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HARVARD MATH 21A - FIRST HOURLY PRACTICE II

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7/8/2010 FIRST HOURLY PRACTICE II 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, 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 ~v = h1, 3, 5i is perpendicular to the plane x + 3y + 5z = 1.2)T FThe set of points which satisfy x2− y2+ z2− 2z + 1 = 0 forms a doublecone.3)T FThe set of points in R3which have distance 1 from a point form a cylinder.4)T FThe surface −x2+ y2+ z2= 1 is called a one-sheeted hyperboloid.5)T FThe two vectors h2, 3, 0i and h6, −4, 5i are orthogonal to each other.6)T FTwo nonzero vectors are parallel if and only if their dot product is 0.7)T FThe cross product is asso ciative: ~u ×(~v × ~w) = (~u ×~v) × ~w.8)T FEvery vector contained in the line ~r(t) = h1 + 2t, 1 + 3t, 1 + 4ti is parallelto the vector (2, 3, 4).9)T FThe linex−12=y−13=z−14hits the plane 2x + 3y + 4z = 9 at a right angle.10)T FTwo planes ax + by + cz = d and ux + vy + wz = e intersect in a line if|ha, b, ci× hu, v, wi| > 0.11)T FThe equations x − 2 = y − 3 = z − 4 describe a line which contains thevector h1, 1, 1i.12)T FIn spherical coordinates, the equation cos(φ) = sin(φ) defines a cone.13)T FA point with spherical coordinates (ρ, θ, φ) = (1, π/2, π/4) has cylindercoordinates (r, θ, z) = (1/√2, π/2, 1/√2).14)T FIf in rectangular coordinates, a point is given by (1, 0, −1), then its sphericalcoordinates are (ρ, θ, φ) = (√2, π/2, 3π/4).15)T FThe volume of a parallelepiped spanned by (1, 0, 0), (0, 1, 0) and (0, 1, 1) isequal to 2.16)T FThe vector projection of the vector (2, 4, 5) onto the vector (1, 1, 0) is(3/2, 3/2, 0).17)T FIf g(x, y, z) = 0 is a surface given implicitly, then ~r(u, v) =hu, v, g(u, v, g(u, v, 1))i is a parametrization of the surface.18)T FIf z = g(x, y) is a graph then ~r(u, v) = hu, v, g(u, v)i is a parameterizationof the surface.19)T FThe distance from the point P = (1, 1, 1) to the x axes is√2.20)T FThe distance between the point P = (1, 1, 1) and the xy plane is√2.Total2Problem 2) (10 points)a) (3 points) Match the contour maps with the corresponding functions f(x, y) of two variables.Enter O if no figure matches. No justifications are needed.I II II I IVEnter I,II,III,IV or O Function f(x, y)f(x, y) = x cos(y)f(x, y) = 3x2+ 4y2f(x, y) = cos(x)Enter I,II,III,IV or O Function f(x, y)f(x, y) = x2− y2f(x, y) = |xy|f(x, y) = |x| − |y|b) (4 points) Match the quadrics with the functions. No justifications are necessary.abcdEnter a,b,c,d here Equationx + y2− z2− 1 = 0y2− z2+ 1 = 0Enter a,b,c,d here Equationx2+ y2− z2+ 1 = 0x2+ y2− z2− 1 = 0c) (3 points) Match the equation with their graphs. No justifications are necessary.A B CEnter A,B,C here Equation Enter A,B,C here Equation Enter A,B,C here Equationz = e−x2−y2z = x cos(x) z = x − y3Problem 3) (10 points)a) (5 points) Surfaces z = f(x, y) which are graphs can be written implicitly as g(x, y, z) = 0,parametrized as ~r(u, v). For example, z = log(xy) is given by g(x, y, z) = 0 with g(x, y, z) =z −log(xy) or parametrized as ~r(u, v) = hu, v, log(uv)i. Complete the following table by fillingin the choices A − J below. No justifications are needed in this problem.z = f(x, y) for g(x, y, z) = 0 ~r(u, v) = hx(u, v), y(u, v), z(u, v)ihv cos(u), v sin(u), vix + y − 2z = 0f(x, y) = x2− y2hcos(u) sin(v), sin (u) sin(v), cos(v)i, v < π/2z − sin(xy) = 0f(x, y) = xA) f(x, y) = x − yB) f(x, y) = x2+ y2C) z − x2− y2D) h1 + u + v, 1 + u − v, uiE) z − x2+ y2.F) f(x, y) =√1 − x2− y2G) hu, v, u2− v2iH) x2+ y2+ z2− 1 = 0I) hu, v, u iJ) z − x = 0b) (5 points)Quantity Check if it depends on parametrization of ~r Is a vectorCurvature of ~r(t)Arc length of ~r(t) from 0 to 1Acceleration of ~r(t)Jerk of ~r(t)Speed of ~r(t)Unit tangent of ~r(t)Normal of ~r(t)Binormal of ~r(t)∇f(~r(t)) ·~r′(t)~r′(t) ×~r′′(t)Problem 4) (10 points)A billard ball starts at A = (1, 1, 0), travels along the vector ~u = h2, −2, 0i to other point Bwhere it bounces off an other ball. It travels from there along the vector ~v = h−3, 4, 0i to athird point C, where it bounces off a wall, rolling along the vector ~w = h1, 1, 0i to its finaldestination D. In other words, you know A,~AB = ~u,~BC = ~v and~CD = ~w.a) (5 points) What are the coordinates of the point D?4b) (5 points) Find the total distance traveled by the ball along the path A, B, C, D.Problem 5) (10 points)a) (5 points) Find the symmetric equation of the line which contains the point P = (3, 4, 1)and the point Q = (5, 5, 5).b) (5 points) What is the equation of the plane perpendicular to the line in a) which passesthrough the point P = (3, 4, 1)?Problem 6) (10 points)We look at a polyhedron which has the shape of a scaled octahedron. Its vertices are A =(1, 1, 0), B = (−1, 1, 0), C = (−1, −1, 0), D = (1, −1, 0), E = (0, 0, 1), F = (0, 0, −1).a) (5 points) Parametrize the line L passing through A,E and the line K passing through B,F.b) (5 points) Find the distance between these two lines L and K.Problem 7) (10 points)The p lane 3x + y + 2z = 6 cuts out a triangle T from the octant x > 0, y > 0, z > 0. Thistriangle as well as the coordinate planes bound a solid E in space.a) (5 points) Find the area of this triangle T = ABC.b) (5 points) The volume of the solid E is known to be 1/6’th of the volume of the parallelepipedspanned by~OA,~OB,~OC. Find the volume of E.5Problem 8) (10 points)a) (2 points) Find the dot product of h3, 4, 5i and h3, 2, 1i.b) (2 points) Find the vector projection of the vector h4, 5, 6i onto the vector h3, 3, 3i.c) (2 points) Do the vectors h1, −1, −1i and h4, −5, 6i form an acute or obtuse …


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HARVARD MATH 21A - FIRST HOURLY PRACTICE II

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