7/8/2010 FIRST HOURLY PRACTICE V Maths 21a, O.Knill, Summer 2010Name:• Start by writing 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. 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 are illegible for the grader can not be given credit.• No notes, bo oks, calculators, comput ers, 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 1011 10Total: 1201Problem 1) (20 points) No justifications are needed.1)T FThe vector ~v = h1, −1, 0i is perpendicular to the line 1 −x = 1 −y = 1 −z.Solution:The line contains the vector h1, 1, 1i which is perpendicular to the vector h1, −1, 0i.2)T FThe length of the vector ~v = h1, 0, −1i is 2.Solution:It is the square root of 2.3)T Fh1, 2, 3i × h2, 4, 6i = h0, 0, 0i.Solution:The vectors are parallel. So, indeed their cross product is the zero vector.4)T FThe distance between a point P and a line through the origin O can not belarger than |OP |.Solution:The distance is the smallest distance between P and any point in the line. So, it is smallerthan the distance between P and O.5)T FThe vectors ~u = h1, −1, 0i and~P Q with P = (1, 1, 1) and Q = (2, 2, 2) areparallel.Solution:They are perp end icular, not parallel6)T FThe surface y2− z2= 1 is an elliptical paraboloid.Solution:It is a cylindrical hyperboloid.27)T FIf two vectors ~v, ~w are orthogonal, then their cross product is the zerovector.Solution:No, their dot product is zero.8)T FThe surface x2+ y2− z2= 2x is called a one-sheeted hyperboloid.Solution:Complete the square to get (x − 1)2+ y2− z2= 1.9)T FThe set of points which have distance 1 from the x-axis is a cylinder.Solution:Yes, it is y2+ z2= 1.10)T FIf in spherical coordinates a point is given by (ρ, θ, φ) = (1, π, π/2), then itsEuclidean coordinates are (x, y, z) = (−1, 0, 0).Solution:Just make a picture.11)T FThe volume of a parallelepiped spanned by (1, 1, 1), (1, 0, 0) and (0, 1, 2) isequal to 1.Solution:The triple scalare product is −1 but the volume is 1.12)T FThe two planes x + y − z = 1 and −2x − 2z + 2z = 5 intersect in a line.Solution:The second equation simplifies to x = −5/2. The planes intersect in a line. The originalintention had been to have the planes parallel (replace -2z by 2z) but we left it like that.313)T FIn cylindrical coordinates, the equation r2= z defines a paraboloid.Solution:Indeed, x2+ y2= z is an elliptical paraboloid.14)T FThe vector projection of the vector (1, 1, 1) onto the vector h0, 2, 0i ish0, 1, 0i.Solution:Use the formula or take the dot product with j.15)T FThe point given in cylindrical coordinates as (r, θ, z) = (1, π/2, 1) is inCartesian coordinates the p oint (x, y, z) = (0, 1, 1).Solution:Just make a picture.16)T FIf z = g(x, y) is a graph then ~r(u, v) = (u, v, g(u, v)) is a parameterizationof the surface.Solution:Graphs are one of the 4 basic surface types we have introduced.17)T FThere is a function f(x, y, z) such that f(x, y, z) = 1 is a hyperboloid andf(x, y, z) = −1 is a paraboloid.Solution:Go through the list of functions which define hyperboloids or paraboloids. Hyperboloidscontain squared quantities x, y, z, paraboloids are graphs and one of the variables appearsonly in a linear fashion.18)T FThe projection of ~v = h1, 1, 1i onto the vector h1, 0, 0i is h√3, 0, 0i.Solution:It is (1, 0, 0).419)T FThe points satisfying (x −1)2−(y −1)2+ (z + 1)2= 1 form a hyperboloidSolution:Indeed, it is a one-sheeted hyperboloid20)T FThe equations x = y = z describe a line which contains th e vector h1, 1, 1i.Solution:Yes, it is a special case of a symmetric equation, where the denominators are a = 1,b = 1and c = 1.TotalProblem 2) (10 points)Match the equations g(x, y, z) = d with the surfaces.I II IIIIV V VI5Enter I,II,III,IV,V,VI here Equationx2+ y2− z2= 0x2− y2− z2= 1x2− y2+ z2= 1x2− y − z2= 1x2− z2= 1x + y2+ z2= 0Solution:Enter I,II,III,IV,V,VI here EquationVIx2+ y2− z2= 0Ix2− y2− z2= 1IIIx2− y2+ z2= 1IVx2− y − z2= 1IIx2− z2− 1Vx + y2+ z2= 0Problem 3) (10 points)Match the functions with their graphs. No justifications are needed.6I IIIII IVV VIEnter I,II,III,IV,V or VI here Equationz = 3x + 5y + 1z = x/(1 + x2+ y2)z = y2− xz = |x|·|y|z = sin(6(x + y))z = y37Solution:Enter I,II,III,IV,V or VI here EquationIIz = 3x + 5y + 1IVz = x/(1 + x2+ y2)Iz = y2− xVIz = |x|·|y|IIIz = sin(6(x + y))Vz = y3Problem 4) (10 points)Au usual, proj~v(~w) is the vector projection of ~w onto the vector ~v and comp~v(~w) is thescalar projection. Compute:a) (4, 5, 1) · (1, −1, 1)b) (−1, 1, 3) × (1, 1, 1)c) (2, 1, 3) · ((3, 4, 5) × (1, 1, 3)).d)~projh1,0,0ih7, 3, 2ie) |h0, 3, 4i| + comph1,0,0ih3, 4, 5iSolution:a) 4 − 5 + 1 = 0 .b)(−2, 4, −2) .c) ((3, 4, 5) × (2, 1, 3)) = (7, −4, −1) and the result is 7 .d) This is a vector projection:(7, 0, 0) .e) 5 + 3 = 8 .Problem 5) (10 points)8We want to fin d the distance between the point P = (5, 0, −5) and the cylinder which hasan axes going through the points A = (1, 1, 1) and B = (0, 2, 1) and radius 1. To do so:a) (4 points) Find first a parameterization ~r (t) = Q + t~v of the line.b) (4 points) Find the distance between P and the line.c) (2 points) Now find the distance between P and the cylinder.Solution:a) ~v =~AB = (h0, 2, 1i − h1, 1, 1i) = h−1, 1, 0i is in the line. Because we can takeQ = A, a parameterization is ~r(t) = h1, 0, −1i + th−1, 1, 0i. We can write it also as~r(t) = h1 − t, t, (−1)i . (note that there are other parametrizations which are ok too).b) We use the distance formula:|~AP ×~v||~v|=|h4, −1, −6i × h−1, 1, 0i||h−1, 1, 0i|=|h6, 6, 3i|√2.The answer is9/√2 .c) Because the distance to the cylinder is 1 less, the final answer is9/√2 − 1 .Problem 6) (10 points)The angle between a line and a plane is defined as π/2 − α, if α is the angle between thenormal vector to the plane and a vector in the line.Find the angle between the plane x + y − z/2 = 1 and the line ~r(t) = h1, 1, 1i + th1, 1, 3i.9Solution:The normal vector to the plane is ~n = h1, 1, −1/2i. We havecos(α) =|h1, 1, 3i · h1, 1, −1/2i|||h1, 1, 3i| · |h1, 1, −1/2i|= 1/(3√11 .So that α = arccos(1/3√11) and the answer is π/2 − arccos(13√11) .Problem
View Full Document
Unlocking...