Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 70 docx

20.7 A Brief Introduction to Vectors 671
20.7 A BRIEF INTRODUCTION TO VECTORS
Questions of displacement and velocity have figured prominently in our studies. When we
talk about displacement, we ask
“how far?” and “in what direction?”
When we talk about velocity, we ask
“how fast?” and “in what direction?”
We’ve been answering the question “in what direction?” by using a plus or a minus sign.
This has restricted the type of motion we describe. Our objects are going up or down; our
boats are headed east or west; our cars travel on long straight highways; our trains are either
inbound or outbound. If we wander along a meandering path, we can report our velocity
only with reference to some benchmark point on the path; we’re either heading toward it or
away from it.
We can free ourselves from this world of dichotomy by using an arrow to indicate
direction. In fact, we can have the arrow answer both the amount and direction questions by
having the length of the arrow indicate “how far” or “how fast” and the direction the arrow
is pointing indicate the direction of the displacement or motion. An arrow used in this way
is called a vector.
Defining Vectors
Definition
A vector has two defining characteristics:

length (or magnitude) and
direction.
A vector is often denoted by

v, a boldface letter with an arrow above it, and represented
by an arrow. The length of

v is denoted by |


v|.
Vectors can be used in a large variety of situations; for example, vectors can be used
to model velocity, force, and displacement. A velocity vector is a vector whose length
represents speed and whose direction indicates the direction of motion. A force vector is
a vector whose magnitude corresponds to the magnitude of the force and whose direction
indicates the direction of application of the force. We can give a similar interpretation for
a displacement vector. We refer to the head of the vector (the tip of the arrow) and the tail
of the vector (the other end). A vector represented by an arrow whose tail is at A = (a
1
, a
2
)
and whose head is at B = (b
1
, b
2
) can be denoted by
−−−→
AB.
672 CHAPTER 20 Trigonometry—Circles and Triangles
vector
head
v
→
tail
v
→
A = (a
1
, a

2
)
B = (b
1
, b
2
)
|b
1
– a
1
|
|b
2
– a
2
|
(i) (ii)
Figure 20.46
Using the Pythagorean Theorem we calculate the length of
−−−→
AB as

(b
2
− a
2
)
2
+ (b

1
− a
1
)
2
.
Two vectors,

u and

v are equivalent if they have the same magnitude and the same direction.
For example, all the vectors drawn in Figure 20.47 are equivalent.
2
1
1
–22
2
–11
x
y
v
→
v
→
v
→
Figure 20.47
The simplest way of determining that these vectors are all equivalent is to notice that
each is constructed (from tail to head) by a displacement of −2 units in the horizontal
direction followed by a displacement of +1 unit in the vertical direction. We say that each

of these vectors has a horizontal component of −2 and a vertical component of +1. The
horizontal and vertical components of a vector together completely determine the vector.
They determine its direction, and its magnitude is given by

(horizontal component)
2
+ (vertical component)
2
.
Horizontal and Vertical Components of Vectors
Consider the vector
−−− →
OP , where O = (0, 0) and P is a point on the unit circle. The length
of
−−− →
OP is 1. Let’s suppose that
−−− →
OP makes an angle of θ with the positive x-axis, where
θ ∈ [0, 2π] and θ is swept out counterclockwise from the positive x-axis. The coordinates
of P are (cos θ, sin θ), where cos θ is the horizontal component of
−−− →
OP and sin θ is the
vertical component of
−−− →
OP .
20.7 A Brief Introduction to Vectors 673
y
x
P
1

1
O
θ
Figure 20.48
Given any vector

v of length L we can position

v so that its tail is at the origin. Its head
lies somewhere on a circle of radius L centered at the origin, as shown in Figure 20.49. Let
θ be the angle

v makes with the positive x-axis, where θ ∈ [0, 2π].
The horizontal component of

v = L cos θ, and
the vertical component of

v = L sin θ.
θ
y
x
L
v
Lsin θ
Lcos θ
→
Figure 20.49
If the tail of a vector is at the origin, the horizontal component is the x-coordinate of the
head and the vertical component is the y-coordinate of the head.

EXERCISE 20.10 D is the displacement vector for a turtle. Its horizontal component is +3 and its vertical
component is −4.
(a) How far has the turtle moved?
(b) If the turtle started at (1, 3), where did it end up?
ANSWERS
(a) 5 units (b) (4, −1)
674 CHAPTER 20 Trigonometry—Circles and Triangles
EXERCISE 20.11

v is the velocity vector at a certain instant for an object in motion. At this instant the object
is traveling southwest at 50 miles per hour. Let due east correspond to the positive x-axis.
What are the horizontal and vertical components of the velocity?
50
50
45°
v
→
horizontal component
vertical component
Figure 20.50
◆
EXAMPLE 20.15 A wheel 60 inches in diameter is oriented vertically and spinning steadily at a rate of 1
revolution every π/2 minutes. There’s a piece of gum stuck on the rim. Let
−−−−→
G(t) be the
position vector pointing from the center of the wheel to the wad of gum. At time t = 0, the
vector
−−−−→
G(t) is pointing in the same direction as the positive x-axis. Find a formula for h(t),
the vertical component of

−−−−→
G(t).
SOLUTION This is simply the Ferris wheel question formulated using different language.
30
t
G(t )
→
30 sin t
Figure 20.51
Since h(0) = 0, a formula of the form 30 sin(Bt) can be used. One revolution is completed
in π/2 minutes, so the period is π/2.
2π
B
= π/2 =>B=4.
h(t) = 30 sin(4t)
◆
The purpose of Example 20.15 is to show that we already know how to find the vertical
and horizontal components of a vector. Given a vector (such as a velocity vector, force
vector, or displacement vector), physicists, engineers, and architects often find it useful to
resolve
15
the vector into its vertical and horizontal components. The next example asks for
the resolution of a force vector into its horizontal and vertical components.
◆
EXAMPLE 20.16 A suitcase is being pulled along a horizontal airport floor. A force of 50 pounds is being
applied at an angle of 35
◦
with the horizontal.
(a) What is the component of the force in the direction of motion?
(b) What is the component of the force perpendicular to the direction of motion?

15
To “ resolve” a vector is to separate it into its constituent parts.
20.7 A Brief Introduction to Vectors 675
SOLUTION (a) Let x = the component of the force in the direction of motion.
cos 35
◦
=
x
50
x = 50 cos 35
◦
x = 50 cos

35
◦
π
180
◦

≈ 40.96
Approximately 40.96 pounds of force are being applied in the direction of motion.
x ≈ 40.96 lbs
35°
50 lbs
y ≈ 28.68 lbs
Figure 20.52
(b) Let y = the component of the force perpendicular to the direction of motion.
sin 35
◦
=

y
50
y = 50 sin 35
◦
y ≈ 28.68
Approximately 28.68 pounds of force are being applied perpendicular to the direction
of motion.
◆
What is the significance of what we have calculated? We’ve resolved the force into
two perpendicular components. If a force of (50 sin 35
◦
) pounds is applied upwards and
simultaneously a force of (50 cos 35
◦
) pounds is applied in the direction of motion, the sum
of these two forces is equivalent to the original force.
◆
EXAMPLE 20.17 Consider the trajectory made by some thrown or launched object. The object could be a
tennis ball, a javelin, a basketball, or a stone. The exact nature of the trajectory is of critical
importance to athletes and marksmen alike. If we neglect air resistance and spin and consider
only the force of gravity, then the trajectory of the object is determined by the initial velocity
of the object, given by the velocity vector

v
0
. Let’s denote the length of

v
0
by v

0
(with no
arrow above it). v
0
is the initial speed. We’ll denote the angle of launch by θ, where θ is in
the interval (0, π/2). For simplicity’s sake, we’ll assume the object is launched from ground
level. The only force acting on the object is the force of gravity, resulting in a downward
acceleration of g meters per second squared.
(a) Find the vertical position of the object at time t.
(b) When will the object hit the ground?
(c) How far away will the object be from its launching spot when it hits the ground?
676 CHAPTER 20 Trigonometry—Circles and Triangles
v
0
v
0
sin θ
θ
→
Figure 20.53
SOLUTION (a) To analyze the vertical position of the object at time t, we can restrict our attention to
the vertical component of velocity. We can express the vertical component of the initial
velocity,

v
0
,asv
0
sin θ.
Let v(t) be the vertical velocity of the object (in meters per second) at time t and let

s(t) be the vertical position (in meters) of the object at time t.
Let t = 0 be the moment of launch.
The vertical acceleration of the object is −g, where g is the gravitational constant;
therefore v

(t) =−g.
Our strategy is as follows:
Using information about acceleration, (the derivative of v(t)), find v(t).
Using information about velocity, (the derivative of s(t)), find s(t).
Let’s guess a function whose derivative is −g; we know that any other such function
differs from it by a constant.
16
The derivative of −gt is −g so
v(t) =−gt + C. We know that v(0) = v
0
sin θ; this information is the initial
condition that allows us to solve for the constant C.
v
0
sin θ =−g·0+C
v
0
sin θ = C (Keep in mind that v
0
sin θ is a constant! Both v
0
and θ are
fixed.)
v(t) =−gt + v
0

sin θ.
The velocity is the derivative of the displacement function; v(t) = s

(t).
s

(t) =−gt + v
0
sin θ
Let’s guess a function whose derivative is −gt + v
0
sin θ; we know that any other such
function differs from it by a constant.
s(t) =−g
t
2
2
+ (v
0
sin θ)t + C
1
We know that s(0) = 0; this information is the
initial condition that allows us to solve for the
constant C
1
.
0 = 0 + 0 + C
1
0 = C
Therefore s(t) =−g

t
2
2
+ (v
0
sin θ)t.
(b) The object hits the ground when s(t) = 0, so we must set s(t) equal to zero.
−g
t
2
2
+ (v
0
sin θ)t = 0 Solve for t.
16
See Appendix C for a proof of the Equal Derivatives Theorem.
20.7 A Brief Introduction to Vectors 677
t(−g
t
2
+ v
0
sin θ)= 0
t = 0or g
t
2
= v
0
sin θ
t = 0or t=

2v
0
sin θ
g
The object hits the ground
2v
0
sin θ
g
seconds after it is launched.
(c) To find the horizontal distance the object will travel we direct our attention to the
horizontal component of velocity. Unlike the vertical component of velocity (which
is affected by the force of gravity), the horizontal component of velocity is constant.
Therefore, the horizontal distance traveled is given by
(the horizontal component of velocity) · (time). The horizontal component of

v
0
is
v
0
cos θ.
v
0
v
0
cos θ
θ
→
Figure 20.54

The time the object travels is the time between launch and when it hits the ground:
2v
0
sin θ
g
seconds.
horizontal distance traveled = v
0
cos θ ·
2v
0
sin θ
g
=
2v
2
0
cos θ sin θ
g
So, for instance, if an object is launched at ground level at an angle of 30
◦
and with an
initial speed of 28 meters per second it will hit the ground after
2(28 m/sec) sin(π/6)
9.8m/sec
2
≈ 2.86
sec. It travels a horizontal distance of (28m/sec) cos(π/6) ·
2(28 m/sec) sin(π/6)
9.8m/sec

2
≈ 69.28
m.
◆
EXERCISE 20.12 Suppose an object is launched from the ground with an initial velocity of 96 feet per second
at an angle of 40
◦
. The gravitational constant is 32 ft/sec
2
. Answer questions (a), (b), and
(c) from the last example by working through all the steps using these concrete numbers.
ANSWERS (a) v(t) =−32t + 96 sin 40
◦
,sos(t) =−16t
2
+ (96 sin 40
◦
)t + 5.
s(t) ≈−16t
2
+ 61.7t + 5 in ft/sec
(b) The object hits the ground after about 3.936 seconds.
(c) ≈ 289.463 feet
Finding the Component of a Vector in an Arbitrary Direction
In many situations it is useful to find the component of a vector not in the horizontal and
vertical directions but in some other direction. For instance, a sailor might be interested
in the component of the wind’s velocity in the direction of his boat’s motion. If an object
678 CHAPTER 20 Trigonometry—Circles and Triangles
is being pulled up an incline we are interested in the component of the force applied in
the direction of the object’s motion. A swimmer swimming in a current is interested in the

component of the force of the water in her direction of motion.
The component of

v in the direction of

u gives a measure of how much of

v is in
the direction of

u. Below we give some examples. (Recall that |

v| denotes the length or
magnitude of

v.)

The component of

v
in the direction of

u

is



|


v| if

u and

v point in the same direction
−|

v| if

u and

v point in the opposite directions.
0if

uand

v are perpendicular.
We denote the component of

v in the direction of

u by v
u
. (It will be a number, not a
vector, so we don’t put an arrow over the v.) To visualize v
u
, begin by placing the tails of
the vectors

v and


u together. Let θ be the angle between

u and

v.We’ll measure θ as an
angle between 0 and π; there is no need to distinguish between its terminal and initial sides.
The direction of

u matters but its length does not. If the angle between

u and

v is acute,
then v
u
is positive; if the angle is obtuse, then

v is not in the direction of

u, and v
u
is negative.
You can visualize what is meant by the component of

v in the direction of

u in the examples
shown in Figure 20.55 by tilting your head so that


u looks horizontal and visualizing the
horizontal component of

v.
v
v
u

→
→→
v
→
v
→
v
→
u
→
u
→
u
→
u
→
→
v
u

→
v

u

→
v
u

→
(i )(ii)(iii)(iv)
(the component of
v in the direction of u)
is negative
Figure 20.55
In the examples shown in Figure 20.56 it is easier to tilt your head so that you’re letting

u be vertical and visualize the vertical component of

v.
v
→
v
→
v
→
→
→
u
→
u
→
u

→
v
u

→
v
u

→
v
u

→
(v)(vi)(vii)
here the component
of v in the direction
of u is negative
Figure 20.56
20.7 A Brief Introduction to Vectors 679
Defining and computing the component of

v in the direction of

u
The component of

v in the direction of

u, v
u

is defined as v
u
=|

v|cos θ,
where θ is the angle between

u and

v.
v
→
v
→
u
→
u
→
→
v
u

→
v
u

→
v
u

→
θ
θ
cos θ =
length of v
→
v
u

→
cos θ =
| v |
Figure 20.57
Observations
The definition of v
u
involves the length of

v but not the length of

u. This is in line with
our earlier remark that the length of

u is irrelevant; only the direction of

u matters.
If θ ∈ [0, π/2), then cos θ is positive; if θ ∈ (π/2, π], then cos θ is negative. This is
exactly what we want.
|


v| cos θ









has a maximum value of |

v| when θ = 0.


u and

v have the
same direction.

has a minimum value of −|

v|when θ = π.


u and

v have
opposite directions.


is zero when θ = π/2. (

u and

v are perpendicular.)
This is in agreement with our intuition.
◆
EXAMPLE 20.18 A suitcase is being pulled up a ramp that makes a 10
◦
angle with the horizontal. A force of
50 pounds is applied at an angle of 35
◦
with the horizontal. What is the component of the
force in the direction of motion?
SOLUTION Let

u be a vector in the direction of motion and

F be the force vector. |

F|=50. The angle
between

u and

F is 25
◦
.
Let x = the component of the force in the direction of motion.
x = 50 cos 25

◦
≈ 45.3
v
→
50
25°
10°
ramp
Figure 20.58
Approximately 45.3 pounds of force are being applied in the direction of motion.
◆
680 CHAPTER 20 Trigonometry—Circles and Triangles
EXERCISE 20.13 The angle between vectors

u and

v is 2π/3. |

u|=3and |

v|=8.
(a) Find the component of

v in the direction of

u. Give an exact answer. Accompany your
answer by a sketch.
(b) Find the component of

u in the direction of


v. Accompany your answer by a sketch.
PROBLEMS FOR SECTION 20.7
1.

v is a vector of length 3. When its tail is at the origin, it makes an angle of 60
◦
with the
positive x-axis. What is the horizontal component of

v? The vertical component of

v?
For Problems 2 through 5, vectors

u and

v are described. Find
(a) the component of

v in the direction of

u.
(b) the component of

u in the direction of

v.
2. |


u|=5, |

v|=7, and the angle between

u and

v is
π
6
.
3. |

u|=5, |

v|=7, and the angle between

u and

v is
π
3
.
4. |

u|=5, |

v|=7, and the angle between

u and


v is
2π
3
.
5.

u is a vector of length 2 directed due south.

v is a vector of length 3 directed northeast.
6. Suppose

v is a vector of length 8 and the component of

v in the direction of

u is 4.
(a) Can the angle between

u and

v be determined? If so, what is it?
(b) Can the direction of

u be determined? If so, what is it?
(c) Can the length of

u be determined? If so, what is it?
7. You’re pushing an object along a table, exerting a force of 10 pounds in the direction
indicated below. What is the component of force in the direction of motion?
direction

of motion
20°
8. A plane is traveling 300 miles per hour. There is a 50-mph wind. The angle between
the velocity vector of the wind and the velocity vector of the plane is 110
◦
.
(a) What is the component of the direction of the plane’s motion?
(b) In the absence of the wind but all else remaining the same, how fast would the
plane be traveling?
9. Suppose an object is launched from a height of 64 feet with an initial velocity of 96
feet per second at an angle of π/6 radians. Assume that the only force acting on the
object is the force of gravity, which results in a downward acceleration of 32 ft/sec.
(a) Find the vertical position of the object at time t.
(b) When will the object hit the ground?