4/18/2006 SECOND HOURLY Math 21a, Spring 2006Name:MWF 10 Samik BasuMWF 10 Joachim KriegerMWF 11 Matt LeingangMWF 11 Veronique GodinTTH 10 Oliver KnillTTH 115 Thomas Lam• Start by printing your name in the above box and checkyour section in the box to the left.• Do not detach pages from this exam packet or unstaplethe packet.• Please write neatly. Answers which are illegible fo r thegrader can not be given credit.• Justify all your answers for problems 4-9. As usual, thepath to the answer is always graded too.• No notes, books, calculators, computers, or other elec-tronic aids can be allowed.• All unspecified functions ar e differentiable as many timesas necessary.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 10Total: 1001Problem 1) True/False questions (20 points)Mark for each of the 20 questions the correct letter. No justifications are needed.1)T FR R RR1 dxdydz is the volume of R.2)T FAt a critical point of a function f, the gradient vector has length 1.3)T FAt a critical point (x, y) of a function f, the tangent plane to the graph off does not exist.4)T FFor any point (x, y) which is not a critical point, there is a unit vector ~u forwhich D~uf(x, y) is nonzero.5)T FIf fxx(0, 0) = 0, D = fxxfy y− f2xy6= 0, and ∇f(0, 0) = h 0, 0i, then ( 0 , 0) isa saddle point.6)T FA continuous function defined on the closed unit disc x2+ y2≤ 1 has anabsolute maximum inside t he disc o r on the boundary.7)T FThe function f(x, y) = x2− y2has a neither a local maximum nor a localminimum at (0, 0).8)T FIf (x, y) is a maximum o f f(x, y) under the constraint g(x, y) = 5 then it isalso a maximum of f (x, y) + g(x, y) under the constraint g(x, y) = 5.9)T FThe functions f (x, y) and g(x, y) = (f(x, y))6always have t he same criticalpoints.10)T FFor f(x, y, z) = x2+ y2+ 2z2, the vector ∇f(1 , 1, 1) is perpendicular to thesurface f (x, y, z) = 4 at the point (1, 1, 1).11)T Ff(x, y) =√16 − x2− y2has both an absolute maximum and an absoluteminimum on its domain of definition.12)T FIf (x0, y0) is a critical point of f(x, y) and fxy(x0, y0) < 0, then (x0, y0) is asaddle point of f.13)T FThe vector ~rv(u, v) of a parameterized surface (u, v) 7→ ~r(u, v) =(x(u, v), y(u, v), z(u, v)) is always perpendicular to the surface.14)T FSuppose f has a maximum value at a point P relative t o the constraintg = 0. If the Lagrange multiplier λ = 0, then P is also a critical point forf without the constraint.15)T FAt a saddle point, all directional derivatives are zero.16)T FThe minimum of f (x, y) under the constraint g(x, y) = 0 is always the sameas the maximum of g(x, y) under the constraint f (x, y) = 0.17)T FAt a local maximum (x0, y0) of f(x, y), one has fy y(x0, y0) ≤ 0.18)T FThe volume of a sphere of radius 1 is equal to the volume under the graphof f(x, y) =√1 − x2− y2inside the unit disc x2+ y2≤ 1.19)T FR10R10(x2+ y2) dxdy = 2/3.20)T FThe function f(x, y) =Rx0Ry0g(u) + g(v) dudv has the critical points (t, t),where t is a root of g.2Problem 2 ) (10 points)Match the regions R with the corresponding double integralR RRf(x, y) dxdy. No justifi-cations are needed for this problem.I IIIII IVEnter I,II,I II,IV here IntegralR10R1√xf(x, y) dydxR10R1x2f(x, y) dydxR10R√x0f(x, y) dydxR10Rx20f(x, y) dydxEnter I,II,III,IV here IntegralR10R1y2f(x, y) dxdyR10Ry20f(x, y) dxdyR10R1√yf(x, y) dxdyR10R√y0f(x, y) dxdy3Problem 3 ) (10 points)Match the solids E with the corresponding triple integralR R REf d V . There is one tripleintegral, f or which no picture of the solid is given. Mark this triple integral with O. Nojustifications are needed for this problem.I IIIII IVEnter O, I,II,III or IV here IntegralR10Rπ/20R1−x2−y20f(x, y, z) rdzdθdrR10Rπ0Rπ/20ρ2sin(φ)f(ρ, θ, φ) dφdθdρR10Rz0Rπ/20f(r, θ, z) r dθdrdzR10Rz0Rπ0f(r, θ, z) r dθdrdzR10Rπ0R10f(r, θ, z) r drdθdz4Problem 4 ) (10 points)Find all t he critical points of the function f(x, y) = xy3−x22−3y22.For each critical point, specify whether it is a local maximum, a local minimum or a saddlepoint and show how you know.Problem 5 ) (10 points)a) (6 points) Find all critical points of f(x, y) = 3xey− e3y− x3and classify them.b) (4 points) Does the function have a absolute maximum or absolute minimum? Makesure to justify also this answer.Problem 6 ) (10 points)We minimize the surface of a roof of height x and width 2x and length L =√2y if thevolume V (x, y) = x2√2y of the roof is fixed and equal to√2. In other words, you have tominimize f(x, y) = 2x2+ 4xy under the constraint g(x, y) = x2y = 1. Solve the problemwith the Lagrange method.Note: this problem can also be solved by substituting one of the variables in the con-straints. If done properly, such a solution needs more work. No credit is given for sucha substitution solution in this problem.xLxProblem 7 ) (10 points)5Evaluate the double integralZ270Z3x1/311 + y4dydx .Problem 8 ) (10 points)The flower type region R is given in polar coordinates as 0 ≤ θ ≤ 2π, 0 ≤ r ≤q|sin(5θ)|.Compute the moment of inertia I =R RRr2dA of this region.RProblem 9 ) (10 points)A solid E in space is given by the inequalities x2+ y2+ z2≤ 1, x2+ y2≥ z2. (In otherwords, the solid is obtained by cutting away the double cone x2+ y2≤ z2from the unitball x2+ y2+ z2≤ 1. ) Compute the triple integralZ Z ZE(x2+ y2+ z2) dxdydz
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