2014fstt - University of Toronto Scarborough Department of...

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University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2014/2015Term Test Solutions1.(a)(i) Supposef:URnR.A pointaUis called alocal (relative)minimumoffif there is an open ballBr(a) such thatf(x)f(a) for allxBr(a).(ii) Those pointsa, in the domain off, at whichfis either not differentiableorD f(a) =Oare calledcritical points.(b) From the lecture notes we haveExtreme Value Theorem. LetDbe a compact set inRnand letf:DRnRbe continuous.Thenfassumes both a (global) maximum and a(global) minimum onD.2.(a)(i)lim(x,y)(0,0)x25xy+y2x2+y2. Evaluating along the liney= 0, we havelim(x,y)(0,0)x25xy+y2x2+y2= limx0x2x2= limx01 = 1, but along the liney=xwehavelim(x,y)(0,0)x25xy+y2x2+y2= limx03x22x2=32negationslash= 1. Hence this limit doesnot exist.(ii)lim(x,y)(0,0)cos(xy)1x2y2.Using another single variable technique, we havelim(x,y)(0,0)cos(xy)1x2y2=lim(x,y)(0,0)(cosxy1) (cosxy+ 1)x2y2(cosxy+ 1)=lim(x,y)(0,0)sin2xyx2y2(cosxy+ 1)=lim(x,y)(0,0)parenleftbiggsin2xy(xy)2parenrightbiggparenleftbigg1cosxy+ 1parenrightbigg=parenleftbigg1parenrightbiggparenleftbigg12parenrightbigg=12.(b) Forfto be continuous at (0,0), we needlim(x,y)(0,0)f(x, y) = 2 =f(0,0).Nowlim(x,y)(0,0)f(x, y) =lim(x,y)(0,0)y4+ 2y2+ 2x2x4x2+y2divide=lim(x,y)(0,0)(2x2+y2) =2 =f(0,0). Hence we conclude thatfis continuous at (0,0).
MATB41HTerm Test Solutionspage23.f(x, y) =yx2+y2.Domain is{(x, y)R2|(x, y)negationslash= (0,0)}.Puttingf(x, y) =cwe haveyx2+y2=c. Forc= 0, the level curve isy= 0.Forcnegationslash= 0,we havey=c(x2+y2)⇐⇒x2+y2yc=0⇐⇒x2+parenleftbiggy12cparenrightbigg2=parenleftbigg12cparenrightbigg2.Theseare circles centered atparenleftbigg0,12cparenrightbiggwith radius12c.

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