11 de Cuong Toan Cao Cap

7/26/2019 11 de Cuong Toan Cao Cap

1/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh 1

Phan 1: Gii tch: Ham nhieu bien, Tch phan boi haiBAI 1: AO HAM - VI PHAN -NG DUNG

0. ao ham r ieng:

0.1 ao ham r ieng bac cao:

0.1 ao ham hon hpnh ly(Schwarz):Giasz/x, z/y z//xy ton tai val ien tuc th coz//yx vaz//xy = z//yx

1. Cc tr tdo ham 2, 3 bin:nh lyveieu kien can: f = f (x ,y) hay f = f (x , y, z) cocc tra phng tai M 0(x 0, y0)(M 0(x 0, y0, z0)) vacac ao ham rieng ton tai ==> M 0(x 0,y0) (M 0(x 0, y0, z0)) laiem dng.

nh l yveieu kien u: Neu M 0(x 0,y0) laiem dng, th cha chacM 0(x 0,y0 ) laiem cc tr, macan phai cothem nhng ieu kien cu theveao ham bac hai nhphng phap sau:Hai bi"n:

1/ //1 0xxH f= > ,// //

2 // //0

xx xy

xy yy

f fH

f f

= > ==> f at cc tieu tai X 0.

2/ //1 0xxH f= < ,// //

2 // //0

xx xy

xy yy

f fH

f f

= > ==> f at cc ai tai X 0.

Ba bi"n:

Cc t ieu: (dng ca) //1 0xxH f= > ,// //

2 // //0

xx xy

yx yy

f fH

f f

= > ,

// // //

// // //3

// // //

0

xx xy xz

yx yy yz

zx zy zz

f f f

H f f f

f f f

= >

Cc ai: (an dau, bat au tam)

//1 0xxH f= < ,

// //

2 // //0

xx xy

yx yy

f fH

f f

= > ,

// // //

// // //3

// // //

0

xx xy xz

yx yy yz

zx zy zz

f f f

H f f f

f f f

= <

ieu kien udng vi phan nhieu b ien tong quat: M 0(x 0, y0, z0)) laiem dng, xet dang v i phan

( )2

0

d f X :

a/ Laxac nh dng ( )( )2 0 0d f X > th f at cc tieu tai 0X .

b/ Laxac nh am ( )( )2 0 0d f X < th f at cc ai tai 0X .

c/ ( )( )2 0 0d f X = th tai 0X cha coket luan. Phai xet vi phan cap 3...Vi phan cap 2: f= f(x,y)

2 22 // 2 // // 22 xy

x yd f f dx f dxdy f dy = + +

Vi phan cap 2: f= f(x,y,z)

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2/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh2

2 2 22 // 2 // 2 // 2 // // //

2 2 2xy xz yz x y z

d f f dx f dy f dz f dxdy f dxdz f dydz = + + + + +

2. Cc tr vng ham 2 bin co1 rang buoc: Cc tr 1 2( , )f f x x = va1 rang buoc 1 2( , ) 0x x =

Ham L agrange: 1 2 1 2 1 2( , , ) ( , ) ( , )x x f x x x x = +

ieu kien can: 1 2( , )f f x x = at cc trtai M 0th M 0thoa 0 = (gi ai hephng tr nh nay etm

M 0 laiem dng cua ham 1 2( , , )x x

).nh l yveieu kien u:i/ H2< 0 th f at cc t ieu.ii/ H2> 0 th f at cc ai.

Trong o:

1 2

1 1 1 2 1 1 1 1 2

2 1 2 2 2 2 1 2 2

// // /// /

/ // // // // //2

/ // // // // //

0 x xx y

x x x x x x x x x x

y x x x x x x x x x

H

= =

ieu kien udng vi phan tong quat cho ham L agr ange :

1 2 1 2 1 2( , , ) ( , ) ( , )x x f x x g x x = + , xet dang vi phan ( )2

0d f X :

a/ Laxac nh dng ( )( )2 0 0d X > th f at cc tieu tai 0X .

b/ Laxac nh am ( )( )2 0 0d X < th f at cc ai tai 0X .

c/ ( )( )2 0 0d X = th tai 0X cha coket luan. Phai xet vi phan cap 3... Ta tnh ham 2 bien: L(x , y, ) = f(x, y) + (x, y)

d2L =2

//

xL dx2+ 2 //xyL dxdy + 2

//

yL dy2. (*)

Chuy: /x dx+/y dy=0, dx

2+ dy2>0

Neu: d2L (x 0,y0) > 0 th (x0,y0) laiem cc tieu.d2L (x 0,y0) < 0 th (x0,y0) laiem cc ai .d2L (x 0,y0) = 0 th (x0,y0) khong laiem cc tr.

Tng tham 3 bien: L (x ,y,z, ) = f(x,y,z) + (x,y ,z)

Th d2L=2

//

xL dx2+

2

//

yL dy2+

2

//

zL dz2+2( //xyL dxdy+

//xzL dxdz+

//yzL dydz)

3. Cc tr vng ham 3 bin co1 rang buoc:Cc tr 1 2 3( , , )f f x x x = varang buoc 1 2 3( , , ) 0g x x x =

Ham L agrange: 1 2 3 1 2 3 1 2 3( , , , ) ( , , ) ( , , )x x x f x x x g x x x = + ieu kien udng vi phan tong quat ham Lagr ange :

1 2 3 1 2 3 1 2 3( , , , ) ( , , ) ( , , )x x x f x x x g x x x = + , xet dang vi phan ( )2

0d f X :

a/ Laxac nh dng ( )( )2 0 0d X > th f at cc tieu tai 0X .

b/ Laxac nh am ( )( )2 0 0d X < th f at cc ai tai 0X .

c/ ( )( )2 0 0d X = th tai 0X cha coket luan. Phai xet vi phan cap 3...



3/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh 3ieu k ien uvenh thc Hess, xet 2 nh thc:i/ H2 0, H3< 0 th f at cc ai .

1 2

1 1 1 2 1 1 1 1 2

2 2 1 2 22 1 2 2

// // //1 2

// // // // //2

1 // // //// //

2

0

x x

x x x x x x x x x

x x x x x x x x x

g g

x x

gH

x

g

x

= =

,

1 2 3

1 1 1 2 1 31 1 1 1 2 1 3

2 2 1 2 2 2 32 1 2 2 2 3

3 3 1 3 2 3 3

3 1 3 2 3 3

// // // //1 2 3

// // //// // // //

13 // // // //// // //

2// // //

// // //

3

0

x x x

x x x x x x x x x x x x x

x x x x x x x x x x x x x

x x x x x x x

x x x x x x

g g g

x x x

g

xH

g

x

g

x

= =

//

4. Cc tr vng ham 3 bin co2 rang buoc: Cc tr 1 2 3( , , )f f x x x = va2 rang buoc

1 1 2 3

2 1 2 3

( , , ) 0

( , , ) 0

g x x x

g x x x

= =

.

1 2 3 1 2 1 2 3 1 1 1 2 3 2 2 1 2 3( , , , , ) ( , , ) ( , , ) ( , , )x x x f x x x g x x x g x x x = + +

ieu k ien uvenh thc Hess, xet 2 nh thc:xet 1 nh thc H 3: H3< 0 th C% va H3> 0 th CT

1 1 1 2 1 3 1 2 3

2 1 2 2 2 3 1 2 3

1 1 2 1 1 1 1 2 1 3 1

1 2 2 2 2 1 2 2 2 3

1 3 2 3 3 1 3 2 3 3

// // // / / /1, 1, 1,

// // // / / /2, 2, 2,

// // // // //3 1,

// // // // //

// // // // //

0 0 0 0

0 0 0 0

x x x x x x

x x x x x x

x x x x x x x x x

x x x x x x x x

x x x x x x x x

g g g

g g g

gH

= =1 1 1 1 2 1 3

2 2 2 1 2 2 2 3

3 3 3 1 3 2 3 3

/ / // // //2,

/ / // // //1, 2,

/ / // // //1, 2,

x x x x x x x

x x x x x x x x

x x x x x x x x

g

g g

g g

Tng tham 3 bien: L (x ,y,z, ) = f(x,y,z) + (x,y ,z)

Th d2

L= 2//

xL dx2

+ 2//

yL dy2

+ 2//

zL dz2

+2//xyL dxdy+2

//xzL dxdz+2

//yzL dydz.

---------------------------------------------------------------------------------------3.3 Cac vdu:

Vdu1:

Tm cac iem ti han cua: ( )2 2 2

, ,4

y zf x y z x

x y z= + + + , x 0, y 0, z 0.

Dung ieu kien cap 2 cua ao ham ephan biet iem M ax vaM in.



4/78


Giai: Giai 3 phng trnh: = =2

/

21 0

4x

yf

x; = =

2/

20

2y

y zf

x y;

= =/2

2 20z

zf

y z

2

2 2

11, ,

2 2

y y z z

x x yy z

= = =

, vay y, x, z cung dau.

Kh&y trong 3 ph'(ng trnh econ phng trnh 2 bien x, z: z8=1==> 1z= .....tm c ung 2 nghiem dng: M (1/2, 1, 1) vaN(-1/2, -1, -1). Tnh:

=2

//

32xx

yf

x = +

2//

3

1 2

2yy

zf

x y = +//

3

2 4zzf

y z

= //22

xy

yf

x, =// 0xzf

=//

2

2yz

zf

y

+ Bay gixet nghiem tai M (1/2, 1, 1):Xet vi phan bac 2:

( )= + + + + +2 // 2 // 2 // 2 // // //2xx yy zz xy xz yz d f f dx f dy f dz f dxdy f dxdz f dydz lamot dang toan phng theo dx ,dy. Ta coma tran:

= = =

// // //

// // //3

// // //

4 2 0

2 3 2

0 2 6

xx xy xz

yx yy yz

zx zy zz

f f f

f f f H

f f f

32 > 0, H1=4 >0,

2

4 28

2 3H

= =

>0

d2f (M ) >0 (DTP xac nh dng) ,dx dy=> M (1/2, 1, 1) cc ti eu.+ Bay gixet nghiem tai N(-1/2, -1, -1)

H1= -4 0

= = =

// // //

// // //3

// // //

4 2 0

2 3 2

0 2 6

xx xy xz

yx yy yz

zx zy zz

f f f

f f f H

f f f

-32 < 0,

Vay d2f (N) < 0 (DTP xac nh am) ,dx dy=> N(-1/2, -1, -1) cc ai .

Cach 2:X et nghiem tai M (1/2, 1, 1):Ta tnh trc tiep t vi phan bac 2:

( )= + + + + +2 // 2 // 2 // 2 // // //2xx yy zz xy xz yz d f f dx f dy f dz f dxdy f dxdz f dydz la:2 2 2 24 3 6 4 4d f dx dy dz dxdy dydz = + + , sau mot sotnh toan :

( )2

22 24 2 4 0

2

dyd f dx dy dz dz

= + + >

==> M (1/2, 1, 1) CT

Xet nghiem tai N(-1/2, -1, -1):Ta tnh trc tiep t vi phan bac 2:



5/78


( )= + + + + +2 // 2 // 2 // 2 // // //2xx yy zz xy xz yz d f f dx f dy f dz f dxdy f dxdz f dydz la:2 2 2 24 3 6 4 4d f dx dy dz dxdy dydz = + + , sau mot sotnh toan :

( )2

22 24 2 4 0

2

dyd f dx dy dz dz

= N(-1/2, -1, -1) C

Vdu2: Xet lai v dutrc ay: Ham: z=f(x,y ) = x

3

+ y

3

3xy Tm tap cac iem dng:z/x= 3x2 3y = 0z/y= 3y2 3x = 0

Giai henay ta co2 iem dng laM 1(1,1) vaM 2(0,0)

Ta co:2

//

xz = 6x, z//xy = -3, 2

//

yz = 6y

V avi phan cap 2: 2 // 2 // 2 //2xx yy xy d f f dx f dy f dxdy = + +

Tai M 1:2 2

2 2 2 96 6 6 6 0

2 2

dx dx d f dx dy dxdy dy

= + = + >

,

nen M 1 laiem cc t ieu.Cothethay bang H1= 6>0, H2= 36 -9 >0

Tai M 2: 2 3d f dxdy = = ? cotheam, cothedng,nen M 2khong laiem cc tr. Cothethay bang H2= -9 0, x vaz bang nhau vacung dau. Giai ra:Tm c ung 2 nghiem dng: M (1, 1/2, 1) vaN(-1, 1/2, -1).

Tnh: //3

4xx

yf

x= //

3yyz

fy

= //3

2zzf

z=

//

2

2xyf

x

= , =// 0xz

f //2

1

2yzf

y

=

+ Bay gixet nghiem tai M (1, 1/2, 1):Xet vi phan bac 2:

( )= + + + + +2 // 2 // 2 // 2 // // //2xx yy zz xy xz yz d f f dx f dy f dz f dxdy f dxdz f dydz lamot dang toan phng theo dx, dy, dz. Ta conh thc ma tran Hess:



6/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh6// // //

// // //3

// // //

2 2 0

2 8 2

0 2 2

xx xy xz

yx yy yz

zx zy zz

f f f

f f f H

f f f

= = =

16 > 0, H1= 2 >0,

2

2 212

2 8H

= =

>0

d2f (M ) >0 (DTP xac nh dng) , ,dx dy dz => M (1, 1/2, 1) cc ti eu.Cach khac: 2 2 2 22 8 2 4 4d f dx dy dz dxdy dydz = + + =

( ) ( )2 22

2 4 2 0dx dy dy dz dy = + + > => M (1, 1/2, 1) cc ti eu.

+ Bay gixet nghiem tai N(-1, 1/2, -1)

H1= -2 < 0, 22 2

122 8

H

= =

> 0

// // //

// // //

3// // //

2 2 0

2 8 20 2 2

xx xy xz

yx yy yz

zx zy zz

f f f

f f f H

f f f

= = = -16 < 0,

Vay d2f (N) < 0 (DTP xac nh am) , ,dx dy dz => N(-1, -1/2, -1) cc ai .

Cach khac: 2 2 2 22 8 2 4 4d f dx dy dz dxdy dydz = =

( ) ( )2 22

2 4 2 0dx dy dy dz dy = + + < => N (-1, -1/2, -1) cc ai .

Vdu5: Tm cc trcua f(x, y, z) =2x+y+3z.

Vi ieu k ien: g(x,y,z) = 2 2 24 2 35x y z+ + =0.+ Cach 1:

Ham Lagrange: ( )2 2 22 3 4 2 35L f g x y z x y z = + = + + + + + /

2 2xL x= + ,/

1 8yL y= + ,/

3 4zL z= + ,/ 2 2 2

4 2 35L x y z = + +

//2xxL = ,

//8yyL = ,

//4zzL = ,

//0L=

//0xyL = ,

//0xzL = ,

//0yzL = ,

//2xL x= ,

//8yL y= ,

//4zL z=

1 21 1 1 1

0 4, , 3, , 4, , 3,

2 4 2 4

L M M = = =

uuur r

cohai iem dng. X et v i phan bac 2 .

( )2 // 2 // 2 // 2 // // //2xx yy zz xy xz yz d L L dx L dy L dz L dxdy L dxdz L dydz = + + + + + lamot dang toan phng theo 3 bi en: dx ,dy ,dz. Chng minh ieu nay tng t nh trng hp 2bien.

Xet vi phan bac 2 tai 11 1

4, , 3,2 4

M =

, f=17.5



7/78


1

2 // 2 // 2 // 2 2 2 21 2 02

xx yy zz M

d L L dx L dy L dz dx dy dz = + + = <

Nen 11 1

4, ,3,2 4

M =

laC.

Xet vi phan bac 2 tai 21 1

4, , 3,2 4

M =

, f= -17.5

2

2 // 2 // 2 // 2 2 2 21 2 02

xx yy zz M

d L L dx L dy L dz dx dy dz = + + = + + >

Nen 21 1

4, , 3,2 4

M =

laCT.

+ Cach 2:Dung nh thc, x et 2 nh thc (n=3,m=1).

Tai 11 1

4, ,3,2 4

M =

, ( )1 0

kkH > , 1, 2,3k m n= + = , that vay:

Ta xet:

/ /

/ // //12 2

/ // //

0 0 8 4

8 0

4 0 2

x y

x xx xy

y yx yy

g g

H g L L

g L L

= =

=136 > 0

/ / /

/ // // // 12

3 / // // //

/ // // //

00 8 4 12

8 0 0

4 0 2 0

12 0 0 1

x y z

x xx xy xz

y yx yy yz

z zx zy zz

g g g

g L L LH

g L L L

g L L L

= =

= -280


8/78


Tai M 1(3 3

,2 2

,3

2

= ), z= 1.299,

Ta xet:

/ /

/2 11 12

/21 22

0 0 1 3

1 3 1 6 3

3 1 3

x y

x

y

H a a

a a

= = =

>0, M 1laC

Tai M 2( 3 3,2 2

, 32

= ), z= -1.299,

Ta xet:

/ /

/2 11 12

/21 22

0 0 1 3

1 3 1 6 3

3 1 3

x y

x

y

H a a

a a

= = =

>0, M 2laCT

Tai M 0( 0 , 0 , bat ky), Ta xet:/ /

/2 11 12

/21 22

00 2 0

2 2 1 8

0 1 2

x y

x

y

H a a

a a

= = = , M 0cha coket luan

Vdu7: Lay v du trc z = xy, thoa (x,y) = 3x+2y 5 = 0L = f + . = xy + [3x+2y 5 ]

Co1 iem dng duy nhat M (5 5 5

, ,6 4 12

= ).

2 2

// //0

x yL L= = , // 1xyL = ,

// /3x xL = = ,

// /2y yL = = .

Tai M(5 5 5

, ,6 4 12

= ), z=25/24,

Ta xet:

/ /

/2 11 12

/21 22

00 3 2

3 0 1 12

2 1 0

x y

x

y

H a a

a a

= = = >0, M laC

---------------------------------------------------------------------------------------Chng 2

TCH PHAN BOI 2 (KEP)BAI 1: BAI TOAN MAU

1.1 Bai toan mau :

Ta nhlai rang, tch phan xac nh hay tch phan n phat xuat tbai toan tnh dien tch vatchphan kep lai xuat phat tbai toan tnh thetch nh sau :

Can tnh thetch cua hnh coay phang lamien D Oxy, ng sinh song song vi truc oz,mat tren lamat cong xac nh bi hamz = z(x,y), vi M (x,y) D. T a chia mien D thanh n phan, di en tch moi phan l a( Si), cot thetchcodien tch ay



9/78


yd

c

a b x

( Si) M i(x i,y i), vay cathetch la:

1 1

( ) ( ) ( ) ( , )n n

n i i i i i i i

V S f M S f x y

= == =

max 0lim

i

nd

V V

= , vi di= dia(Si) = ban knh cua Si

chuy: max di0 n

1.2 nh ngha :Gii han tren neu ton tai hu han (khong phuthuoc vao cach chia mien D) c goi latch phan

kep cua ham z = f (x,y ) tren mien D, khi ota noi f khatch tren mien D vakyhieu nola( , )

D

f x y dx dy

Nhan xet :1) Theo nh ngha neu f (x ,y) = 1 (x,y) D th 1 ( )

D

dxdy S D = (dien tch mien D).

2) f (x ,y) > 0, l ien tuc (x,y) D th f(x,y) V( )D

dxdy D = l athetch hnh trucocac ng

sinh song song vi Oz, hai ay gii han bi z= 0, z= f (x,y).1.3 Tnh chat cua tch phan kep :

Cac tnh chat cua tch phan kep cung giong nh tnh chat cua t ch phan xac nh.Tnh chat 1 :f l ien tuc trong D th f khatch tren DTnh chat 2 :cotnh tuyen tnh

( )

,

D D D

D D

f g dxdy f dxdy gdxdy

Kf dxdy K f dxdy K R

+ = +

=

Tnh chat 3 :neu D = D1D2vaD1D2= th

1 2D D D

f f f= +

BAI 2:PHNG PHAP TNH TCH PHAN KEP

2.1 Trong hetr uc Descar tes.

q M ien D lahnh chnhat :D = (x,y) : a x b, c y d =[a,b][ c,d]

( , ) ( ( ( , ) )

d b

D c a

f x y dxdy f x y dx dy = =

= ( ( , ) )b d

a c

f x y dy dx

V ay vi mien l ahnh chnhat ta cothehoan v can .V du: D = [0,1][1,2]

1 22 2 2 2

0 1

( ) ( ( ) )dx

D

I x y dxdy x y dy = + = +



10/78


y4 B

O 4 8 x

-2

A

-4

dc

a b x

2

1

( )( )

y xy x

1( )x y 2( )x y x

y

y=1/xD1

D211/2

1 2 x

y=x

1 132 2

0 0

2 7 8( ) ( )

3 1 3 3

yx y dx x dx = + = + = , hoac

2 1 2 23 32 2 2 2

1 0 1 1

1 21 8I ( ( ) )dy ( ) ( )

3 0 3 3 3 1 3

x y yx y dx xy dy y dy = + = + = + = + =

q M ien D lamien bat ky:

y y

D = (x,y) : a x b, y1(x )y y2(x) =[ a,b][ y1(x), y2(x)]2

1

( )

( )

( , ) ( , )

y y xb

D a y y x

f x y dxdy dx f x y dy

=

==

D = (x,y) : c y d, x1(y )x x2(y) =[ x1(y), x2(y)][c,d]2

1

( )

( )

( , ) ( , )

x x yy d

D y c x x y

f x y dxdy dy f x y dx

==

= ==

Neu D lamien phc tap th phai phan D ra thanh nhng mien n gian nh tren

V du1 :tnh2

2D

xI dxdy

y=

D=D1D2

Vi D gii han bi 3 ng sau :1

2 , ,x y y x x

= = =

cach 1 :2 2

211

9

4

y x

yx

xI dx dy

y

=

=

= =

cach 2 :

2 2D D

I= + v D = D1D2vaD1D2=

2

1

2 2 2 2

21 1

8 5

3 6 6

y x

D y x y

x ydy dx

yy

= =

= =

= = =

1

1

2

1 2 24

21 12

2

8 1 8 15

3 12 3 12

x

Dx

y

x

dy dx y yy

=

=

= = + =

5 8 15 9

6 3 12 4I = + =

V du2 :TnhD

I xydxdy = ,

D gii han bi2 ng y = x 4 vay2= 2x,Tnh giao iem ta co:



11/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh 11giao iem : A (2,-2), B(8,4)

22

444 4 2 4 6

3 2

2 22

2

41 8( ) ( 8 ) 90

2 2 4 3 24 2

yx y

yy

x

x y yI dy xydx y dy y y

+= +

=

= = = + + =

Cach 2 :oi can danh cho ban oc coi nh bai tap.V du3 :Tnh thetch khoi, c gii han bi cac mat z= 4-x-y va

z 0, y=x2, y1.

zE

4

y1 4

F

4x

V 2=31 1

2

0 0

68(4 ) (4 ) (8 2 )

15

x yy y

D y yx y

x y dxdy dy x y dx y y dy

=+= =

= ==

= = =

Cach 2: oi can khac:2

21

2

1

68(4 )

15

yx

x y x

V dx x y dy

==

= =

= =

V du4 :Tnh thetch mien V, c gii han bi cac ngz= 4-x-y va z 0, y=x2, y1.Danh cho H oc Sinh k iem tra chi ti et. T nh gi ao iem ta cohai gi ao i em :

1 17 1 17, ,

2 2F Ex x

+ = = thetch mien V can tnh:

V= V 1 - V 2, (V 2laket quatren), ta can tnh V 1l athetch coay laparabol y=x2.

V 1=2

1

4

(4 ) (4 )F

E

x x y x

V x x y x

x y dxdy dx x y dy

= = +

= =

= =

=

1 1742

2 3

1 17

2

7 289( 4 8 ) 17

2 2 60

x

x

xx x x dx

+=

=

+ + + =

2.2 Phng phap oi bien :

Xet ( , )D

I f x y dxdy = , bay giat van eoi bien trong tch phan k ep cog mi so vi tch phan

n ahoc, muc ch la: v l ay tch phan tren mien Dxy quaphc tap nen ta bien v emien D uv ngian nh cac hnh chnhat chang han, luc olay tch phan rat n gian.



12/78


v5 B C

D/

2

A D

1 3 u

y B114

12 C1

D

5 A13 D1

4 7 8 11 x

u

v

(u,v)

uvD

F

1F

(x,y)

xyD

f

x

g f F= o

0

Dxy (x,y) (u,v) Duv phep bien oi nay 1 1 nen nocophep bien oi ngc :F(u,v)=(x,y) vaF1(x ,y)=(u,v ), vaco:

F(Duv)=Dxy (D xy laanh cua Duv)

luc oham sof: Dxy R, t ch phan cua ham sof(x,y ) tren Dxy , c oi bi en nh sau:

[ ]( )

( , )( , ) ( , ), ( , )

( , )xy uv uv D F D D

D x yf x y dxdy f x u v y u v dudv

D u v== =

= ( ) ( )0 ( , ). det ( , ) ( , ). det ( , )uv uv

F F

D D

f F u v J u v dudv g u v J u v dudv =

trong oJF laJacobi cua phep bien oi .

( )/ /

/ // /

/ /

( , ) 1 1det ( , )

( , )( , )

( , )

u vF

x yu v

x y

x xD x yJ u v

D u vD u v u uy yD x y

v v

= = = =

V du1 : Hnh chnhat A (1,2), B(1,5), C(3,5), D(3,2), c bien hnh qua phep bien oi tuyen tnhcomatran tng ng la:

2 1

1 3G

= , tm dien tch A 1B1C1D1bien hnh qua PBTTT. G tren. X em hnh di.Dedang kiem chng qua PBTT. G ta co A 1(4,5), B1(7,14), C1(11,12), D1(8,3) , dien tchABCD=2.3=6. qua lien hetoa o:

2 1 ( , )

1 3 ( , )

x u u D x yG J

y v v D u v

= = = =

2 1det( ) 7

1 3J G= = =

;

Vay dien t ch mi la:

/ /

( , ) 1. . 7 7 6 42

D D D D

f x y dxdy dxdy J dudv dudv = = = = =



13/78


1

/2

34

5

67

8

910

/3

/4

/5

/6

/7

/8

/9

2

Ta cothekiem chng ket quaqua cong thc hnh giai tch nh sau:

S= ( )4

1 11

4 8 11 7 41 1

2 2 5 3 12 14 5i i i i

i

x y y x + +=

= =

=1

2[(4.3-5.8)+(8.12-3.11)+(11.14-12.7)+(7.5-14.4)]=

84

2=42.

Ta cothegai bang cong thc bang ex cel, hay V inacal -570M S4 8 11 7 4

5 3 12 14 5

-28 63 70 -21 84

V du2 : Hnh ch s nhoco10 toa o: iem 10 trung iem 1 nh sau:TT 1 2 3 4 5 6 7 8 9 10

x 1 2 4 3 4 2 0 2 2 1

y 0 -2 2 2 3 4 3 2 1 0

c bien hnh qua phep bien oi tuyen t nh coma tran tng ng la:2 1

1 3

G

=

, tm dien tch S ln qua bien hnh PBTTT. G tren. X em hnh.

Giai:

TT 1 2 3 4 5 6 7 8 9 10

x 1 2 4 3 4 2 0 2 2 1

y 0 -2 2 2 3 4 3 2 1 0

1 1i i i i x y y x + + -2 12 2 1 10 6 -6 -2 -1 20

Dien tch: ( )9

1 11

1 120 10

2 2i i i i

i

s x y y x + +=

= = = (gai excel, hang 3)

2 1 ( , )

1 3 ( , )

x u u D x yG J

y v v D u v

= = = =

2 1det( ) 7

1 3J G= = =

;




14/78


1C 2C

1H

/D

D

u

v

x

y

/ /

( , ) 1. . 7 7 10 70

D D D D

f x y dxdy dxdy J dudv dudv = = = = = .

Cach khac: Qua phep bien hnh ta tm cac toa ocua hnh S ln mh sau (nh hnh vetren) . Donhan hai ma tran sau ta conh cua hnh S l a:

2 1

1 3

1 2 4 3 4 2 0 2 2 1

0 -2 2 2 3 4 3 2 1 0

=

2 2 10 8 11 8 3 6 5 2-1 -8 2 3 5 10 9 4 1 -1 V ay toa ohnh S ln la:

TT 1/ 2/ 3/ 4/ 5/ 6/ 7/ 8/ 9/ 10/ x 2 2 10 8 11 8 3 6 5 2

y -1 -8 2 3 5 10 9 4 1 -1

1 1i i i i x y y x + + -14 84 14 7 70 42 -42 -14 -7 140

Dien tch: ( )9

1 11

1 1140 70

2 2i i i i

i

S x y y x + +=

= = = (gai excel, hang 3)

V du3 : Hnh tron ( ) ( )2 2

1 : 2 1 1C u v + = , c bien hnh qua phep bien oi tuyen t nh co

matran tng ng la:2 1

1 3G

=

, tm dien tch hnh el l ip labien hnh qua

PBTTT. G tren.


( )/ /

1( , ) 1. . 7 7 1 . 7

D D D D

f x y dxdy dxdy J dudv dudv = = = = =

Cach 2: V ong tron 1C bi en thanh el l ip 2C qua PBTT: X em hnh tren.

( )

( )

13

27

3 12

7

u x yx u v

y u vv x y

= = + = + = +

, 2 2210 2 5

: 2 4 049 49 49

C x xy x y + + =

Sau PB trc giao ta a vedang ellip : 2 249

5 495

Y X+ = hay

( )

2 2

2 21

7 5

5

X Y+ =

, th di en tch la:7

5 75

S ab = = = .



15/78


v y1 G(D/)=D

1 G

D/0 1 x

1 u

V du4 : M ien D/lahnh tam giac O(0,0), A (2,0) , B(0,2), c bien hnh qua phep bien oi phi

tuyen G: (x,y)=G(u,v)= (u+v, u2-v). Tnh tch phan cua ham:1

( , )1 4 4

f x yx y

=+ +

, tren mien

bien hnh G(D/).

v G y2 G(D/ )=D

D/2 u x

==

+=

=

),(),(

vuDyxD

Jvu

vu

v

uG

y

x2

=1 1

det( ) (1 2 )2 1

J G uu

= = = +

;

Vay :

/ 2( )

1

( , ) 1 4( ) 4( )DD G Df x y dxdy dxdy

u v u v = = =+ + +

=/

1.

2 1D

J dudvu+ / /

1 2 11. 2 2 2

1 2 2D D

ududv dudv

u

+= = = =

+

V du5 : M ien D/laphan t hnh tron n vtrong mp Ouv , c bien hnh qua phep bien oi phi

tuyen G: (x,y )=G(u,v)=(u2-v2, 2uv ). T nh tch phan cua ham:2 2

1( , )f x y

x y

=+

, tren mien bien

hnh G(D/).M ot phan t v ong tron qua phep bien oi G trthanh mot na vong tron nh hnh ve

2 2 ( , )

( , )2

x u D x yu vG J

y v D u v uv

= = = =

2 22 2det( ) 4( )

2 2

u vJ G u v

v u

= = = + ; Vay :

/ /2 22 2 2 2 2

( )

1 1( , ) . .

( ) 4DD G D D

f x y dxdy dxdy J dudv u vu v u v =

= = =+ +

/ /

2 2 2

2 2

.14 4 1. 4

4D D

u vdudv dudv

u v

+= = = =

+

Cach khac:1

2 20 0

1 1r

D r

dxdy rdr d rx y

= =

= == =

+



16/78


2

1

u

v= u

v

v= -u

O x O 1 3 u

y y= 2x y= x 2

2

3x y

x y

=

=

D

D/

v

1

2

V du6 :D

I xydxdy = D l amien cong gii han bi bon ng

y2= x, y = x, y2= 3x, y= 2x

oi bien at2

,y y

u vx x

= = , rorang ta congay tgii han mien D la:

1 u 3 va1 v 2 va Jacobi cophep bien oi nay la:3

2 42

2

2

( , ) 1 1

( , )( , ) 2

( , )

1

D x y x u J

D u vD u v y vy y

D x y xx

y

xx

= = = = =

neu lam phc tap hn ta cothetnh:2 3

2 4

2

1 2

( , ),

( , ) 1

u

u u D x y u v vy x

v D u v u v v

v v

= = = =

Vay'

3 23

2 4 7

1 1

105. . .

32D D

u u u dv I xydxdy dudv u du

vv v v= = = =

V du7 :M ien Q gii han bi 4 i em (0,1), (0,2), (2,0) va (1,0),

hay tnh I=

( )( )

y x

y x

Q

e dxdy

+

Giai :at ( )u y x= , ( )v y x= +

( )

( ) ( )( )

, 1 1 1

1 1,, 2, 1 1

D x y

D u vD u v

D x y

= = =

M ien l ay tch phan oxy,c bien thanh mien ouv nh hnh ve.Nen

I=

( )( ) 1

2xy uv

y xu

y x v

Q Q

e dxdy e dudv

+ = =

y2

2 x



17/78


y B/

4 y=x2

G(D/) =D

O 2 x

A/

- y= -

x

v2 A

D/

2 B u

-1 0 1 2 3 4 5 6-1

0

1

2

3

4

( )2

1

1

1 3

2 4

v u v uv

v u v

dv e du e e

= =

= == =

V du8 :Tnh dien tch mien D gii han biy2= px, y2= qx, x2= ay, x2= by vi 0 < p


18/78


-2 -1 0 1 2 3 4 5 6-1.5

-1

-0.5

0

0.5

1

1.5

2 - =

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3-1

0

1

2

3

4

5

6

7

8

9

H1-->

H2

B(u=2, v=0) bien thanh B/(x=2, y=4) , oan thang A B bien thanh ng thang x=2, oan OB(v=0) bi en thanh cung OB/parabol y=x2, oan OA (u=0) bien thanh dng OA/ y= -x.

Vay S=

22

0

y xx

x y x

dx dy

==

= =

=14

3

Tv dunay, ta cothetnh lai v du4 trc:

( )

2

/

2 2

0 0( )

1 1( , ) 2 1 2

2 21 4 4

y xx x

x y x x D G D

dxdyf x y dxdy x dx

x y

== =

= = ==

= = + = + +

V du10 : Tnh dien t ch gii han bi 2 ng sau:

( )

( )

2 21 : 2 8 8 16 2 0

2 : 18 19 0

x y xy x

y x

+ =

+ =.

T(1) ta cotheviet lai: ( )2 2

2 2 4 32 34 2x y y v u = + = , v

at 2 4u x y= , 32 34v y= + , thevao (2) ta co: v=2u+4, T nh J=1

32

vay' 2

2 2 4

1 2

1 1 9

32 32 32

u v u

D uD v u

S dxdy dudv du dv

= = +

= =

= = = = .

Giai :

V du11 : Tnh dien tch gii han bi 2 ng

sau: ( )

( )

2 2

12

1 : 3 12 12 12 8 1 0

2 :

x y xy x y

y

+ + + =

= .

Gi ai :

Dung phng phap L agrange ta a (1) ve ( )2

3 2 2 16 11x y y = +

at2 2

16 11

u x y

v y

= = +

==> 23u v= , a (2) ve 3v= ,( )

( ) ( )( )

, 1 1 1

, 1 2, 16

, 0 16

D x yJ

D u vD u v

D x y

= = = =

, giao

iem: 23u v= , va 3v=



19/78


lau=-1, u=1, vay S= ( )1

2

1

1 13 3 4

16 4

u

u

J u du

=

= = =

V du6 : Tnh tch phan suy rong:2

xI e dx

= = , ta cothexem

( )2 2

2 22

y xx yx y

x y y x

I e dx e dy e dxdy

= = +

= = = =

= =

, mien lay tch phan ay lavong

tron tam O, ban knh lavohan R=

( )2 2 2220 0

y x rx y r

y x r

I e dxdy e rdr d

= = = = +

= = = == = =

0

22

u u

u

e

=

=

= =

I = , tay suy ra, neu ta tnh tien th :

( )2

x ye dx

=

V du7 : Tnh2 2

a xI e dx

a

= = , do oi bien mot t.

V du8 : Tnh( )2 2 2 3

3

x y xy

y x

I e dxdy

+

= == =

( ) 22 2 34 23

yx y xy

y x y

I e dxdy e dy

+

= = == = = ,

Xem v dusau.

V du9 : Tnh ( )2 2

2

3 3

4

x y xy

x y

I e dxdy A

+

= == = = = ,

do oi bien mot t vedang toan phng

( )2 2 2 2112

11

2

x Tx y xy x y xy x y

yX AX

+ = + = =

( ) ( )2 2 2 2 22 2

2 4

y yx y xy x xy y x y + = = + , vadet(A )=

3

4A =

BAI 3: NG DUNG TCH PHAN KEPTnh dien tch vathetch laxuat phat cua tch phan kep . ay ta xet cac ng dung khac, tnh cacac trng vat l ytren dien tch v a thetch.



20/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh203.1 Momen quan tnh cua tam phang :

M at okhoi lng tai iem (x,y ) laham (x,y ) th cac momen quan tnh theo cac truc chnh

la: 2( , ) ,OxD

J x y y dxdy = 2( , )OyD

J x y x dxdy =

Vay momen quan t nh cua vat thekhong thetr iet t ieu.

oi vi truc ( )Oz Oxy th: 2 2( , )( )OzD

J x y x y dxdy = +

oi vi vat thea giac: Nh: iem ( )1 1, , , 1,n i i i M M M x y i n + =

( ) ( )2 2

1 1 1 11

1

12

i n

Ox i i i i i i i i i D

J x y x y y y y y y dxdy =

+ + + +=

= + =

( ) ( )2 2

1 1 1 11

1

12

i n

Oy i i i i i i i i i D

J x y x y x x x x x dxdy =

+ + + +=

= + =

oi vi truc ( )Oz Oxy th: 2 2OzD

J x y dxdy = + =

( ) ( ) ( )2 2

1 1 1 1 1 11

1

12

i n

i i i i i i i i i i i i i

x y x y x x x x y y y y =

+ + + + + +=

= + + +

V du1 :tm Jycua hnh D gii han bi y2= 1 x, x = 0, y = 0 vi mat okhoi l ng l a(x, y) = y11 1 3 4

2 2

0 0 0

1 1 1. (1 ) ( )

2 2 3 4 24

y x

y

y

x xJ dx y x dy x x dx

=

== = = =

V du2 :tnh Jxcua cacioi t,cophng tr nh l a:r = a(1 + cos ), (x,y) = 1

ta co: 2 ,xD

J y dxdy = oi sang toa occ ta cocos

sin

x r

y r

=

=,

(1 cos )2 2(1 cos )2 2 2 4

00 0 0

1sin . sin

4

r aa

x

r

J r rdr r d

= +=+

= == = =

242 4 4

0

21sin (1 cos ) .

4 32

ad a

= + =

V du4 :Tm momen quan tnh cua hnh vanh khan co2 ban knh a R< a) oi vi truc ox trong mat phang (H3)b) oi v i t ruc oz thang goc vi mat phang (H4).Giai :

a/ 2

xy

ox

D

J y dxdy = =

x

ya

2a

M(x,y)

M(x,y)x

z

O O

H3 H4PDF created with pdfFactory Pro trial version www.pdffactory.com


21/78


= ( )2

2 2

0

sin

r R

r a

d r rdr

= =

= = =

=4 4

12 2 4

4 R a =

4 4

4R a

=

= ( )2 2 4mR a+ , ( )2 2m R a=

b/ ( )2 2xy

oz

D

J x y dxdy = + = ( )2 2

0

r R

r a

d r r dr

= =

= = =

4 4

2R a

=

=( )2 2 2mR a+

Chuy: K hi a=0 th v du3 latrng hp r ieng cua v du4.V du5 :Tm momen quan tnh cua hnh vuong cocanh 2R a) oi vi truc oz thang goc vi mat phang tai i em O. (H5)b) oi vi truc ox nam trong mat phang ( H6) , vatruc x///Ox qua A .c) oi vi t ruc oz thang goc vi mat phang (H7).d) oi vi truc oz tai iem C (gia canh) thang goc vi mat phang (H5).Giai :a/ Hnh (H5): ay ta lay mat okhoi lng =1.

( )2 2xy

oz

D

J x y dxdy = + = ( )22

2 2

0 0

y Rx R

x y

y x dxdy

==

= =+ =

=4

83R =

243

mR , vi ( )2 2

2 2m R R= = latrong lng cua a vuong.

Cach khac: toa occ:2

2 coscos

B

Rx R r r

= = =

2

xy

oy

D

J r rdrd = =

2

cos43

0 0

2

R

r

r

d r dr

==

= =

=44 84

3 32 RR = =

243

mR ,

b/ Hnh (H6): 2OA R= , BA R= , v oi xng nen

22

xy

ox

D

J y dxdy = =22

2

0

y R x y R

y x y

y dxdy

= = +

= = =

4

3R =

2

6mR , vi ( )

2 22 2m R R= = latrong lng cua

a vuong. oi vi truc x/ta co:

M(x,y)

M(x,y)x

z

B O

H6 H7

O

H5

A

B

A x/

R

2RO

2R H8

A

B

C

O

C



22/78


( )4 4

/

22 2 4 73 3

2 2R ROxOxJ J a F R R R = + = + = + = (nh lyHuygen)

c/Hnh (H7, H8): 22

ABR

R OC= =

( )2 2

xy

oz

D

J x y dxdy = + = ( )2 2

2 2

0 0

4

R Rx y

x y

y x dxdy

= =

= =

+ =

=4 4

4 26 3R R=

2

3mR , vi ( )

2 22 2m R R= = latrong lng cua a vuong.

d/ Sinh vien tgiai, hnh H8.==> H6 quay denh hnh H7 va hnh H5 kho quay nhat. V ata cotheso sanh squay cua hnhvuong v ahnh tron qua v du3 vav du5.3.2 Momen tnh vatrong tam :

Tnh ngha cua bieu thc momen quan tnh, ta nhan ra rang momen quan tnh bieu hien tnh ynoi tai cua chat iem cokhoi lng (x,y ) > 0 khi chuyen ong quay quanh mot truc nao o, nhngban than nocha bieu dien c scan bang tnh hoc oi vi truc ora sao ? mc okhathi khi

quay quanh truc ora sao? vav vay mot khai ni em venolamomen tnh vatoa otrong tam cnh ngha nh sau :M omen tnh oi vi truc x la ( , )x

D

S y x y dxdy =

M omen tnh oi vi truc y la ( , )yD

S x x y dxdy =

Toa otrong tam C : ycS

xS

= , xcS

yS

=

Vi (x,y ) mat okhoi lng tai iem (x,y ) va

S= ( , )D

x y dxdy ladi en t ch mien D.

oi vi vat thea giac:

( ) ( )1 1 11

1

6

i n

Ox i i i i i i i D

S x y x y y y ydxdy =

+ + +=

= + =

( ) ( )1 1 11

1

6

i n

Oy i i i i i i i D

S x y x y x x xdxdy =

+ + +=

= + =

Nh: iem ( )1 1, , , 1,n i i i M M M x y i n + =

That vay theo nh lytrung bnh ta co:

( , ) ( , ) . y

y C C C

D D

SS x x y dxdy x x y dxdy x S x

S= = = =

( , ) ( , ) . xx C C C D D

SS y x y dxdy y x y dxdy y S y

S= = = =

Nhan xet :

y

c

H9



23/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh 23M omen tnh cotheam, dng, momen tnh oi vi 1 truc nao o matriet tieu th truc ogoi

latruc trung hoa, giao cua 2 truc trung hoa bat kyhng nao, cung chnh latoa otrong tam.

That vay goi (H9) latruc trung hoa th

Cong thc oi truc ta co: Cx x x= + , ( )y C CD D D D

S xdxdy x x dxdy x dxdy x dxdy = = + = + =

0C yD D D

x S x dxdy S x dxdy x dxdy S = + = + = = Vamomen quan tnh oi vi truc l anhonhat. That vay Cx x x= +

( )2

2 22Oy C C C

D D D D

J x x dxdy x dxdy x dxdy x x dxdy = + = + + =

= 2 20 .C CD

x dxdy J x S J J + + = + . pcm. vaco 2 .Oy CJ x S J = +

Vaay chnh lanh lyHuyghen.

M omen quan t nh l uon l uon dng. V amomen quan tnh oi vi truc trung hoa lanhonhat, so

vi cac truc / / vi truc trung hoa.

V du1 :tm trong tam cua 1 tam giac

ong chat nh hnh veben.

ta cophng trnh ng thang 1x y

a b+ = , dien tch S =

1

2ab

(1 )2 3

00 0 0

(1 )2 3

xy b

a aaa

y

D

x bx b x S xdxdy xdx dy xb dx

a a

=

= = = = =

=2 2 2

2 3 6

ba ba ba = . V ay

21 1

6 2 3

yc

S bax ab a

S= = = ,

tng tta co1

3

xc

Sy b

S= = .

M ot lan na ta tm lai c ket quamatrong hnh s cap abiet .V du2 :tm toa otrong tam cua hnh phang ong chat gii han bi 2ng cong sau y2= 4x + 4 vay2= -2x + 4 .

Dien tch

2

2

22 2

21

4

yx

D yx

S dxdy dy dx

= +

=

= = =

o a

y

b

x



24/78


A1= A9 A2

A8

cy

2 2 2

0

4 42 ( )

2 4

y ydy

= =

22

0

32 3 8

4y dy

= =

yS

2

2

4

2 2 224

0 04

4

33 32 2 2 4 32

yx

y y

y yyx

y

dy xdx y dy

== =

= ==

= = +

yS

22 2 534

0 0

3 33 16(3 ) 3

2 16 2 80 5

y yyy dy y = + = + =

2

5

yc

Sx

S= = , yc= 0 larorang do tnh chat oi xng

V du3 :T m toa otrong tam vamomen quan tnh oi vi truc t rung hoa cua no, cua hnh phangong chat cotoa onh hnh ve. Co8 iem, iem th9 trung vi i em thnhat. A 1(1,0) , A 2(2,0) ,

A3(2,3), A4(3,3), A5(3,4), A6(0,4), A 7(0,3), A8(1,3), A9(1,0).a) Ta ap cong thc vi 9 iem (tnh bang V inacal -570M S).

( ) ( )9

1 1 11

1 9015

6 6

i

x i i i i i i i

S x y x y y y =

+ + +=

= + = = ,

(Cach bam may chnh xac, xin xem sach V inacal -570M S cung tac gia)

Tnh theo tch phan:

1 2

9 2115

2 2x

D D

S ydxdy ydxdy = + = + = , do tach:

Hnh chnhat di:

1

32

1 0

9

2

yx

D x y

ydxdy dx ydy

==

= =

= =

Hnh chnhat tren:

2

43

0 3

21

2

yx

D x y

ydxdy dx ydy

==

= == =

Dien tch hnh S= 3 1 3 1 6 + = hay ta cothedung cong thc:

tt x yMomen

tnh

MomenquantnhA1 1 0

A2 2 0 0 0

A3 2 3 18 54A4 3 3 -18 -81

A5 3 4 21 111

A6 0 4 96 576

A7 0 3 0 0

A8 1 3 -18 -81

A9 1 0 -9 -27

90 552

Bang A 46

x

y

2

2

2

-1

2 4 4y x= +

2 2 4y x= + 2

5



25/78


( )8

1 11

1

2

i

i i i i i

S x y x y =

+ +=

= 1 2 2 3 3 0 0 1 11 12

62 0 0 3 3 4 4 3 3 0 2

= = = .

Vay15

6

xc

Sy

S= = = 2.5 ay tung ocua truc tung hoa nh hnh ve.

Thc ra ta cothetm trong tam cua hnh bang cach lay iem gia oan noi hai trong tam hnh chnhat tren vadi bang nhau, ta cung thu ngay c ket quatren cy = 2.5. V i phng phap nay ta

cothetm trong tam bat kyhnh nao, bang cach phan hnh ra nhieu manh, mamoi manh ta abiettrong tam cua chung, lan lt tm trong tam noi 2 manh mot, theo ttrong cua dien tch 2 manh o,vacthe....ay ladam cau uc san ch T, c lap ghep ethanh mat cau trong cac cong trnh cau ng.Khi cotai trong th phan tren truc trung hoa cua dam sebnen, ngc l ai phan di truc trung hoacua dam sebkeo.b) oi vi momen quan tnh /v truc Ox, ta ap cong thc vi 9 iem nh cot cuoi bang tren, tnhbang excel , hay tnh bang V inacal-570M S.

( ) ( )8

21 1 1 1

1

1 55246

12 12

i

x i i i i i i i i i

J x y x y y y y y =

+ + + +=

= + = = ,

Tnh theo tch phan:

1 2

2 29 37 46x

D D

J y dxdy y dxdy = + = + = , do

Hnh chnhat di:

1

322 2

1 0

9

yx

D x y

y dxdy dx y dy

==

= == =

Hnh chnhat tren:

2

432 2

0 3

37

yx

D x y

y dxdy dx y dy

==

= == =

c) etnh momen quan tnh oi vi truc trung hoa , ta cothedung nh lyHuygentrong vat l yve

oi truc trong cong thc tnh momen quan tnh:

46= 2xJ J a F = +2

5 1506 46 8.5

2 4J J

= + = =

etnh trc tiep ta dung cong thc oi truc: 2.5y Y a Y y = + = , vasau ota tnh tng t

nh bang A tren mot lan na, ta co:

y x 2.5Y y= Momentnh

Momenquan tnhA1 0 1 -2.5

A2 0 2 -2.5 -12.5 46.875

A3 3 2 0.5 -12 31.5

A4 3 3 0.5 -0.5 -0.375A5 4 3 1.5 6 9.75

A6 4 0 1.5 13.5 30.375

A7 3 0 0.5 0 0

A8 3 1 0.5 -0.5 -0.375

A9 0 1 -2.5 6 -15.75

0 102

Bang B 8.5

Xem bang ta thay lai ket qua: Momen tnh oi vi th tr iet tieu, va



26/78


cy

A

B

DC

( ) ( )8

21 1 1 1

1

1 1028.5

12 12

i

i i i i i i i i i

J x y x y y y y y =

+ + + +=

= + = =

V du4 :Tm toa otrong tam vamomen quan tnh oi vi truc trung hoa cua notheo hnh vecotoa onh hnh sau: A1(0,0), A2(3,0), A3(2,3), A4(3,3), A 5(4,4), A6(-1,4), A 7(0,3), A8(1,3).

Sinh vien tgiai. 2768

6412

x

D

J y dxdy = = = , S=10

130 130 13

6 6 10 6x cS y= = =

=2.1666

(Cach bam may phan nay, x in xem sachV inacal -570M S cung tac gia)

V du5 :Tm toa otrong tam vamomen quan tnh oi v i t ruc Oy theo hnh vecotoa onh hnhsau: B (2,3), D(5,7) , A (-1,5), B(2,3). Nghi em lai bang cong thc gi ai tch.Dien tch S:

( )3

1 11

1 189

2 2

i

i i i i i

S x y x y =

+ +=

= = =

M omen tnh:

( ) ( )3

1 1 11

1

6

270456

i

x i i i i i i i

S x y x y y y =

+ + +=

= + =

= =

( ) ( )3

1 1 11

1 10818

6 6

i

y i i i i i i i

S x y x y x x =

+ + +=

= + = =

Vay45

9

xc

Sy

S= = = 5,

18

9

yc

Sx

S= = = 2

( ) ( )3

21 1 1 1

1

1 59449.5

12 12

i

Oy i i i i i i i i i

J x y x y x x x x =

+ + + +=

= + = =

Ta chnghi em lai momen quan t nh /Oy: ta chi a thanh 2 vung

( )

162 23

2 2 2

2 131 1

3

271

4

xyx x

xABC x x y

x dxdy dx x dy x x dx

+== =

+= ==

= = + =

( )

16

5 532 2 2

4 12 2

3

1715

4

xy

x x

xBDC x x y

x dxdy dx x dy x x dx

+=

= =

+= ==

= = + =



27/78


0 2 4 6 8 10 1 2

-8

-6

-4

-2

0

2

4

6

8

10

H1

-3 -2 -1 0 1 2 3 4-2

-1

0

1

2

3

4

5

H2

Vay27 171

49.54

OyJ += = . Ta cotruc BC=qua trong tam nen ta tm momen quan t nh J (oi

truc v i x=2) : ta van chi a thanh 2 vung

[ ] [ ]

16

2 32 2

2 131

3

2 2

xy

x

xABC x y

x dxdy dx x dy

+==

+= =

= = [ ] ( )2

2

1

272 1

4

x

x

x x dx

=

= + =

[ ] [ ]

165 3

2 2

4 12

3

2 2

xy

x

xBDC x y

x dxdy dx x dy

+==

+= =

= =

[ ] ( )5

2

2

272 5

4

x

x

x x dx

=

== + = . V ay

27 27

4 4J= + = 13.5. Cothetm:

2 249.5 2 .9 49.5 36OyJ J a F J J = = + = + = = 13.5

Vay momen quan tnh oi vi truc t rung hoa l abenhat.

V du6 :Tm toa otrong tam cua hai hnh: hnh goc vahnh bi en hnh, mata axet v du2, muc2.2, cocac toa otheo bang:

TT 1 2 3 4 5 6 7 8 9 10

x 1 2 4 3 4 2 0 2 2 1

y 0 -2 2 2 3 4 3 2 1 0

c bien hnh qua phep bien oi tuyen tnh coma tran tng ng la:2 1

1 3G

=

Nhan ma tran G vi ma tran 2,10M tren ta c ma tran toa obien hnh nh sau (H1):

x 2 2 10 8 11 8 3 6 5 2y -1 -8 2 3 5 10 9 4 1 -1

Dung cac cong thc tren etm toa otrong tam 2 hnh H1:

trong tam hnh goc:

34

15

23

15

, trong tam cua hnh bien:

91

15

7

3

--> H1



28/78


0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

A

B

C

Chuyta co:

34 91

2 1 15 15

1 3 23 7

15 3

=

? V aneu lay:31

2 2

3 12 2

G

=

lam phep bien hnh: ngha lanhan

ma tran G vi ma tran 2,10M tren ta c ma tran toa obien hnh nh sau (H2):

x 0.5 2.732 0.267949 -0.232 -0.598 -2.4641 -2.598 -0.732 0.13397 0.5y 0.866 0.732 4.464102 3.5980 4.9641 3.7320 1.5 2.7320 2.23205 0.8660

Dung cac cong thc tren etm toa otrong tam 2 hnh H2:

Trong tam hnh goc v an la:

34

15

23

15

, trong tam cua hnh bien:-0.194572

2.729657

--> H2, ta van co

312 2

3 12 2

34

-0.19457215

23 2.72965715

=

Nhan xet: M a tran2 1

1 3G

=

ban au lama tran cua phep bien oi af f in nen hnh bien sebmeo

mo, con ma tran31

2 2

3 12 2

G

=

sau lacua phep quay (hnh bien khong meo mo) cua cothe, nen

toa otrong tam c ginguy en qua phep quay, cac thao tac tnh toan tren, cothecai at thuattoan bang excel.

V du7 : Tnh hai tch phan: I=D

xdxdy ,

2

D

J x dxdy = D latam giac cocac nh la

A(2;3), B(7;2), C(4;5).

Giai : Xem hnh

I1 =17

5 5

14

2 x

y xx

x y

xdx dy

= +=

= = + =

( )

43 262

5 5 2

x

x

x x=

=

=8

I2 =17

5 5

97

4 x

y xx

x y

xdx dy

= +=

= = + = ( )

73 24 14

5 5 4

x

xx x

=

= + =18

Vay I =8+18=26



29/78


A

B C

D

E

Cach 2: T2 bieu thc I , J muon tnh, ochnh lamomen tnh, momen quan tnh cua tam gi ac /v

truc Oy. Nen ay ta cotheap dung cong thc:D

xdxdy = ( ) ( )3

1 1 11

1

6

i

Oy i i i i i i i

S x y x y x x =

+ + +=

= +

TT 1 2 3 4x 2 7 4 2

y 3 2 5 3

Xem cach bam may nhanh V inacal-570M S (cung tac gia) cho cong thc tren, ta co

I=1

1566

D

xdxdy= = 26, tng t:1

1206

D

ydxdy= = 20.

Tnh 2

D

J x dxdy = . bang cach tach 2 tch phan:

J1 =175 5

142

2 x

y xx

x y

x dx dy

= +=

= = +

= ( )4

2 6 125

2

136

5

xx

x

x dx

=

=

=

J2=17

5 5

972

4 x

y xx

x y

x dx dy

= +=

= = + = ( )

72 4 28

54

459

5

xx

x

x dx

= +

== =>J=

136 459

5 5+ =

=119Cach 2: N hcong thc momen quan tnh:

2

D

J x dxdy = = ( ) ( )2

1 1 1 11

1

12

i n

Oy i i i i i i i i i

J x y x y x x x x =

+ + + +=

= +

Bam may tong tren:1

142812

OyJ = = 119, tng t:1

82812

OxJ = = 69

V du8 :M ien Q gii han bi 4 iem (0, -2), (0, -4), (-2,0) va (-1,0),

hay tnh I= ( ) ( )2

2

Q

y x x y dxdy + +

Giai :at ( )2u y x= + , ( )v x y= + ,

( )

( ) ( )

( )

, 1 1

12 1,,

, 1 1

D x y

D u vD u v

D x y

= = =

M ien bien hnh lamien hnh thang nho:

( ) ( )2 / 2 2 4

2 2

4 4

3 3722

8 5

v u

Q v u

uI y x x y dxdy u vdudv du

=

=

= + + = = =

Cach 2: Tnh trc tiep, ta phan ra hai mien D1= AB C, D2=ACDE



30/78


( ) ( )1

012

1

2 2 4

3562

15

yx

D x y x

I dxdy y x x y dxdy

==

= =

= = + + =

( ) ( )2

2 202

2

1 2 4

1522

3

y xx

D x y x

I dxdy y x x y dxdy

= =

= =

= = + + =

1 2356 152 372

15 3 5

I I I

= + = = ,

tnh chi tiet danh cho sinh vien.V du8 : M ien Q gii han bi 4 iem(-2, 2), (0, 2), (3, 0) va (5/2, -1),

hay tnh I=3

2

3Q

xy dxdy

+

Giai:

at2

3

xu y= + , v y= (cotheat tuy y)

( )( ) ( )

( )

23

, 1 1 3,, 21

,0 1

D x yD u vD u v

D x y

= = = . M ien bien hnh lamien hnh thang nhocotoa otheo s(onhan

ma tr)n (bang nhan):

52

2 2 23 3 3

2 0 3

2 2 0 1

1 2 2

0 1 2 2 0 1

2 43 3

3 2 23

2 0

3 2 3 1832

2 3 2 135v

v

Q u

xI y dxdy u dudv

=

+ =

= + = =

V du9: M ien Q gii han bi 4 iem (-2, 0), (1, 2), (4, 1) va (-1/2, -2),

hay tnh I= ( )3

3 2

Q

y x dxdy

Giai:

( )3

3 23 2 27

3Q Q

xI y x dxdy y dxdy

= =

at2

3

xu y= , v y= (cotheat tuy y)

( )

( ) ( )

( )

23

, 1 1 3

,, 21

,0 1

D x y

D u vD u v

D x y

= = =



31/78


M ien bien hnh lahnh thang nho(bang nhan):

12

5 52 4 43 3 3 3 3

2 1 4

0 2 1 2

1

0 1 0 2 1 2

4 143 3

5 83 3

3 03

3

2 3 0

2 3 317727 27.3 2 20

u v

Q u v

xI y dxdy u dudv

+ =

+ = = = =

BAI TAP1/ Tnh cac tch phan sau:

a/1 3

0 2

( ( ) )I x y ydy dx = va3 1

2 0

( ( ) )J x y ydx dy = .

so sanh I vaJ

b/ Cho

1

0 0

( ( ) )

x

K x y ydy dx = va1

0 0

( ( ) )

x

L x y ydx dy = .

So sanh K vaL , to cho ket luan.

c/ Chng minh :1

0 0

( )

x

x y ydydx =1 1

0

( ) x y

y

x y yd d

2/ Hoan vcan tch phan sau:

a/2

2 4

2

( , )

x

dx f x y dy

b/

23

1 0

( , )

y

dy f x y dx c/ln

1 0

( , )

xe

dx f x y dy

3/ Tnh cac tch phan sau:a) 2 2(cos sin )

D

x y dxdy + , D lahnh vuong: 04

x

, 04

y

b) ln .D

x y dxdy D lahnh chnhat 0 4x , 1 y e

c)3

.( )D

dxdy

x y+ D xac nh bi 1x , 1y , 3x y+

d) 2( ) .D

x y x dxdy D gii han bi cac ng y = x2, x = y2

4/ Tnh cac tch phan sau:a) 3( ) ( ) .

D

x y x y dxdy + D gii han bi:

x + y = 1, x y = 1, x + y = 3, x y = -1b) (2 ) .

D

x y dxdy D gii han bi :

x + y = 1, x + y = 2, 2x y = 1, 2x y = 3c) .

D

xydxdy D gii han bi :



32/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh32 xy = 1, xy = 2, y = x, y = 3x (x > 0, y > 0 )

d) 2 2 .D

x y dxdy + D gii han bi :

x2+ y 2= a2, x2+ y 2=4a2 ( a > 0 )

e)2 2

.1D

dxdy

x y+ + D gii han bi 21y x= vaox

f) 2 .D

xy dxdy D gii han bi cac ng tron:x2+ ( y 1)2= 1, va x2+y 2 4y = 0

g)2 2

.x y

D

e dxdy + D lamien x2+ y 2a2

h)2 2

2 21 .

D

x ydxdy

a b D laEll ip

2 2

2 21

x y

a b+ =

5/ Tnh dien tch phang gii han bi cac ng :a) x2+ y2= 2x, x2+ y2= 4x, y = x, y = 0

b) (x 2y + 3 )2+ ( 3x + 4y 1 )2= 100c) x2= ay , x2= by , y2= x , y2= x ( 0 < a < b, 0 < < )d) x = 4y y2 , x + y = 6

e) x2+ y2= 1 va x2+ y 2=2

3x

6/ Tnh thetch gii han bi cac mat :a) y = 1 + x2, z = 3x , y = 5 , z = 0 nam trong phan tam thnhat.b) x2+ y2= 2 , z = 4 x2 y2 , z = 0c) x2+ y2=a2, x2+ z2= a2

d) x = 0 , y = 0 , z = 0 , 11 1 2x y z+ + =

e) z = x 2+ y 2 , z = x + y.7/ Tnh momen quan tnh cua :

a/ H nh vanh khan cocac ban knh d, D, d < D, = 1 ( tkhoi ) .oi vi tam cua no

oi vi ng knh cua nob/ Tm momen quan tnh cua hnh vuong canh a oi vi truc i quanh cua no, trc giao oi vi mat phang cua hnh v uong , = 1c/ T m momen quan tnh cua mien gii han xy = 4, x + y = 5,

= 1, oi vi ng thang y = x.8/ X ac nh trong tam cac ban phang ong chat gii han bi cac ng :

a)2 2

125 9

x y+ = , 1

5 3

x y+ =

b) y2=x , x2= y

c) 22y x x= , y = 09/ Tm momen quan tnh cua hnh

vuong ben vi truc Ox :10/ Tm dien tch giao bi 2 ng:

H8

A

B

C

Ox



33/78


A

B

CD

0 1 2 3 4 5-2

-1

0

1

2

3

4

5

2

2

2 12 8

31

4 2 4

y x x

y yx

= + = + +

( lam tieu luan)

11/ Tnh ( )2 22 3 2

y xx y xy

y x

I e dxdy

= = +

= ==

12/ Trong xay dng, Ngi ta dung mayknh veo khoang cach vagoc xoay (v duau tien ng tai iem 1,ta sexac nh c khoang cach12 , ng 2 ta cothem 23 vagoc

nh hng dng ( )21,23 ) .. cuoicung ng lai 1 ta cothem

goc nh hng dng ( )1,11,12 nhhnh ve. Hay mohnh cong thc tnhdien tch hnh nay vi cac k hoangcach vagoc nh hnh vevacho v duroi dung may tnh V inacal etnh.(Bai nay danh cho Sv lam tieu l uan).

13/ Cho 5 iem A 1(2,-2), A 2(4,2), A 3(3, 2),A 4(4,3), A 5(2,4) . Bay gi tnh tien5 iem nay thanh B1, B 2, B3, B4, B5

vi vecto ( )1,0.5a = . Tm toa o

trong tam cua a giac kn

10 iem tren nh hnh ve.S=(43/12, 43/24)

14/ Cho hnh phang bat ky, l ay mot iemM trong hnh phang o. Hay tm mot truc qua i qua iem M , sao cho momen quan tnh cua vat(hnh phang) oi vi truc olanhonhat.(Cothelatruc qua M vai qua trong tam cua hnh phang ?)

1

2

3

4

5

6

7

8

9

10

11



34/78


Phan 2:ai so: M a tran, nh thc, Hephng trnh, K GV T, Anh xa tuyen tnh

CHNG 1MA TRAN-HEPHNG TR NH TUYEN TNH

BAI 1: MA TRAN

1.1 nh ngha:M ot ma tran mn l amot bang chnhat gom mn phan t nh sau: =

11 12 1

21 22 2,

1 2

....

....

.....

....

n

nm n

m m mn

a a a

a a aA M

a a a

(R)

Ta cocac kyhi eu: A=[ aij ] i = 1, , m, j = 1, , nHai ma tran A vaB bang nhau, ghi laA = B, neu chung cocung dang kch thc vacac phan ttng ng bang nhau.

Trong trng hp m=n goi lama tran vuong, vaneu trong ma tran vuong macotat caphan teubang 0, ngoai trcac phan ttren ng cheo chnh bang 1, th ta goi olama tran n v.M a tran chuyen vA Tc xay dng tma tran A , bang cach oi hang thanh cot cua A .

In=

1 0 0 0

0 1 0 0

0

0 0 1

M O

L

,

=

11 21 1

12 22 2,

1 2 ,

....

....

.....

....

m

mTn m

n n m n

a a a

a a aA M

a a a

(R)

M a tran tam giac tren, di:

11 12 13 1

22 21 2

33 3

0

0 0

0 0 0 0

M

M

M

M M M O

n

n

n

nn

a a a a

a a a

a a

a

,

11

21 22

31 32 33

1 2 3

0 0 0

0 0

0

0

M

M

M

M M M O

Mn n n nn

a

a a

a a a

a a a a

V du 1: Tm x,y,z,w sao cho:+ +

=

2 3 5

1 4

x y z w

x y z w

Ta c hephng tr nh:x + y = 3; x y = 1; 2z + w = 5; z - w = 4Nghiem cua helax = 2, y = 1, z = 3, w = -1.

V du2: hi en nhien ta co( ) =T

TA A

1.2 Cac phep toan tren ai soma tran:

Cho A, B ,C, 0 M m,n, , R Cong: C=A+B [cij ]= [aij ]+[ bij ] Nhan vohng: A = [ aij ] M m,n, R.

V et cua ma tran vuong nA M , ( )1

n

i ii

tr A a

==

Ta cocac tnh chat sau:



35/78


1jb

2jb

3jb

4jb

1ia

2ia

3ia

4ia

mpA

pnB

mnC ijc

(I ) (A + B) + C =A + (B + C)(II) A + 0 = A(III) A + (-A) = 0(IV) A + B = B + A(V ) (A + B ) = A + B(VI) (+ ) A = A + A(VII) (. ) A = (A )

(VII I ) 1. A = AVay tap cac ma tran M m,n(R) cohesothc, lamot K GV T tren R.

Nhan ma tran vi ma tran:Gias[aij ] va[bij ] lacac ma tran sao cho socot cua A th bang sodong cua B ngha laA m,pvaBp,n.Khi otch cua ma tran A vaB, ghi laA B lama tran m x n masohang cij thu c bang cach nhanvohng vet dong thi cua A vi vet cot thj cua B , cuthe:

== + + + =1 1 2 2

1

...p

i j i j i j i p pj ik kj k

c a b a b a b a b

S onhan ma t ran:

V du1: S onhan ba ma tran:[ A (2,2)* B(2,3) ] * C(3,2)= D (2,2) nh hnh ben di:

= = =

0 21 1 1 1 1

, , 1 12 1 2 1 1

1 1

A B C

0 2

1 1 1 1 1

2 1 1 1 1

1 1 1 2 2 4 2

2 1 4 1 1 2 10

,

ma tran cuoi D(2,2)=

4 2

2 10.



36/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh36Ta cotheviet oan maPascal tch 3 ma tran

[A (m,n)* B(n,p) ]* C(p,q)= D(m,q) nh sau:K hai bao dimension A (m,n), B(n,p), C(p,q), D(m,q)

for i=1, mdo begin

for j=1,q do begin

D(i,j)=0;for k=1,p do begin

for kk=1,n do begin

D(i,j)= D(i,j)+ A(i ,kk) * B (kk,k) * C(k,j);end

endend

end

V du2:1 0 0 0

0, 00 0 1 0

C D

= =

= =

0 00

0 0CD ,

=

0 0

1 0DC

Tay ta thay rang, trong ma tran neu coCD=0 (C=0 hay D=0).

V dukhac

0 0 2 1 2 4 0 0 0

0 2 1 0, 0 0 0 0 0 0 0

0 1 1 0 0 0 0 0 0

C D CD

= = =

.

V du3: ac biet nhan 2 ma tran hang vacot, ket qualasothc:

[ ]1 2 L nh h h

1

2

n

c

c

c

= + + + + 1 1 2 2 3 3 L n nh c h c h c h c R

giong nh tch vohng chnh tac abiet lp 12 phothong.

Tnh chat tch ma tr an:1/ AIn= InA vi A M n,n, InM n,n (ma tran n vcap n)2/ A (BC)=(AB )C3/ A(B+C)=AB+AC, (B+C)A=BA +CA4/ Tong quat ta khong coAB=BA .V du4: Lay lai v du tren ekiem tra s onhan A (BC)

= = =

0 2

1 1

0 2 1 11 1 1 1 1

, , 1 1 , 1 1 1 2 42 1 2 1 1

1 1 2 1 1 2 2

1 1 4 2

2 1 2 10

A B C



37/78


ma tran cuoi A (B C)= D(2,2) =

4 2

2 10so sanh vi v du1.

Luy tha ma tran: A M n(R), ta ky hieu nh sau:

=0A I , =1A A , =2A AA , = 1m mA A A .

V du5:

=

0 1 0

0 0 1

0 0 0

A ,

=

2

0 0 1

0 0 0

0 0 0

A , = =

3

0 0 0

0 0 0 0

0 0 0

A

V du6: Cho ma tr)n A vuong c*p 2007 maphan t&+dong th,i l a( )1 i

i , th ph.n t&+ dong 2

cot 3 cua A 2la:

( ) ( ) ( ) ( )2007

1

2 1 2 1 2 3 4 ........ 2005 2006 2007i

i = + + + + =

= ( )2006

2 1 2007 20082

=

nh ly: Gias A, B M n(R), vagiao hoan c, nghialaA B=BA. K hi o:

1/ ( ) =m m mAB A B .

2/ ( )( ) = + + + +1 2 3 2 1Lm m m m m m A B A B A A B A B B .

3/ ( )=

=+ =

0

i mm i m i i

mi

A B C A B . =!

!( )!

im

mC

i m i

nh ly: A Tlama tran c thiet lap tA bang cach oi dong thanh cot vay: A M m,n(R), vaA TM n,m(R) ). Ta cocac tnh chat sau:

i/ ( ) =T

T

A A .

ii/ ( )+ = +T T T

A B A B .

i i i / ( ) =T T T

AB B A .

BAI 2: HEPHNG TR NH TUYEN TNH2.1 nh ngha:

Hephng tr nh tuyen tnh lahephng tr nh codangtong quat (m n):

+ + + =

+ + + = = + + + =

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

(1) .

K

K

L L L L L L L L L L

K

n n

n nm n

m m mn n m

a x a x a x b

a x a x a x b A x b

a x a x a x b

Trong o:



38/78


=

11 12 1

21 22 2

1 2

(2)

L

L

L O O L

L O O L

L

n

n

m n

m m mn

a a a

a a a

A

a a a

lama tran hesocua (1),

= =

1 1

2 21 1;

L Ln m

n m

x b

x bx b

x b

[ ] +

= = =

11 1 1

( 1)

1

(3)

L

L L L L

L

a

m n

m mn m

a a b

A A b A

a a b

,

lama tran mrong cua he(1).Henay com phng trnh tuyen tnh, n an sox1, .., xn.Hec goi lathuan nhatneu cac hang sob1, ., bmtat cabang 0.2.2 Nghiem cua hephng trnh tuyen tnh:

Bo (x1, x2, x3,.....xn) goi l anghiem neu chung thoa man he(1). Gi ai he(1) latm tat cabonghiem tren.

Hai hephng trnh tuyen tnh goi l atng ng neu chung cung chung mot tap nghiem. Chuyhethuan nhat l uon cot nhat mot nghiem tam thng la(0,0,...,0).

2.3 Cac phep bien oi s cap (PBSC) tren dong cua ma tran:Phep bien oi s cap tren dong (PBSC) cua ma tran gom:

1/ oi cho2 dong cho nhau.

2/ Nhan mot dong vi mot hesokhac khong.3/ Cong vao mot dong vi mot dong khac sau khi a nhan vi mot hang so.

A M m,n(R). M ot PBSC e tren ma tran A ec ma tran A/, ta kyhieu /e

A A .

Nhan xet: Neu e laPBSC bien A thanh A/, th cung coPBSC e/esao cho /

/ eA A .

nh nghia:A, B M m,n(R) c goi latng ng dong, khi A coc tB qua hu han motsoPBSC. L uc ota k hieu A ~ B .

nh nghi a: M ot ma tran v uong S c goi lama tran s cap k hi comot PB SC sao cho

eI S. K hi ota v iet S = e(I).

V du1: S=

1 0 2

0 1 0

0 0 1

lama tran s cap v +1 32d dI S

M enh e: A M m,n(R) (khong can vuong). e laM ot PBSC. Khi o:

/eA A M m,nvae

m m mI S M th: A/=SA.

V du2: A=

1 1 2

4 3 2

1 2 3

, +1 32 /d dA A =

3 5 8

4 3 2

1 2 3



39/78


S1=

1 0 2

0 1 0

0 0 1

lama tran s cap v 1 32 1d d

I S+

A/ =

3 5 8

4 3 2

1 2 3

=S1A=

1 0 2

0 1 0

0 0 1

1 1 2

4 3 2

1 2 3

. T iep tuc, hoan v dong 2, 3:

2 3/ //

3 5 8 3 5 8

4 3 2 1 2 3

1 2 3 4 3 2

d dA A

= =

, 2 3 2d d

I S

S2=

1 0 0

0 0 1

0 1 0

. V ay //1 0 0 1 0 2 1 1 2

0 0 1 0 1 0 4 3 2

0 1 0 0 0 1 1 2 3

A

=

=S2S1A

V du3:A =1 1 2

1 2 3

, +1 32 /d dA A =3 5 8

1 2 3

S1= 1 20 1

lama tran s cap v 1 32 1d dI S+ . Hay th&lai:

A/ =3 5 8

1 2 3

=S1A=1 2

0 1

1 1 2

1 2 3

Hequa: A, B M m,n(R). A ~ B khi vachkhi ton tai cac ma tran s cap S1, S2, S3,...., Skcap msao cho: B= SkSk-1.... S2. S1. A

Do 31 21 2 3ke ee e

kA A A A A B =L

ta coA 1= S1A , A 2= S2A 1, A 3= S3A 2,..... . . ., B=A k= SkA k-1,suy ra pcm.

Chuy: Neu e l aPBSC tren cotA M m,n(R) (khong can vuong). e l aM ot PBSC tren cot. K hi o:

/eA A M m,nvae

m m mI S M th: A/=A S.

V du: 1 2 1 /1 2 1 2

3 4 1 4

c cA A

+ = =

1 2 12

1 0 1 0

0 1 1 1

c cI S

+ = = .

Hay thlai: /1 2 1 2 1 0

.

1 4 3 4 1 1

A A S

= = =

2.4 nh lyvecac phep bien oi s cap:nh ly: Sau hu han cac phep bien oi s cap tren he(1) ta thu c hetng ng.Chng minh nh l ynay khan gian.V du1:

He

+ = + + = + + =

1 2 3 4

1 2 3 4

1 2 3 4

3 2 3 1 3 1 2 3

3 5 9 4 8, 3 5 9 4 8

4 3 5 7 144 3 5 7 14

L

x x x x

x x x x

x x x x



40/78


1 3 1 2 31 3 1 2 3

1 20 9 9 1 2 0 1 1

9 90 14 12 10 17

76 1810 0 2

9 9

Vay heau tng ng:

+ =

= =

1 2 3 4

2 3 4

3 4

3 2 3

9 9 2

76 1812

9 9

x x x x

x x x

x x

vata tm nghiem tren hen gian nay dedang hn nhieu.V ay sau hu han cac phep bien oi s cap tren tran A m,,n, ta thu c ma tran tng ng, ma

dang cua ma tran sau cung A/codang bac thang ( nhng hesokhac zero tao thanh bac thang rutgon ).

nh ngha ma tran bac thang: R M m,,n(R) c goi la ma tran bac thang rut gon neu cac ieukien sau ay c thoa:i/ Cac dong zero (neu co) phai ben di cac dong khac zero (neu co).i i / Phan tau tien khac 0 tren cac dong khac zero (neu co) laso1, vaben di cot cua somot naytat caeu bang 0.

V du: ma tran bac thang codang nh sau:

1

0 1

0 0 1

0 0 0

0 0 0 0

M

M

M

M M

M

nh ly:M oi ma tran eu tng ng dong vi ma tran codang bac thang rut gon duy nhat.

V du2:A =

0 2 4 1 4 5 1 0 3

1 4 5 0 2 4 0 1 2

3 1 7 0 5 10 0 0 0

0 5 10 0 11 22 0 0 0

2 3 0 0 5 10 0 0 0

=B

B laladang ma tran bac thang rut gon tng ng vi A .

Bay gineu goi x, y , z l anghiem cua

0 2 4 0

1 4 5 0

3 1 7 0

0 5 10 0

2 3 0 0

x y z

x y z

x y z

x y z

x y z

+ =

+ = + + = + =

+ + =

th chung cung langhiem cua

1 3 0

1 2 0

x z

y z

+ = =

vanghiem nay la:

3

2

x t

y t

z t

= = =

.

Luc nay ta nh ngha hang ma tran nh sau:



41/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh 41nh ngha hang cua ma t ran: Sohang khac zero cua ma tran dang bac thang rut gon cua matran A c goi lahang cua ma tran A . K yhieu lar(A), oi khi con k hieu l arank(A ).Ngi ta con nh ngha hang cua ma tran l abac ln nhat cua mot nh thc con nao o khac zero(c trch ra tA, k hang, k cot bat k yvamoi nh thc cap k+1 tri eu bang 0) . Ta cothechng minh hai nh ngha nay latng ung.

V du3:A =

2 1 4 5 6

0 3 2 3 0

0 0 4 1 5

0 0 0 0 0

, cohang la3, v ay lama tran bac thang cha rut gon, vamoi ma tran

bac 4 eu bang 0, v cha hang tht toan bang 0. V dunay cho ta thay 2 nh ngha vehang trenlatng ng.

M enh e: A M m,,n(R) khi o:i/ 0 /r(A) /min (m,n).ii / Neu A ~B th r(A ) =r(B) .iii/ r(A) =0 A=0.

iv/ r(A)=r(AT

)2.5 nh lyKronecker-Capell i :nh ly: He =.m nA x b (1), trong o: [ ]A A b = , ta cocac k et luan:

a/ Ta cohoac r(A ) = r(A/) hoac r(A/)= r(A ) +1.b/ (1) conghi em khi vachkhi r(A )= r(A/), cu the:

Neu r(A )= r(A/)= n, (n l asoan) th heconghi em duy nhat. Con neu r(A )= r(A/) < n, (n lasoan) th hecovosonghiem.V i soan tdo lan - r(A )

c/ r(A ) < r(A/)= r(A ) +1, hevonghiem.

V du1: Da vao nh lyK ronecker-Capell i , xet vtr tng oi cua 4 mat phang trong R3: chrocautruc nghiem cua hephng trnh neu co:

+ + + = + + + = + + + = + + + =

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

0(1)

0(2)

0(3)

(4) 0

a x b y c z d

a x b y c z d

a x b y c z d

a x b y c z d

,

at A =

1 1 1

2 2 2

3 3 3

4 4 4

a b c

a b c

a b c

a b c

, [ ]

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

|

a b c d

a b c d A b

a b c d

a b c d

=

Do soan t hn sophng trnh nen khong bao gicor(A)= r([A |b] )=4. Vay chcon cac trng hp sau:1/ r(A )= r([A |b])=3, ! nghiem lamot iem (x0, y0, z0).2/ r(A )=3 < r([A |b] )=4, vonghiem. Hesuy bien thanh ba mat phang (mp) cat nhau, va comp (4)// vi mot trong ba mp tren.3/ r(A )=2 = r([A |b] )=2, voso nghiem. Hesuy bien thanh hai mat phang cat nhau. Nghiem namtren ng thang giao tuyen.



42/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh424/ r(A )=2 < r([A |b] )=3, vonghiem. Hesuy bi en thanh hai mat phang cat nhau, vaco mot mp //

vi mot trong hai mp tren.5/ r(A )=2 = r([A |b] )=4, vonghi em. Hesuy bien thanh hai mat phang cat nhau, con hai mp // vimot trong hai mp tren.6/ r(A )=1 = r([A |b] )=1, vosonghi em. Hesuy bi en thanh mot mat duy nhat, con ba mp kia trungvi mp tren.7/ r(A )=1 < r([A |b] )=2, vonghi em. Hesuy bi en thanh mot mat duy nhat, va co mot mp // vi mp

tren.8/ r(A )=1 < r([A |b] )=3, vonghiem. Hesuy bien thanh mot mat duy nhat, va cohai mp // vi mptren.9/ r(A )=1 < r([A |b] )=4, vonghi em. Hesuy bien thanh mot mat duy nhat, va coba mp // vi mptren.

V du2: Giai he:

+ = + = + + =

1 2 3 4

1 2 3 4

1 2 3

2 2

3 6 4 2 4

2 2 0

x x x x

x x x x

x x x

Ax=b (1)

Ta x et cap 3:

1 2 1 1 2 1

1 23 6 4 3 6 4 1 03 6

1 2 2 0 0 1

= = =

Xet nh thc cap 2:

2 10

6 4, nen rank(A )=2. Xet ma tran m rong A/ ta co:

1 2 2 1 2 21 2

3 6 4 3 6 4 2 03 6

1 2 0 0 0 2

= = =

,

=> r(A/)=2=r. V ay heconghi em (co2= n -r = 4-2 bac tdo)

= + = + 1 3 2 4

1 3 2 4

2 23 4 4 6 2x x x x

x x x x ==> x1=2x2-2x 4+ 4 x3= -x4+2,

Cach 2:

1 2 1 1 2 1 2 0 2 4

3 6 4 2 4 ... 0 0 1 1 2

1 2 2 0 0 0 0 0 0 0

=> ket qua.

Nhan xet: ( ) ( ) ( ) ( )(0) (1) (2)

1 2 3 4 2 4, , , 4,0,2,0 2,1,0,0 2,0, 1,1

x x x

x x x x x x = + + E55555F E55555F E5555555F

, ta

co ( )0x laN0cua Ax=b, con (1)x , (2 )x laN0cua A x=0, hay kiem chng!V du3: Gi ai vabien luan hephng tr nh theo tham som:

+ + = + + = + + = + + =

1

1

1

x y z m

x y mz

x my z

mx y z

2

1 1 11 1 1

0 0 1 11 1 1

0 1 0 11 1 1

1 1 1 0 1 1 1

mm

m mm

m mm

m m m m

Neu m-1=0 th



43/78


1 1 1 1

0 0 0 0

0 0 0 0

0 0 0 0

Hecovosonghiem

Neu m-1 0 th tiep tuc bien oi.1 1 1 1 0 1 1 1 0 0 2

0 1 0 1 0 1 0 1 0 1 0 1...0 0 1 1 0 0 1 1 0 0 1 1

0 1 1 1 0 0 0 3 0 0 0 3

m m m

m m m

+ +

+ + +

Tom lai : m 1 vam -3 hevo nghi em.m = -3 heduy nhat nghi em x = y =z = -1.m =1 hevosonghi em.

V du4: Giai vabien luan hephng trnh theo cac thong soa,b:1

(2 1) 3 1

( 3) 1

bx ay z

b x ay z

bx ay b z

+ + = + + =

+ + + =

Giai:

1 1

3 2 1 1

3 1

z y x

a b

a b

b a b

+

=>

1 1

0 2 1 2

0 ( 2) ( 2) 2

a b

a b

a b b b b

+

a/ b -2: =>1 1

0 2 1 2

0 1

a b

a b

a b

+

=>

1 1

0 1

0 0 1 0

a b

a b

b

+

=>

1 0 0 0

0 1

0 0 1 0

a b

b

+

b 1, a 0: => N0= (0, 1/a, 0) duy nhat nghiem.

b =1 =>

y x

1 1 1

0 1 1

0 0 0 0

z

a

a

=>

x y

1 0 0 0

0 1 1

0 0 0 0

z

a

, r(A)=2=r(A/)VosoN0

, k hong gian nghiem lang thang.

b/ b = -2: =>

y x

1 2 1

0 2 1 2

0 0 0 0

z

a

a

=>

x y

1 2 1

0 1 2 2

0 0 0 0

z

a

a

, r(A)=2=r(A/)VosoN0, k hong gian nghiem lang thang.Tom lai :b 1, b -2, a 0: => N0= (0, 1/a, 0) duy nhat nghi em.b 1, b -2, a = 0: => VoN0.



44/78

ecng On thi cao hoc, TS.GVC Nguyen Phu V inh44b = 1, VosoN0, khong gian nghiem lang thang..

b = -2, VosoN0, k hong gian nghiem lang thang.

That ra ta co 32 8= trng hp, rut ve 6 trng hp sau, nhng thu vechco4 trng hp tomtat tren:

1/ 0, 1, 2a b b= 2/ -- 1b= 3/ -- 2b= 4/ 0, 1, 2a b b 5/ -- 1b= 6/ -- 2b=

Cach khac:

1

2 1 3

3

b a

A b a

b a b

= +

( xem chng 2, tnh nh thc)

2

2

1 11 2

2 1 3 1 2 02 2

3 2 2 0

b a b a b

b a b a a b b b

b a b b b a ab

= =

+

( ) ( )( )2det 2 2 1A b a ab a a b b = + = + . Ta thecac gi atrb=1,

b= -2, a=0 vao phng trnh valyluan tng tnh tren ta secoket quatng t.

V du5: Gi ai, bien l uan hephng tr nh theo tham so:2

1x y z

x y z

x y z

+ + =

+ + =

+ + =

Giai:

2 2

1 1 1 1 1

1 1 1 1 1

1 1 1 1

--> 2 2

2

1 1

0 1 1 1

0 1 1

+

a/ 1:

1 1

0 1 1 1

0 1 1

+ +

-->

1 1

0 1 1

0 1 1 1

+ +

-->

-->2

1 1

0 1 1

0 0 2 ( 1)

+ +

va-2: => N0 duy nhat nghiem:( )

211 1

, ,2 2 2

x y z

++ = = = + + +

= -2: --> V oN0

b/ = 1:--> 2 2

2

1 1 1 1 1 1

0 1 1 1 0 0 0 0

0 0 0 00 1 1

+ =

N0lamp.



45/78


Cach khac: ( ) ( ) ( )23

1 1

1 1 det 3 2 1 2

1 1

A A

= = + = +

= 1:-->1 1 1 1 1 1 1 1

1 1 1 1 0 0 0 0

1 1 1 1 0 0 0 0

=

coN0lamp.

= -2:-->1 2 1 2 1 2 1 2 1 2 1 2

2 1 1 1 0 3 3 3 0 3 3 3

1 1 2 4 0 3 3 6 0 0 0 3

--> VoN0.-2 va1, duy nhat nghiem.2.6 Phng phap khGauss:

egiai hephng trnh tuyen tnh, ta dung cac phep bien oi s cap, a hevemot hemitng ng codang bac thang, sau odung nh lyCronecker-Capel l i exem heconghiem haykhong? Neu cota tm tap nghiem bang cach thetdi len tren egiai.

V du 1: Giai hephng tr nh:+ + =

+ + = + + =

1 2 3

1 2 3

1 2 3

2 1

2 5 6

4 2 2

x x x

x x x

x x x

, ap phep bien oi s cap vao, ta co:

+ + = + + = + + =

1 2 3

1 2 3

1 2 3

2 1

2 5 6

4 2 2

x x x

x x x

x x x

+ + = = + + =

1 2 3

2 3

2 3

2 1

4

6 3 3

x x x

x x

x x

+ + = = + =

1 2 3

2 3

3

2 1

4

9 21

x x x

x x

x

Ta thay ngay r(A )= r(A/)= r=3=n (coduy nhat nghiem),

thetdi len tren ta coheduy nhat nghiem: x= 5 70; ;3 3 .

Trong quatrnh bien oi ta chtnh toan tren cac hesonen bat au tay, etien ta chbieu dienma tran mrong cho cac phep bien oi s cap mathoi.V du 2: Giai hephng tr nh:

+ = + + = + =

1 2 3

1 2 3

1 2 3

2 3

4 2 1

2 3 4 2

Documents

11 de Cuong Toan Cao Cap