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University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2013/2014Solutions #51. A normal vector to the plane 4x−8y−z= 3 is (4,−8,−1)and a normal vector to the tangent plane to the graphoff(x,y) =x2+y2−1 is (2x,2y,−1).For the planesto be parallel we need (2x,2y,−1) =k(4,−8,−1), forsomek. From the third component we see thatk= 1, sowe havex= 2 andy=−4.The tangent plane passesthrough (2,−4,f(2,−4)) = (2,−4,19) and has equation4x−8y−z= 21.2x-4y19z2.(a) We havef(0,0) = 0, so we can compute the partial derivatives from the defini-tion.∂f∂x(0,0) = limh→0f(h,0)−f(0,0)h= limh→00−0h= 0∂f∂y(0,0) = limh→0f(0,h)−f(0,0)h= limh→00−0h= 0(b) Recall that for a functionfto be differentiable at a pointa, we needlimx→abardblf(x)−f(a)−D f(a)(x−a)bardblbardblx−abardbl= 0.Hereweneedtoevaluatelim(x,y)→(0,0)|x1/3y1/3−0−(0,0)·(x,y)|bardbl(x,y)bardbl=lim(x,y)→(0,0)|x1/3y1/3|radicalbigx2+y2. If we approach along the liney=x, withx >0, this limitreduces to limx→0+x2/3√2x2= limx→0+x2/3x√2= limx→0+1x1/3√2=∞. Since this limit is not0,fis not differentiable at (0,0).3. The rate of change in depth is the directional derivative.(a) The depth will increase most rapidly in the direction of the gradient; i.e., indirection∇D(1,−2). Now∇D= (−6xy2,−6x2y) so the rubber duck swims indirection∇D(1,−2) = (−24,12).(b) The depth will stay constant if the duck stays on the level set which passes through(1,−2). Since the gradient is orthogonal to level sets it can proceed in direction(1,2) or direction (−1,−2) since±1 (1,2)·∇D(1,−2) =±1 (1,2)·(−24,12) = 0.
MATB41HSolutions # 5page24.(a) Letx=u2−v2andy=v2−u2. Theng(u,v) =f(x,y) and the Chain Rulecan be applied to give∂g∂u=∂f∂x∂x∂u+∂f∂y∂y∂u=∂f∂x(2u)+∂f∂y(−2u)and∂g∂v=∂f∂x∂x∂v+∂f∂y∂y