trn min D ca n th lim

() = c ngha l, vi mi

e > 0, tn ti s > 0 sao cho, nu x D v 0 < |x a| < th |f(x) L| < e.

Ch rng nu n = 1 th x = x v a = a, nh ngha 5 chnh l nh ngha v gii hn i

vi hm mt bin. Trong trng hp n = 2, ta c

x = x, y, a = a, b, v | | = ( ) + ( ),

nh ngha 5 tr thnh nh ngha 1. Nu n = 3 th x = x, y, z, a = a, b, c, v nh ngha 5

tr thnh nh ngha v gii hn ca hm ba bin.

Trong mi trng hp ta u vit lim

() = ().

2.3. o hm ring

2.3.1. Khi nim v o hm ring

Vo mt ngy nng, m cao lm cho chng ta ngh rng nhit cao hn nhit thc

ca n, trong khi trong khng kh rt kh, chng ta cm nhn nhit thp hn ch th ca nhit

k. Dch v Thi tit Quc gia (National Weather Service) a ra cc ch s nhit (cn gi

l ch s nhit - m, hoc ch s m mt s nc) m t tc ng kt hp ca nhit

v m. Ch s nhit I l nhit khng kh cm nhn c khi nhit thc t l T v

m tng i l H. V vy, I l mt hm ca T v H, v chng ta c th vit I = f(T, H).

Bng di y ca cc gi tr ca I c trch t mt bng c bin son bi cc Dch v Thi

tit Quc gia.

Nu chng ta tp trung vo ct c nh du ca bng, tng ng vi m tng i

ca H = 70%, chng ta coi ch s nhit nh l hm mt bin T i vi gi tr c nh ca H. Ta

vit g(T) = f (T, 70). Sau g (T) m t cch thc ch s nhit I tng ln khi nhit thc t T

tng, tng ng vi m l 70%. o hm ca g khi T = 96oF l tc thay i ca I i vi

T khi T = 96oF:

(96) = lim

()()

= lim

(,)(,)

Chng ta c th xp x g'(96) bng cch s dng cc gi tr trong Bng 1 vi h = 2 v -2

(96) ()()

=

(,)(,)

=

= 4

(96) ()()

=

(,)(,)

=

= 3.5

n Ng Minh o hm ring

Trang 16

Ly trung bnh cng hai gi tr ny, ta c th ni rng o hm g'(96) xp x bng 3.75.

Ngha l, khi nhit thc t l 96oF v m tng i l 70%, nhit biu kin tng khong

3.75oF so vi mi tng ca nhit thc t.

Gi chng ta xem xt dng c nh du trong Bng 1, tng ng vi nhit c nh

T = 96oF. Cc s trn dng l cc gi tr ca hm G(H) = f(96, H), chng m t cch thc ch

s nhit tng ln khi m m tng i tng, trong khi nhit thc t T = 96oF. o hm

ca hm ny khi H = 70% l tc bin thin ca I i vi H khi H = 70%:

(70) = lim

()()

= lim

(,)(,)

Chng ta c th xp x G'(70) bng cch t h = 5 v -5:

(70) ()()

=

(,)(,)

=

= 1

(70) ()()

=

(,)(,)

=

= 0.8

Ly gi tr trung bnh, ta c c lng G'(70) 0.9. iu ny ni ln rng, khi nhit

l 96oF v m tng i l 70%, ch s nhit tng khong 0.9oF i vi mi phn trm tng

ca nhit tng i.

Tng qut, nu f l hm ca hai bin x v y, gi s c nh y = b - const v cho x bin

i. Khi ta c hm mt bin g(x) = f(x, b). Nu g c o hm ti a th ta gi n l o hm

ring ca f theo bin x ti (a, b) v k hiu fx(a, b). V vy

[1] fx(a, b) = g'(a) vi g(x) = f(x, b)

Theo nh ngha o hm ring ta c () = lim

()()

, v vy [1] tr thnh

[2] (, ) = lim

(,)(,)

Tng t, o hm ring ca f theo y ti (a, b), k hiu fy(a, b), nhn c bng cch c

nh x = a v tnh o hm ti b ca hm mt bin G(y) = f(a, y):

[3] (, ) = lim

(,)(,)

Vi cc k hiu ny ca cc o hm ring, ta c th vit tc thay i ca ch s nhit

I theo nhit thc t T v m tng i H khi T = 96oF v H = 70% nh sau:

fT(96, 70) 3.75 fH(96, 70) 0.9

Gi chng ta coi im (a, b) thay i, fx v fy tr thnh hm hai bin.

[4] Nu f l hm hai bin, cc o hm ring ca n fx v fy c xc nh nh sau:

(, ) = lim

(,)(,)

(, ) = lim

(,)(,)

C nhiu k hiu i vi cc o hm ring, v d, nu z = f(x, y) th

(, ) = =

=

(, ) =

= = =

Quy tc tnh cc o hm ring:

1. tnh fx, coi y l khng i v o hm f(x, y) theo x.

2. tnh fy, coi x l khng i v o hm f(x, y) theo y.

n Ng Minh o hm ring

Trang 17

V d 1 Cho f(x, y) = x3 + x2y3 2y2, tm fx(2, 1) v fy(2, 1).

Li gii Gi y c nh v o hm theo x, ta nhn c fx(x, y) = 3x2 + 2xy3, v vy

fx(2, 1) = 3.22 + 2.2.13 = 16

Gi x c nh v o hm theo y, ta nhn c fy(x, y) = 3x2y2 - 4y, v vy

fy(2, 1) = 3.22.12 4.1 = 8

2.3.2. ngha ca cc o hm ring

a ra ngha ca cc o hm ring, chng ta nh li rng phng trnh z = f(x, y)

m t mt mt cong S. Nu f(a, b) = c th im P(a, b, c) thuc S. C nh y = b, chng ta

hn ch s ch ca ta trn ng cong C1 l giao ca mt phng y = b vi S. Ni khc i, C1

l giao tuyn ca S trn mt phng y = b. Tng t th, mt phng x = a giao vi S theo ng

cong C2. C hai ng cong C1 v C2 u i qua im P (xem Hnh 1).

Ch rng ng cong C1 l th ca hm g(x) = f(x, b),

v th dc ca tip tuyn ca n T1 ti P l g'(a) = fx(a, b).

ng cong C2 l th ca hm G(y) = f(a, y), v th dc ca

tip tuyn ca n T2 ti P l G'(b) = fy(a, b).

V th cc o hm ring fx(a, b) v fy(a, b) c th hiu l

dc ca cc ng tip tuyn ti P(a, b, c) ca cc giao tuyn

C1 v C2 ca S vi cc mt phng y = b v x = a.

V d 2 Cho f(x, y) = 4 x2 2y2, tm fx(1, 1) v fy(1, 1) ri gii thch ngha.

Li gii V fx(x, y) = -2x nn fx(1, 1) = -2, v fy(x, y) = -4y nn fy(1, 1) = -4.

th ca f l paraboloid z = 4 x2 2y2 v giao ca mt phng y = 1 vi n l parabola

z = 2 x2, y = 1. (ng cong C1 trong Hnh 2). dc ca tip tuyn ca parabola ti im

(1, 1, 1) l fx(1, 1) = -2. Tng t, ng cong C2 l giao ca paraboloid vi mt phng x = 1,

l mt parabola z = 3 2y2, x = 1. dc ca tip tuyn ti (1, 1, 1) l fy(1, 1) = -4 (Hnh 3).

Hnh 4 m t my tnh v tng ng vi Hnh 2. Phn (a) biu th mt phng y = 1 giao

vi mt cong theo giao tuyn v phn (b) m t C1 v T1. Chng ta s dng cc phng trnh

vc t r(t) = t, 1, 2 t2 cho C1 v r(t) = 1 + t, 1, 1 2t cho T1. Tng t, Hnh 5 tng ng

vi Hnh 3.

n Ng Minh o hm ring

Trang 18

V d 3 Cho f(x, y) = sin

, tnh

v

.

Li gii S dng quy tc Chain i vi hm mt bin, ta c

= cos

= cos

= cos

= cos

()

V d 4 Tm z/x v z/y nu z c xc nh n nh l hm ca x, y theo phng

trnh x3 + y3 + z3 + 6xyz = 1.

Li gii tm z/x, chng ta o hm hm n theo x, coi y nh hng s:

3 + 3

+ 6 + 6

= 0

Gii ra ta c

=

. Tng t,

=

2.3.3. Hm nhiu hn hai bin

Cc o hm ring c th c nh ngha cho cc hm nhiu hn hai bin. V d, nu f

l hm ba bin x, y v z th o hm ring theo x c nh ngha l

(, , ) = lim

(,,)(,,)

Nu w = f(x, y, z) th fx = w/x c th xem l tc thay i ca w theo x khi y v z

khng i. Nhng chng ta khng th gii thch hnh hc bi v th ca f nm trong khng

gian bn chiu.

n Ng Minh o hm ring

Trang 19

Tng qut, nu u l hm ca n bin, u = f(x1, x2, ..., xn) th o hm ring theo bin th i

s l

= lim

(, , , , + , , , ) (, , , , , , , )

v chng ta cng vit

=

= = =

V d 5 Tm fx, fy v fz nu f(x, y, z) = exylnz.

Li gii Gi y v z khng i v o hm theo x ta c fx = yexylnz.

Tng t, fy = xexylnz v fz = exy/z

2.3.4. o hm cp cao

Nu f l hm hai bin th cc o hm ring fx v fy cng l hm hai bin, v vy chng ta

c th ly o hm ring ca chng v gi l cc o hm ring cp hai ca f.

Nu z = f(x, y), chng ta s dng cc hiu sau:

() = = =

=

=

() = = =

=

=

= = =

=

=

= = =

=

=

V th k hiu fxy (hay 2f/yx) c ngha l u tin ly o hm theo x, sau ly o

hm theo y, trong khi fyx th o li th t.

V d 6 Tnh cc o hm ring cp hai ca f(x, y) = x3 + x2y3 2y2

Li gii Trong V d 1 chng ta tm c fx(x, y) = 3x2 + 2xy3, fy(x, y) = 3x2y2 4y

V vy =

(3 + 2) = 6 + 2 =

(3 + 2) = 6

=

(3 4) = 6 =

(3 4) = 6 4

Hnh 7 cho thy th ca hm f trong V d 6 v cc th ca cc o hm ring cp mt v

cp hai vi -2 x 2, -2 y 2. Ch rng cc th ny ph hp vi cch gii thch ca

chng ta, fx v fy l dc ca ng tip tuyn vi cc chu tuyn ca th ca f. V d,

th ca gim nu chng ta bt u ti (0, -2) v di chuyn theo hng dng ca trc x. iu

ny c phn nh bi gi tr m ca fx. Bn nn so snh cc th ca fyx v fyy vi th ca

fy xem cc mi quan h.

n Ng Minh o hm ring

Trang 20

Ch rng trong V d 6, fxy = fyx. y khng phi l s trng hp. N ch ra rng cc

o hm ring hn hp fxy v fyx l bng nhau trong hu ht cc hm chng ta gp trong thc

hnh. nh l sau y, c pht hin bi nh ton hc ngi Php Alexis Clairaut (1713-

1765), cho ra iu kin c th khng nh fxy = fyx.

nh l Clairaut Gi s f xc nh trn min D cha im (a, b). Nu cc o hm ring

cp hai fxy v fyx cng tn ti v lin tc trn D th fxy(a, b) = fyx(a, b).

Cc o hm ring cp 3 hoc cao hn cng c th xc nh. V d

= =

=

v s dng nh l Clairaut c th chng minh rng fxyy = fyxy = fyyx nu cc hm ny cng tn

ti v lin tc.

V d 7 Tnh fxxyz nu f(x, y, z) = sin(3x + yz).

Li gii fx = 3cos(3x + z) fxx = -9sin(3x + yz)

fxxy = -9zcos(3x + yz) fxxyz = -9cos(3x + yz) + 9yzsin(3x + yz)

2.4. Mt phng tip din v xp x tuyn tnh

Mt trong nhng tng quan trng nht trong cc php tnh gii tch hm mt bin l

khi chng ta phng to mt im trn th ca mt hm kh vi, th tr nn khng th phn

bit so vi ng tip tuyn ca n v chng ta c th xp x hm bi mt hm tuyn tnh.

y chng ta pht trin tng tng t trong ba chiu. Khi chng ta phng to mt im trn

mt mt cong l th ca mt hm hai bin kh vi, mt cong trng ging mt mt phng (mt

phng tip tuyn ca n) v chng ta c th hm bi mt hm tuyn tnh hai bin. Chng ta

cng c th m rng tng ny i vi hm nhiu bin hn.

2.4.1. Mt phng tip din

Gi s rng mt S c phng trnh z = f(x, y), trong f c cc o hm ring cp mt

lin tc, v gi s P(x0, y0, z0) l mt im trn S. Nh phn trc, gi s C1 v C2 l cc giao

tuyn ca cc mt phng y = y0 v x = x0 vi mt S, hi im P nm trn c C1 v C2.

Gi s T1 v T2 l cc ng tip tuyn ca C1 v C2 ti P,

khi mt phng tip din (tangent plane) ca S ti P l mt phng

cha c hai tip tuyn T1 v T2 (Xem Hnh 1).

Trong phn 2.6 chng ta thy rng nu C l ng cong bt

k nm trong S v i qua P, th tip tuyn ca n ti P nm trn

mt phng tip din. Do ta c th xem rng tip din ca S ti

n Ng Minh o hm ring

Trang 21

P cha tt c cc tip tuyn ti P ca cc ng cong nm trn S v i qua P.

Chng ta bit rng phng trnh ca mt phng i qua P(x0, y0, z0) c dng

A(x x0) + B(y y0) + C(z z0) = 0

Chia hai v cho C v t a = -A/C, b = -B/C, ta vit li di dng

[1] z z0 = a(x x0) + b(y y0)

Nu phng trnh [1] m t tip din ti P th giao ca n vi mt phng y = y0 s l

ng tip tuyn T1. t y = y0 vo phng trnh [1] ta nhn c

z z0 = a(x x0) khi y = y0

v chng ta bit rng y l phng trnh ca ng thng vi dc a. Nhng trong mc 2.3

ta bit rng dc ca tip tuyn T1 l fx(x0, y0), v th a = fx(x0, y0).

Tng t, t x = x0 vo phng trnh [1], ta nhn c z z0 = b(y y0), chnh l

tip tuyn T2, vy b = fy(x0, y0).

nh ngha 2 Gi s f c cc o hm ring lin tc. Phng trnh tip din ca mt cong

z = f(x, y) ti im P(x0, y0, z0) l z z0 = fx(x0, y0)(x x0) + fy(x0, y0) (y y0)

V d 1 Tm mt phng tip din ca paraboloid elliptic z = 2x2 + y2 ti im (1, 1, 3).

Li gii Gi s f(x, y) = 2x2 + y2. Khi

fx(x, y) = 4x fx(1, 1) = 4 fy(x, y) = 2y fy(1, 1) = 2

Theo nh ngha 2, phng trnh mt phng tip din ti (1, 1, 3) l

z 3 = 4(x 1) + 2(y 1) hay 4x + 2y z 3 = 0.

Hnh 2(a) m t paraboloid elliptic v mt phng tip din ti P(1, 1, 3) ni n trong V

d 1. Cc hnh (b) v (c) l phng to ti im (1, 1, 3). Ch rng cng phng to th th cng

phng v cng ging vi mt phng tip din ca.

Trn Hnh 3 chng ta khng nh thm v iu , bng cch phng to im (1, 1) trn

bn ng ng mc ca hm f(x, y) = 2x2 + y2. Ch rng cng phng to th cc ng

mc nhn nh cc ng song song, l c trng ca ng thng.

n Ng Minh o hm ring

Trang 22

2.4.2. Xp x tuyn tnh

Trong V d 1, chng ta thy rng phng trnh ca tip din ca th ca hm

f(x, y) = 2x2 + y2 ti im (1, 1, 3) l z = 4x + 2y 3. V vy, cc chng c trn Hnh 2 v Hnh

3 cho thy, hm tuyn tnh hai bin L(x, y) = 4x + 2y 3 l xp x tt nht ca f(x, y) khi (x, y)

gn (1, 1). Hm L c gi l tuyn tnh ha ca f ti (1, 1) v s xp x f(x, y) 4x + 2y 3

c gi l xp x tuyn tnh hoc xp x tip din ca f ti (1, 1).

V d, ti im (1.1, 0.95) xp x tuyn tnh s cho

f(1.1, 0.95) 4(1.1) + 2(0.95) 3 = 3.3

kh gn vi gi tr thc s ca f(1.1, 0.95) = 2(1.1)2 + (0.95)2 = 3.3225. Nhng nu ta ly im

xa im (1, 1) hn, v d im (2, 3), chng ta khng cn nhn c xp x tt. Tht vy, ta c

L(2, 3) = 11, trong khi f(2, 3) = 17.

Tng qut, phng trnh tip din ca th ca hm f hai

bin ti im (a, b, f(a, b)) l

z = f(a, b) + fx(a, b)(x a) + fy(a, b)(y b)

Hm tuyn tnh c th l tip din l hm

[3] L(x, y) = f(a, b) + fx(a, b)(x a) + fy(a, b)(y b)

c gi l tuyn tnh ha ca f ti (a, b), v xp x

[4] f(x, y) f(a, b) + fx(a, b)(x a) + fy(a, b)(y b)

c gi l xp x tuyn tnh hoc xp x tip din ca f ti (a, b).

Chng ta xc nh c tip din ca mt cong z = f(x, y) ti ni m f c cc o hm

ring cp 1 lin tc. iu g s xy ra nu fx v fy khng lin tc? Hnh 4 l th ca mt hm

nh vy, n c phng trnh

(, ) =

(, ) (0,0)

0 (, ) = (0,0)

C th kim tra rng ti (0, 0), cc o hm ring tn ti, fx(0, 0) = fy(0, 0) nhng khng

lin tc. Xp x tuyn tnh s l f(x, y) = 0, nhng thc t f(x, y) = ti tt c cc im trn

ng thng y = x. V vy vn xy ra tnh trng xu ngay c khi cc o hm ring ca hm hai

bin tn ti. khc phc, ta a ra khi nim kh vi i vi hm hai bin.

Nh li rng, vi hm mt bin y = f(x), nu x thay i t a ti a + x, th s gia ca y l

y = f(x + ) f(x)

Nu f kh vi ti a th

[5] y = f '(a)x + ex vi e 0 khi x 0

By gi ta xt hm hai bin z = f(x, y), v gi s x thay i t a ti a + x, cn y thay i

t b n b + y. Khi s gia tng ng ca z l

[6] z = f(a + x, b + y) f(a, b)

Nh vy s gia z biu th s thay i gi tr ca f khi (x, y) thay i t (a, b) ti (a + x,

b + y). Tng t [5], ta nh ngha tnh kh vi ca hm hai bin nh sau

[7] nh ngha Nu z = f(x, y) th f l kh vi ti (a, b) nu z c th biu din di dng

n Ng Minh o hm ring

Trang 23

z = fx(a, b)x + fy(a, b)y + e1x + e2y

trong e1 v e2 cng dn v 0 khi (x, y) dn v (0, 0).

nh ngha 7 ni ln rng, hm kh vi s c xp x tt khi (x, y) gn (a, b). Ni khc i,

mt phng tip din xp x tt nht vi th ti tip im.

[8] nh l Nu cc o hm ring fx v fy tn ti trong ln cn ca (a, b) v lin tc ti

(a, b) th kh vi ti (a, b).

V d 2 Chng t rng f(x, y) = xexy kh vi ti (1, 0) v tm tuyn tnh ha ca n ti .

Sau s dng tm xp x f(1.1, -0.1).

Li gii Cc o hm ring l fx(x, y) = exy(1 + xy) fx(1, 0) = 1

fy(x, y) = x2exy fy(1, 0) = 1

C fx v fy l cc hm lin tc, v vy f kh vi theo nh l 8.

Tuyn tnh ha l

L(x, y) = f(1, 0) + fx(1, 0)(x 1) + fy(1, 0)(y 0) = x + y

Xp x l xexy x + y, vy f(1.1, -0.1) 1.1 0.1 = 1.

So snh vi gi tr thc, f(1.1, -0.1) = 1.1e-0.11 0.98542.

Hnh 5 th hin th ca f v tuyn tnh ha L trong V d 2.

V d 3 u phn 2.3 chng ta ni n ch s nhit I l hm ca nhit thc t v

m tng i H, v a ra bng cc gi tr t Dch v Thi tit Quc gia

Tm xp x tuyn tnh ca ch s nhit I = f(T, H) khi T gn 96oF v H gn 70%. S dng

n c lng ch s nhit khi nhit l 97oF v m tng i l 72%.

Li gii T bng ta c f(96, 70) = 125. Trong phn 2.3 ta c c lng fT(96, 70)

3.75 v fH(96, 70) 0.9. V vy xp x tuyn tnh l

f(T, H) f(96, 70) + fT(96, 70)(T 96) + fH(96, 70)(H 70)

125 + 3.75(T 96) + 0.9(H 70)

c bit, f(97, 72) 125 + 3.75(1) + 0.9(2) = 130.55

V vy, khi T = 97oF v H = 72% th ch s nhit l I 131oF

2.4.3. Vi phn

Vi hm kh vi mt bin y = f(x), chng ta xem vi phn dx l bin c lp, tc l dx c

th nhn bt c gi tr thc no. Vi phn ca y c nh ngha l

[9] dy = f '(x)dx.

n Ng Minh o hm ring

Trang 24

Hnh 6 m t mi quan h gia s gia y v vi phn

dy: y biu th s thay i theo chiu cao ca ng cong

y = f(x), cn dy biu th s thay i theo chiu cao ca

ng tip tuyn khi x thay i mt lng dx = x.

i vi hm kh vi hai bin z = f(x, y), chng ta

xem cc vi phn dx v dy l cc bin c lp. Khi vi

phn dz c gi l vi phn ton phn (total differential),

c xc nh:

[10] = (, ) + (, ) =

+

i khi s dng df thay cho dz.

Nu chng ta t

dx = x = x a v dy = y = y b

trong phng trnh 10, th vi phn ca z l

dz = fx(a, b)(x a) + fy(a, b)(y b)

V vy, theo k hiu ca cc vi phn,

xp x tuyn tnh [4] c th vit nh sau

f(x, y) f(a, b) + dz

Hnh 7 tng ng vi Hnh 6 trong

khng gian ba chiu, m t ngha hnh hc ca vi phn dz v s gia z: dz biu th s thay

i theo chiu cao ca mt phng tip din, trong khi z bi th s thay i theo chiu cao

ca mt cong z = f(z, y) khi (x, y) thay i t (a, b) n (a + x, b + y).

V d 4

(a) Cho z = f(x, y) = x2 + 3xy y2, tm vi phn dz.

(b) Cho x thay i t 2 ti 2.05 v y thay i t 3 ti 2.96, so snh cc gi tr z v dz.

Li gii

(a) T nh ngha 10 ta c

=

+

= (2 + 3) + (3 2)

(b) t x = 2, dx = x = 0.05, y = 3, dy = y = -0.04, ta c

dz = [2(2)+ 3(3)]0.05 + [3(2) 2(3)](-0.04) = 0.65

S gia ca z l

z = f(2.05,2.96) f(2, 3) = [(2.05)2 + 3(2.05)(2.96) (2.96)2] [22 + 3(2)(3) 32] = 0.6449

Ch rng z dz nhng d tnh hn.

V d 5 Bn knh c s v chiu cao ca hnh nn trn c xc nh tng ng l 10cm

v 25cm, vi sai s 0.1cm trong mi gi tr o. S dng vi phn c lng sai s ln nht

khi tnh ton th tch ca hnh nn.

Li gii Th tch ca hnh nn vi bn knh c s r v chiu cao h l V = pr2h/3. V vy

vi phn ca V l

n Ng Minh o hm ring

Trang 25

=

+

=

+

Bi v mi sai s l 0.1cm, ta c |r| 0.1, |hr| 0.1 cng vi r = 10, h = 25. Do

=500p

3(0.1) +

100p

3(0.1) = 20p

V vy li ln nht khi tnh th tch hnh nn l 20pcm3 63 cm3.

2.4.4. Cc hm ba hoc nhiu bin

Cc khi nim xp x tuyn tnh, tnh kh vi v vi phn c th c m rng cho hm

nhiu hn hai bin. S kh vi c m rng t nh ngha 7. Xp x tuyn tnh l

f(x, y, z) f(a, b, c) + fx(a, b, c)(x a) + fy(a, b, c)(y b) + fz(a, b, c)(z c)

v tuyn tnh ha L(x, y, z) l v phi ca biu thc ny.

Nu w = f(x, y, z) th s gia ca w l w = f(x + x, y + y, z + z) f(x, y, z).

Vi phn dw c xc nh theo cng thc =

+

+

V d 6 Kch thc ca khi hp ch nht c cc s o l 75cm, 60cm v 40cm cng

mt sai s l 0.2cm. S dng vi phn c lng sai s ln nht c th khi th tch ca hp

c o vi o .

Li gii Nu cc kch thc ca hp l x, y v z th th tch ca n l V = xyz, v vy

=

+

+

= + +

Ta cho |x| 0.2, |y| 0.2, |z| 0.2. c lng sai s ln nht, chng ta s dng

dx = dy = dz = 0.2 v x = 75, y = 60. z = 40:

V dV = (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2) = 1980

Nh vy, ch vi sai s 0.2cm trn mi s o dn n sai s xp x 1980cm3 khi tnh

th tch. iu xem ra c v l sai s ln, nhng n ch chim 1% s o th tch ca hp.

2.5. Quy tc dy chuyn

Nh li quy tc dy chuyn ca hm mt bin i vi hm hp: Nu y = f(x) v x = g(t)

l cc hm kh vi th y l hm kh vi gin tip theo t, v

[1]

=

i vi hm nhiu bin, quy tc dy chuyn c mt s dng, mi dng a ra mt quy

tc tnh vi phn ca hm hp. Dng th nht (nh l 2) cp ti trng hp z = f(x, y) v

cc bin x v y l cc hm theo t. iu c ngha l z ph thuc gin tip vo t, z = f(g(t),

h(t)), v quy tc dy chuyn cho ra dng vi phn z nh l hm ca t. Chng ta gi thit rng f

l kh vi (nh ngha 2.4.7). Nh li rng y l trng hp fx v fy lin tc (nh l 2.4.8).

[2] Quy tc dy chuyn th nht Gi s rng z = f(x, y) l hm kh vi theo cc bin x v y,

trong x = g(t) v y = h(t) l hai hm kh vi theo t. Khi z l hm kh vi theo t v

=

+

Chng minh S thay i ca t ti t sinh ra cc s thay i ca x ti x v y ti y. V th

dn ti s thay i ca z ti z v t nh ngha 2.4.7 ta c

n Ng Minh o hm ring

Trang 26

=

+

+ +

trong c e1 v e2 cng dn v 0 khi (x, y) (0, 0). [Nu cc hm e1 v e2 khng xc nh

tai (0, 0), chng ta cn nh ngha chng bng 0 ti .] Chia c hai v cho t ta c

=

+

+

+

Nu cho t 0 th x = g(t + t) g(t) 0 v g l hm kh vi nn n lin tc. Tng

t, y 0. iu c ngha l e1 0 v e2 0, v vy

= lim

=

lim

+

lim

+ lim

lim

+ lim

lim

=

+

+ 0.

+ 0.

=

+

Bi v chng ta thng vit

thay cho

nn c th vit li quy tc dy chuyn di

dng sau:

=

+

V d 1 Cho z = x2y + 3xy4 vi x = sin2t v y = cost. Tm dz/dt khi t = 0.

Li gii Ta c z/x = 2xy + 3y4, z/y = z2 + 12xy3,

dx/dt = 2cos2t, dy/dt = -sint

Theo quy tc dy chuyn, dz/dt = (2xy + 3y4)(2cos2t) + (z2 + 12xy3)(-sint)

Tc l khng cn thit thay x v y bi cc biu thc theo t. D thy rng khi t = 0 ta c x

= 0 v y = 1. V vy ti t = 0 th dz/dt = (0 + 3)(2cos0) + (0 + 0)(-sin0) = 6

Cc o hm trong V d 1 c th hiu l tc thay i ca z

theo t khi im (x, y) di chuyn dc theo ng cong C vi phng

trnh tham s x = sin2t, y = cost. (Xem Hnh 1.) c bit khi t = 0,

im (x, y) l (0, 1) v dz/dt = 6 l tc ca s tng khi di chuyn

dc theo ng cong C qua im (0, 1). C th, nu z = T(x, y) = x2y

+ 3xy4 biu th nhit ti im (x, y) th hm hp z = T(sin2t, cost)

biu th nhit ti im trn C v o hm dz/dt biu th thay i

ca nhit dc theo C.

V d 2 p lc (pressure) P(kilopascals, 9.8692326*10-3 atmosphere), th tch V(liters)

v nhit T(kelvins, 0 Kelvin = 273.15 Celsius) ca mt phn t gam (mol) kh l tng lin

quan vi nhau theo cng thc PV = 8.31T. Tm tc m p lc thay i khi nhit l 300K

v tng vi tc 0.1K/s, th tch V bng 100L v tng vi tc 0.2L/s.

Li gii Nu t biu th khong thi gian theo giy th ti thi im tc thi ta c T =

300, dT/dt = 0.1, V = 100, dV/dt = 0.2. Bi v P = 8.31T/V nn

=

+

=

.

.

= .

(0.1)

.()

(0.2) = 0.04155

p lc gim khong 0.042 kPa/s.

By gi ta xt trng hp z = f(x, y) nhng c x v y u l hm ca hai bin s v t:

n Ng Minh o hm ring

Trang 27

x = g(s, t), y = h(s, t)

Do z gin tip l hm ca s v t nn chng ta c th tm z/s v z/t. Nh li rng

Nh li rng khi tnh z/t ta gi s c nh. p dng nh l 2 ta nhn c

=

+

Tng t, ta tm c z/s. V th ta c dng th hai ca quy tc dy chuyn

[3] Quy tc dy chuyn th hai: Gi s rng z = f(x, y) l hm kh vi theo hai bin x v y,

trong x = g(s, t) v y = h(s, t) l cc hm kh vi lin tc theo s v t. Khi

=

+

=

+

V d 3 Cho z = exsiny, vi x = st2 v y = s2t. Tm z/s v z/t.

Li gii p dng quy tc dy chuyn th 2, ta nhn c

= ()() + ()(2) =

sin() + 2

cos ()

= ()(2) + ()() = 2

sin() +

cos ()

Dng 2 ca quy tc dy chuyn cha ba dng bin: s v t l cc bin c lp, x v y c

gi l cc bin trung gian, z l bin ph thuc.

Mt cch hiu qu nh quy tc dy chuyn l v s cy

(tree diagram) trn Hnh 2. Chng ta v cc nhnh bt u t bin

c lp z ti cc bin trung gian x v y biu th rng z l hm

ca x v y. Sau v cc nhnh t x v y ti cc bin c lp s v

t. Trn mi nhnh chng ta vit cc o hm ring tng ng.

tm z/s, chng la ly tch ca cc o hm ring dc theo ng

t z ti s.

By gi ta xt trng hp tng qut, bin ph thuc u l hm ca n bin trung gian x1, x2,

..., xn, mi bin trung gian li l hm ca m bin c t1, t2, ..., tm. Vic chng minh tng t nh

trng hp 1.

[4] Quy tc dy chuyn tng qut Gi s u l hm kh vi ca n bin x1, x2,..., xn v mi xj l

hm kh vi ca m bin t1, t2, ..., tm. Khi u l hm ca t1, t2, ..., tm v

`

=

+

+ +

vi mi i = 1, 2, ..., m.

V d 4 Vit ra quy tc dy chuyn cho trng hp w = f(x, y, z, t)

v x = x(u, v), y = y(u, v), z = z(u, v) v t = t(u, v).

Li gii Chng ta p dng nh l 4 vi n = 4 v m = 2. Hnh 3 biu th s cy.

Mc d chng ta cha vit cc o hm ring trn cc nhnh, nhng c hiu l, nu mt

nhnh dn t y ti u th o hm ring i vi nhnh l

y/u. Vi s tr gip ca s cy, chng ta c th vit

cc biu thc cn thit:

=

+

+

+

=

+

+

+

n Ng Minh o hm ring

Trang 28

V d 5 Cho u = x4y + y2z3, vi x = rset, y = rs2e-t, z = r2ssint.

Tm gi tr ca u/s khi r = 2, s = 1, t = 0.

Li gii Vi s tr gip ca s cy trn Hnh 4, ta c

=

+

+

=

= (4x3y)(ret) + (x4 + 2yz3)(2rse-t) + (3y2z2)(r2sint)

Khi r = 2, s = 1 v t = 0 ta c x = 2, y = 2 v z = 0.

V vy u/s = (64)(2) + (16)(4) + (0)(0) = 192.

2.5.1. o hm hm n

Gi s rng t phng trnh dng F(x, y) = 0 xc nh y nh l hm ca x, tc l y = f(x)

vi F(x, f(x)) = 0 vi mi x trn min xc nh ca f. Nu F kh vi, ta p dng dng th nht

ca quy tc dy chuyn, o hm hai v phng trnh F(x, y) = 0 theo bin x. V c x v y u

l hm ca x, ta nhn c

+

= 0

Nhng dx/dx = 1, nu F/y 0 ta gii ra c

[6]

=

=

nhn c phng trnh ny chng ta gi thit rng F(x, y) = 0 xc nh hm n

ca y theo x. nh l hm n cho ta iu kin m qua gi thit ca chng ta l hp l: Nu f

c xc nh trong ln cn ca (a, b), y F(a, b) = 0, Fy(a, b) 0, Fx v Fy l cc hm lin

tc trong ln cn , th phng trnh F(x, y) = 0 xc nh hm n y theo x trong ln cn ca (a,

b) v o hm ca hm ny c cho bi phng trnh 6.

V d 8 Tm y' nu x3 + y3 = 6xy.

Li gii Phng trnh cho c th vit l F(x, y) = x3 + y3 6xy = 0, v vy phng

trnh 6 cho ra

=

=

=

.

Gi s z = f(x, y) l hm n c cho bi dng F(x, y, z) = 0, tc l F(x, y, f(x, y)) = 0

vi mi (x, y) thuc min xc nh ca f. Nu F v f kh vi, ta s dng quy tc dy chuyn vi

phng trnh ny

+

+

= 0

Nhng v

=

() = 1 v

=

() = 0 nn

+

= 0

Nu F/z 0, ta nhn c z/x nh trong cng thc 7. Tng t, ta cng xy dng

c cng thc tnh z/y.

[7]

=

=

=

=

Thm mt dng na ca nh l hm n ch ra iu kin gi thit ca chng ta tha

mn: Nu F c xc nh bn trong mt cu cha (a, b, c), ti F(a, b, c) = 0, Fz(a, b, c) 0

n Ng Minh o hm ring

Trang 29

v cc o hm ring Fx, Fy, Fz lin tc trn mt cu, th phng trnh F(x, y, z) = 0 xc nh z

l hm ca x v y trong ln cn ca (a, b, c) v hm ny kh vi, vi cc o hm ring c

tnh theo cng thc 7.

V d 9 Tm z/x v z/y nu x3 + y3 + z3 + 6xyz = 1.

Li gii Gi s F(x, y, z) = x3 + y3 + z3 + 6xyz 1 = 0. T phng trnh 7 ta c

=

=

2.6. o hm theo hng v vc t gradient

Bn thi tit trn Hnh 1 biu th bn ng mc

ca hm nhit T(x, y) ca cc bang California v

Nevada ti 3:00pm trong mt ngy ca thng Mi. Cc

ng mc, hay ng nhit (isothermals), kt ni cc vng

c cng nhit . o hm ring Tx ti mt a phng nh

Reno l tc thay i ca nhit i vi khong cch

nu chng ta di chuyn xung pha ng t Reno, Ty l tc

thay i ca nhit nu chng ta di chuyn xung pha

bc. Nhng s th no nu chng ta mun bit tc thay

i ca nhit khi chng ta di chuyn v pha ng Nam

(n Las Vegas), hoc trong mt s hng khc? Trong phn ny chng ta gii thiu mt dng

o hm, c gi l o hm theo hng, cho php chng ta tm tc thay i ca mt hm

hai hay nhiu bin theo bt k hng no.

2.6.1. o hm theo hng

Nh li rng cc o hm ring fx v fy c nh ngha nh sau

v biu th tc thay i ca z theo hng x v y, tc l hng ca cc vc t n v i v j.

n Ng Minh o hm ring

Trang 30

Gi s rng ta mun tm tc thay i ca z ti (x0, y0) theo hng vc t n v bt

k u = a, b. (Xem Hnh 2.) lm iu chng ta xt mt S vi phng trnh z = f(x, y) v

gi s z0 = f(x0, y0), khi im P(x0, y0, z0) thuc mt S. Mt phng i qua P v song song vi

u giao vi S theo ng cong C. (Xem Hnh 3.) dc ca ng tip tuyn ca C ti im P

chnh l tc thay i ca z theo hng u.

Nu Q(x, y, z) l mt im khc nm trn C, P' v Q' l cc hnh chiu ca P v Q ln

mt phng xy th vc t song song vi u v = = , vi hng s h no .

Bi v x x0 = ha, y y0 = hb nn x = x0 + ha, y = y0 + hb v

=

=

( , )(, )

Chuyn qua gii hn khi h 0, ta nhn c tc thay i ca z (theo khong cch)

theo hng u, v c gi l o hm theo hng ca f theo hng ca u.

[2] nh ngha o hm theo hng ca f ti (x0, y0) theo hng ca vc t n v

u = a, b (nu gii hn tn ti) l

(, ) = lim

( + , + ) (, )

Bng cch so snh nh ngha 2 vi phng trnh [1], ta thy rng nu u = i = 1, 0 th

Dif = fx v nu u = j = 0, 1 th Djf = fy. Ni khc i, cc o hm ring ca f theo x v y l cc

trng hp c bit ca o hm theo hng.

V d 1 S dng bn thi tit trn Hnh 1 c lng gi tr ca o hm theo hng

ca hm nhit ti Reno theo hng ng Nam.

Li gii Vc t n v nh hng ng Nam l = ( )/2 , nhng chng ta khng

mun dng biu thc ny. Ta bt u v mt

ng thng qua Reno hng ng Nam

(Xem Hnh 4). Chng ta xp x o hm theo

hng DuT bi gi tr trung bnh ca tc

thay i ca nhit gia cc im m ng

thng ny ct cc ng ng mc T = 50 v

T = 60. Nhit ti im ng Nam ca Reno

l T = 60oF v nhit ti im Ty Bc ca

Reno l T = 50oF. Khong cch gia cc im

c xem l khong 75 dm. V vy tc thay

i ca nhit theo hng ng Nam l

0.13/

Khi tnh o hm theo hng ca hm c xc nh bi cng thc, chng ta hay dng

nh l sau.

[3] nh l Nu f l hm kh vi ca x v y th f c o hm theo mi hng ca vc t n

v u = a, b, v Duf(x, y) = fx(x, y)a + fy(x, y)b.

n Ng Minh o hm ring

Trang 31

Chng minh Nu nh ngha hm g mt bin h, g(h) = f(x0 + ha, y0 + hb), th theo nh ngha

ca o hm ta c

[4] (0) = lim

()()

= lim

( , )(, )

= (, )

Mt khc, ta c th vit g(h) = f(x, y), trong x = x0 + ah, y = y0 + bh, nn theo quy tc

dy chuyn

() =

+

= fx(x, y)a + fy(x, y)b

t h = 0 th x = x0 v y = y0, v

[5] g'(0) = fx(x0, y0)a + fy(x0, y0)b

So snh phng trnh 4 v phng trnh 5 ta thy rng

Duf(x0, y0) = fx(x0, y0)a + fy(x0, y0)b

V d 2 Tm o hm theo hng Duf(x, y) nu f(x, y) = x3 3xy + 4y2 v u l vc t n

v c cho bi gc = p/6. Hi rng Duf(1, 2) l g?

Li gii Cng thc 6 cho ra Duf(x, y) = fx(x, y)cosp/6 + fy(x, y)sinp/6

= (3 3)

+ ( 3 + 8)

=

33 3 + 3 33

Do

(1,2) =

33(1) 3(1) + 3 33(2) =

o hm theo hng Duf(1, 2) biu th tc thay i ca

z theo hng u. l dc ca tip tuyn ca ng cong l

giao ca mt cong z = x3 3xy + 4y2 vi mt phng nm ngang

i qua im (1, 2, 0) theo hng ca u (Xem Hnh 5).

2.6.2. Vc t gradient

Theo nh l 3, o hm theo hng ca hm kh vi c th vit di dng tch v hng

ca hai vc t:

[7] Duf(x, y) = fx(x, y)a + fy(x, y)b = fx(x, y), fy(x, y)a, b = fx(x, y), fy(x, y)u

Vc t u tin trong tch v hng khng ch xut hin trong vic tnh o hm theo

hng m cn trong mt s ng cnh khc. V vy ta t cho n mt ci tn c bit (gradient

ca f) v k hiu c bit ( grad f, hoc f, c l "del f").

[8] nh ngha Nu f l hm ca hai bin x v y th gradient ca f l vc t f c xc

nh bi

(, ) = (, ), (, )=

+

V d 3 Nu f(x, y) sinx + exy th

(, ) = , = , v (1,2) = 2,0

Vi k hiu ny ca vc t gradient, ta c th vit phng trnh 7 di dng

n Ng Minh o hm ring

Trang 32

[9] Duf(x, y) = f(x, y)u

iu ny th hin o hm theo hng ca vc t n v u l chiu v hng ca vc t

gradient ln u.

V d 4 Tm o hm theo hng ca hm f(x, y) = x2y3 4y ti im (2, -10 theo hng

ca vc t v = 2i + 5j.

Li gii Trc ht ta tnh vc t gradient ti (2, -1):

f(x, y) = 2xy3i + (3x2y2 4)j f(2, -1) = -4i + 8j

Ch rng v khng phi l vc t n v, nhng v |v| = 29 nn vc t n v theo hng

v l

=

||=

+

V vy theo phng trnh 9 ta c

(2, 1) = (2, 1) = ( 4 + 8)

+

= ()()()()

=

Vc t gradient f(2,-1) trong V d 4 c minh ha trong

Hnh 6. C vc t v l hng ca o hm theo hng. C hai vc

t u xp chng ln cc ng mc ca th ca f.

2.6.3. Hm ba bin

Chng ta c th nh ngha o hm theo hng ca hm ba bin mt cch tng t. Mt

ln na, Duf(x, y, z) c th hiu l tc thay i ca hm theo hng ca vc t n v u.

[10] nh ngha o hm theo hng ca f ti (x0, y0, z0) theo hng ca vc t n v

u = a, b, c l (nu gii hn tn ti):

(, , ) = lim

( , , )(, , )

Nu s dng k hiu vc t th chng ta c th vit c hai nh ngha (2 v 10) ca o

hm theo hng trong dng chung

[11] () = lim

( )()

trong x0 = x0, y0 nu n = 2 v x0 = x0, y0, z0 nu n = 3. iu l hp l bi v phng

trnh vc t ca ng thng i qua x0 theo hng ca vc t u l x = x0 + tu v do f(x0 +

tu) th hin gi tr ca f ti mt im trn ng ny.

Nu f(x, y, z) l kh vi v u = a, b, c th vi phng php s dng chng minh

nh l 3, ta c th s dng chng t rng

[12] Duf(x, y, z) = fx(x, y, z)a + fy(x, y, z)b + fz(x, y, z)c

Vi hm ba bin f, vc t gradient l f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z)

hoc dng ngn gn

[13] = , , =

+

+

V th, ging nh vi hm hai bin, cng thc [12] c th vit li di dng

[14] Duf(x, y, z) = f(x, y, z)u

n Ng Minh o hm ring

Trang 33

V d 5 Cho f(x, y, z) = xsinyz. (a) tm gradient ca f v (b) tm o hm theo hng ca

f ti (1, 3, 0) theo hng ca v = i + 2j k.

Li gii (a) Gradient ca f l

(, , ) = (, , ), (, , ), (, , )= , ,

(b) Ti (1, 3, 0) ta c f(1, 3, 0) = 0, 0, 3. Vc t n v theo hng v l

=

+

Do phng trnh 14 cho ra

Duf(1, 3, 0) = f(1, 3, 0)u = 3

+

=

2.6.4. Cc i ca o hm theo hng

Gi s ta c hm hai hoc ba bin v ta xt mi kh nng ca o hm theo hng ca f

ti mt im cho. Chng a ra cc tc thay i ca f theo mi hng. Ta c th t cu

hi: Hng no th f thay i nhanh nht v nh th no th tc thay i t gi tr ln nht?

Cu tr li chnh l nh l sau.

[15] nh l Gi s f l hm kh vi hai hoc ba bin. Gi tr ln nht ca o hm theo hng

Duf(x) l |f(x)| v n xy ra khi u c cng hng vi vc t gradient f(x).

Chng minh T phng trnh 9 hoc 14 ta c

Duf = f u = |f(x)||u|cos = |f|cos

trong l gc gia f v u. Gi tr ln nht ca cos l 1 v n xy ra khi = 0. Do gi

tr ln nht ca Duf l |f| v n xy ra khi = 0, tc l khi u cng hng vi f.

V d 6

(a) Cho f(x, y) = xey, tm tc thay i ca f ti im P(2, 0) theo hng t P ti Q(1/2,2).

(b) Theo hng no th f t tc thay i ln nht? Gi tr ln nht ny l g?

Li gii

(a) Truocs ht ta tnh gradient ca f

f(x, y) = fx, fy = ey, xy f(2, 0) = 1, 2

Vc t n v theo hng ca = 1.5, 2 l =

,

, v vy tc thay i ca

f theo hng t P ti Q l

(2,0) = (2,0) = 1,2

,

= 1

(b) Theo nh l 15, f tng nhanh nht theo hng ca vc t gradient (2,0) = 1,2.

Tc thay i ln nht l |(2,0)| = |1,2| = 5.

n Ng Minh o hm ring

Trang 34

Ti (2, 0) hm trong V d 6 tng nhanh nht theo hng ca vc t gradient (2,0) =

1,2. Hnh 7 ni ln rng vc t ny dng nh vung gc vi ng mc i qua (2, 0).

Hnh 8 m t th ca f v vc t gradient.

V d 7 Gi s rng nhit ti im (x, y, z) trong khng gian c cho bi

T(x, y, z) = 80/(1 + x2 + 2y2 + 3z2), trong T l thang o oC v th nguyn ca x, y, z l mt.

Theo hng no th nhit ti im (1, 1, -2) tng nhanh nht? Gi tr ln nht bng bao nhiu?

Li gii Vc t gradient ca T l

=

+

+

=

=

( x 2 3)

Ti im (1, 1, -2) vc t gradient l

(1,1, 2) =

()

()

Theo nh l 15, nhit tng nhanh nht theo hng ca vc t gradient, hoc tng

ng, theo hng ca vc t 2 + 6, hoc vc t n v ( 2 + 6)/41.

Tc nhanh nht ca s tng l ln ca vc t gradient:

|(1,1, 2)| =

| 2 + 6| =

41 4 (/ )

2.6.5. Mt phng tip din ca mt mc

Gi s S l mt cong vi phng trnh F(z, y, z) = k, tc l mt mc ca hm ba bin F.

Gi s P(x0, y0, z0) l im thuc S v C l ng cong bt k thuc S v i qua P. Nh li t

mc 1.1, ng cong C c m t bi hm vc t lin tc r(t) = x(t), y(t), z(t). Gi s t0 l

gi tr ca tham s ng vi im P, tc l r(t0) = x0, y0, z0. Boeir v C thuc S nn mi im

(x(t), y(t), z(t)) phi tha mn phng trnh ca S, tc l

[16] F(x(t), y(t), z(t)) = k

Nu x, y v z l accs hm kh vi theo t v F cng kh vi th chunga ta c th s dng quy

tc dy chuyn ly o hm hai v ca phng trnh 16:

[17]

+

+

= 0

Nhng F = Fx, Fy, Fz v r'(t) = x'(t), y'(t), z'(t), phng trnh 17 c th vit di dng

tch v hng nh sau F r'(t) = 0

Trng hp ring, khi t = t0 ta c r(t0) = x0, y0, z0, v th

[18] F(x0, y0, z0) r'(t0) = 0

Phng trnh 18 ni ln rng vc t gradient ti P, F(x0,

y0, z0), vung gc vi vc t tip tuyn r'(t0) ca mi ng cong

C thuc S m i qua P. (Xem Hnh 9.) Nu F(x0, y0, z0) 0,

mt cch t nhin nh ngha mt phng tip din ca mt

mc F(x, y, z) = k ti P(x0, y0, z0) chnh l mt phng i qua P

v c vc t php tuyn F(x0, y0, z0). Chng ta c th vit

phng trnh ca mt phng tip din nh sau

n Ng Minh o hm ring

Trang 35

[19] Fx(x0, y0, z0)(x x0) + Fy(x0, y0, z0)(y y0) + Fz(x0, y0, z0)(z z0) = 0

ng php tuyn ca S ti P l ng i qua P v vung gc vi mt phng tip din.

Hng ca php tuyn l xc nh bi vc t gradient F(x0, y0, z0), v th phng trnh i

xng (symmetric) ca n l

[20]

(,,)=

(,,)=

(,,)

Trong trng hp ring, phng trnh ca mt cong S c dng z = f(x, y) (tc S l th

ca hm hai bin), chng ta c th vit li di dng F(x, y, z) = f(x, y) z = 0, v S l mt mc

(vi k = 0) ca F. V vy

Fx(x0, y0, z0) = fx(x0, y0) Fy(x0, y0, z0) = fx(x0, y0) Fz(x0, y0, z0) = -1

v phng trnh 19 tr thnh fx(x0, y0)(x x0) + fy(x0, y0)(y y0) (z z0) = 0

n tng ng vi phng trnh 2.4.2. Do , nh ngha tng qut ny ca mt phng tip

din l ph hp vi nh ngha c trong trng hp c bit trong phn 2.4.

V d 8 Tm cc phng trnh ca mt phng tip din v vc t php tuyn ti im (-

2, 1, -3) ca ellipsoid

+ +

= 3

Li gii Ellipsoid l mt mc (vi k = 3) ca hm F(x, y, z) =

+ +

. Do ta c

(, , ) =

(, , ) = 2 (, , ) =

( 2,1, 3) = 1 ( 2,1, 3) = 2 ( 2,1, 3) =

T phng trnh 19 suy ra phng trnh tip din ti ( 2,1, 3) l

-1(x + 2) + 2(y 1)

(x + 3) = 0, hay 3x 6y + 2z + 18 = 0.

Theo phng trnh 20, phng trnh i xng ca php tuyn l

=

Hnh 10 biu th ellipsoid, mt phng tip din v php tuyn trong V d 8.

2.6.6. Tm quan trng ca vc t gradient

Chng ta tm tt mt s vai tr quan trng ca vc t gradient. Trc ht chng ta xem

xt hm hai bin f v im P(x0, y0, z0) thuc min xc nh ca n. Mt mt, t nh l 15 ta

bit rng vc t gradient F(x0, y0, z0) a ra hng m f tng nhanh nht. Mt khc, ta bit

rng f(x0, y0, z0) trc giao vi mt mc S ca f ti P. (Tham kho Hnh 9.) Theo trc gic, hai

tnh cht ny tng i ph hp bi khi ta di chuyn ri xa P trn mt mc, gi tr ca f khng

h thay i. V vy, n c v hp l rng nu chng ta di chuyn theo hng vung gc, chng

ta nhn c s tng ti a.

Tng t nh th, chng ta xem xt hm hai bin f v im P(x0, y0) thuc min xc nh

ca n. Ln na, vc t gradient f(x0, y0) xc nh hng tng nhanh nht ca f. Tng t nh

cch xem xt mt phng tip din, chng t rng f(x0, y0) vung gc vi ng mc f(x, y) =

k m i qua P. Mt ln na, trc gic ng tin bi cc gi tr ca f khng i khi chng ta di

chuyn dc theo ng cong (Xem Hnh 11.)

n Ng Minh o hm ring

Trang 36

Nu chng ta xem xt bn a hnh ca mt ngn i v gi s f (x, y) th hin cho

chiu cao trn mc nc bin ti im c ta (x, y), th mt ng cong dc nht c th

c v trong Hnh 12 bng cch lm cho n vung gc vi tt c cc ng ng mc.

H thng my tnh i s c cc lnh v mu cc vc t gradient. Mi vc t gradient

f(a, b) c v bt u ti im (a, b). Hnh 13 biu th trng vc t gradient i vi hm

f(x, y) = x2 y2 chng ln bn ng mc ca f. Theo d kin, cc vc t gradient ch ra im

"ln dc" v l vung gc vi cc ng mc.

2.7. Cc gi tr ln nht v nh nht

2.7.1. Cc i a phng v cc tiu a phng

Nh bit i vi hm mt bin, mt trong nhng ng dng ca o hm l tm cc gi

tr cc i v cc tiu, gi chung l cc tr (extreme values). Trong phn ny chng ta xem xt

s s dng cc o hm ring xc nh cc gi tr cc tr ca hm hai bin. c bit, trong

V d 6 chng ta nht nh s thy bng cch no cc i ha th tch ca khi hp khng

c np vi cng mt din tch ca tm ba cc tng.

Hy xem cc nh li (hill) v lm (valley) trn th

ca f trong Hnh 1. C hai im (a, b) m f c cc i

a phng, tc l f(a, b) l ln hn cc gi tr f(x, y) gn n.

Gi tr ln nht trong hai gi tr ny l cc i tuyt i.

Tng t, f c hai cc tiu a phng, wor f(a, b) l nh

hn mi gi tr gn n, Gi tr nh hn trong hai gi tr l

cc tiu tuyt i.

[1] nh ngha Hm hai bin c cc i a phng ti (a, b) nu f(x, y) f(a, b) khi

(x, y) gn (a, b). [Ngha l f(x, y) f(a, b) vi mi (x, y) trong hnh trn tm (a, b).] S f(a, b)

c gi l gi tr cc i a phng. Nu f(x, y) f(a, b) khi (x, y) gn (a, b) th f c cc tiu

a phng ti (a, b) v f(a, b) l gi tr ca tiu a phng.

Nu cc bt ng thc trong nh ngha 1 ng cho mi im (x, y) trong min xc nh

ca f th f c cc i tuyt i (hoc cc tiu tuyt i) ti (a, b).

[2] nh l Nu f c cc i hoc cc tiu a phng ti (a, b) v cc o hm ring cp

mt tn ti th fx(a, b) = 0 v fy(a, b) = 0.

Chng minh Gi s g(x) = f(x, b). Nu f c cc i a phng (hoc cc tiu) ti (a, b) th g

c cc i a phng (hoc cc tiu) ti a, v th g'(a) = 0 theo nh l Fermat. Nhng g'(a) =

fx(a, b) nn fx(a, b) = 0. Tng t, ta nhn c fy(a, b) = 0.

n Ng Minh o hm ring

Trang 37

Nu chng ta t fx(a, b) = 0 v fy(a, b) = 0 vo phng trnh ca mt phng tip din

(Phng trnh 2.4.2), ta nhn c z = z0. V vy ngha hnh hc ca nh l 2 l nu th

ca f c mt phng tip din ti im cc tr th mt phng tip din nm ngang.

im (a, b) c gi l im ti hn (critical) hay im dng (stationary) ca f nu fx(a,

b) = 0 v fy(a, b) = 0, hoc mt trong hai o hm ring khng tn ti. nh l 2 ni ln rng,

nu f c cc tr a phng ti (a, b) th (a, b) l im ti hn ca f. Tuy nhin, nh vi hm

mt bin, khng phi tt c cc im ti hn u l im cc tr. Ti im ti hn, hm c th

c cc tiu a phng hoc cc i a phng hoc khng.

V d 1 Gi s f(x, y) = x2 + y2 2x 6y + 14. Khi fx(x, y) = 2x 2, fy(x, y) = 2y 6

Cc o hm ring ny bng 0 khi x = 1 v y = 3, v vy

ch c duy nht im ti hn (1, 3). a v dng bnh phng,

ta c f(x, y) = 4 + (x 1)2 + (y 3)2. Bi v (x 1)2 0 v (y

3)2 0, ta c f(x, y) 4 vi mi gi tr ca x v y. V vy

f(1, 3) = 4 l cc tiu a phng v thc t n l cc tiu tuyt

i ca f. iu c th c khng nh hnh hc t th

ca f, l mt paraboloid elliptic vi nh (1, 3, 4) trong

Hnh 2.

V d 2 Tm cc gi tr cc tr ca f(x, y) = y2 x2.

Li gii T fx = -2x v fy = 2y, duy nht im ti hn l (0, 0). Ch rng i vi cc

im trn trc x ta c y = 0, nn f(x, y) = -x2 < 0 (nu x 0).

Tuy nhin, vi cc im trn trc y ta c x = 0 nn f(x, y) =

y2 > 0 (nu y 0). V vy mi hnh trn tm (0, 0) u cha

nhng im lm cho f dng v cha nhng im lm cho f

m. V vy f(0, 0) = 0 khng th l cc tr ca f, v vy f

khng c cc tr.

V d 2 minh

ha cho s kin hm

khng t cc tr ti im ti hn. Hnh 3 m t khi no

th c th. th ca f l mt paraboloid hyperbolic z =

y2 x2, n c mt phng tip din nm ngang (z = 0) ti

gc ta . Ta c th thy rng f(0, 0) = 0 l cc i theo

hng ca trc x nhng l cc tiu theo hng ca trc

y. Gn gc ta th c hnh dng ci yn nga

(saddle), v vy (0, 0) c gi l im yn nga.

Mt ng i trong ni cng c hnh dng ca yn

nga. Nh bc nh ca s hnh thnh a cht minh ha,

vi nhng ngi leo ni theo mt hng th im yn

n Ng Minh o hm ring

Trang 38

nga l im thp nht trn tuyn ng ca h, trong khi i vi nhng ngi i du lch theo

mt hng khc im yn nga l im cao nht.

Chng ta cn xc nh khi no mt hm c gi tr cc tr ti im ti hn. nh l sau

y, c chng minh cui phn ny, tng t nh i vi hm mt bin.

[3] nh l Gi s cc o hm ring cp hai ca f lin tc trn mt hnh trn tm (a, b), v

gi s rng fx(a, b) = 0 v fy(a, b) = 0, tc l, (a, b) l im ti hn ca f. Gi s

D = D(a, b) = fxx(a, b) fyy(a, b) [fxy(a, b)]2

(a) Nu D > 0 v fxx(a, b) > 0 th f(a, b) l cc tiu a phng

(b) Nu D > 0 v fxx(a, b) < 0 th f(a, b) l cc i a phng

(c) Nu D < 0 th f(a, b) khng l cc tr a phng

Ch 1 Trong trng hp (c) im (a, b) c gi l im dng ca f v th ca f bt

cho qua vi mt phng tip din ca n ti (a, b).

Ch 2 Nu D = 0, chng ta khng khng nh c g, f c th c hoc khng cc tr

a phng, hoc (a, b) l im dng ca f.

Ch 3 d nh cng thc ca D, ta vit n di dng nh thc

=

=

V d 3 Tm cc gi tr cc i a phng v cc tiu a phng v cc im yn nga

ca f(x, y) = x4 + y4 4xy + 1.

Li gii Trc ht chng ta xc nh cc im ti hn: fx = 4x3 4y fy = 4y3 4x

t cc o hm ring bng 0 ta nhn c cc phng trnh

x3 y = 0 v y3 x = 0.

gii h ny ta thay y = x3 t phng trnh u tin vo phng trnh th hai.

Ta c

0 = x9 x = x(x8 1) = x(x4 1)(x4 + 1) = x(x2 1)(x2 + 1)(x4 + 1)

v vy cc nghim thc l x = 0, x = 1, x = -1. Ba im ti hn l (0, 0), (1, 1) v (-1, -1).

Tip theo chng ta tnh cc o hm ring cp hai v D(x, y):

fxx = 12x2 fxy = -4 fyy = 12y2 D(x, y) = 144x2y2 16

Bi v D(0, 0) = -16 < 0, dn ti trng hp (c) ca

nh l 3 nn gc ta l im yn nga, tc l f khng c

cc tr a phng ti (0, 0).

Bi v D(1, 1) = 128 > 0 v fxx(1, 1) = 12 > 0, t trng

hp (a) ca nh l 3 suy ra f(1, 1) = -1 l cc tiu a

phng.

Tng t, ta c D(-1, -10 = 128 > 0 v fxx(-1, -1) = 12

> 0 nn f(-1, -1) = -1 cng l cc tiu a phng. th ca

f c th hin Hnh 4.

n Ng Minh o hm ring

Trang 39

Mt bn ng mc ca hm f trong V d 3 c

th hin trong Hnh 5. Cc ng mc gn (1, 1) v (-1, -1)

l hnh bu dc v ch ra rng khi chng ta di chuyn ra khi

(1, 1) hoc (-1, -1) theo bt k hng no th cc gi tr ca f

tng ln. Mt khc, cc ng mc gn (0, 0), ging cc

hyperbola. Chng ni ln rng khi chng ta di chuyn ra khi

gc ta ( gi tr ca f l 1), cc gi tr ca f gim theo

mt s hng nhng tng theo cc hng khc. Do , bn

ng mc cho thy s hin din ca cc cc tiu v im

yn nga m chng ta tm thy trong V d 3.

V d 4 Tm v phn loi cc im ti hn ca hm

f(x, y) = 10x2y 5x2 4y2 x4 2y4

T tm im cao nht trn th ca f.

Li gii Cc o hm ring cp mt l fx = 20xy 10x 4x3 v fy = 10x2 8y 8y3

V th, tm cc im ti hn ta cn gii cc phng trnh

[4] 2x(10y 5 2x2) = 0

[5] 5x2 4y 4y3 = 0

T phng trnh 4 ta thy x = 0 hoc 10y 5 2x2 = 0.

Trong trng hp th nht (x = 0), phng trnh 5 tr thnh

4y(1 + y2) = 0, v th y = 0 v ta c im ti hn (0, 0).

Trong trng hp th 2 (10y 5 2x2 = 0) ta c

[6] x2 = 5y 2.5

v th vo phng trnh 5 ta c 25y 2.5 4y 4y3 = 0. Ta phi gii phng trnh bc ba

[7] 4y3 21y + 12.5 = 0

S dng my tnh v th ca hm g(y) = 4y3 21y + 12.5 nh Hnh 6, ta thy

phng trnh 7 c ba nghim thc: y -2.5452 y 0.6468 y 1.8984

(Chng ta c th s dng phng php Newton tm cc nghim xp x ny).

T phng trnh 6 gi tr x tng ng c cho bi cng thc = 5 2.5

Nu y -2.5452 th x khng c nghim thc. Nu y 0.6468 th x 0.8567. Nu y

1.8984 th x 2.6442. V th chng ta c tt c 5 im ti hn, c phn tch trong biu

sau. Tt c c lm trn ti 2 ch s thp phn.

n Ng Minh o hm ring

Trang 40

Hnh 7 v Hnh 8 l hai gc nhn th ca f v ta thy mt cong m dn xung pha

di. [iu ny c th nhn thy t biu thc ca f(x, y): Hng thc chim u th l x4 2y2

khi |x| v |y| ln]. So snh cc gi tr ca f ti cc im cc i a phng, ta thy rng gi tr

cc i tuyt i ca f l f(2.64) 8.50. Ni khc i, cc im cao nht trn th ca f l

(2.64, 1.90, 8.50)

Hnh 9 l bn ng mc ca hm f trong V d 4. Nm im ti hn c th hin bi

cc chm trn.

V d 5 Tm khong cch t im (1, 0, -2) ti mt phng x + 2y + z = 4.

Li gii Khong cch t im bt k (x, y, z) ti im (1, 0, -2) l

= ( 1) + + ( + 2)

nhng nu (x, y, z) thuc mt phng x + 2y + z = 4 th z = 4 x 2y v v vy ta c

= ( 1) + + (6 2)

Chng ta c th cc tiu d theo biu thc n gin hn

= (, ) = ( 1) + + (6 2)

Bng cch gii cc phng trnh

fx = 2(x 1) 2(6 x 2y) = 4x + 4y 14 = 0

fy = 2y 4(6 x 2y) = 4x + 10y 24 = 0

chng ta tm ng cong duy nht im ti hn

,. Bi v fxx = 4, fyy = 4 v fxy = 10,

chng ta c D(x, y0 = fxxfyy (fxy)2 = 24 > 0 v fxx > 0, nn theo nh l 3 f c cc tiu a

phng ti

,. Bng trc gic chng ta c th thy rng cc tiu a phng ny thc t l

cc tiu tuyt i bi v phi c mt im trn mt phng cho l gn im (1, 0, -2) nht.

Nu =

v =

th =

+

+

=

6.

V d 6 Mt khi hp ch nht khng np c to nn t 12m2 ba cc tng.

Tm th tch cc i c th ca n.

Li gii Gi s di, rng v chiu cao ca hp (theo m) l x, y v z, nh trn Hnh

10. Khi th tch ca hp l V = xyz.

Chng ta c th xem V l hm ca hai bin x v y bng

cch s dng s kin rng din tch ca bn mt bn v y di

ca hp l 2xz + 2yz + xy = 12.

T y gii ra z ta c = (12 )/[2( + )], v th

biu thc ca V l =

()=

()

n Ng Minh o hm ring

Trang 41

Chng ta tnh cc o hm ring

=

()

=

()

Nu V t cc i th V/x = V/y = 0, nhng x = 0 hoc y = 0 s cho V = 0, nn chng

ta cn gii cc phng trnh 12 2xy x2 = 0 12 2xy y2 = 0

iu bao hm x2 = y2 v do x = y. (Ch rng x v y cn phi dng.) Nu chng

ta t x = y vo mt trong hai phng trnh ta nhn c 12 3x2 = 0, gi ra c x = 2, dn

n y = 2 v z = [12 (2)(2)]/[2(2 + 2)] = 1.

Chng ta cn s dng nh l 3 chng t rng iu dn n gi tr cc i ca V,

hoc chng ta c th lp lun n gin t l tnh ca bi ton rng l gi tr cc i tuyt i

ca th tch khi x = 2, y = 2 v z = 1. Vy V = 4, tc l th tch cc i ca hp l 4m3.

2.7.2. Cc tr tuyt i

Vi hm mt bin, nh l v gi tr cc tr ni rng nu f lin tc trn min ng [a, b],

th f c gi tr cc tiu tuyt i v gi tr cc i tuyt i. Cc gi tr ny c th l gi tr ti

hn hoc gi tr ti cc mt a v b.

Hm hai bin cng c tnh trng tng t nh th. Cng

nh on ng cha cc im mt, tp ng trong R2 l tp

cha tt c cc im bin ca n. [im bin ca D l im (a,

b) sao cho bt k hnh trn tm (a, b) u cha nhng im

thuc D cng nh khng thuc D.]

V d hnh trn D = {(x, y) | x2 + y2 1} bao gm tt c

cc im trn v trong ng trn x2 + y2 = 1, l tp ng bi

v n cha tt c cc im bin ca n (l nhng im trn

ng trn x2 + y2 = 1). Nhng nu ch mt im trn bin ny

b b i th tp khng cn l ng na. (Xem Hnh 11.)

[8] nh l cc tr i vi hm hai bin Nu f lin tc trn min ng gii ni D trong R2

th f t gi tr cc i tuyt i f(x1, y1) v gi tr cc tiu tuyt i f(x2, y2) ti cc im

(x1, y1) v (x2, y2) trong D.

tm cc gi tr cc tr theo nh l 8, chng ta ch rng, t nh l 2, nu f c cc

tr ti (x1, y1) th (x1, x2) hoc l im ti hn hoc l im bin ca D. V vy chng ta c quy

tc sau y.

[9] tm cc tr tuyt i ca hm lin tc f trn min ng gii ni D:

1. Tm cc gi tr ca f ti cc im ti hn ca n trn min D.

2. Tm cc gi tr cc tr ca f trn bin ca D.

3. Gi tr ln nht trong cc gi tr trn chnh l gi tr cc i tuyt i, gi tr nh nht

trong cc gi tr trn chnh l gi tr cc tiu tuyt i.

V d 7 Tm cc tr tuyt i ca hm

f(x, y) = x2 2xy + 2y trn min ch nht D = {(x, y) | 0 x 3, 0 y }.

Li gii V f l a thc, lin tc trn min ng gii ni nn theo nh l 8, c c cc

i tuyt i v cc tiu tuyt i. Theo bc 1 trong [9] chng ta tm cc im ti hn, xy ra

n Ng Minh o hm ring

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khi fx = 2x 2y = 0 v fy = -2x + 2 = 0, v th c duy nht im ti hn l (1, 1), v gi tr l

f(1, 1) = 1.

Ti bc th 2, chng ta xem xt cc gi tr ca f trn bin

ca D, gm bn on thng l L1, L2, L3 v L4 nh trn Hnh

12. Trn L1 ta c y = 0 v f(x, 0) = x2 , 0 x 3. y l hm

tng theo x, nn gi tr cc tiu ca n l f(0, 0) = 0 v gi tr

cc i ca n l f(3, 0) = 9. Trn L2 ta c x = 3 v f(3, y) = 9

4y, 0 y 2. y l hm gim theo y, nn gi tr cc tiu ca

n l f(3, 2) = 1 v gi tr cc i ca n l f(3, 0) = 9. Trn L3

ta c y = 2 v f(x, 2) = x2 4x + 4, 0 x 3.

D thy f(x, 2) = (x 2)2 nn gi tr cc tiu ca n l

f(2, 2) = 0 v gi tr cc i ca n l f(0, 2) = 4.

Cui cng, trn L4 ta c x = 0 v f(0, y) = 2y, 0 y 2.

D thy gi tr cc tiu ca n l f(0, 0) = 0 v gi tr cc i

ca n l f(0, 2) = 4. V th trn bin, hm c gi tr cc tiu l

0 v gi tr cc i l 9.

Ti bc th 3, chng ta so snh cc gi tr vi gi tr ti im ti hn f(1, 1) = 1, suy

ra rng gi tr cc tiu tuyt i l f(0, 0) = f(2, 2) = 0 v gi tr cc i tuyt i l f(3, 0) = 9.

Hnh 13 m t th ca f.

Chng ta kt thc phn ny bng s chng minh phn u tin ca nh l 3. Phn (b)

chng minh tng t.

Chng minh nh l 3, phn (a) Chng ta tnh cc o hm ring cp 1 v cp 2 ca f theo

hng u = h, k. Theo nh l 2.6.3, Duf = fxh + fyk. p dng nh l ny ln hai ta c

[10] = () =

() +

() = + + +

= + 2 +

= +

+

Chng ta c fxx(a, b) > 0 v D(a, b) > 0. Nhng fxx v D = l cc hm lin

tc, v vy tn ti hnh trn B tm (a, b) bn knh > 0 sao cho fxx(x, y) > 0 v D(x, y) > 0 vi

mi (x, y) thuc B. Do t phng trnh 10 ta suy ra rng (, ) > 0 vi mi (x, y) thuc

B. iu c ngha l nu ng cong C l giao ca th ca f vi mt phng nm ngang i

qua P(a, b, f(a, b)) theo hng u th C lm ln trn mt khong c di 2. iu ny ng vi

mi hng u, v vy nu hn ch trn B, th ca f nm pha trn mt phng ngang ti P. V

vy f(x, y) f(a, b) vi mi (x, y) thuc B, ngha l f(a, b) l cc tiu a phng.

2.8. Nhn t Lagrange

2.8.1. Phng php nhn t Lagrange

Trong V d 6 phn 2.7, chng ta lm cc i hm th tch V = xyz chu rng buc

(constraint) 2xz + 2yz + xy = 12, l iu kin din tch (khng k np) bng 12m2. Trong

n Ng Minh o hm ring

Trang 43

phn ny chng ta trnh by phng php ca Lagrange cc i hoc cc tiu mt hm tng

qut f(x, y, z) chu rng buc dng g(x, y, z) = k.

Rt d gii thch c s hnh hc ca phng php

Lagrange i vi hm hai bin. V th chng ta bt u tm cc

gi tr c tr ca f(x, y) chu rng buc dng g(x, y) = k. Ni

khc i, chng ta tm cc gi tr cc tr ca f(x, y) khi im (x,

y) b hn ch trn ng mc g(x, y) = k. Hnh 1 m t ng

cong ny cng vi mt vi ng mc ca f, chng c phng

trnh f(x, y) = c, y c = 7, 8, 9, 10 v 11. Vic cc i f(x,

y) chu rng buc g(x, y) = k l tm gi tr ln nht ca c sao cho ng mc f(x, y) = c tip xc

vi g(x, y) = k. iu c minh ha trn Hnh 1, rng iu xy ra khi cc ng cong c

tip tuyn chung. iu c ngha l cc ng php tuyn ti im (x0, y0), ni chng chm

nhau, l nh nhau. V vy cc vc t gradient l song song, tc l f(x0, y0) = lg(x0, y0) vi

hng s l no .

Cch lp lun nh th cng p dng c cho bi ton tm cc tr ca f(x, y, z) chu rng buc

g(x, y, z) = k. V th im (x, y, z) b hn ch trn mt mc S vi phng trnh g(x, y, z) = k. Thay

cho cc ng mc trong Hnh 1, chng ta xem xt cc mt mc f(x, y, z) = c v lp lun rng nu

cc tr ca f l f(x0, y0, z0) = c th mt mc f(x, y, z) = c l tip din ca mt mc g(x, y, z) = k, v

v th cc vc t gradient tng ng l song song vi nhau.

Lp lun trc quan c th pht biu chnh xc nh sau. Gi s rng hm f c cc tr

ti im P(x0, y0, z0) trn mt cong S v gi s C l ng cong vi phng trnh vc t l

r(t) = x(t), y(t), z(t) nm trn S v i qua P. Nu t0 l gi tr tham s tng ng vi im P th

r(t0) = x0, y0, z0. Hm hp h(t) = f(x(t), y(t), z(t)) biu th cc gi tr m f c th trn ng

cong C. V f c cc tr ti (x0, y0, z0), dn n h c cc tr ti t0, nn h'(t0) = 0. Nhng nu f kh

vi, chng ta s dng quy tc dy chuyn vit

0 = h'(t0) = fz(x0, y0, z0)x'(t0) + fy(x0, y0, z0)y'(t0) + fz(x0, y0, z0)z'(t0) = f(x0, y0, z0) r'(t0)

iu chng t rng vc t gradient f(x0, y0, z0) trc giao vi vc t tip tuyn r'(t0)

vi mi ng cong C. Nh chng ta bit trong phn 2.6, rng vc t gradient ca g, f(x0,

y0, z0), trc giao v r'(t0) i vi mi ng cong nh th. (Xem phng trnh 2.6.18.) iu

ny c ngha rng cc vc t gradient f(x0, y0, z0) v g(x0, y0, z0) song song vi nhau. V vy

nu g(x0, y0, z0) 0, tn ti s l sao cho

[1] f(x0, y0, z0) = lg(x0, y0, z0)

S l trong phng trnh 1 ng cong gi l nhn t Lagrang (Lagrange multiplier). Th

tc da trn phng trnh 1 nh sau.

Phng php nhn t Lagrange tm cc tr ca f(x, y, z) chu rng buc g(x, y, z) = k

[gi s rng cc cc tr tn ti v g 0 trn mt cong g(x, y, z) = k]:

(a) Tm tt c cc gi tr ca x, y, z v l sao cho f(x0, y0, z0) = lg(x0, y0, z0)

v g(x, y, z) = 0

n Ng Minh o hm ring

Trang 44

(b) nh gi f ti tt c cc im (x, y, z) nhn c t bc (a). Gi tr ln nht trong chng

l gi tr ln nht ca f, gi tr nh nht trong chng l gi tr nh nht ca f.

Nu chng ta vit phng trnh vc t f = lg dng cc thnh phn, th cc phng

trnh trong bc (a) tr thnh

fx = lgx fy = lgy fz = lgz g(x, y, z) = k

y l h bn phng trnh vi bn n s x, y, z v l, nhng khng nht thit tm c th

cc gi tr ca l.

i vi cc hm hai bin, phng php nhn t Lagrange tng t nh phng php

trnh by. tm cc cc tr ca f(x, y) chu rng buc g(x, y) = k, ta tm cc gi tr x, y v l

sao cho f(x, y) = lg(x, y) v g(x, y) = k.

iu tng ng vi gii h ba phng trnh ba n

fx = lgx fy = lgy g(x, y) = k

Minh ha u tin ca chng ta v phng php Lagrange l xem xt li cc vn c

a ra trong V d 6 ti mc 2.7.

V d 1 Mt khi hp ch nht khng np c lm t 12m2 ba cc tng. Tm th tch

ln nht ca hp.

Li gii Nh trong V d 6 phn 2.7, chng ta gi s x, y v z tng ng l s o di,

rng v chiu cao ca hp (cm). Sau chng ta mun lm cc i V = xyz chu rng buc

g(x, y, z) = 2xz + 2yz + xy = 12. S dng phng php nhn t Lagrange, ta tm cc gi tr x,

y, z v l sao cho V = lg v g(x, y, z) = 12. iu cho ra cc phng trnh

Vx = lgx Vy = lgy Vz = lgz 2xz + 2yz + xy = 12

hay

[2] yz = l(2z + y)

[3] xz = l(2z + x)

[4] xy = l(2x + 2y)

[5] 2xz + 2yz + xy = 12

Khng c quy tc tng qut gii h phng trnh ny, nn i khi cn n su kho

lo. Trong v d ny chng ta nhn ln lt [2] vi x, [3] vi y v [4] vi z, dn n cc v tri

ca cc phng trnh nh nhau. Lm nh vy ta c

[6] xyz = l(2xz + xy)

[7] xyz = l(2yz + xy)

[8] xyz = l(2xz + 2yz)

Chng ta thy rng l 0, bi v nu l = 0 th t [2], [3] v [4] ko theo yz = xz = x y =

0, iu ny mu thun vi [5]. Do , t [6] v [7] ta c 2xz + xy = 2yz + xy, dn n xz = yz.

Nhng z 0 (v z = 0 s cho V = 0), vy x = y. T [7] v [8] ta c 2yz + xy = 2xz + 2yz, dn

n 2xz = xy v do x 0 nn y = 2z. t x = y = 2z vo [5] ta nhn c 12z2 = 12, hay z2 = 1.

V x, y v z dng nn ta c z = 1, do x = 2 v y = 2. Ph hp vi kt qu trong 2.7.

V d 2 Tm cc cc tr ca hm f(x, y) = x2 + 2y2 trn ng trn x2 + y2 = 1.

n Ng Minh o hm ring

Trang 45

Li gii Chng ta hi cc tr ca f chu rng buc g(x, y) = x2 + y2 = 1. S dng

phng php nhn t Lagrange, chng ta gii phng trnh f = lg

v g(x, y) = 1 c vit li nh sau:

fx = lgx fy = lgy g(x, y) = 1

c th l

[9] 2x = 2xl

[10] 4y = 2yl

[11] x2 + y2 = 1

T [9] ta c x = 0 hoc l = 1. Nu x = 0 th t [11] suy ra y = 1.

Nu l = 1 th t [10] suy ra y = 0, v vy t [11] cho ra x = 1.

Do f c th c cc tr ti cc im (0, 1), (0, -1), (1, 0) v (-1,

0). Lng gi f ti bn im ny ta tm c

f(0, 1) = 2 f(0, -1) = 2 f(1, 0) = 1 f(-1, 0) = 1

Do gi tr ln nht ca f trn ng trn x2 + y2 = 1 l

f(0, 1) = 2 v gi tr nh nht l f(1, 0) = 1. Kim tra trn Hnh

2, chng ta thy cc gi tr l ph hp.

Hnh 3 m t v phng din hnh hc ca V d 2. Cc cc tr ca f (x, y) = x2 + 2y2

tng ng vi cc ng mc m tip xc vi ng trn.

V d 3 Tm cc gi tr cc tr ca f(x, y) = x2 + 2y2 trn hnh trn x2 + y2 1.

Li gii Theo th tc trong 2.7.9, chng ta so snh cc gi tr ca f ti cc im ti hn

vi cc gi tr trn bin. Bi v fx = 2x v fy = 4y nn ch duy nht im ti hn l (0, 0). Chng

ta so snh gi tr ca f ti im ny vi cc cc tr trn bin t V d 2:

f(0, 0) = 0 f(1, 0) = 1 f(0, 1) = 2

Do , trn min hnh trn x2 + y2 1, gi tr ln nht ca f l f(0, 1) = 2 v gi tr nh

nht l f(0, 0) = 0.

V d 4 Tm cc im trn mt cu x2 + y2 + z2 = 4 m khong cch t ti im (3, 1,

-1) l gn nht v xa nht.

Li gii Khong cch t im (x, y, z) ti im (3, 1, -1) l

= ( 3) + ( 1) + ( + 1)

nhng n gin, ta tm gi tr ln nht v gi tr nh nht ca bnh phng khong cch:

= ( 3) + ( 1) + ( + 1)

Theo phng php nhn t Lagrange, ta gii f = lg, g = 4.

Dn n

[12] 2(x 3) = 2xl

[13] 2(y 1) = 2yl

[14] 2(z + 1) = 2zl

[15] x2 + y2 + z2 = 4

n Ng Minh o hm ring

Trang 46

Cch n gin nht gii cc phng trnh ny l gii x, y v z theo l t [12], [13] v

[14], sau thay cc gi tr vo [15]. T [12] ta c