Transcript Slide 1

VECTORS
The study of vectors is closely related to the study of such
physical properties as force, motion, velocity, and other related
topics.
Vectors allow us to model certain characteristics of these
phenomena with numbers that tell us their magnitude and
direction.
SCALAR QUATITIES:
Measurements involving such things as time, area, volume,
energy, and temperature are called scalar measurements
because each can be described adequately using their
magnitude alone (with the appropriate units).
27 ft3 adequately describes the volume of a cube with side 3
ft. 980 F adequately describes the temperature of a person.
27ft3 and 980 F are called scalars.
Some properties such as force, velocity, and displacement
require both magnitude and direction to be described
completely. These quantities are called vector quantities.
Example: You are driving due north at 45 miles per hour. The
magnitude is the speed, 45 miles per hour. The direction of
motion is due north.
NOTATION AND GEOMETRY OF VECTORS
Two airplanes travel at 400 mph on a parallel course and in the
same direction. This situation can be modeled using directed
line segments.
B
D
A
A
C
The directed segments are drawn parallel with arrowheads
pointing the same way to indicate direction of flight, while
making them the same length indicates that the velocities
are the same. The length of the vector models the
magnitude of the velocity, while the arrowhead indicates
the direction of travel.
Vectors are named using the initial and
terminal points that define them as in AB and CD
Or with a bold, small case letter such as u or
v. We may also write them as
u. or v
Q
P
The magnitude of the directed line segment PQ is its length.
We indicate this by PQ . PQ is the distance from point P
to point Q.
Reference Angle θr :
For any angle θ in standard position, the acute angle θr
formed by the terminal side and the x-axis is called the
reference angle for θ.
Find the reference angle, θr , for each of the following
angles.
a) θ = 3450
b) θ = -1350
c) 5800
a) θr = 150
b) θr = 450
c) θr = 400
Example
Show that u=v.
y

Vectors that are equal have the
same magnitude and direction.
Use the distance formula to show
that u and v have the same
magnitude.


(2,3)
u

x





(2, 1)





d = (x 2  x1 )2  ( y 2  y1 )2

v

u  [0  (3)]2  [3  (3)]2 
13

(4, 4)
v  (3  0)2  (6  0)2
 13
Thus u and v have the same magnitude: u  v
One way to show that u and v have the same direction is to
find the slopes of the lines on which they lie.
y 2  y1
m
x2  x1
Verify to show that each vector has a slope of 3/2.
POSITION VECTORS
For a vector v with initial point (x1, y1) and terminal point
(x2, y2), the position vector for v is
v = x 2 - x1 , y 2 - y1 ,
an equivalent vector with initial point (0,0) and terminal point
(x2 – x1, y2 – y1).
The vector in component form is denoted as a, b ,
where a is the horizontal component and b is the
vertical component.
Find the position vector for
vector u and graph it.
(3, 7)
(-2, 3)
u
7
3
(-5, -4)
The position vector for u is:
v = -2 - (-5), 3 - (-4) = 3, 7
The position vector v has
(0,0) as its initial point, and
(3, 7) as its terminal point.
For a position vector v = ‹a, b› shown below at left and
angle θr, observe the following:
The magnitude of vector v =
vertical component:
b
sinθr 
v
horizontal component:
cosθr 
a
v
a b
2
2
b v sinθr
a  v cosθr
y
b
tanθr 
a
v
b
θr
a
x
b
θr  tan  
a
1
Finding the Magnitude and Direction Angle of a Vector
Example: For v1  2.5,  6 and v 2 3 3, 3 ,
a. Find their magnitudes.
b. Graph each vector and name the quadrant where
located.
c. Find the angle θ for each vector (round to tenths
of a degree).
a. v1 
b. v 2 
 2.5   6
2
3 3 
2
2
 6.25  36  42.25  6.5
 3  27  9  36  6
2
 6 
c. For v1 : θr  tan1 


2.5


 tan1  2.4  67.40
 3
 3 
1
0

tan

30
d. For v 2 : θr  tan 



3 3 
 3 
1
v 2  3 3, 3 in QI.
3 3, 3
v1  2.5,  6 is located in QIII. Wh
300
2.5,  6
Find the Horizontal and Vertical Components of
a Vector.
The vector v = a, b is in QIII, has a magnitude of v = 21, and forms
an angle of 250 with the negative x-axis. Graph the vector and find its
horizontal and vertical components.
y
Note: θr = 25⁰, therefore θ = 1800 + 250 = 2050.
Horizontal Component:
-19
x
250
-8.9
v  21
a = v cos  21cos2050   19
Vertical Component:
b  v sin
b  21sin2050   8.9
Vector Addition:
Addition of vectors using the “tail – to – tip” method.
Shift one vector (without changing its direction) so
that its tail (initial point) is at the tip (terminal point)
of the other vector.
Given vectors u and v
y
x
u
v
Tail of u to tip of v
y
Tail of v to tip of u
y
x
x
u
u+v
v
v
u
u+v
v
u
Vector Addition: Add vectors v
and w.
w
v +w
Terminal
point of w
w
v
v
Initial
point of v
APPLICATION OF VECTORS:
A common example of a vector quantity is force.
Other vector quantities that appear in engineering
mechanics are moment, displacement, velocity,
and acceleration.
FORCE: The effect of one physical body on
another physical body. The force effect between
two bodies can be interpreted as a ‘push’ or ‘pull’
of one of the bodies on the other body.
RESULTANT FORCE: When two or more forces
are added to obtain a single force, it produces the
same effect as the original system of forces. This
single vector is called the sum, or the resultant
force of the original system of forces.
y
FR
F1
F2
x
The resultant force, FR ,
is the sum of F1 and F2 .
COMPONENTS OF A FORCE:
Two or more forces acting on a particle may be replaced
by a single force which has the same effect on the
particle.
Conversely, a single force F acting on a particle may be
replaced by two or more forces which, together, have the
same effect on the particle. These forces are called the
components of the original force F.
RECTANGULAR COMPONENTS OF A FORCE.
Often it is desirable to resolve a force into two components
which are perpendicular to each other. In the figure below,
the force F has been resolved into a component Fx and a
component Fy. Fx and Fy are called rectangular components.
y
Fx = Fcosθ
Fy = Fsinθ
tanθ =
Fy
F
θ
0
x
Fx
Fy
Fx
 Fy 
θ = tan  
 Fx 
-1
Example: A force of 800 N is exerted on a bolt A as shown.
Determine the horizontal and vertical components of the force.
Note: θR = 350; What is θ ?
F = 800 N
Fy
F = 800 N
350
θ = 1450
A
350
Fx
A
Fx = Fcos1450 = 800cos1450 = -655 N
Fy = Fsin1450 = 800sin145˚ = 459 N
We write F in the form F = -(655N)i + (459 N) j
Vectors in the Rectangular
Coordinate System
The i and j Unit Vectors.
Vector i is the unit vector (vector of length 1) whose
direction is along the positive x-axis. Vector j is the unit
vector whose direction is along the positive y-axis.
y
Vectors in the rectangular
coordinate system can be
represented in terms of i and j.
1
j
i
1
x
P = (a, b)
b
v = ai + bj
0
a
x
A unit vector is defined to be a vector whose magnitude is
one. In many applications, it is useful to find the unit
vector that has the same direction as a given vector.
For any nonzero vector v, the vector
v
is the unit vector that has the
v
same direction as v. To find this vector, divide v by its magnitude.
Example: Find the unit vector in the same direction
as v = 5i – 12j. Then verify that the vector has
magnitude 1.
v  52  (12)2  25  144  169  13
5i  12 j
5
12


u
j
13
13
13
v
v
5 2
12 2
25
144
169
Verify : ( )  ( )  (
) (
) 
 11
13
13
169
169
169
Sketch the vector and find its magnitude.
v = -3i + 4j
a = -3 and b = 4.
(-3, 4)
v=5
Example
Sketch the vector v=-2i+j and
find its magnitude.
y




x













Let v be a vector from initial point P1 ( 2, 2)
Example
to terminal pont P2 (2,3). Write v in terms
of i and j.
y




x













Operations with Vectors in Terms of i
and j
Vector Subtraction:
The difference of two vectors, u – v is defined as
u – v = u + (-v):
The terminal point of u coincides with the initial point of -v
y
x
y
x
u
v
u
Example
If v=2i-3j and w=-i+4j find each of the following:
a.
v+w
b. v-w
c. 2v
d. -3w
e. 2v-3w
Unit Vectors
Example
Find the unit vector in the same direction as v=-3i-4j.
Then verify that the vector has the magnitude of 1.
Writing a Vector in Terms of Its
Magnitude and Direction
Example
The wind is blowing at 16 mph in the direction of N600 E.
Express its velocity as a vector v in terms of i and j.
Application
Example
Two forces F1 and F2 , of magnitude 5 lbs and 12 lbs,
respectively act on an object. The direction of F1 is
N200 E and the direction of F2 is N750 E. Find the
magnitude and the direction of the resultant force.
Express the magnitude to the nearest hundredth of a
pound and the direction angle to the nearest tenth
of a degree.
If u=- 3i+4j, v=2i-j find 2u+v
(a) -i+3j
(b) 4i-j
(c) 5i+3j
(d) -4i+7j
Find the magnitude of the vector v=-3i+4j
(a)
(c)
5
6
7
(d)
8
(b)
Converting from Rectangular Coordinates to Polar
Coordinates.
a) Radians.
b) Polar Coordinates.
A) RADIAN MEASURE
We measure angles by determining the amount
of rotation from the initial side to the terminal
side. Two units of measurement for angles are
degrees and radians.
Radian Measure
An angle whose vertex is at the center of a circle
is called a central angle. A central angle
intercepts the arc of the circle from the initial side
to the terminal side. A positive central angle that
intercepts an arc of the circle of length equal to the
radius of the circle has a measure of 1 radian.
r
ө
r
How many radians are there in a
circle?
6.28
How many degrees are there in
360
1 radian?
o
6.28
 57.3
RELATIONSHIP BETWEEN DEGREES AND
RADIANS:
360˚ = 2π radians
Degrees to radians: θ
180˚ = π radians
o
 π 
= θ
 radians
 180 
 180 
Radians to degrees: θ radians = θ 
 degrees
 π 
Convert each angle from degrees to radians.
a) 30˚
b) 90˚
c) -225˚
d) 55˚
π
a) 30  30
 radians
180 6

b)90
π

radians
2
5π
c)- 2 25  radians
4

55
d) 55  55

 0.96radians
180 180
Convert each angle in radians to degrees.
π
a)
radians
3
3π
b) = radians
4

 180
o
a)
radians 
 60
3
3 
3
3 180
b) - radians =  -135 o
4
4 
c) 2 radians  2
180

 114.6o
c) 2.5 radians
The foundation of the polar coordinate system is
a horizontal ray that extends to the right. This ray
is called the polar axis. The endpoint of the polar
axis is called the pole. A point P in the polar
coordinate system is designated by an ordered
pair of numbers (r, θ).
P = (r, θ)
r
θ
polar axis
pole
r is the directed distance form the
pole to point P ( positive, negative,
or zero).
θ is angle from the pole to
P (in degrees or radians).
Plotting Points in Polar Coordinates.
To plot the point P(r, θ ), go a distance of
r at 00 then move θ0 along a circle of
radius r.
If r > 0, plot a point at that location. If r < 0, the
point is plotted on a circle of the same radius,
but 180 in the opposite direction.
0
Plot each point (r, θ)
a) A(3, 450)
C
b) B(-5, 1350)
A
B
c) C(-3, -π/6)
CONVERTING BETWEEN POLAR AND
RECTANGULAR FORMS
CONVERTING FROM POLAR TO RECTANGULAR
COORDINATES.
To convert the polar coordinates (r, θ) of a point to
rectangular coordinates (x, y), use the equations
x = rcosθ
and
y = rsinθ
Convert the polar coordinates of each point to its
rectangular coordinates.
a) (2, -30⁰ )
b) (-4, π/3)
3
a) x = rcos(-30⁰)  2(
) 3
2
y  2sin(30 )  2(1/ 2)   1
The rectangular coordinates of (2,  30 ) are( 3,-1)
b) x= -4cos(π/3) = -4(1/2) = -2
3
)  2 3
y= -4 sin(π/3) = 4(
2
The rectangular coordinates of (-4,

3
) are(-2, - 2 3 )
CONVERTING FROM RECTANGULAR TO
POLAR COORDINATES:
To convert the rectangular coordinates (x, y) of a point to
polar coordinates:
1) Find the quadrant in which the given point (x, y) lies.
2) Use r =
x 2  y 2 to find r .
y
3) Find  by using tan 
and choose  so that it lies in the
x
same quadrant as the point (x, y ).
Find the polar coordinates (r, θ) of the point P with r > 0
and 0 ≤ θ ≤ 2π, whose rectangular coordinates are
(x, y) = (1, 3)
The point is in quadrant 2.
r  ( 1)2  ( 3)2 
 3
tanθ = 
 1 


4 2
r  tan1 3  60

2
  180  60  120 or 120 (
)
180
3
The required polar coordinates are (2, 2π/3)
10.1 Three-Dimensional Coordinate Systems
Distance Formula in Three Dimensions
The distance PP
1 2 between the points P1(x1, y1, z1 )and P2 (x 2 , y 2 , z2 ) is
2
2
2
PP

(
x

x
)

(
y

y
)

(
z

z
)
1 2
2
1
2
1
2
1
Equation of a sphere
An equation of a sphere with center C(h, k, l) and radius r is
(x – h)2 + (y – k)2 + (z – l)2
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
Copyright © 2011 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
R(0, 0, 12)
Q(4, 0, 12)
P(4, 9, 12)