Math 3C — Homework #11
Solutions
Laney College, Spring 2014
Fred Bourgoin
Section 17.1
4. Find a parametrization for the curve.
Solution. The easiest thing to do is to start with the usual parameterization of the unit circle: x = cos θ, y = sin θ. Luckily that gives
the correct orientation (counterclockwise). Now, shift the center by
adding 1 to y: x = cos θ, y = 1 + sin θ. Finally, set the limits for θ so
that it starts and ends where it should:
x = cos θ ,
y = 1 + sin θ ,
π ≤ θ ≤ 2π .
Other parameterizations are possible.
10. Find parametric equations for the line in the direction of 5 j + 2 k and
through (5, −1, 1).
Solution. Call the point P and the vector v . Then a parameterization
−−→
is r(t) = OP + t v , or
r(t) = 5i − j + k + t(5 j + 2 k) .
17. Find parametric equations for the line intersecting the x-axis at x = 3
and the z-axis at z = −5.
Solution. Since we’re given two points, (3, 0, 0) and (0, 0, −5), we can
make the vector v = 3i + 5 k and proceed as above:
r(t) = 3i + t(3i + 5 k) .
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22. Find a parametrization for the circle of radius 2 parallel to the xyplane, centered at the point (0, 0, 1), and traversed counterclockwise
when viewed from below.
Solution. How about this?
r(t) = 2 cos ti + 2 sin t j + k ,
0 ≤ t ≤ 2π .
31. Find a parametrization for the curve y = 4 − 5x4 through the point
(0, 4, 4), parallel to the xy-plane.
Solution. Parameterize the curve in the xy-plane, then shift it up 4
units:
r(t) = ti + (4 − 5t4 ) j + 4 k .
√
40. Find a parametric equation for the graph of y = x from (1, 1) to
(16, 4).
√
Solution. r(t) = ti + t j, 1 ≤ t ≤ 16.
52. Find an equation for the plane containing the point (2, 3, 4) and the
line x = 1 + 2t, y = 3 − t, z = 4 + t.
Solution. A vector in the direction of the line (and thus parallel to
the plane) is v = 2i − j + k. A point on the line is (1, 3, 4). With
the other point, we make the vector w
= i, also parallel to the plane.
Now, v × w
= j + k is perpendicular to the plane, so
0(x − 2) + 1(y − 3) + 1(z − 4) = 0 ,
or
z = 7−y.
Section 17.2
In Exercises 9 and 12, find the velocity v (t) and the speed v (t). Find any
times at which the particle stops.
7. x = 3t2 , y = t3 + 1
Solution. Since r(t) = 3t2i + (t3 + 1) j,
v (t) = r (t) = 6ti + 3t2j and v (t) = (6t)2 + (3t2 )2 = 3|t| 4 + t2 .
The particle stops when v (t) = 0; that is, when t = 0.
12. x = 3 sin2 t, y = cos t − 1, z = t2
Solution. Since r(t) = (3 sin2 t)i + (cos t − 1) j + t2 k,
v (t) = (6 sin t cos t)i − (sin t) j + 2t k
and v (t) = 36 sin2 t cos2 t + sin2 t + 4t2 .
2
The particle stops when t = 0 (by inspection because there is no way
to solve the equation algebraically).
22. A particle passes through the point P = (5, 4, −2) at time t = 4,
moving with constant velocity v = 2i − 3 j + k. Find a parametric
equation for its motion.
Solution. Proceed as usual, but delay the motion by 4 seconds:
r(t) = 5i + 4 j − 2 k + (t − 4)(2i − 3 j + k) .
29. A stone is thrown from a rooftop at time t = 0 seconds. Its position
at time t is given by
r(t) = 10ti − 5t j + (6.4 − 4.9t2 ) k .
The origin is at the base of the building, which is standing on flat
ground. Distance is measured in meters. The vector i points east, j
points north, and k points up.
(a) How high is the rooftop above the ground?
Solution. Since the stone is thrown from the rooftop, that’s
where it is when t = 0: r(0) = 6.4 k. The rooftop is 6.4 m above
the ground.
(b) At what time does the stone hit the ground?
Solution. The stone hits the ground when the k component is
0:
=⇒
t = ± 6.4/4.9 = ±8/7 ≈ ±1.14 s.
6.4 − 4.9t2 = 0
Of course, only the positive answer makes sense.
(c) How fast is the stone moving when it hits the ground?
Solution. First, first the velocity:
v (t) = r (t) = 10i − 5 j + (−9.8t) k .
√
Then find the speed: v (t) = 125 + 96.04 t2 . So, when the
stone hits the ground, its speed is
v (8/7) = 125 + 96.04 (8/7)2 ≈ 15.8 m/s.
(d) Where does the stone hit the ground?
40 Solution. r(8/7) = 80
7 i − 7 j.
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(e) What is the stone’s acceleration when it hits the ground?
Solution. The acceleration vector is
a(t) = v (t) = −9.8 k .
The acceleration is always 9.8 m/s2 (because the stone is subject
to only one force: that of gravity).
35. The base of a 20-meter tower is at the origin; the base of a 20-meter tree
is at (0, 20, 0). The ground is flat and the z-axis points upward. The
following parametric equations describe the motion of six projectiles
each launched at t = 0 in seconds.
(I) r(t) = (20 + t2 ) k
(II) r(t) = 2t2j + 2t2k
←−
mistake in the homework
←−
mistake in the homework
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(III) r(t) = 20 i + 20 j + (20 − t ) k
(IV) r(t) = 2t j + (20 − t2 ) k
(V) r(t) = (20 − 2t)i + 2t j + (20 − t) k
(VI) r(t) = ti + t j + t k
(a) Which projectile is launched from the top of the tower and goes
downward? When and where does it hit the ground?
Solution. This is (IV) since r(0) = 20 k and the k component
decreases as t increases. The√projectile hits the ground√when
2 = 0; that is, when t = 2 5. It hits the ground at r (2 5) =
20−t
√
4 5 j.
(b) Which projectile hits the top of the tree? When? From where is
it launched?
√
Solution. This is (II)
√ because r( 10) = 20 j + 20 k. Clearly, this
happens when t = 10. The projectile is launched from r(0) = 0,
the base of the tower.
(c) Which projectile is not launched from somewhere on the tower
and hits the tree? Where and when does it hit the tree?
Solution. This is (V) because r(0) = 20i + 20 k, not on the
tower; and r(10) = 20 j + 10 k, which is 10 m up the tree when
t = 10 s.
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Section 17.3
20. For each description of a vector field in (a)–(d), choose one or more of
the vector fields I–IX.
(a) Pointing radially outward, increasing in length away from the
origin.
Solution. III.
(b) Pointing in a circular direction around the origin, remaining the
same length.
Solution. II.
(c) Pointing towards the origin, increasing in length farther from the
origin.
Solution. IV.
(d) Pointing clockwise around the origin.
Solution. II and V.
I.
IV.
VII.
xi + y j
x2 + y 2
II.
−y i + x j
x2 + y 2
III.
r
−r
V.
−y i + x j
VI.
y i − x j
r
r3
IX.
−
y i + x j
VII.
r
r3
21. Each vector field in Figures (I)–(IV) represents the force on a particle
at different points in space as a result of another particle at the origin.
Match up the vector fields with the descriptions below.
(a) A repulsive force whose magnitude decreases as distance increases,
such as between electric charges of the same sign.
Solution. (IV).
(b) A repulsive force whose magnitude increases as distance increases.
Solution. (III).
(c) An attractive force whose magnitude decreases as distance increases, such as gravity.
Solution. (I).
(d) An attractive force whose magnitude increases as distance increases.
Solution. (II).
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Section 17.4
16. Match the vector fields (a)–(f) with their flow lines (I)–(VI). Put arrows on the flow lines indicating the direction of flow.
Solution.
(a) (III).
(b) (I).
(c) (II).
(d) (V).
(e) (VI).
(f) (IV).
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Section 17.5
7. Describe in words the object parameterized by x = 3 cos θ, y = 2 sin θ,
z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 7.
Solution. This is a elliptical cylinder centered around the z-axis,
resting on the xy-plane and 7 units high.
17. Parameterize the plane through (5, 1, 4) with normal vector i+2 j+3 k.
Solution. We need to find two non-parallel vectors in the plane, that
is, two vectors perpendicular to i + 2 j + 3 k. How about v = 2i − j
and w
= 3i − k? Then the plane can be parameterized by
r(s, t) = 5i + j + 4 k + s(2i − j) + t(3i − k) .
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28. Find parametric equations for the sphere (x−a)2 +(y −b)2 +(z −c)2 =
r2.
Solution. This is a sphere of radius r centered at (a, b, c). Let’s use
spherical coordinates, remembering to shift the center:
r(φ, θ) = (a + r sin φ cos θ)i + (b + r sin φ sin θ) j + (r cos φ) k .
30. Find parametric equations for the cone x2 + y 2 = z 2 .
Solution. One possibility:
r(t, θ) = (t cos θ)i + (t sin θ) j + t k .
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