MATH 265 Section A
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Spring 2005
EXAM 1
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1. (30 points) Let ~u =< 2, 1, −5 > and ~v =< 6, −2, 3 >. Calculate
(a) 2~u − 3~v
(b) A unit vector in the direction of ~v
(c) ~u · ~v
(d) ~u × ~v
(e) The angle between ~u and ~v
(f) The projection of ~v onto ~u
Solution: a) < −14, 8, −19 >, b) < 6/7, −2/7, 3/7 >,
c) −5, d) < −7, −36, −10 >,
e) cos θ = − 7√530 , so θ ≈ 1.7, f) < −1/3, −1/6, 5/6 >
2. (15 points) Find the equation of the plane containing the three points
P = (1, 2, 1), Q = (2, −4, 5) and R = (3, 0, −1).
Solution: 2x + y + z = 5
3. (20 points) Let ~r(t) =< t, 2t, t2 >
(a) Find the velocity, acceleration and unit tangent vectors.
(b) Show that the curve passes through the point P0 = (2, 4, 4) and find
the tangent line to the curve at P0 .
< 1, 2, 2t >
Solution: a) ~v (t) =< 1, 2, 2t >, ~a(t) =< 0, 0, 2 >, T~ (t) = √
5 + 4t2
y−4 z−4
b) ~r(2) =< 2, 4, 4 >; x − 2 =
=
2
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MATH 265 Section A
Spring 2005
EXAM 1
1
1
4. (10 points) Compute the curvature of y = at the point (2, ). What
x
2
is the radius of curvature at this point?
16
173/2
Solution: κ = 3/2 and radius of convergence is
16
17
5. (10 points) A line passes through P0 = (1, 0, −2) perpendicular to the
plane 3x − y + 5z = 1.
(a) Find the equation of the line in parametric form.
(b) Find the coordinates of the point at which the line intersects the xy
plane.
Solution: a) x = 1 + 3t, y = −t, z = −2 + 5t, b) (11/5, −2/5, 0)
6. (15 points) The vector function ~r(t) =< t−sin t, 1−cos t > is the position
vector for a cycloid.
(a) Find a point on the curve at which the speed is zero.
(b) Find a point on the curve at which the tangent vector is horizontal.
(c) What is a unit normal vector at the point you found in part b)?
Solution: a) (0, 0) (or (2π, 0), (4π, 0) etc.), b) (π, 2) (or (3π, 2), (5π, 2)
etc.), c) (0, ±1)
Some points to emphasize: Question asks for the point on the curve,
not the value of t; the zero vector has no well defined direction, so
in part b) it is necessary to find points where the first component
is nonzero; a zero speed therefore generally means that there is no
~ is the hard way to do part c), the easy
tangent vector; computing N
way is just to realize that if the tangent vector is horizontal then any
normal vector is vertical.
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