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HARVARD MATH 21A - SECOND HOURLY

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11/15/2005, SECOND HOURLY Math 21a, Fall 2005Name:MWF9 Ivan PetrakievMWF10 Oliver KnillMWF10 Thomas LamMWF10 Micha el ScheinMWF10 Teru YoshidaMWF11 Andrew DittmerMWF11 Chen-Yu ChiMWF12 Kathy PaurTTh10 Valentino TosattiTTh11.5 Kai-Wen LanTTh11.5 Jeng-Daw Yu• Please mark the box to the left which lists your section.• Do not detach pages from this exam packet or unstaplethe packet.• Show your work. Answers without reasoning can not begiven credit.• Please write neatly. Answers which the grader can notread will not receive credit except for the True/False andmultiple choice problems.• No notes, books, calculators, computers, o r o ther elec-tronic aids can be used.• All unspecified functions mentioned in this exam are as-sumed to be smooth: you can differentiate a s many timesas you want with respect to any variables.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 10Total: 100Problem 1) True/False questions (20 points)Mark for each of the 20 questions the correct letter. No justifications are needed.1)T FIf ∇f(x, y) 6= h0, 0i at a given point (x0, y0), there exists a unit vector ~u forwhich D~uf(x0, y0) is zero.2)T FIf fxx(0, 0) = 0, D 6= 0, and ∇f(0, 0) = h0, 0i, then (0, 0) is a saddle point.3)T FThe surface described in spherical coordinates by the equation ρ cos(φ) =ρ2sin2(φ) is an elliptic paraboloid.4)T FThe function f(x, t) = x + t satisfies the heat equation ft= fxx.5)T Ff(x, y) = 3x2y − y3is a solution of the Laplace equation fxx+ fyy= 0.6)T FA smooth function defined on the closed unit disc x2+y2≤ 1 has an absolutemaximum in this disc (including the boundary).7)T FA surface defined in cylindrical coordinates by the equation g(r, θ, z) = 0 isalways a surface of revolution.8)T FThe function f( x, y) = x2− y2has a neither a local maximum nor a localminimum at (0, 0 ).9)T FThe functions f(x, y) and g(x, y) = (f (x, y))4always have the 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 FIf f(x, y) = c and fx6= 0, thendxdy= fy(x, y)/fx(x, y).12)T Ff(x, y) =√16 − x2− y2has both an absolute maximum and a n absoluteminimum on its domain of definition.13)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.14)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.15)T FThe directional derivative D~vf is a vector perpendicular to ~v.16)T FSuppose f has a maximum value at a point P relative to the constraintg = 0. If the Lagrange multiplier λ = 0, then P is also a critical point forf without the constraint.17)T FAt a saddle point, all directional derivatives are zero.18)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) = 0.19)T FThe function f(t, x, y) = y sin(x−t) satisfies the partial differentia l equationftt= fxx+ fyy.20)T FAt a local maximum (x0, y0) of f (x, y), o ne has fyy(x0, y0) ≤ 0.Problem 2) (10 points)Match the pa rametric surfaces with their parameterization. No justifications are neededin this problem.I IIIIIIV VEnter I,II,III,IV here Parameterization(u, v) 7→ (cos(u) sin(v), sin(u) sin(v), 5 cos(v))(u, v) 7→ (u2, v2, u2− v2)(u, v) 7→ (cos(u) sin(v), sin(u) sin(v), 5 sin(v))(u, v) 7→ (u, v, u2− v2)(u, v) 7→ (u, v, eu sin(v))Problem 3) (10 points)Consider the following differential equations:A) Laplace equation fxx+ fyy= 0B) Wave equation fxx= fyyC) Poisson equation fxx+ fyy= 4D) Heat equation fx= fyyE) Transport equation fx= fyGiven the functions g(x, y) = sin(x + y) and h(x, y) = x2+ y2. Which of the partialdifferential equations A,B,C,D,E do they satisfy?Equation g is a solution g is not a solution h is a solution h is not a solutionA)B)C)D)E)Problem 4) (10 points)xy01234556677-1-2-3-4-5-6-7-8RSTUVWXYa) (4 points) Circle the point at which t he magnitude of the gradient vector ∇f is greatest.Mark exactly one point. Justify your answer.R S T U V W X Yb) (3 points) Circle t he points at which the partial derivative fxis strictly positive. Markany number of points on this question. Justify your answers.R S T U V W X Yc) (3 points) We know that the directional derivative in the direction (1, 1)/√2 is zero atone of the following points. Which one? Mark exactly one point on this question.R S T U V W X YProblem 5) (10 points)Find all the critical points of the function f(x, y) =x22+3y22− xy3.For each critical point, specify if it is a local maximum, a local minimum or a saddle pointand show how you know.Problem 6) (10 points)A beach wind protection is manufactured as follows. There is a rectangular floor ACBDof length a and width b. A pole of height c is located at the corner C and perpendicularto the ground surface. The top point P of the pole forms with the corners A and C onetriangle and with the corners B and C an other triangle. The total material has a fixedarea of g(a, b, c) = ab + ac/2 + bc/2 = 12 square meters. For which dimensions a, b, c is thevolume f(a, b, c) = abc/6 of the tetrahedral protected by this configuration maximal?Problem 7) (10 points)A spaceship approaches its base B = (0, 0, −π/2) along the pathr(t) = (sin2(t), 1 − cos(t), −π/2 − t).The base is prot ected by a fo r ce shield given by the equation x2+ 2y2+ z2/π2= 3. Attime t = −π/2, the spaceship passes through the shield.a) (5 points) At that time, does the ship pass through the shield at a right angle to theshield?b) (5 points) The fo r ce shield is generated by a power station located at the point (0, 0, 0).In the moment when the spaceship is passing through the shield, what is the rate of changeof the distance from the spaceship to the power station?Problem 8) (10 points)Given the functionf(x, y) =q105 − 2x2− 3y2.a) (4 points) Use the technique of linear approximation at the point (1, 1) to estimatef(1.01, 0.9).b) (3 point s) Find a unit vector pointing in the direction at (1, 1) where the functiondecreases fastest.c) (3 points) Find the tangent line to the curve√105 − 2x2− 3y2= 10 at the point (1, 1).Problem 9) (10 points)Let S be the surface of revolution for which the distance r to the z-a xis is g(z) = ez.a) (3 points) Find a parameterization of S.b) (3 points) Find an implicit equation f(x, y, z) = c which describes this surface.c) (4 points) Find the


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