7/8/2010 FIRST HOURLY PRACTICE I Maths 21a, O.Knill, Summer 2010Name:• 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 i on a l paper, write your nam eon it.• Do not det ach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which are i l l egi b l e for the grader can not be given credit.• No n ot 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 1010 10Total: 1101Problem 1) (20 points) No justifications are needed.1)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.2)T FThe triple scalar product between three vectors is zero if and only if two ofthe vectors are parallel.Solution:They can be nonparallel but in the same plane.3)T FThere are two vectors ~v an d ~w so th at the dot product ~v · ~w is equal to thelength of t he cross product |~v × ~w|.Solution:Take two vectors which make an angle of 45 degrees. Then sin(θ) = cos(θ).4)T FThe distance between two spheres of radius 1 whose centers have distan ce10 is 8.Solution:The connection between the centers is also the connection between the nea r est points onthe sphere.5)T FIf two vectors ~v and ~w are both parallel and perpendicular, then one of thevectors must be the zero vector.Solution:If ~v = λ ~w, then ~v · ~w = |~v|~w| = 0 implies that one must be empty.6)T FThe curvature κ(~r(t)) is always smaller or equal than the length |~r′′(t)| ofthe acceleration vector ~r′′(t).2Solution:If you drive along a circl e very slowly the acceler at i on is small but the curvature is thesame.7)T FThe curve ~r(t) = hcos(t) sin(t), sin(t) sin(t), c os( t ) i is located on a sphere.Solution:Check x2+ y2+ z2= 1.8)T FThe s u r face x2+ y2+ z2= 2z is a sphere.Solution:Complete the square to see that it is indeed a sphere centered at (0, 0, 1) with radius 1.9)T FThe l ength of the vector h4, 2, 4i is an integer.Solution:It i s the square root of 42+ 22+ 42which is 6.10)T FThe curvature of th e curve h2 cos(t3), 2 sin(t3), 1i is constant 2.Solution:It i s 1/2.Total3Problem 2) (10 points) No justifications are needed in this problem.a) ( 2 points) Match contour maps with functions f(x, y). Enter O, where no match.I II II IFunction f(x, y) = Enter I,II,II Ix2+ y2x2− y2x2− yx − y2b) ( 3 points) Match the graphs with the functions f(x, y). Enter O, where no match.I II II IFunction f(x, y) = Enter I,II,IIIx − y|x| − yx2− y2x2y2x − y3c) (2 points) Match the curves with their parametrizations ~r(t). Enter O, where no match.I II II ICurve ~r(t) = Enter I,II,II Iht, t2iht4, 1 + 2 t4ih−t sin(t), t cos(t)ihsin(t), cos(t)id) ( 3 points) Match level surfaces with definition g(x, y, z) = 0. Enter O, where no match.I II II IFunction g(x, y, z) = Ent er I,II,IIIx2+ y2− z2x2− y2− 1x2− y2− zx2+ y2− zx − y + z4Solution:a) II,I,III,Ob) O , III,I,II,Oc) I,O , III,IId) I ,III,II,O,OProblem 3) (10 points) No justifications are needed in this problem3a) (5 points) Matching traces with surfaces.xy-trace yz-tracexz-tracexy-trace yz-tracexz-traceA BC DThe figures above show the xy-trace,(the intersection of the surface withthe xy-plane), the yz-trace (the in-tersection of the surface with the yz-plane), and the xz-trace (the intersec-tion of the surface with the xz-plane).Match the following equ at i ons withthe traces. No justifications required.Enter A,B,C,D,E,F here Equationx2+ y2− (z − 1/3)2= 0x2− y2+ z = 0x2+ y2− z2− 1 = 0x2+ y2− z = 1Solution:II I,II,I,OO,III,O,I,III,O III,IIO,III,I,II,O3b) (5 points) Matching parametrized surfaces.5I II II I IVMatch the para-metric surfaceswith their param-eterization. Nojustifications areneeded.Enter I,II,III, IV her e Parametrization~r(u, v) = hu, v, v2− u2i~r(u, v) = hcos(u) sin(v), 2 sin(u) sin(v), cos(v)i~r(u, v) = h(v2+ 1) cos(u), (v2+ 1) sin(u), vi~r(u, v) = hu, 3, viSolution:a) DABCb) I ,IV, II,IIIProblem 4) (10 points)We want to find t h e distance between the lines x = y = z and (x−1)/2 = (y −2)/3 = (z −4)/4.a) ( 4 points) Find a parametrization for each of the two lin e s.b) ( 6 points) Find the distance between the two lines.Solution:a)~r(t) = ht, t, ti~R(t) = h 1 + 2t, 2 + 3t, 4 + 4ti.b) ~n = h1, 1, 1i × h2, 3, 4i = h1, −2, 1i. The connection vector between the first line andsecond line is~P Q = h1, 2, 4i. Project it onto ~n to get|h1, 2, 4i · h1, −2, 1i||h1, −2, 1i|=1√6.6Problem 5) (10 points)At the independence day celebration on July 4, 2010 in Boston, two rockets were laun chedat t he same time. Their paths follow parabola:~r(t) = ht, t, 5 − t2i ,~R(t) = h2 − t, t, 4 + t − t2i .a) ( 3 points) They collide at some time t = t0. Fin d this time.b) ( 4 points) Compute the two velocity vectors ~r′(t) and~R′(t) at t = t0.c) (3 points) Determin e the cos of the angle of intersection between the curves at the impactpoint.Solution:a) The time ist = 1 as can be seen by solving ~r(t) =~R(t). (Just look at the firstcomponent for example.)b) The velocity vectors are~r′(t) = h1, 1, −2ti ,~R′(t) = h−1, 1, 1 − 2ti .Therefore,~r′(1) =h1, 1, −2i an d~R′(1) =h−1, 1, −1i .c) The impact angle is the angle between of the velocity vectors at the point ~r(1) =~R(1).The co s of the angle is~r′(1) ·~R′(1)|~r′(1)| · |~R′(1)|= 2/(√6√3) ,which simplifies to√2/3 .Problem 6) (10 points)An octahedron has 4 vertices A = (−1, −1, 0), B = (1, −1, 0), C = (1, 1, 0), D = (−1, 1, 0) inthe xy plane. Two oth er vertices are at E = ( 0, 0, a) and F = (0, 0, −a).a) (4 points) For which positive value of a is the distance between A and F equal to 2 and thesolid a regular octahedron?b) ( 6 points) Find the distance between A and th e line connectin g the points B and F .7Solution:a) 2 + a2= 4 shows a =√2.b) | h − 1, −1, −ai × h−2, 0, 0i| /√2 + a2=√12/2 =√3 .Problem 7) (10 points)Let ~v = h3, 4, 5i, ~w = h1, 1, 1i. Compute the following expressions:a) ( 2 points) the area of the parallelogram spanned by ~v and ~w,b) ( 2 points) the vector (~v × ~w) × ~w,c) (2 points) the scalar ~v · ~w,d) ( 2 points) the vector Proj~v(~w),e) (2 points) cos(α), where α is the angle
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