8/11/2011 FINAL EXAM PRACTICE VI Maths 21a, O. Knill, Summer 2011Name:• Start by printing your name in the above box.• Try to answer each question on the same page as the question is asked. If needed, usethe back or the next empty page for work.• Do not detach pages from this exam packet or un s t ap l e the packet.• Please try to write nea t ly. Answers which are illegible for the grader ca n not be givencredit.• No not es, books, calculators, computers, or other electronic aids are allowed.• Problems 1-3 do not r equ i r e any justifications. For the rest of the p r ob l em s you have toshow your work. Even correct answers without derivation can not be given credit.• You h ave 180 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 10Total: 1501Problem 1) ( 20 points)1)T FThe qu ad r a ti c surface −x2+ y2+ z2= −5 is a one-sheeted hyperboloid.2)T F|~u×~v| < |~u·~v| implies that the angle α between u and v satisfies |α | < π/4.3)T FR30R2π0r sin(θ) dθ dr is the area of a disc of radius 3.4)T FIf a vector field~F (x, y, z) satisfies curl(F )(x, y, z) = 0 for all points (x, y, z)in sp a ce, then~F is a gradient field.5)T FThe acce ler a t io n of a parameterized curve ~r(t) = hx(t), y(t), z(t)i is zero ifthe cu r ve ~r(t) is a line.6)T FThe curvature of the curve ~r(t) = h3 + sin(t) , t, t2i is half of the curvatureof th e curve ~s(t) = h6 + 2 sin(t), 2t, 2t2i.7)T FThe cur ve ~r(t) = hsin(t), t, cos(t)i for t ∈ [0, π] is a helix.8)T FIf a functi on u(t, x) is a solution of th e partial differential equation utx= 0,then it is constant.9)T FThe unit tangent vector~T of a curve is always perpendicular to the a ccel -eration vector.10)T FLet (x0, y0) b e the maximum of f(x, y) under the constraint g(x, y) = 1.Then t h e gradient of g at (x0, y0) is perpendicular to the gradient of f at(x0, y0).11)T FAt a critical point for which fxx> 0, the discriminant D determines whetherthe point is a local maximum or a local m i n i mum.12)T FIf a vector field~F (x, y) is a gradient field, then for any closed curve C theline integralRC~F ·~dr is zer o.13)T FIf C is part of a level curve of a function f(x, y) and~F = hfx, fyi is thegradient field of f, thenRC~F ·~dr = 0.14)T FThe gradient of the divergence of a vector field~F (x, y, z) = ∇f(x, y, z) isalways the zero vector field.15)T FThe line integral of the vector field~F (x, y, z) = hx, y, zi along a l ine segmentfrom ( 0, 0, 0) to (1, 1, 1) is 1.16)T FIf~F (x, y) = hx2− y, xi and C : ~r(t) = hqcos(t),qsin(t)i par am e te r ize s theboundary of the region R : x4+ y4≤ 1, thenRC~F ·~dr is twice the area ofR.17)T FThe flux of the vector field~F (x, y, z) = h0, 0, zi through the boundary S ofa solid toru s E is equal to the volume the torus.18)T FIf~F is a vector field in space and S is the boundary of a cube then the fluxof~curl(~F ) thr ough S is 0.19)T FIf div(~F )(x, y, z) = 0 for all (x, y, z) and S is a half sphere then the flux of~F throu g h S is zero.20)T FIf the cur l of a vector field is zero everywhere, then its divergence is zeroeverywhere too.2Problem 2) (1 0 po ints)a) Mat ch the following contour surface maps with t h e functio n s f(x, y, z)I IIIII IVEnter I,II,III,IV here Functionf(x, y, z) = −x2+ y2+ zf(x, y, z) = x2+ y2+ z2f(x, y, z) = −x2− y2+ zf(x, y, z) = −x2− y2+ z2b) Mat ch the following parametrized surfaces with their definitions3I IIIII IVEnter I,II,III,IV here Function~r(u, v) = hu − v, u + 2v, 2u + 3v sin( uv)i~r(u, v) = hcos(u) sin(v), 4 sin(u) sin(v), 3 cos(v)i~r(u, v) = h(v4− v2+ 1) cos(u), (v4− v2+ 1) sin(u), vi~r(u, v) = hu, v, sin ( uv)iProblem 3) (1 0 po ints)No justifications are required in this p r oblem. The first p i ct u r e sh ows a gr ad i ent vectorfield~F (x, y) = ∇f(x, y).4ABCThe critical points o f f(x, y) are called A, Band C. What can you say about the natureof these three critical points? Which one is alocal max, which a local min, which a saddle.point local max local min saddleABCRP →Q~F ·~dr denotes the l i n e integral of~Falong a strai ght line path from P to Q.statement True FalseRA→B~F~dr ≥ 0RA→C~F~dr ≥RA→B~F~drThe second picture again shows an oth er gradient vector field~F = ∇f(x, y) of a differentfunction f(x, y).ACBDWe want t o identify th e maximum of f(x, y)subject to th e constraint g(x, y) = x2+ y2=1. The soluti on s of the Lagrange equ at i on sin this case are labeled A, B, C , D. At whichpoint on the circle if f maximal?point maximumABCDRγ~F ·~dr denotes the line integral of~F alongthe circle γ : x2+ y2= 1, oriented counterclockwise.Rγ~F ·~r> 0 < 0 = 0Check if true:Problem 4) (1 0 po ints)a) Mat ch the following vector fields in space with the corresponding formulas:5I IIIII IVEnter I,II,III,IV here Vector Fie ld~F (x, y, z) = hy, −x, 0i~F (x, y, z) = hx, y, zi~F (x, y, z) = h0, 0, 1i~F (x, y, z) = h−x, −2y, −zi6b) Choose from the following words to complete the following table: ”arc length formula”,”surface area formula”, ”chain rule”, ”volume of parall el epiped”, ”area of p a r al l elo gr am ” ,”Consequence of Clairot theorem”, ”Fubini Theorem”, ”line integral”, ”Flux integral”,”vector projecti on ”, ”scalar projection”, ”partial derivative”:Formula Name of formula or rule or theoremP~v(~w) = ~w~v· ~w| ~w|2R RR|~ru×~rv| dudvddtf(~r(t ) ) = ∇ f (~r(t)) ·~r′(t)curl(grad(f)) =~0Rba|~r′(t)| dt|~u · (~v × ~w)|Problem 5) (1 0 po ints)The poi nt P = (1, 2, 2) is mir r or ed at the plane which contains the points A = (0, 0, 0), B =(4, 0, 2) an d C = (2, 2, −1). The mirror point is called Q.a) (3 point) Find a normal vector to the plane which has len gt h 1.b) (4 points) Find the distance of the point P to the plane.c) (3 po ints) Find the coordinates of the point Q.Problem 6) (1 0 po ints)The tip of Cape Cod near Provincetown is full of dunes and a nice place to relax. Theelevation at position (x, y) isf(x, y) = x3/12 − xy − 2y − y2− x .7Find a ll crit i cal points and decide whether they are maxima, minima or saddle points.Problem 7) (1 0 po ints)The roman surface is defined as …
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