2009 Solution - University of Toronto Scarborough...

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University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2009/2010Term Test Solutions1.(a) For a functionf:R2R, the equation of the tangent plane at (a, b) isz=f(a, b) +∂f∂x(a, b) (xa) +∂f∂y(a, b) (yb).Here we havefx= 3x26y,fx(1,2) =9;fy=6x+3y2,fy(1,2) = 6 andf(1,2) =3. Hence the equationof the tangent plane isz=3+(9)(x1)+(6)(y2) =39x+9+6y12 =9x+ 6y6, which can be rewritten as 9x6y+z=6.(b) Since (0.99,2.01) is “near” (1,2) we can use our work from part (a).Hencethe linear approximation is given byT1(x, y) =3 + (9)(x1) + (6)(y2).Evaluating at (0.99,2.01) we haveT1(0.99,2.01) =3+(9)(0.991)+(6)(2.012) =3 + (9)(0.01) + (6)(0.01) =3 + 0.09 + 0.06 =2.85.2.(a)(i)lim(x,y)(2,0)(x2)2(x2)2+y2. If we restrict to the linex= 2, the limit reduces tolimy0parenleftbigg0y2parenrightbigg= 0. On the other hand, if we restrict to the liney=x2, thelimit reduces to limx2parenleftbigg(x2)2(x2)2+ (x2)2parenrightbigg= limx2parenleftbigg12parenrightbigg=12. Hence this limitdoes not exist.(ii)lim(x,y)(0,0)sin(xy)bardbl(x, y)bardbl=lim(x,y)(0,0)sin(xy)radicalbigx2+y2. If we restrict to the liney=x,the limit reduces to limx0sin 02x2= 0. On the other hand, if we restrict to theliney=x, the limit reduces tolimx0+sin(2x)2x2=limx0+2 sin(2x)2x=2.Hence this limit does not exist.(b)f(x, y) =xyx2+y2,if (x, y)negationslash= (0,0)0,if (x, y) = (0,0).Forfto be continuous at (0,0), weneedlim(x,y)(0,0)f(x, y) = 0 =f(0,0). Evaluating the limit along the liney=x,we have limx0parenleftbiggx22x2parenrightbigg=12. Sincef(0,0) = 0negationslash=12, we can conclude thatfis notcontinuous at (0,0).
MATB41HTerm Test Solutionspage23.f(x, y) =x2x+y+ 1.

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Term
Fall
Professor
moore
Tags
Math, Calculus, lim, Level set, Orientability, Term Test Solutions

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