7/14/2011 FIRST HOURLY PRACTICE II 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 additional paper, write your nameon it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which a re illegi b l e for the grader can not be given credit.• No notes, boo ks, calcu l at or s, comp u t er s , 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 the 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 that the dot product ~v · ~w is equal to thelength of the cross product |~v × ~w|.Solution:Take two vectors which make an angle of 45 degrees. Then si n ( θ) = cos(θ).4)T FThe distance between two spheres of radius 1 whose centers have distanc e10 is 8.Solution:The connection between t h e centers is also the connection between t h e nearest points onthe sp here.5)T FIf two vectors ~v and ~w are both parallel and perpendicula r , 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 a lways smaller or equal than the length |~r′′(t)| ofthe acce ler a t io n vector ~r′′(t).2Solution:If you drive along a cir cl e very slowly the accel er at i on is small but the curvature is thesame.7)T FThe cu r ve ~r ( t ) = hcos(t) sin(t), sin(t) sin(t), cos(t)i is located on a sphere.Solution:Check x2+ y2+ z2= 1.8)T FThe su r face x2+ y2+ z2= 2z is a sphere.Solution:Complete the square t o see that it is indeed a sphere centered at (0 , 0, 1) with radius 1.9)T FThe length of the vector h4, 2, 4i is an integer.Solution:It is the square root of 42+ 22+ 42which is 6.10)T FThe cu r vature of the curve h2 cos(t3), 2 sin(t3), 1i is co n st ant 2.Solution:It is 1/2.11)T FThe graph of the funct i on f(x, y) = x2−y2is called an elliptic paraboloid.Solution:It is a hyperbolic paraboloid.12)T FThe equ a t io n φ = 3π/2 in spherical coordinates defines a p l an e .3Solution:φ = 3π/2 is out of range13)T FThe vector h1, 2, 3i is perpendicular to the plane 2x + 4y + 6z = 4.Solution:The vector h2, 4, 6i is the normal vector.14)T FThe cr oss product between h2, 3, 1i and h1, 1, 1i is 6.Solution:It is the dot product which is 6. The cross product is a vector.15)T FThe curve ~r ( t ) = hcos(t), t2, sin(t)i, 1 ≤ t ≤ 9 and the curve ~r(t) =hcos(t2), t4, sin(t2)i, 1 ≤ t ≤ 3 have th e same length.Solution:This i s a change of parametrization16)T FIf a stone falls for 3 seconds from height z = h to the ground z = 0 withgravitational acceleration −10 then the height is 30 meters.Solution:The t im e and height is related by 5t2= h so that the height is about 45 met er s.17)T FThe point (1, −1, 1) has the spherical coordinates the form (ρ, θ, φ) =(√3, π/4, π/4).Solution:Apply the t r an s for m at i on formulas. We have θ = −pi/4.18)T FThe point (1, −1, 1) has the cylindrical coordinates the form (r, θ, z) =(√3, π/4, 1).4Solution:Apply the t r an s for m at i on formulas. The correct answer is (√2, −π/4, 1).19)T FThe distance between two parallel li n es in space is the dist a n ce of a pointon one line to the other line.Solution:This i s only true for parallel lines.20)T FFor two nonzero arbitrary vectors ~v and ~w the identity Proj~v(~v × ~w) =~0holds.Solution:Make a change of variables.Total5Problem 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 IIIFunction f(x, y) = Enter I,II,IIIx2+ y2x2− y2x2− yx − y2b) ( 3 points) Match the graphs with the fu n ct i ons f(x, y). Enter O, where no match.I II IIIFunction 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 IIICurve ~r(t) = Enter I,II,IIIht, t2iht4, 1 + 2t4ih−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 IIIFunction g( x, y, z) = Enter I,II,IIIx2+ y2− z2x2− y2− 1x2− y2− zx2+ y2− zx − y + z6Solution: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-pl an e ), 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 equat i ons withthe t r aces . 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:III,II,I,OO,III,O,I,III,O III,IIO,III,I,I I, O73b) ( 5 points) Matching parametrized surfaces.I II III IVMatch the para-metric surfaceswith their param-eterization. Nojustifications areneeded.Enter I,II,III,IV here 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 the 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 es.b) ( 6 points) Find the distance between the two lines.Solution:a)~r(t) = ht, t, ti~R(t) = h1 + 2t, 2 + 3t, 4 + 4ti.b) ~n = h1, 1, 1i × h2, 3, 4i = h1, −2, 1i . The connection vector between t he fir st line andsecond line i s~P Q = h1, 2, 4i. Project it onto ~n to get|h1, 2, 4i · h1, −2, 1i ||h1, −2, 1i|=1√6.8Problem 5) (10 points)At the independence day celebrat i on on July 4, 2010 in Boston, two rockets were launchedat the same time. Their paths follow parabola:~r(t) = ht, t, 5 −
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