# 2015fstt - University of Toronto Scarborough Department of...

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University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2015/2016Term 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) From the lecture notes we haveChain Rule. Letf:URnRmandg:VRmRkbe given functionssuch thatf[U]Vso thatgfis defined. LetaRnandb=f(a)Rm.Iffis differentiable ataandgis differentiable atb, thengfis differentiableataandD(gf) (a) = [Dg(b)] [Df(a)].2.(a)(i)lim(x,y)(0,0)x2+ 2xy+y2x2+y2. Evaluating along the liney= 0, we have limx0x2x2= 1,but along the liney=x, we have limx0x2+ 2x2+x2x2+x2= limx04x22x2= 2. Since1negationslash= 2,lim(x,y)(0,0)x2+ 2xy+y2x2+y2does not exist.(ii)lim(x,y)(0,0)x2+ 2xy+y2x+y=lim(x,y)(0,0)(x+y)2x+y=lim(x,y)(0,0)(x+y) = 0.(b) Forfto be continuous we needlim(x,y)(1,0)f(x, y) = 0 =f(1,0). Evaluating thelimit along the linex= 1, we have limy0yyy2= 0, but along the liney=x1,we have limx1(x1)2(x1)2+ (x1)2= limx1(x1)22 (x1)2=12. Hence the limit does notexist and, consequently,fcan not be continuous at (1,0).
MATB41HTerm Test Solutionspage23.f(x, y) =1radicalbig(y1)(yx2).(a)(i) The domainDis given byD=braceleftbig(x, y)R2|(y1)(yx2)>0bracerightbig=braceleftbig(x, y)R2|y >1 andy > x2bracerightbigbraceleftbig(x, y)R2|y <1 andy < x2bracerightbig.

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Term
Fall
Professor
moore
Tags
Math, Continuous function