Mathematics 2210−1,2
EXAM I SOLUTIONS
K. M. Golden
(20 points)
Fall 2015
1. (a) Let ~u = (3, 1) and ~v = (1, 2). Find ~u + 2~v and illustrate the scalar
multiplication and vector addition with a diagram in the plane,
showing ~u, ~v , 2~v and the resultant vector ~u + 2~v .
Answer: ~u + 2~v = (3, 1) + (2, 4) = (5, 5). The resultant vector
points from the origin to the head of 2~v , after translating 2~v so
that its tail is located at the head of ~u.
(b) Let ~a = (1, 1). Find a vector ~b which is orthogonal to ~a, and
verify that they are perpendicular.
Answer: ~b = (t, −t) for any t ∈ R, since (1, 1) · (t, −t) = t − t = 0
for any t ∈ R.
(20 points)
2. Let ~u = (2, 1) and ~v = (1, 2). Calculate the following. Be sure to show
all of your work.
(a) ~u · ~v
Answer: (2)(1)+(1)(2) = 4.
(b) ~u × ~v
Answer: 0i + 0j + 3k
(c) Find the angle θ between ~u and ~v .
Answer: θ = arccos( 54 ), 36.87 degrees or .6435 radians
~ = (0, 0), P~ =
(d) Determine the area of the triangle with vertices O
~ = (1, 2).
(2, 1), and Q
Answer: Area = 12 ||(2, 1, 0) × (1, 2, 0)|| = 21 (3) = 32 .
(10 points)
3. Consider the following vectors: ~a = (1, 1), ~b = (−1, 1).
1
(a) Multiply each vector by an appropriate scalar in order to normalize its length (that is, make each vector unit length), so that the
set {~e1 , ~e2 } forms an orthonormal basis for R2 .
Answer: ~e1 = ( √12 , √12 ), ~e2 = (− √12 , √12 )
(b) Expand the vector (2, 4) as a linear combination of the new orthonormal basis vectors {~e1 , ~e2 }.
Answer: (2, 4) =
(15 points)
√6 ~
e
2 1
+
√2 ~
e
2 2
4. Find the equation of the plane which contains the vectors ~u = (2, 1, −1)
and ~v = (0, 3, −1) and the point (0, 1, 0).
Answer: 2x + 2(y − 1) + 6z = 0 (or 2x + 2y + 6z = 2)
(20 points)
5. Projectile Motion. Determine the position ~r(t) = (x(t), y(t)) of a
projectile fired from the origin with an initial speed of 50 f/s at an angle of 45◦ . Be sure to show all your work, not just the final formulas.
(Hint: start with the the fact that the acceleration due to gravity is
~a(t) = −32j, then integrate Newton’s second law twice.)
50
t, −16t2 +
Answer: ~r(t) = ( √
2
(15 points)
50
√
t)
2
6. The position of a particle in motion in the plane at time t is given by:
~r(t) = a cos(t) i + a sin(t) j.
(a) Find T~ (t), the unit tangent vector to ~r(t), and determine the
speed of the particle as a function of time.
Answer: T~ (t) =
~
r0 (t)
||~
r0 (t)||
sin(t),a cos(t))
= √(−a
2
2
2
2
a sin (t)+a cos (t)
= (− sin(t), cos(t)).
Speed is ||~r0 (t)|| = a, as we found above in our calculation of T~ (t).
(b) Find the curvature κ at any point ~r(t), t ∈ R, along the planar
curve determined by the particle’s motion.
Answer: κ =
||T~ 0 (t)||
||~
r0 (t)
=
||(− cos(t),− sin(t))||
a
2
= a1 .
(c) Find a parameterization for the unit helix in R3 , and sketch its
trajectory for 0 ≤ t ≤ 4π.
Answer: α
~ (t) = (cos(t), sin(t), t)
3