11 Limiti coordinate polari

Pagina 1

QuestecurvesonounfasciodiMini: Rate= ESTREM1(pueearcizioDeterminaregliestremidifluy)=rbyalvindoP(x,y)== m4+4g-1=0. 10/05/831)x byI=>6-(R4 dae- 2 eminf(x,y)=D x= acosO0-0,25] 1- 3 Ey=bsinc p(x,y):Eede02[0,25) f(x0),y(0))=41058+9sinO= g(0) g(0)= 4(1-sin)+9sind= 4-4sin+9sind 90)= - 8einOca8+9ca,0 Ot[0,24] 10)-80inO+9)50 cos05,000V3T2- 8sin8+ 97,0en8=98WOt[0,2]0 4a 2π I 0= 42Atodimax & stodiminP(0,3)P(0,-3) 0= 42π) f(0,3)=9 =-9OfMax = 9 f(0,- 3)= - 9 Caystd 2)Determinaregliestremirelatinassolutidi · f(xy)= 4x(x2y2)-3x4y2vincolatiallacurva d(x,y)=x- y21=0 fluy)-3°(R2)d(y):ry=14y==x bluy)nonèuncompatto->nonpassoapplicareweierstrass Simi· yu-ChatgereFesa e re ·yagusterrazt)totee g(x)= - 2(x- 44) = 48x(4,- 48)4(42)- 44)- flny)vincolataa P(ny)ammettemanimoandutopara 18= f(2,0)= - Iminopercheplay)è& (y)-Pmimitatorinferiormen- 2)(appello11712019)SiconsideriHe 1= xeylafunzionefluymere Determinareimassimiminimirelati/andutinelarchig diragges1centratoin(1P).D= ((xy)+R?(x- 1)4y2= 13Iancompatto.Exendof(x,y)A conteniaaDpercheprotode ⑳ . funzionicontinue,alloraperWeiert(P) esistemaxomif(x,y)tDI(xy)tD P=DoUSDConideroparatamentoDoeID Doaperto:ft6Y(D)=>ciceco,wcandoilteamdi Fermot,Ipunticandidatiadescemax/minitraquelliche annullomoilgradientexf(y)= (0,0)fu=exy+xeny"(-2x)= = e- (x4y2)(1-20 fy=xe - xity2)(2y)= - Rage- (**) (4)(1-20)=0 e=IYEA(YE2,0)OKAtDoe- I+yz)(-2xy)=0 Iy=0 B(- 42,)H fun=y(+electy2)(-(in)BCDo " fuy=fyx= iyut-- que-tyY-Ry(-2y)e-cty)=fyy= -que- xyz)(1+ 2y