Sol - Exam2 2009 - Spring.pdf - Math 18 Solutions to and...

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Math 18Solutions toand Comments onSecond ExaminationSpring, 2009Given the long faces I saw as you guys left the exam, I expected a disaster. But this was nothing ofthe sort. Grades ranged from 63 to a single 100 (I’m impressed!), with 4 each in the range 90-99,80-89, and 63-79. The median was 85; if I were to scale, I’d have to pull the numbers down. So Iguess all my cursing at previous classes had an effect. Good job!1.(a) (10 points)Give each definition below (you may assume the function is definedeverywhere):i. The functionf(x, y) iscontinuousat (x, y) = (x0, y0) iflim(x,y)(x0,y0)f(x, y) =f(x0, y0) ;equivalently, if wheneverxkis a sequence withxk6=x0for allkwhich converges tox0, thenf(xk) converges tof(x0, y0).ii. The functionf(x, y) isdifferentiableat (x, y) = (x0, y0) if there exists an affinefunctionT(x, y) which has first-order contact withfat (x0, y0), that is,lim|f(x, y)-T(x, y)
(b)Is the functionf(x, y) =(x2yx2+y2for (x, y)
(i) continuous (10 points), and (ii) differentiable (10 points)according to yourdefinition? You may use theorems from class or the book to justify your answer:i.Continuity: Automatic except at the origin (rational functions are continuouswhere defined). To establish continuity at the origin, we need to show thatf(x, y)0 as (x, y)(0,0). The easiest way to do this is the polar trick: setx=rcosθy=rsinθ
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thenf(r, θ) =r3cos2θsinθr2=rcos2θsinθ0asr0.Almost everyone got this right, using the polar trick. One person tried to useL’Hˆopital’s rule to find a limit. This applies only to functions of one variable; and inparticular it is not clear which derivatives you are supposed to take when looking at alimit of a rational function of two variables when numerator and denominator both goto zero. I wasn’t as insistent as I should have been that you said something aboutcontinuity at points other than the origin.ii.Differentiability:Again, automatic except at the origin. Since the function iszero along each axis, the two partial derivatives at the origin are both zero. Thusthe only candidate for the affine functionT(x, y) isT(x, y) =f(0,0) + 0(x-0) + 0(y-0) = 0. Let us check for contact withfat theorigin:f(x, y)-T(x, y)k(x, y)k=x2yx2+y2-0px2

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