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HARVARD MATH 21A - First practice exam second hourly

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FIRST PRACTICE EXAM SECOND HOURLY Math 21a, Spring 2003Name:MWF10 Ken ChungMWF10 Weiyang QiuMWF11 Oliver KnillTTh10 Mark LucianovicTTh11.5 Ciprian Manolescu• Start by printing your name in the above box andcheck your section in the box to the left.• Do not detach pages from this exam packet or un-staple the packet.• Please write neatly. Answers which are illegible forthe grader can not be given credit.• No notes, books, calculators, computers, or otherelectronic aids can be allowed.• You have 90 minutes time to complete your work.• The hourly exam itself will have space for work oneach page. This space is excluded here in order tosave printing resources.1 202 103 104 105 106 107 108 109 10Total: 100Problem 1) TF questions (20 points) Circle the correct letter. No justifications are needed.The score for this question is the number of correct answersT FAt a local maximum (x0, y0) of f(x, y), one has fyy(x0, y0) ≥ 0.T FIf R is the region bounded by x2+ 4y2= 1 thenR RRxy4dxdy < 0.T FThe gradient h2x, 2yi is perpendicular to the surface z = x2+ y2.T FThe equation f(x, y) = k implicitly defines x as a function of y anddxdy=∂f∂y/∂f∂x.T Ff(x, y) =q(16 − x2− y2) has both an absolute maximum and anabsolute minimum on its domain of definition.T FIf (x0, y0) is a critical point of f(x, y) under the constraint g(x, y) =0, and fxy(x0, y0) < 0, then (x0, y0) is a saddle point.T FThe vector ru(u, v) of a parameterized surface (u, v) 7→ r(u, v) =(x(u, v), y(u, v), z(u, v)) is normal to the surface.T FThe identityR10R√1−x20(x2+ y2) dydx =R10Rπ/20r2dθdr holds.T Ff(x, y) and g(x, y) = f (x2, y2) have the same critical points.T FIf f(x, t) satisfies the Laplace equation fxx+ ftt= 0 and simulta-neously the wave equation fxx= ftt, then f(x, t) = ax + bt + c.T FEvery smooth function satisfies the partial differential equationfxxyy= fxyxy.T FThe function (x4−y4) has has neither a local maximum nor a localminimum at (0, 0).T FR10Rπ/20r dθdr = π/4.T FAt a saddle point, the directional derivative is zero for two differentvectors u, v.T FIt is possible to find a function of two variables which has no max-imum and no minimum.T FThe value of the function f(x, y) = exy at (0.001, −0.001) can bylinear approximation be estimated as −0.001.T FFor any function f(x, y, z) and any unit vectors u, v, one has theidentity Du×vf(x, y, z) = Duf(x, y, z)Dvf(x, y, z).T FGiven 2 arbitrary points in the plane, there is a function f(x, y)which has these points as critical points and no other critical points.T FThe maximum of f(x, y) under the constraint g(x, y) = 0 is thesame as the maximum of g(x, y) under the constraint f (x, y) = 0.T FAssume (x0, y0) is a critical point of f(x, y) and fxxfyy−f2xy6= 0 atthis point. Let T be the tangent plane of the surface S = {f(x, y)−z = 0} at P = (x0, y0, f(x0, y0)). If the intersection of T with S isa single point, then (x0, y0) is a local max or local min.2Problem 2) (10 points)Match the parametric surfaces S = ~r(R) with the corresponding surface integralR RSdS =R RR|~ru×~rv| dudv. No justifications are needed.I IIIII IVEnter I,II,III,IV here Surface integralR RR|~ru×~rv| dudv =R10R10√1 + 4u2+ 4v2dvduR RR|~ru×~rv| dudv =R10R10√3 dvduR RR|~ru×~rv| dudv =R2π0Rπ0sin(v)q1 + cos(v)2dvduR RR|~ru×~rv| dudv =R2π0Rπ0sin(v) dvduProblem 3) (10 points)Find all the critical points of the function f(x, y) = xy(4 −x2−y2). Are they maxima, minimaor saddle points?3Problem 4) (10 points)Let f(x, y) = e(x−y)so that f(log(2), log(2)) = 1. Find the equation for the tangent plane tothe graph of f at (log(2), log(2)) and use it to estimate f(log(2) + 0.1, log(2) + 0.04).Problem 5) (10 points)FindR R RRz2dV , where R of the solid obtained by intersecting {1 ≤ x2+ y2+ z2≤ 4} withthe double cone {z2≥ x2+ y2}.Problem 6) (10 points)A can is a cylinder with a circular base. Its surface area (top, bottom and sides) is 300π cm2.What is the maximum possible volume of such a can?Problem 7) (10 points)EvaluateR20R√4−x20xy5x2+y2dy dx.Problem 8) (10 points)a) Find the area of the region D enclosed by the lines x = ±2 and the parabolas y = 1 + x2,y = −1 − x2.b) Find the integral of f(x, y) = y2on the same region as in a). (The result can be interpretedas a moment of inertia).Problem 9) (10 points)Let T (u, v) = (v cos(u), 2v sin(u)) = (x, y).a) Find the image S = T (R) of the rectangle R = {0 ≤ u ≤ π, 0 ≤ v ≤ 1} under the map Tand find its area using the formula for the change of variables.b) Write the integralR40R(y/2)+1y/2(2x+y)2dxdy using uv coordinates with a change of variablesT (u, v) = (x, y) = (u + v, 2v) and evaluate that


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