Transcript Document
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-1
Chapter 10: Applications of Trigonometry and Vectors 10.1 The Law of Sines 10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex Numbers 10.5 Powers and Roots of Complex Numbers 10.6 Polar Equations and Graphs 10.7 More Parametric Equations
Slide 10.3-2
Copyright © 2011 Pearson Education, Inc.
10.3 Vectors and Their Applications • Basic Terminology – A
scalar
is a magnitude • E.g. 45 pounds – A
vector quantity
is a magnitude with direction • E.g. 50 mph east • Vectors – Represented in boldface type or with an arrow over the letters • E.g.
OP
, and OP represent the vector
OP
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Slide 10.3-3
10.3 Basic Terminology • Vector
OP
– First letter represents the
initial point
– Second letter represents the
terminal point
– Vector
OP
and vector
PO
are not the same vectors. They have the same magnitude, but in opposite directions.
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Slide 10.3-4
10.3 Basic Terminology • • Two vectors are equal if and only if they both have the same magnitude and direction.
1.
2.
The sum of two vectors
A
and
B
Place the initial point of
B
at the terminal point of
A
. The vector with the same initial point as
A
and the same terminal point as
B
is the sum
A
+
B
.
The
parallelogram rule
: place the vectors so that their initial points coincide. Then complete the parallelogram as shown in the figure.
Figure 22 pg 10-47
Slide 10.3-5
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10.3 Basic Terminology • Vectors are commutative:
A
+
B
=
B
+
A
, and the sum
A
+
B
is called the
resultant
of
A
and
B
.
• For every vector
v
there is a vector –
v
such that
v
+ (–
v
) =
0
, the
zero vector
.
• The
scalar product
of a real number (scalar)
k
and a vector
u
is the vector k·u , which has magnitude |
k
| times the magnitude of
u
. If
k
< 0, then k·u is in the opposite direction as
u
.
Slide 10.3-6
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10.3 Algebraic Interpretation of Vectors • A vector with its initial point at the origin is called a
position vector
.
• A position vector
u
as
u
with endpoint (
a
,
b
) is written =
a
,
b
, where
a
is called the
horizontal component
and
b
is called the
vertical component
of
u
.
Slide 10.3-7
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10.3 Algebraic Interpretation of Vectors
Magnitude and Direction Angle of a Vector
a, b
The magnitude (length) of vector
u
=
a
,
b
is given by
u
a
2
b
2 .
The direction angle satisfies tan
a
,
a
0.
Slide 10.3-8
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10.3 Finding the Magnitude and Direction Angle
Example
Find the magnitude and direction angle for
u
= 3, –2 .
Analytic Solution u
3 2 ( 2 ) 2 13 tan
y x
2 3 2 3 tan 1 2 3 33 .
7 Vector
u
has a positive
x
-component and a negative
y-
component, placing
u
yields the direction of in quadrant IV. Adding 360° 33 .
7 360 326 .
3 .
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Slide 10.3-9
10.3 Finding Horizontal and Vertical Components
Horizontal and Vertical Components
The horizontal and vertical components, respectively, of a vector
u
are given by having magnitude |
u
| and direction angle
a
= |
u
| cos and
b
= |
u
| sin .
That is,
u
=
a
,
b
= |
u
| cos , |
u
| sin .
.
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Slide 10.3-10
10.3 Finding Horizontal and Vertical Components
Example
From the figure, the horizontal component is
a
= 25.0 cos 41.7° 18.7. The vertical component is
b
= 25.0 sin 41.7° 16.6.
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Slide 10.3-11
10.3 Finding the Magnitude of a Resultant
Example
Two forces of 15 and 22 newtons act on a point in the plane. If the angle between the forces is 100°, find the magnitude of the resultant force.
Solution
From the figure, the angles of the parallelogram adjacent to angle
P
each measure 80º, since they are supplementary to angle
P
. The resultant force divides the parallelogram into two triangles. Use the law of cosines on either triangle.
|
v
| 2 = 15 2 + 22 2 –2(15)(22) cos 80º |
v
| 24 newtons Copyright © 2011 Pearson Education, Inc.
Slide 10.3-12
10.3 Some Properties of a Parallelogram • Properties of Parallelograms 1. A parallelogram is a quadrilateral whose opposite sides are parallel.
2. The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles are supplementary.
3. The diagonals of a parallelogram bisect each other but do not necessarily bisect the angles.
Slide 10.3-13
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m
10.3 Operations with Vectors
, n
, and p
.
Slide 10.3-14
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10.3 Operations with Vectors
Vector Operations
For any real numbers
a
,
b
,
c
,
d
, and
k
,
a
,
b
c
,
d
a
c
,
b
d k
a
,
b
ka
,
kb
.
a
,
b
If
a
c
,
d a
1 ,
a
2 , then
a
,
b
a
c
,
d
a
1 ,
a
2 .
a
c
,
b
d
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Slide 10.3-15
10.3 Performing Vector Operations
Example
Let
u
= –2, 1 and the following: (a)
u v
+ = 4, 3 . Find each of
v
, (b) –2
u
, (c) 4
u
– 3
v
.
Solution
(a)
u
+
v
= –2, 1 + 4, 3 = –2 + 4, 1 + 3 = 2, 4 (b) –2
u
= –2 · –2, 1 = –2(–2), –2(1) = 4, –2 (c) 4
u
– 3
v
= 4 · –2, 1 – 3 · 4, 3 = 4(–2) – 3(4), 4(1) – 3(3) = –8 – 12, 4 –9 = –20, –5 Copyright © 2011 Pearson Education, Inc.
Slide 10.3-16
10.3 The Unit Vector • A
unit vector
is a vector that has magnitude 1.
• Two very useful unit vectors
i
are defined as = 1, 0 and
j
= 0, 1 .
i
,
j
If
v
Forms for Unit Vectors =
a
,
b
, then
v
= ai + bj .
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Slide 10.3-17
10.3 Dot Product
Dot Product
The
dot product v
=
c
,
d
of two vectors is denoted
u
·
v
, read “
u u
=
a
,
b
dot
v,
and ” and given by
u
·
v
=
ac + bd
.
Example
Find the dot product 2, 3 · 4, –1 .
Solution
2, 3 · 4, –1 = 2(4) + 3(–1) = 8 – 3 = 5
Slide 10.3-18
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10.3 Dot Product
Properties of the Dot Product
For all vectors
u
,
v
, and
w
and real numbers
k
, a)
u
·
v
=
v
·
u
b)
u
· (
v
+
w
) =
u
·
v
+
u
·
w
c) (
u
+
v
) ·
w
=
u
·
w
+
v
·
w
d) ( ku ) ·
v
=
k
(
u
·
v
) =
u
· ( kv ) e)
0
·
u
=
0
f)
u
·
u
= |
u
| 2 .
Slide 10.3-19
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10.3 Dot Product
Geometric Interpretation of the Dot Product
If
u
is the angle between the two nonzero vectors and
v
, where 0º 180º, then
u
v
u v
cos or cos
u
v u v
.
Slide 10.3-20
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10.3 Finding the Angle Between Two Vectors
Example Solution
Find the angle between the two vectors
u
= 3, 4 and
v
= 2, 1 .
cos
u
v u v
3 , 4 3 , 4 2 , 1 2 , 1 3 ( 2 ) 4 ( 2 ) 9 16 4 1 5 10 5 .
894427191 Therefore, 26.57º.
NOTE If
a
·
b
= 0, then cos = 0 and = 90º. Thus
a
and
b
are perpendicular or
orthogonal vectors.
Slide 10.3-21
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10.3 Applying Vectors to a Navigation Problem
Example
A plane with an airspeed of 192 mph is headed on a bearing of 121º. A north wind is blowing (from north to south) at 15.9 mph. Find the groundspeed and the actual bearing of the plane.
Solution
Let
|x|
be groundspeed. We must find angle . Angle
AOC
= 121º.
Find
|x|
using the law of cosines .
x
2 192 2 15 .
9 2 2 ( 192 )( 15 .
9 ) cos 121 40 , 261 sin 15 .
9 sin sin 121 200 .
7 .
06792320
x
200 .
7 mph 3 .
89 Copyright © 2011 Pearson Education, Inc.
Slide 10.3-22
10.3 Finding a Required Force
Example
Find the force required to pull a wagon weighing 50 lbs up a ramp inclined at 20º to the horizontal.
(Assume no friction.)
Solution
The vertical 50 lb force
BA
represents the force of gravity.
BA
is the sum of the vectors
BC
and –
AC
. Vector
BC
represents the force with which the weight pushes against the ramp. Vector
BF
represents the force required to pull the weight up the ramp. Since
BF
and
AC
are equal, |
AC
| gives the magnitude of the required force.
sin20 Copyright © 2011 Pearson Education, Inc.
AC 50
AC
50sin20 2017 lbs
Slide 10.3-23