7/14/2011 FIRST HOURLY PRACTICE I Maths 21a, O.Knill, Summer 2011Name:• Start by writing your name in the above box.• Try to answer each question on the sam e page as the question is asked. If need ed , usethe back or the next empty page for work. If you need addi t io n al paper, write your nam eon i t .• Do not detach pages from this exam p acket or unst a p l e the packet.• Please write neatly. Answers which are i l leg i b l e for the grader can not be given credit.• No no t es, books, calculators, computers, or other electronic aids can be allowed.• You have exactly 90 minutes to complete your work.1 202 103 104 105 106 107 108 109 10Total: 1001Problem 1) (20 points) No justifications are needed.1)T FThe l en g th of the vector h1, 1, 1i is equal to 3.Solution:The l en g th is the square root of 3.2)T FAny two distinct points A, B in space determine a unique lin e which containsthese two points.Solution:The two points allow to give a parametrization A + t~AB3)T FFor any two non-intersecting li n es L, K, there is exactly one point P whichhas equal d i st ance to both li n es.Solution:There are many points which have e qu a l distance. Points with this property form a planeas d escr i bed in a main problem below.4)T FThe graph of f (x, y) is a surface in space which is equal to th e level surfaceg(x, y, z) = 0 of g(x, y, z) = f(x, y) − z.Solution:It is the level surface to c = 0.5)T FThe graph of the function f(x, y) = x2−y2is called an elliptic paraboloid.Solution:Nope, it is a hyperbolic p a ra boloid. An elliptic one would have a plus sign f(x, y) =x2+ y2.6)T FThe eq u at i on ρ cos(θ) = 1 in spherical coordinates defines a plane.2Solution:In spherical coordinates, the equation ρ cos(φ) i s z = 1 but ρ cos(θ) = 1 is not a pla n e.7)T FThe vector h1, 2, 3i is perpendicular to the plane x + 2y + 3z = 4.Solution:The vector h1, 2, 3i is ind e ed a normal vector.8)T FThe cr os s product between the vectors h1, 2, 3i and h1, 1, 1i is 6.Solution:It is the dot product which is 6. The cross product is a vector.9)T FThe two parametrized curves ~r(t) = ht, t2, t6i, 0 ≤ t ≤ 1 and~R(t) =ht2, t4, t6i, 0 ≤ t ≤ 1 have the same arc len g t h .Solution:This is a change of parametrization but the powers do not match.10)T FThe point (1, 0, 1) ha s the spherical coordinates (ρ, θ, φ) = (√2, 0, π/4).Solution:Apply the t r a n sfor m a ti o n formulas. We have θ = −π/4.11)T FThe distance between two parallel planes is the distance of any point on oneplane to the other plane.Solution:Note t h a t this is only true for parallel lines.12)T FProj~w(~v × ~w) =~0holds for al l nonzer o vectors ~v, ~w.3Solution:The vector (~v × ~w) is projected onto a vector perpendicular to it.13)T FThe vector projection of h2, 3, 4i onto h1, 0, 0i is h2, 0, 0i.Solution:Apply t h e formula. Because the vector on which we project has length 1, the result is thedot product times this vector.14)T FThe triple scalar product ~u · (~v × ~w) between three vectors ~u,~v, ~w is zero ifand only if two or more of the 3 vectors are parallel.Solution:They c an be nonparallel but in the same plane.15)T FThere are two vectors ~v and ~w so tha t the dot product ~v · ~w is equal to thelength of t h e cross product |~v × ~w|.Solution:Take two vectors which make an angle of 45 degrees. Then sin(θ) = cos(θ) and the len g t hof the cross product is equal to the dot product. Examples are ~v = h1, 0, 0i, ~w = h1, 1, 0i .16)T FTwo cylinders of radius 1 whose axes are lines L, K have distance d(L, K) =3 have d i st an ce 2.Solution:The d i st ance is 1 = 3 − 1 − 1. We h ave to subtr a ct the two radii of the cylinders.17)T FThere are two unit vectors ~v and ~w that are both parallel and perpendicular.Solution:The cr os s and dot product have to be zero simultaneously which is not possible.418)T FAssuming the curvature to exist for all time, the curvature κ (~r(t)) is alwayssmaller than or equal to |~r′′(t)|/|~r′|2.Solution:Use the formula and th e fact that the cross product in norm is smaller or equal than theproduct |r′′(t)| · |r′(t)|.19)T FThe cu r ve ~r(t) = hcos(t) sin(t), sin(t) sin(t) , sin(t)i is located on a sp here.Solution:Check x2+ y2+ z2= 1 is false. With a cos(t) in the end, it would be true.20)T FThe su r fa ce x2+ y2+ z2= 4z − 3 is a sphere of radius 1.Solution:Complete the square.Total5Problem 2) (10 points) No justifications are needed in this problem.a) ( 2 points) Match the graphs of the functio n s f(x, y). Enter O, if there is no match.I II IIIFunction f(x, y) = Enter O,I,II or IIIx2||x| − |y||x2+ y2x/(1 + y2)b) (3 points) Match the space curves with their parametrizations ~r(t). Enter O, if there is n omatch.I II IIIParametrization ~r(t) = O, I, II,III~r(t) = ht cos(t), t sin(t), ti~r(t) = hcos(t), sin(t), ti~r(t) = hsin(t), cos(t), 0i~r(t) = hsin(3t), cos(2t), cos(t)ic) (2 points) Match the functions g with the level sur face g(x, y, z) = 1. Enter O, where nomatch.I II IIIFunction g(x, y, z) = 1 O, I,II,IIIg(x, y, z) = x2− y2− z = 1g(x, y, z) = x − y2= 1g(x, y, z) = x2− y2+ z2= 1g(x, y, z) = x2+ y2+ z2= 1d) ( 3 points) Match the surface with the parametrization. Enter O, where no match.I II IIIFunction g(x, y, z) = O,I,II,III~r(s, t) = ht, s, tsi~r(s, t) = ht2+ s2, s, ti~r(s, t) = ht, t cos(s), t sin(s)i~r(s, t) = ht, s, tiSolution:a) III,I,0,IIb) III,I,0,IIc) III,II,I,0d) III,0,II,I6Problem 3) (10 points) No justifications are needed in this problem.a) ( 2 points) Translate from polar to Cartesian coordinates or back:Polar coordinates (θ, r) = Cartesian coordinates (x , y) =(π/2, 1)(1, 1)b) ( 2 points) Tr an s la t e from spherical to Cartesian coordinates or back:Spherical coordinates (θ, φ, ρ) = Cartesian coordinates (x, y, z) =(π/2, π/2, 1)(1, 1, 1)c) (3 points) Match the curves given in polar coordinates. Enter O, if there is no match.I II IIISurface Enter I,II,II, Or = π/4r = −θθ = π/4r = θd) ( 3 points) Match the surfaces given in spherical coordinates. Enter O, if there is n o match.I II IIISurface Enter I,II,III,Oρ = π/4φ = π/4θ = φθ = π/47Solution:Polar coordinates (θ, r) = Cartesian coordinates (x , y) =(π/2, 1) (0, 1)(π/4,√2) (1, 1)Spherical coordinates (θ, φ, ρ) = Cartesian coordinates (x, y, z) =(π/2, π/2, …
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