Math 101C Ch. 11 Test
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1. Find the equation of the line in vector form r(t) that passes through (−2, −4, 1) and (5, −1, 1).
Solution:
r(t) = h−2 + 7t, −4 + 3t, 1i
2. Find a vector of length 7 in the direction of h6, 8i.
Solution:
7h6, 8i
h21, 28i
√
=
5
62 + 82
3. Find an expression for the angle between the vectors h2, 3, −1i and h−2, 1, 5i.
Solution: Using u = h2, 3, −1i and v = h−2, 1, 5i,
θ = arccos
u·v
−6
= arccos √ √
||u|| ||v||
14 30
4. (10 points) Find a vector orthogonal to the vectors h2, 3, −1i and h−2, 1, 5i.
Solution:
h2, 3, −1i × h−2, 1, 5i = h16, −8, 8i
5. (10 points) Determine the intersection of the line x = 1 + 4t, y = 2 − t, z = −1 + t and the plane
2x + 3y − z = 17.
Solution:
2(1 + 4t) + 3(2 − t) − (−1 + t) = 17 −→ t = 2 which gives the point (9, 0, 1).
6. (10 points) Determine the intersection point of the lines
x = 1 + 4t, y = 2 + 3t, z = −1 + t
x = t, y = 2 + t, z = 4 + 2t
and
Solution: Using the parameter s for the second equation and solving gives t = −1 and s = −3. This
give the point (−3, −1, −2).
7. (10 points) If a(t) = e2t , 3t2 and v(0) = h3, 4i, find v(t).
Solution:
v(t) =
Z D
e2t , 3t2
E
dt
1 2t
=
e + C1 , t3 + C2
2
1 2t 5 3
=
e + , t +4
2
2
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Math 101C
Ch. 11 Test
Page 2 of 3
8. (10 points) Given the equation of the cycloid
r(t) = h2πt − sin 2πt, 1 − cos 2πti
1
.
find an expression for T
4
Solution: Starting with r0 (t) = h2π − 2π cos 2πt, 2π sin 2πti. we have r0 (1/4) = h2π, 2π i. This gives
r0 (t)
h1, 1i
T(1/4) = 0
= √ .
||r (t)||
2
9. (10 points) A golfball is hit east down a fairway with an initial velocity of h50, 0, 30i m/s. A crosswind
blowing to the south produces an acceleration of −0.8m/s2 . Find the position function for the ball.
Solution: (11.7.39)
a(t) = h0, −0.8, −9.8i
v(t) =
r(t) =
Z
Z
a(t) dt = h50, −0.8t, 30 − 9.8ti
D
E
v(t) dt = 50t, −0.4t2 , 30t − 4.9t2
10. (10 points) A satellite is launched from the surface of the earth and follows the path
r(t) = 3 cos 6πt, 3 sin 6πt, 50(1 − e−t
Write the function for the speed of the satellite and determine the speed as time approaches infinity.
Solution: The velocity of the satellite is given by
v(t) = −18π sin 6πt, 18π cos 6πt, 50e−t
The speed is given by
||v(t)|| =
q
(18π )2 + 2500e−2t
As t −→ ∞, ||v(t)|| = 18π.
11. (10 points) mark each question as “True” or “False”.
(a) Torque is a vector. True
(b) Work is a vector. True
False
False
(c) Given two non-zero vectors u and v, u × v = v × u. True
False
(d) Given two non-zero vectors u and v, u × v gives the area of the parallelogram formed by the
vectors. True False
(e) As a particle follows a downward path, its speed is negative. True
False
12. (25 points) Sketch the graphs of the following. Be sure to draw an arrow on the curve to indicate the
direction of the path.
π
π
(a) r(t) = hcos t, sin ti on 0 ≤ t ≤ 1
2
2
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Math 101C
Ch. 11 Test
Page 3 of 3
Solution: This gives a quarter-circle starting at (1, 0) and ending at (0, 1).
π
π
t, − cos ti on 0 ≤ t ≤ 1
2
2
Solution: This gives a quarter-circle starting at (0, −1) and ending at (−1, 0).
(b) r(t) = h− sin
π
π
t, 3 sin t, ti on 0 ≤ t ≤ 8
2
2
Solution: This is a spiral, rotating counterclockwise, passing through (3, 0, 0), (0, 3, 1), (−3, 0, 2),
(0, −3, 3) and ending at (3, 0, 4) after one rotation. The second rotation ends at (3, 0, 8).
(c) r(t) = h3 cos
π
π
t, 3 sin t, 8 − e−t i on 0 ≤ t ≤ 8
2
2
Solution: This is a spiral, rotating counterclockwise, that makes two revolutions. The first revolutions starts at (3, 0, 7) and the the second rotation ends at (3, 0, 8). The second revolution should
be more closely spaced vertically than the first revolution.
(d) r(t) = h3 cos
π
π
t, sin t, 3i on 0 ≤ t ≤ 8
2
2
Solution: This is a circle with radius 1, centered around the z axis, and in the plane z = 3. Note
that it makes two revolutions.
(e) r(t) = hcos
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