2010 Solution - University of Toronto Scarborough...

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University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2010/2011Term Test Solutions1.(a) From the lecture notes we haveLetf:URnRkbe a given function. We say thatfisdifferentiable ataUif the partial derivatives offexist ataand iflimxabardblf(x)f(a)Df(a) (xa)bardblbardblxabardbl= 0,whereDf(a) is thek×nmatrixparenleftbigg∂fi∂xjparenrightbiggevaluated ata.Df(a) is called thederivative offata.(b)(i) LetARn. A pointaAis called aninterior point ofAifBr(a)A, forsomer >0.(Br(a) is an open ball of radiusrcentered ata.)(ii) Supposef:URnR. A pointaUis called alocal (relative) minimumoffif there is an open ballBr(a) such thatf(x)f(a) for allxBr(a).2.(a)lim(x,y)(0,0)x yy2x+y=lim(x,y)(0,0)y(xy)x+y=lim(x,y)(0,0)y(xy)(x+y)x+y=lim(x,y)(0,0)y(xy)= 0.(b) If (x, y)negationslash= (0,0),f(x, y) =x3+ 2x2+ 2xy2+ 4y2x2+ 2y2. Since rational functions arecontinuous on their domains,f(x, y) is continuous for (x, y)negationslash= (0,0).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)x3+ 2x2+ 2xy2+ 4y2x2+ 2y2divide=lim(x,y)(0,0)(x+ 2) =2negationslash=2 =f(0,0). Hence we conclude thatf(x, y) is not continuous at (0,0).
MATB41HTerm Test Solutionspage23.f(x, y) =x2+y21(x+ 1)2+y2.Domain isbraceleftbig(x, y)R2|(x, y)negationslash= (1,0)bracerightbig.Puttingf(x, y) =cwe havex2+y21(x+ 1)2+y2=c⇐⇒x2+y21 =cx2+ 2cx+c+cy2⇐⇒(1c)x2+ (1c)y22cx=c+ 1.Forc= 1, we get the linex=1.Forcnegationslash=1, we get(1c)parenleftbiggx22c1cx+c2(1c)2parenrightbigg+(1c)y2=11c⇐⇒parenleftbiggxc1cparenrightbigg2+y2=parenleftbigg11cparenrightbigg2.

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