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HARVARD MATH 21A - First practice exam second hourly

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11/15/2005, FIRST PRACTICE SECOND HOURLY Math 21a, Fall 2005Name:MWF9 Ivan PetrakievMWF10 Oliver KnillMWF10 Thomas LamMWF10 Michael ScheinMWF10 Teru YoshidaMWF11 Andrew DittmerMWF11 Chen-Yu ChiMWF12 Ka t hy 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 notbe given credit, except for the TF and multiple choiceproblems.• Please write neatly. Answers which the grader can notread will not receive credit.• No notes, books, calculators, computers, or o ther elec-tronic aids can be used.• All unspecified functions mentioned in this exam are as-sumed to be smooth: you can differentiate as 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 1010 10Total: 110Problem 1) TF questions (20 points)Mark for each of the 2 0 questions the correct letter. No justifications are needed.1)T FAny function of three variables f(x, y, z) satisfies the partial differentialequation fxyz+ fyzx= 2fzxy.2)T FIf fx(x, y) = fy(x, y) for all x, y, t hen f(x, y) is a constant.3)T F(1, 1) is a local maximum of the function f(x, y) = x2y − x + cos(y).4)T FIf f is a smooth function of two variables, then the number of critical pointsof f inside the unit disc is finite.5)T FThe value of the function f(x, y) = sin(−x + 2y) at (0.0 01, −0.002) can bylinear approximation be estimated as −0.003.6)T FIf (1, 1) is a critical po int for the function f(x, y) then (1, 1) is also a criticalpoint for the function g(x, y) = f (x2, y2).7)T FThere is no function f (x, y, z) of three variables, for which every point onthe unit sphere is a critical point.8)T FIf (x0, y0) is a maximum of f (x, y) under the constraint g(x, y) = g(x0, y0),then (x0, y0) is a maximum of g(x, y) under the constraint f(x, y) =f(x0, y0).9)T FIf ~u is a unit vector tangent at (x, y, z) to the level surface of f(x, y, z) thenDuf(x, y, z) = 0.10)T FIn cylindrical coordinates (r, θ, z), the surface z = r2describes a cone.11)T FThe function u (x, t) = x2/2 + t satisfies the heat equation ut= uxx.12)T FThe vector ~ru− ~rvis tangent to the surface parameterized by ~r(u, v) =(x(u, v), y(u, v), z(u, v)).13)T FIf (1, 1, 1) is a maximum of f under the constraints g(x, y, z) = c, h(x, y, z) =d, and the Lagrange multipliers satisfy λ = 0, µ = 0, then (1, 1, 1) is a criticalpoint of f.14)T FIf (0, 0) is a critical point of f(x, y) and the discriminant D is zero butfxx(0, 0) > 0 then (0, 0) can not be a local maximum.15)T FLet (x0, y0) be a saddle point of f(x, y). For any unit vector ~u, there arepoints arbitrarily close to (x0, y0) for which ∇f is parallel to ~u.16)T FA function f(x, y) on the plane f or which the absolute minimum and theabsolute maximum are the same must be constant.17)T FThe sign of the Lagrange multiplier tells whether the critical point of f(x, y)constrained to g(x, y) = 0 is a local maximum or a local minimum.18)T FThe gradient of a function f (x, y, z) is tangent to the level surfaces of f19)T FThe point (0, 1) is a local minimum of the function x3+ (sin(y − 1))2.20)T FIf Duf(x, y, z) = 0 for all unit vectors u, then (x, y, z) is a critical point.Space for workProblem 2) (10 points)Match t he parametric surfaces with their parameterization. No justification is needed.I IIIII IVEnter I,II,III,IV here Parameterization(u, v) 7→ (u, v, u + v)(u, v) 7→ (u, v, sin(uv))(u, v) 7→ (0.2 + u(1 − u2)) cos(v), (0.2 + u(1 − u2)) sin(v), u)(u, v) 7→ (u3, (u − v)2, v)Space for workProblem 3) (10 points)Use the technique of linear approximation to estimate f(log(2)+0.001, 0.006) for f(x, y) =e2x−y. (Here, log means the natural logarithm).Space for workProblem 4) (10 points)Consider the equationf(x, y) = 2y3+ x2y2− 4xy + x4= 0It defines a curve, which you can see in the picture. Nearthe point x = 1, y = 1, the f unction can be written asa graph y = y(x). Find the slope of tha t graph at thepoint (1 , 1).-2 -1.5 -1 -0.5 0.5 1x-2-1.5-1-0.50.51ySpace for workProblem 5) (10 points)Find and classify all the critical points of the function f (x, y) = xy(4 − x2− y2).Space for workProblem 6) (10 points)Let f(x, y) = e(x−y)so that f(log(2), log(2)) = 1. Find the equation for the tangent planeto the graph of f at the point (log(2), log(2)) and use it to estimate f (log(2) +0.1, log(2) +0.004).Remark: As usual on the college life a s well asin galaxy life, log denotes t he natural logarithm.There is no particular reason (except maybe athigh school) to think, an other base is natural,except f or the fact that we earthlings have 10 fin-gers. I’m sure you can read this word of wisdomsomewhere in a ”manual to the galaxy”.Space for workProblem 7) (10 points)Consider the graph of the function h(x, y) = e−3x−y+ 4.a) (2 points) Find a function g(x, y, z) of three variables such that this surface is the levelset of g.b) (2 points) Find a vector normal to the tangent plane of this surface at (x, y, z).c) (2 points) Is this tangent plane ever horizontal? Why or why not?d) (3 points) Give an equation for the tangent plane at (0, 0).Space for workProblem 8) (10 points)Minimize the function E(x, y, z) =k28m(1x2+1y2+1z2) under the constraint xyz = 8, wherek2and m are constants.Remark. In quantum mechanics, E is the ground state energy of a particle in a box withdimensions x, y, z. The constant k is usually denoted by ¯h and called the Planck constant.Space for workProblem 9) (10 points)Assume F (x, y) = g(x2+ y2), where g is a function of one var ia ble. Find Fxx(1, 2) +Fyy(1, 2), given that g′(5) = 3 and g′′(5) = 7.Space for workProblem 10) (10 points)Let g(x, y) be the distance to the curve x2+ 2y2+ y4/10 = 1.. Show that g is a solutionof the pa rt ia l differential equationf2x+ f2y= 1outside the curve.Hint: here no computations are needed. The shape o f the curve is pretty much irrelevant.What does the PDE say about the gradient ∇f?Remark: This example just needs thought. Use it as a ”pillow problem” that is thinkabout it before going to sleep. By the way, the PDE is called eiconal equation. Itdescribes wave fronts in optics.Space for


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