11/21/2002 SECOND HOURLY, PRACTICE EXAM Math 21aName:MWF9 Sasha BravermanMWF10 Ken ChungMWF10 Jake RasmussenMWF10 WeiYang QuiMWF10 Spiro KarigiannisMWF11 Vivek MohtaMWF11 Jake RasmussenMWF12 Ken ChungTTH10 Oliver KnillTTH11 Daniel Goroff• Start by printing your name in the above box and checkyour section in the box to the left.• Try to answer each question on the same page as thequestion is asked. If needed, use the back or next emptypage for work. If you need additional paper, write yourname on it.• Do not detach pages from this exam packet or unstaplethe packet.• Please write neatly. Answers which are illegible for thegrader can not be given credit. Justify your answers.• No notes, books, calculators, computers or other elec-tronic aids are allowed.• You have 90 minutes time to complete your work.1 802 303 404 405 406 407 408 40Total: 350Problem 1) TF questions (80 points) Circle the correct letter. No justifications are needed.T 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.x 4 =2Problem 2) (30 points)Match the parametric surfaces with their parameterization. No justification is needed.I IIIII IVEnter I,II,III,IV here Parameterization(u, v) 7→ (u cos(v), u sin(v), u2cos(u)/(u + 1))(u, v) 7→ (u, v + u, |u − v|)(u, v) 7→ ((u − sin(u)) cos(v), (u − cos(u)) sin(v), u)(u, v) 7→ (u, v, u2− v2)Problem 3) (40 points)Find all the critical points of the function f(x, y) = xy(4 − x2− y2). Are they maxima, minimaor saddle points?3Problem 4) (40 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) (40 points)f is a function which depends on x and y, where x = u3v and y = u2v2. When (u, v) = (1, 1)∂f∂x= −5,∂f∂v= 9. What is∂f∂u?Problem 6) (40 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) (40 points)EvaluateR20R√4−x20xy5x2+y2dy dx.Problem 8) (40 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
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