text: # M 408D Lecture 2: HW 02-solutions
  
  60 views13 pages
  Published on 16 Feb 2020
  School
  UT-Austin
  Department
  Mathematics
  Course
  M 408D
  Professor
  oliver (smo998) – HW 02 – rusin – (52850) 1
  This print-out should have 20 questions.
  Multiple-choice questions may continue on
  the next column or page – ﬁnd all choices
  001 10.0 points
  If ais a vector parallel to the xy-plane and
  bis a vector parallel to k, determine ka×bk
  when kak= 3 and kbk= 2.
  1. ka×bk= 0
  2. ka×bk=3
  3. ka×bk= 3
  4. ka×bk= 6 correct
  5. ka×bk=6
  6. ka×bk=32
  7. ka×bk= 32
  Explanation:
  For vectors aand b,
  ka×bk=kakkbksin θ
  when the angle between them is θ, 0 θ < π.
  But θ=π/2 in the case when ais parallel
  to the xy-plane and bis parallel to kbe-
  cause kis then perpendicular to the xy-plane.
  Consequently, for the given vectors,
  ka×bk= 6 .
  keywords: cross product, length, angle,
  002 10.0 points
  A line passes through the point P(2,3,3)
  and is perpendicular to the plane
  x+ 4y+ 3z= 6 .
  At what point Qdoes intersect the xy-
  plane?
  1. Q(3,7,0)
  2. Q(3,0,1)
  3. Q(1,1,0) correct
  4. Q(7,0,3)
  5. Q(1,0,7)
  6. Q(1,1,0)
  Explanation:
  Since the xy-plane is given by z= 0, we
  have to ﬁnd an equation for and then set
  z= 0.
  Now a line passing through a point
  P(a, b, c) and having direction vector vis
  given parametrically by
  r(t) = a+tv,a=ha, b, c i.
  But for , its direction vector will be parallel
  to the normal to the plane
  x+ 4y+ 3z= 6 .
  Thus
  a=h2,3,3i,v=h1,4,3i,
  and so
  r(t) = h2 + t, 3 + 4t, 3 + 3ti.
  In this case, z= 0 when t=1. Conse-
  quently, intersects the xy-plane at
  Q(1,1,0) .
  keywords: line, parametric equations, direc-
  tion vector, point on line, intercept, coordi-
  nate plane
  003 10.0 points
  Which one of the following shaded-regions
  in the plane consists of all points whose polar
  coordinates satisfy the inequalities
  0r < 3,1
  6πθ7
  12π?
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  oliver (smo998) – HW 02 – rusin – (52850) 2
  1.
  2 4
  2.
  2 4
  3.
  2 4
  4.
  2 4
  correct
  5.
  2 4
  6.
  2 4
  Explanation:
  Using the deﬁnition of polar coordinates
  (r, θ), we see that the region deﬁned by the
  inequalities
  0r < 3,1
  6πθ7
  12π
  is
  2 4
  keywords: polar coordinates, inequalities, po-
  lar graph,
  004 10.0 points
  Find a polar equation for the curve given
  by the Cartesian equation
  4y2=x .
  1. r= 4 csc θtan θ
  2. 4r= sec θtan θ
  3. 4r= sec θcot θ
  4. 4r= csc θcot θcorrect
  5. r= 4 csc θcot θ
  6. r= 4 sec θtan θ
  Explanation:
  We have to substitute for x, y in
  4y2=x
  using the relations
  x=rcos θ , y =rsin θ .
  In this case the Cartesian equation becomes
  4r2sin2θ=rcos θ .
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  oliver (smo998) – HW 02 – rusin – (52850) 3
  Consequently, the polar form of the equation
  is
  4r= csc θcot θ .
  005 10.0 points
  Use the graph in Cartesian coordinates
  r
  θ
  π2π
  of ras a function of θto determine which
  one of the following is the graph of the corre-
  sponding polar function?
  1.
  2.
  3.
  4.
  5. correct
  6.
  Explanation:
  As the Cartesian graph of rnever crosses
  the θ-axis, there is no value of θin [0,2π]
  at which r= 0; in particular, therefore, the
  polar graph never passes through the origin.
  This eliminates 3 of the graphs, leaving only
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