7/14/2011 FIRST HOURLY PRACTICE IV Maths 21a, O.Knill, Summer 2011Name:• Start by printing 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, books, calculators, comput er s, or other electronic aids can be allowed.• You have 90 minutes time to complete your work (the actual exam wil l have only 10questions)1 202 103 104 105 106 107 108 109 1010 1011 10Total: 1201Problem 1) (20 points)Circle for each of the 20 questions the correct letter. No justification s are needed.1)T FThe set of points in the plane which sati sfy x2−y2= −10 is a curve calledhyperbola.2)T FThe length of the sum of two vectors in space is always the sum of thelength of t h e vectors.3)T FFor any three vectors ~u, ~v, ~w, the identity ~u · (~v × ~w) = ( ~w ×~v) · ~u ho l d s.4)T FThe set of points which satisfy x2+ 2x + y2− z2= 0 is a cone.5)T FIf P, Q, R are 3 different points in space that don’t l i e in a line, then~P Q×~RQis a vector orthogonal to the plane containing P, Q, R.6)T FThe line ~r(t) = (1 + 2t, 1 + 3t, 1 + 4t) hits the plane 2x + 3y + 4z = 9 at aright angle.7)T FA surface which is given as r = sin(z) in cylindrical coordinates stays thesame wh en we rotate it around the y axis.8)T FFor any two vectors, ~v × ~w = ~w ×~v.9)T FIf |~v × ~w| = 0 for all vectors ~w, then ~v =~0.10)T FIf ~u and ~v are or t h o gonal vectors, then (~u ×~v) × ~u is parallel to ~v.11)T FEvery vector contained in the line ~r(t) = (1 + 2t, 1 + 3t, 1 + 4t ) is parallelto t h e vector (1, 1, 1).12)T FIf in spherical coordinates a point is given by (ρ, θ, φ) = (2, π/2 , π/2), thenits r ect a n gular coordinates are (x, y, z) = (0, 2, 0).13)T FThe set of points which satisfy x2− 2y2− 3z2= 0 form an ellipsoid.14)T FIf ~v × ~w = (0, 0, 0), then ~v = ~w.15)T FThe set of points in R3which have distance 1 from a line form a cylinder.16)T FIf in rectangular coordinates, a point is given by (1, 0, 1), t hen its sphericalcoordinates are (ρ, θ, φ) = (√2, π/2, −π/2).17)T FIn sp h e r ica l coordinates, the equation cos(θ) = sin(θ) defines the planex − y = 0.18)T FFor any three vectors ~a,~b and ~c, we always have (~a ×~b) ·~c = −(~a ×~c) ·~b.19)T FIf |~v × ~w| = 0 then ~v = 0 or ~w = 0.220)T FTwo nonzero vectors are parallel if and only if their cross product is~0.3Problem 2a) (5 points)Match t he surfaces with their parameteri zat i on ~r(u, v) or the implici t description g(x, y, z) = 0 .Note that o n e of the surfaces is not represented by a formula. No justifications are needed inthis problem.I II IIIIV V VIEnter I,II,III,IV,V,VI here Equation or Parameterization~r(u, v) = ((2 + sin(u)) cos(v), (2 + sin(u)) sin(v), cos(u))~r(u, v) = (v, v − u, u + v)~r(u, v) = (u2, vu, v)x2− y2+ z2− 1 = 0~r(u, v) = (cos(u) s in(v) , cos(v), sin(u) sin(v))4Solution:Enter I,II,III,IV,V,VI here Equation or ParameterizationII~r(u, v) = ((1 + sin(u)) cos(v), (1 + sin(u)) sin(v), cos(u))IV~r(u, v) = (v, v − u, u + v)I~r(u, v) = (u2, vu, v)IIIx2− y2+ z2− 1 = 0V~r(u, v) = (cos(u) s in(v) , cos(v), sin(u) sin( v))Problem 2b) (5 points)Match the contour maps with the corresponding functions f (x, y) of two variables. No justifi-cations are n e ed ed .I II IIIIV V VI5Enter I,II,III,IV,V or VI here Function f(x, y)f(x, y) = sin(x)f(x, y) = x2+ 2y2f(x, y) = |x| + |y|f(x, y) = sin(x) cos(y)f(x, y) = xe−x2−y2f(x, y) = x2/(x2+ y2)Solution:Enter I,II,I II,IV,V or VI here Function f(x, y)IIIf(x, y) = sin(x)IIf(x, y) = x2+ 2y2Vf(x, y) = |x| + |y|IVf(x, y) = sin(x) cos(y)If(x, y) = xe−x2−y2VIf(x, y) = x2/(x2+ y2)Problem 3) (10 points)Match the equation with their graphs and justify briefly your choice:I II6III IVEnter I,II,III,IV here Equation Short Justificationz = sin(3x) cos(5y )z = cos(y2)z = log(x)z = x/(x2+ y2)Solution:Enter I,II,III,IV here Equation Short JustificationIIIz = sin(3x) cos(5y ) two traces show wavesIIz = cos(y2) no x dependence, periodic in yIz = log(x) no y dependence, monotone in xIVz = x/(x2+ y2) singular at (x,y)=(0,0)Problem 4) Distances (10 points)Let L be the linex = 1 + 2t, y = − 3t , z = tand let S be the plane x + y + z = 2.a) Verify that L and S have no intersections.b) Com p ute the distance between the line L and plane S.Hint. Just take any point P on the line and compute the distance from the line to the p la n e.Solution:The vector v = ( 2, −3, 1) is in the line. It i s is n or m al to the normal vector (1, 1, 1) of theplane.The point P = (1, 0, 0) is on the line. The point Q = (1, 1, 0) is on the plan e. Thedistance is the scalar projection of P Q onto the nor m al vector (1, 1, 1) which is (0, 1, 0) ·(1, 1, 1)/√3 = 1/√37Problem 5) (10 points)a) ( 5 points) Find the ar ea of the parallelogram P QSR with cornersP = (0, 0, 0), Q = (1, 1, 1), R = (1, 1, 0), S = (2, 2, 1) .b) (5 points) Find the volume of the pyramid which has as the base t h e parallelogram P QRSand has a fifth vertex at T = ( 3 , 4, 3).8Solution:a) Th e area is |(Q − P ) × (R − P )| =√2.b) The distance h = |w · (u × v)|/|w| of T to the plane containing thepoints P, Q, S, R and use the volume formula V = Ah/3, where A = |u × v|is the area of the parallelogram. The volume of the pyramid is1/3 .Remark. The parallelepiped spanned by ~u =(1, 1, 1),~v = (2, 2, 1) and ~w = (3, 4, 3) has volume 1. Thepyramid has volume 1/3 of this volume because one canchop the pyramid into two tetraedra of the same volumeand each tetrahedron has volume 1/6. One can take 4tetrahedra spanned by u, v, w as well as one of doublevolume spanned by ( u + v, u + w, v + w) to build theparallelepiped. The volume of the pyramid is therefore1/3 .The following pictur e shows that a t et r a h ed r o n …
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