PANEPISTHMIO DUTIKHS ATTIKHS SQOLH MHQANIKWN TMHMA POLITIKWN MHQANIKWN ANWTERA MAJHMATIKA II SUNARTHSEIS POLLWN METABLHTWN h Seirˆ Ask sewn Akrìtata pragmatik n sunart sewn 1. Na brejoôn ta topikˆ akrìtata twn sunart sewn: (a) f(x, y) = 1 x 4xy + 9y + 3x 14y + 1. 1
(b) g(x, y) = x 3 + y 3 1(x + y).. Na deiqjeð ìti h sunˆrthsh f(x, y) = x 4 + 16y 4 (x y) èqei trða kritikˆ shmeða kai na brejeð h fôsh touc. Eidikˆ gia to (0, 0) na deiqjeð ìti den eðnai shmeðo topikoô akrotˆtou. 3. Na brejoôn ta topikˆ akrìtata thc sunˆrthshc f : R R, me f(x, y) = 1 x y + e x + e y. 4. Na brejoôn ta topikˆ akrìtata thc sunˆrthshc f : R R, me f(x, y) = x y 4 x 4 y, x 0, y 0. 5. Na brejeð sto xy-epðpedo èna shmeðo M(x, y) tètoio ste to ˆjroisma twn tetrag nwn twn apostˆsewn tou apì tic eujeðec x = 0, y = 0, x y + 1 = 0 na eðnai to elˆqisto dunatì.
6. Na prosdiorðsete ta topikˆ akrìtata thc sunˆrthshc f : R R pou orðzetai me ton tôpo f(x, y) = sin x + sin y + sin(x + y), an 0 < x < π kai 0 < y < π. 7. Na prosdiorðsete èna shmeðo P (x 0, y 0 ) tou kartesianoô epipèdou (Π) to opoðo èqei thn idiìthta ìti to ˆjroisma twn tetrag nwn twn apostˆsewn tou apì ta tèssera dosmèna shmeða na eðnai to elˆqisto. A(x 1, y 1 ), B(x, y ), Γ(x 3, y 3 ) kai (x 4, y 4 ) tou (Π) 8. Na brejeð h apìstash tou shmeðou P (1, 0, ) apì to epðpedo (Π) me exðswsh x + y + z = 4. 9. Na prosdioristoôn treic jetikoð pragmatikoð arijmoð x, y, z twn opoðwn to ˆjroisma eðnai 1 kai to ginìmenì touc lambˆnei mègisth tim. 10. H jermokrasða T sta shmeða enìc q rou eðnai T (x, y, z) = 400xyz. BreÐte th mègisth kai thn elˆqisth jermokrasða pˆnw sth sfaðra x + y + z = 1. 3
11. Na brejeð h elˆqisth apìstash thc kwnik c epifˆneiac apì thn arq twn axìnwn O(0, 0, 0). z = (x 1) + (y 1) 1. 'Ena sqoinð m kouc 1 cm dipl netai ètsi ste na sqhmatisteð èna trapèzio qwrðc ˆnw bˆsh, ìpwc sto sq ma. Na brejeð to mègisto embadì. y x 1 x x ϑ y u O x 13. Na upologisjoôn oi diastˆseic enìc kibwtðou epifˆneiac 500 cm mègisthc qwrhtikìthtac. 4
14. Na breðte thn elˆqisth apìstash thc arq c twn axìnwn O(0, 0, 0) apì thn epifˆneia z = xy + 4. 15. Na breðte thn elˆqisth apìstash thc arq c twn axìnwn O(0, 0, 0) apì to epðpedo x + 3y z = 1. 16. Qreiˆzetai na kataskeuˆsoume anoiktì kib tio me ìgko 1000 cm 3. Na brejoôn oi diastˆseic ste to ulikì pou ja qreiasteð na eðnai elˆqisto. 5
17. Na prosdiorðsete ta topikˆ akrìtata thc sunˆrthshc f : R R pou orðzetai me ton tôpo f(x, y) = e y 8x +x 4. Anaplhrwt c Kajhght c : Dr. Pappˆc G. Alèxandroc 6
. Lumènec Ask seic 7
Jèma 1 : Na brejoôn ta topikˆ akrìtata twn sunart sewn: (a) (b) f(x, y) = 1 x 4xy + 9y + 3x 14y + 1. g(x, y) = x 3 + y 3 1(x + y). LÔsh (a) H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh (akèraio polu numo). B ma 1 o = BrÐskoume ta stˆsima shmeða. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x (x, y) = 0 f x (x, y) = 0 'Omwc y (x, y) = 0 f y (x, y) = 0 f x = x 4y + 3 (1) f y = 4x + 18y 14 () To sôsthma twn exis sewn grˆfetai: f x (x, y) = 0 f y (x, y) = 0 x 4y + 3 = 0 (3) 4x + 18y 14 = 0 (4) Apì th lôsh tou sust matoc twn exis sewn (3), (4) prokôptei wc stˆsimo shmeðo to x = 1, y = 1 dhlad to (1, 1). B ma o = BrÐskoume ta topikˆ akrìtata. Apì parag gish twn (1), () prokôptoun: f xx = 1, f xy = f yx = 4, f yy = 18. pou eðnai stajerèc. 'Ara sth jèsh (1, 1) èqw: = f xx f xy f yx 'Ara sth jèsh (1, 1) èqw topikì akrìtato. f yy = 1 4 4 18 = > 0. 8
B ma 3 o = ProsdiorÐsoume to eðdoc akrotˆtou (topikì mègisto topikì elˆqisto). Epeid èqw topikì elˆqisto sto (1, 1). f xx = 1 > 0, B ma 4 o = ProsdiorÐsoume thn tim f(x 0, y 0 ). H tim tou eðnai f min = f(1, 1) = 1 1 4 1 1 + 9 1 + 3 1 14 1 + 1 = 5. (b) B ma 1 o = BrÐskoume ta stˆsima shmeða. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn 'Omwc g x (x, y) = 0 g y (x, y) = 0 g x (x, y) = 0 g y (x, y) = 0 g x = 0 3x 1 = 0 (5) g y = 0 3y 1 = 0 (6) To sôsthma twn exis sewn grˆfetai: g x (x, y) = 0 g y (x, y) = 0 3x 1 = 0 3y 1 = 0 Apì th lôsh tou sust matoc prokôptei ìti ta stˆsima shmeða eðnai: (, ), (, ), (, ), (, ). 9
B ma o = BrÐskoume ta topikˆ akrìtata. Epiplèon brðskoume: kai h orðzousa: = g xx g xx = 6x, g xy = 0 = g yx, g yy = 6y. g yx g xy g yy = 6x 0 0 6y = 36xy = (x, y). Sto shmeðo (, ) èqw (, ) = 36 = 144 > 0. 'Ara upˆrqei topikì akrìtato sth jèsh (, ). Sto shmeðo (, ) èqw (, ) = 36 ( ) < 0. 'Ara ed den upˆrqei topikì akrìtato. Sto shmeðo (, ) èqw (, ) = 36 ( ) < 0, ˆra oôte sto shmeðo autì upˆrqei topikì akrìtato. Sto shmeðo (, ) èqw 'Ara sth jèsh (, ) upˆrqei topikì akrìtato. (, ) = 36 ( ) ( ) = 144 > 0. B ma 3 o = ProsdiorÐsoume to eðdoc akrotˆtou (topikì mègisto topikì elˆqisto). Epeid èqw topikì elˆqisto sto (, ). Epeid èqw topikì mègisto sto (, ). g xx (, ) = 6 = 1 > 0 g xx (, ) = 6 ( ) < 0, B ma 4 o = ProsdiorÐsoume thn tim g(x 0, y 0 ). g(, ) = 3 + 3 1( + ) = 3. g(, ) = ( ) 3 + ( ) 3 1( ) = 3. 10
Jèma : Na deiqjeð ìti h sunˆrthsh f(x, y) = x 4 + 16y 4 (x y) èqei trða kritikˆ shmeða kai na brejeð h fôsh touc. Eidikˆ gia to (0, 0) na deiqjeð ìti den eðnai shmeðo topikoô akrotˆtou. LUSH x = 4x3 4(x y) = 0 y = 64y3 4(x y)( ) = 0 Prosjètoume thn pr th exðswsh sthn deôterh kai prokôptei x 3 x + y = 0 x 3 + 8y 3 = 0 x 3 + 8y 3 = 0 x3 x + y = 0 x3 x + y = 0 8y 3 + x y = 0 x 3 x + y = 0 x 3 + (y) 3 = 0 x 3 x + y = 0 (x + y) (x xy + 4y ) = 0 x 3 x + y = 0 x = y x 3 x = 0 x = y x = 0 x = x = x = y 'Ara èqoume ta ex c stˆsima shmeða: ( ),, ( ),, (0, 0). 11
f x = 4x 3 4(x y) = 4x 3 4x + 8y f xx = 1x 4 f x = 4x 3 4x + 8y f xy = 8 f y = 64y 3 4(x y)( ) = 64y 3 + 8x 16y f yy = 19y 16 = 8 ( 4y ) f y = 64y 3 + 8x 16y f yx = 8. = f xx f yx f xy f yy = 1x 4 8 8 8 ( 4y ) 'Ara = 64 ( 3x 1 ) ( 1y 1 ) 64. = 64 ( 3x 1 ) ( 1y 1 ) 64 kai f xx = 1x 4. 'Elegqoi : i. Gia to ( ), èqoume ( ), = 64 4 > 0 kai f xx = 1 4 > 0 ˆra sth jèsh ( ), èqoume topikì elˆqisto. ii. Gia to (, ) èqoume = 64 4 > 0 kai f xx ( ), = 1 4 > 0 ˆra sth jèsh (, ) (, ) èqoume topikì elˆqisto. 1
iii. Gia to (0, 0) èqoume opìte pˆme ston orismì: = 64 64 = 0 Orismìc : 'Estw h sunˆrthsh f(x, y) orismènh se ènan tìpo τ. Lème ìti èqoume topikì mègisto (antðstoiqa topikì elˆqisto) sth jèsh (x 0, y 0 ) tou tìpou τ, ìtan upˆrqei perioq B ((x 0, y 0 ), δ) ètsi ste (x, y) B ((x 0, y 0 ), δ), na isqôei h sqèsh f(x, y) f(x 0, y 0 ) 0 (antðstoiqa f(x, y) f(x 0, y 0 ) 0) efarmìzoume to Mejodologikì sqìlio. Mejodologikì sqìlio : JewroÔme th diaforˆ A = f(x, y) f(x 0, y 0 ) thc opoðac exetˆzoume to prìshmo gôro apì to (x 0, y 0 ). Sun jwc gia (x 0, y 0 ) èqoume to (0, 0). i. An mèsw tautot twn (p.q. tetrˆgwna) deðxoume ìti se perioq gôrw apì to (x 0, y 0 ) isqôei A 0 A 0, tìte èqoume akrìtato. ii. Epilègoume kampôlec pou pernˆne apì to (0, 0), (p.q. y = λx, y = λx ). (a) Gia y = x èqoume f(x, y) = f(x, x), opìte ja eðnai A = f(x, x) f(0, 0) kai exetˆzoume to prìshmo gôrw apì to 0 wc proc x. (b) Gia y = λx èqoume f(x, y) = f(x, λx) me katˆllhlo λ, opìte A = f(x, λx) f(0, 0) kai exetˆzoume to prìshmo gôrw apì to 0 wc proc x. An sto α) β) h diaforˆ A paðrnei jetikèc kai arnhtikèc timèc, tìte den èqoume akrìtato. 13
A = f(x, y) f(0, 0) = x 4 + 16y 4 (x y). Gia x = y A = 16y 4 + 16y 4 = 3y 4, ˆra ìtan to y paðrnei timèc sthn perioq tou mhdenìc kai epð thc eujeðac x = y, èqoume A > 0. (1) Gia x = y A = 16y 4 + 16y 4 3y = 3y ( y 1 ). -1 0 1 + - - + 'Ara ìtan to y paðrnei timèc sthn perioq tou mhdenìc kai epð thc eujeðac x = y, èqoume A < 0. () Apì tic sqèseic (1) kai () sumperaðnoume ìti den upˆrqei akrìtato sth jèsh (0, 0). 14
Jèma 5 : Na brejeð sto xy-epðpedo èna shmeðo M(x, y) tètoio ste to ˆjroisma twn tetrag nwn twn apostˆsewn tou apì tic eujeðec na eðnai to elˆqisto dunatì. x = 0, y = 0, x y + 1 = 0 LUSH y x y + 1 = 0 1 d 3 M(x, y) d 1 d O x Gia to shmeðo M(x, y) tou XOY epipèdou, to ˆjroisma twn tetrag nwn apì tic eujeðec eðnai: Epomènwc zhtˆme to elˆqisto thc sunˆrthshc Upologismìc twn pijan n akrotˆtwn: x = 0 d 1 = x y = 0 d = y ( ) x y + 1 x y + 1 = 0 d 3 = = 1 1 + 1 (x y + 1). f(x, y) = y + x + 1 (x y + 1). EpÐshc x = 0 y = 0 = f xx f yx f xy f yy 3x y = 1 x + 3y = 1 x 0 = 1 4 y 0 = 1 4 = 3 1 1 3 = 8 > 0 kai f xx = 3 > 0. 'Ara sto shmeðo ( 1 4, 1 4) èqoume elˆqisto. 15
Jèma 6 : Na prosdiorðsete ta topikˆ akrìtata thc sunˆrthshc f : R R pou orðzetai me ton tôpo f(x, y) = sin x + sin y + sin(x + y), an 0 < x < π kai 0 < y < π. LUSH H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh sto R. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x = 0 f x = 0 cos x + cos(x + y) = 0 y = 0 f y = 0 cos y + cos(x + y) = 0 cos x = cos y. (1) Apì thn (1) kai ton periorismì 0 < x < π kai 0 < y < π sunepˆgetai ìti 'Etsi apì to sôsthma sunepˆgetai h exðswsh x = y. cos x + cos x = 0. () H trigwnometrik exðswsh () grˆfetai sth morf cos x + cos x 1 = 0, dhl. cos x + cos x 1 = 0. (3) H lôseic thc deuterobˆjmiac exðswshc (3) wc proc cos x eðnai cos x = 1, cos x = 1. Allˆ 0 < x < π, epomènwc h tim cos x = 1 aporrðptetai. Sunep c h monadik epitrept tim tou cos x sto diˆsthma (0, π) eðnai 1. H lôsh thc exðswshc eðnai 'Omwc x = y, sunep c cos x = 1 sto (0, π) x = π 3. x = π 3 kai y = π 3. 'Ara h pijan jèsh akrotˆtou thc f eðnai to shmeðo ( π 3, π 3 ). 16
= f xx f yx ( π f xx = sin x sin(x + y) f xx 3, π ) = sin π 3 3 sin π 3 3 3 = = 3. ( π f yy = sin y sin(x + y) f yy 3, π ) = sin π 3 3 sin π 3 3 3 = = 3. ( π f xy = sin(x + y) = f yx f xy 3, π ) = sin π 3 ( π 3 3 = = f yx 3, π ). 3 = f xx f xy f yx f yy = 3 3 3 3 = 3 3 4 = 9 4 > 0. 'Ara sth jèsh ( π 3, π 3 ) èqw topikì akrìtato. f xy f yy. Epeid ( π f xx 3, π ) = 3 < 0, 3 h sunˆrthsh f parousiˆzei topikì mègisto sto shmeðo ( π 3, π 3 ). H mègisth tim thc f sto orjog nio pou orðzetai me 0 < x < π kai 0 < y < π eðnai ( π f 3, π ) 3 = sin π 3 + sin π 3 + sin π 3 = 3 sin π 3 = 3 3. 17
Jèma 7 : Na prosdiorðsete èna shmeðo P (x 0, y 0 ) tou kartesianoô epipèdou (Π) to opoðo èqei thn idiìthta ìti to ˆjroisma twn tetrag nwn twn apostˆsewn tou apì ta tèssera dosmèna shmeða na eðnai to elˆqisto. A(x 1, y 1 ), B(x, y ), Γ(x 3, y 3 ) kai (x 4, y 4 ) tou (Π) LUSH To ˆjroisma twn tetrag nwn twn apostˆsewn enìc shmeðou P (x, y) tou (Π) apì ta shmeða A(x 1, y 1 ), B(x, y ), Γ(x 3, y 3 ) kai (x 4, y 4 ) eðnai [ (x x1 ) + (y y 1 ) ] + [ (x x ) + (y y ) ] + [ (x x 3 ) + (y y 3 ) ] + [ (x x 4 ) + (y y 4 ) ]. Sunep c eðnai fanerì ìti prèpei na prosdiorðsoume to elˆqisto thc sunˆrthshc pou orðzetai me ton tôpo f : R R f(x, y) = (x x 1 ) + (y y 1 ) + (x x ) + (y y ) + (x x 3 ) + (y y 3 ) + (x x 4 ) + (y y 4 ). H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh sto R wc poluwnumik sunˆrthsh me pedðo orismoô R. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x = 0 f x = 0 (x x 1 ) + (x x ) + (x x 3 ) + (x x 4 ) = 0 y = 0 f y = 0 (y y 1 ) + (y y ) + (y y 3 ) + (y y 4 ) = 0 4x = x 1 + x + x 3 + x 4 x 0 = x1+x+x3+x4 4 4y = y 1 + y + y 3 + y 4 y 0 = y 1+y +y 3 +y 4 4 Gia ton prosdiorismì tou eðdouc tou krðsimou shmeðou thc f ja upologðsoume tic deôterec merikèc parag gouc thc f sto shmeðo (x 0, y 0 ). EÐnai = f xx f xy f yx f yy = 8 0 0 8 = 64 > 0 (x 0, y 0 ) = 64 > 0. 'Ara sth jèsh (x 0, y 0 ) èqw topikì akrìtato. Epeid f xx (x 0, y 0 ) = 8 > 0 h sunˆrthsh f parousiˆzei topikì elˆqisto sto shmeðo (x 0, y 0 ). Sunep c to zhtoômeno shmeðo eðnai to ( x1 + x + x 3 + x 4 P 4, ) y 1 + y + y 3 + y 4 4. 18
Jèma 8 : Na brejeð h apìstash tou shmeðou P (1, 0, ) apì to epðpedo (Π) me exðswsh x + y + z = 4. LUSH H apìstash enìc tuqìntoc shmeðou A(x, y, z) tou R 3 apì to shmeðo P (1, 0, ) dðnetai me ton tôpo d = d(a, P ) = (x 1) + (y 0) + (z + ). (1) 'Omwc to shmeðo A(x, y, z) an kei sto epðpedo (Π), sunep c z = 4 x y. Epomènwc h (1) grˆfetai d = (x 1) + y + (6 x y). () JewroÔme th sunˆrthsh pou orðzetai me ton tôpo To prìblhma elaqistopoðhshc thc sunˆrthshc () eðnai isodônamo me to prìblhma elaqistopoðhshc thc sunˆrthshc (3). f(x, y) = (x 1) + y + (6 x y). (3) H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh sto R wc poluwnumik sunˆrthsh me pedðo orismoô R. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x = 0 y = 0 f x = 0 f y = 0 (x 1) (6 x y) = 0 y + (6 x y)( ) = 0 to opoðo èqei monadik lôsh 4x + 4y 14 = 0 4x + 10y 4 = 0 x + y 7 = 0 x + 5y 1 = 0 (x, y) = ( ) 11 6, 5. 3 x + y = 7 x + 5y = 1 19
Gia ton prosdiorismì tou eðdouc tou krðsimou shmeðou thc f ja upologðsoume tic deôterec merikèc parag gouc thc f sto shmeðo (x 0, y 0 ). EÐnai = f ( ) xx f xy f yx f yy = 4 4 11 4 10 = 4 > 0 6, 5 = 4 > 0. 3 'Ara sth jèsh ( 11 6, 5 3) èqw topikì akrìtato. Epeid ( ) 11 f xx 6, 5 = 4 > 0 3 h sunˆrthsh f parousiˆzei topikì elˆqisto sto shmeðo ( 11 6, 5 3). 'Etsi h elˆqisth apìstash tou shmeðou P (1, 0, ) apì to epðpedo (Π) eðnai d = = (11 ) 6 1 + (5 ) + = 5 6 6 6. ( ) 5 + 3 ( ) ( 5 + 6 11 3 6 10 ) = 3 ( ) 5 = 6 0
Jèma 9 : Na prosdioristoôn treic jetikoð pragmatikoð arijmoð x, y, z twn opoðwn to ˆjroisma eðnai 1 kai to ginìmenì touc lambˆnei mègisth tim. LUSH Upojètoume ìti x, y, z eðnai oi zhtoômenh jetikoð pragmatikoð arijmoð gia touc opoðouc x + y + z = 1. (1) 'Estw h sunˆrthsh f : R R pou orðzetai me ton tôpo Apì thn (1) lambˆnoume f(x, y, z) = xyz. () z = 1 x y. (3) Apì tic sqèseic () kai (3) orðzoume thn sunˆrthsh g : R R me ton tôpo dhl. f(x, y) = xy(1 x y) f(x, y) = 1xy x y xy. (4) H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh sto R wc poluwnumik sunˆrthsh me pedðo orismoô R. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x = 0 f x = 0 1xy xy y = 0 y = 0 f y = 0 1x x xy = 0 To sôsthma èqei tic lôseic (x, y) = (0, 0) kai (x, y) = (4, 4). To shmeðo (0, 0) profan c aporrðptetai. Epomènwc ja exetˆsoume an sto shmeðo (4, 4) h sunˆrthsh f parousiˆzei akrìtato. Gia ton prosdiorismì tou eðdouc tou krðsimou shmeðou thc f ja upologðsoume tic deôterec merikèc parag gouc thc f sto shmeðo (x 0, y 0 ). EÐnai 'Ara f xx = y f xx (4, 4) = 8 f xy = 1 x y f xy (4, 4) = 4 f yx = 1 x y f yx (4, 4) = 4 f yy = x f xx (4, 4) = 8 = f xx f yx f xy f yy = 8 4 4 8 = 48 > 0. 'Ara sth jèsh (4, 4) èqw topikì akrìtato. 1
Epeid f xx (4, 4) = 8 < 0 h sunˆrthsh f parousiˆzei topikì mègisto sto shmeðo (4, 4). Apì thn (3) prokôptei z = 1 4 4 = 4. Epomènwc oi zhtoômenoi jetikoð pragmatikoð arijmoð eðnai x = 4, y = 4, z = 4.
Jèma 11 : Na brejeð h elˆqisth apìstash thc kwnik c epifˆneiac apì thn arq twn axìnwn O(0, 0, 0). z = (x 1) + (y 1) LUSH H apìstash enìc tuqìntoc shmeðou A(x, y, z) thc dosmènhc epifˆneiac apì thn arq twn axìnwn O(0, 0, 0) dðnetai me ton tôpo d = d(o, A) = x + y + z. (1) 'Omwc to shmeðo A(x, y, z) an kei sthn epifˆneia, sunep c z = (x 1) + (y 1). Epomènwc h (1) grˆfetai d = x + y + (x 1) + (y 1). () JewroÔme th sunˆrthsh pou orðzetai me ton tôpo To prìblhma elaqistopoðhshc thc sunˆrthshc () eðnai isodônamo me to prìblhma elaqistopoðhshc thc sunˆrthshc (3). f(x, y) = x + y + (x 1) + (y 1). (3) H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh sto R wc poluwnumik sunˆrthsh me pedðo orismoô R. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn x = 0 y = 0 f x = 0 f y = 0 x + (x 1) = 0 y + (y 1) = 0 x + x = 0 y + y = 0 4x = 4y = x = 1 y = 1 to opoðo èqei monadik lôsh (x, y) = ( ) 1, 1. 3
Gia ton prosdiorismì tou eðdouc tou krðsimou shmeðou thc f ja upologðsoume tic deôterec merikèc parag gouc thc f sto shmeðo (x 0, y 0 ). EÐnai = f xx f xy = 4 0 0 4 = 16 > 0. f yx 'Ara sth jèsh ( 1, 1 ) èqw topikì akrìtato. Epeid f yy ( ) 1 f xx, 1 = 4 > 0 h sunˆrthsh f parousiˆzei topikì elˆqisto sto shmeðo ( 1, 1 ). 'Etsi h elˆqisth apìstash thc kwnik c epifˆneiac apì thn arq twn axìnwn O(0, 0, 0) eðnai: d = x + y + (x 1) + (y 1) = (1 ) ( ) ( ) ( ) 1 1 1 = + + 1 + 1 = (1 ) ( ) ( 1 = + + 1 ( + ) 1 = ) ( ) 1 = 4 = = 1 = = 1. 4
Jèma 1 : 'Ena sqoinð m kouc 1 cm dipl netai ètsi ste na sqhmatisteð èna trapèzio qwrðc ˆnw bˆsh, ìpwc sto sq ma. Na brejeð to mègisto embadì. y x 1 x x ϑ y u O x To embadìn tou trapezðou eðnai Allˆ Epomènwc, E = LUSH (B + β) β = 1 x, y = x cos θ, υ = x sin θ, B = 1 x + x cos θ + x cos θ = 1 x + x cos θ. E = (B + β) υ = = (1 x + x cos θ) + (1 x) x sin ϑ = = 4 4x + x cos θ x sin ϑ = = (1 x + x cos θ) x sin ϑ = = (1 x + x cos θ) x sin θ = υ = 1x sin θ x sin θ + x sin θ cos θ E(x, θ) = 1x sin θ x sin θ + x sin θ cos θ. 5
EÐnai: E(x, θ) = 1x cos θ x cos θ + x cos θ x sin θ θ E(x, θ) = 1 sin θ 4x sin θ + x sin θ cos θ. x Oi anagkaðec sunj kec gia akrìtata dðnoun to sôsthma: x ( 1 cos θ x cos θ + x cos θ x sin θ ) = 0 To sôsthma dðnei tic lôseic sin θ (6 x + x cos θ) = 0. M 1 ( θ = 60 0, x = 4 ), M ( θ = 0 0, x = 0 ). Oi timèc θ = 0 0 kai x = 0 aporrðptontai. Oi timèc eðnai dunatèc kai dðnoun to mègisto embadìn. θ = 60 0 kai x = 4 6
Jèma 16 : Qreiˆzetai na kataskeuˆsoume anoiktì kib tio me ìgko 1000 cm 3. Na brejoôn oi diastˆseic ste to ulikì pou ja qreiasteð na eðnai elˆqisto. LUSH 'Etsw x, y, z oi diastˆseic tou kibwtðou. ElaqistopoÐhsh ulikoô shmaðnei elaqistopoðhsh thc epifˆneiac twn tessˆrwn pleur n kai tou pˆtou, dhlad thc èkfrashc E πιϕ = f(x, y, z) = xy + xz + yz. Epeid o ìgkoc eðnai prokôptei ìti opìte h sunˆrthsh èqei th morf 1000 cm 3 = xyz z = 1000 xy f(x, y) = xy + x 1000 xy, + y 1000 xy f(x, y) = xy + 000 y + 000 x. Aut eðnai h sunˆrthsh pou prèpei na elaqistopoihjeð. H sunˆrthsh f(x, y) eðnai dôo forèc suneq c paragwgðsimh. 7
B ma 1 o = BrÐskoume ta stˆsima shmeða. Ta stˆsima shmeða (upoy fièc jèseic akrotˆtwn) eðnai lôseic tou sust matoc twn exis sewn 'Omwc x (x, y) = 0 y (x, y) = 0 f x (x, y) = 0 f y (x, y) = 0 f x = y 000 x (1) f y = x 000 y () To sôsthma twn exis sewn grˆfetai: f x (x, y) = 0 f y (x, y) = 0 y 000 x = 0 (3) x 000 y = 0 (4) Apì th lôsh tou sust matoc twn exis sewn (3), (4) prokôptei x = 0 y = 0 x = y. Gia to prìblhma oi timèc x = 0 kai y = 0 aporrðptontai, en apì th sqèsh x = y prokôptei x = y = (000) 1 3 3 = 10 'Ara to stˆsimo shmeðo eðnai to (10 3, 10 3 ). B ma o = BrÐskoume ta topikˆ akrìtata. Apì parag gish twn (1), () prokôptoun: f xx = 4000 x 3, f xy = f yx = 1, f yy = 4000 y 3. 'Ara sth jèsh (10 3, 10 3 ) èqw: = f xx f yx f xy f yy 4000 = x 1 3 4000 1 y 3 = 4000 x 3 4000 y 3 1. 'Ara sth jèsh (10 3, 10 3 ) èqw topikì akrìtato. (10 3,10 3 ) > 0. 8
B ma 3 o = ProsdiorÐsoume to eðdoc akrotˆtou (topikì mègisto topikì elˆqisto). Epeid f xx (10 3, 10 3 ) = 4000 (10 3 ) = 4000 3 1000 = > 0, èqw topikì elˆqisto sto ( 10 3, 10 3 ). Ara èqoume topikì elˆqisto ìtan oi diastˆseic tou kibwtðou eðnai: x = 10 3, y = 10 3, z = 1000 xy = 1000 (10 3 )(10 3 ) = 1000 100 3 = 3 10 = 10 3 3 3 = 10 3 = 5 3. 9