Post on 31-Dec-2015
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10.2 Vectors and Vector Value Functions
Quantities that we measure that have magnitude but not direction are called scalars.
Quantities such as force, displacement or velocity that have direction as well as magnitude are represented by directed line segments.
A
B
initialpoint
terminalpoint
AB��������������
The length is AB��������������
A
B
initialpoint
terminalpoint
AB��������������
A vector is represented by a directed line segment.
Vectors are equal if they have the same length and direction (same slope).
A vector is in standard position if the initial point is at the origin.
x
y
1 2,v v
The component form of this vector is: 1 2,v vv
A vector is in standard position if the initial point is at the origin.
x
y
1 2,v v
The component form of this vector is: 1 2,v vv
The magnitude (length) of 1 2,v vv is:2 2
1 2v v v
P
Q
(-3,4)
(-5,2)
The component form of
PQ��������������
is: 2, 2 v
v(-2,-2) 2 2
2 2 v
8
2 2
If 1v Then v is a unit vector.
0,0 is the zero vector and has no direction.
Vector Operations:
1 2 1 2Let , , , , a scalar (real number).u u v v k u v
1 2 1 2 1 1 2 2, , ,u u v v u v u v u v
(Add the components.)
1 2 1 2 1 1 2 2, , ,u u v v u v u v u v
(Subtract the components.)
Vector Operations:
Scalar Multiplication:1 2,k ku kuu
Negative (opposite): 1 21 ,u u u u
v
vu
u
u+vu + v is the resultant vector.
(Parallelogram law of addition)
The dot product (also called inner product) is defined as:
1 1 2 2cos u v u v u v u v
Read “u dot v”
Example:
3,4 5,2 3 5 4 2 23
The angle between two vectors is given by:
1 1 1 2 2cosu v u v
u v
The dot product (also called inner product) is defined as:
1 1 2 2cos u v u v u v u v
This could be substituted in the formula for the angle between vectors (or solved for theta) to give:
1cos
u v
u v
Find the angle between vectors u and v:
2,3 , 2,5 u v
1cos
u v
u v
Example:
1 2,3 2,5cos
2,3 2,5
1 11cos
13 29
55.5
Application: Example 7
A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
N
E
Application Example
A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
N
Eu
Application: Example 7
A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
N
E
v
u
60o
Application: Example 7
A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
N
E
v
u
We need to find the magnitude and direction of the resultant vector u + v.
u+v
N
E
v
u
The component forms of u and v are:
u+v
500,0u
70cos60 ,70sin 60v
500
70
35,35 3v
Therefore: 535,35 3 u v
538.4 22535 35 3 u v
and: 1 35 3tan
535 6.5
N
E
The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5o north of east.
538.4
6.5o
We can describe the position of a moving particle by a vector, r(t) (position vector).
tr
If we separate r(t) into horizontal and vertical components,
we can express r(t) as a linear combination of standard unit vectors i <1, 0> and j <0, 1>.
t f t g t r i j f t i
g t j
In three dimensions the component form becomes:
f g ht t t t r i j k
Most of the rules for the calculus of vectors are the same :
)( ),( )(:VectorPosition tytxtr
dt
dy
dt
dxt , )(:VectorVelocity v
2
2
2
2
, )(:Vectoron Acceleratidt
yd
dt
xdta
b
a
b
a
dttvdttv )( ,)( :Vectornt Displaceme 11
The exceptions???:
22:direction) no hasit (b/c Speed ba v
b
a
22
21 ))(())(()(:Traveled Distance
b
a
dttvtvdttv
Example 7:
A particle moves in an elliptical path so that its position at any time
t ≥ 0 is given by <4 sin t, 2 cos t>.
)cos2(),sin4(dt
d t)(Velocity t
dt
dtv
a.) Find the velocity and acceleration vectors.
)sin2,cos4 t)(Velocity tt v
)sin2(),cos4(dt
d t)(on Accelerati t
dt
dt a
tt cos2,sin4 t)(on Accelerati a
Example 7:
A particle moves in an elliptical path so that its position at any time
t ≥ 0 is given by <4 sin t, 2 cos t>.
4sin2,
4cos4 )4/(Velocity
v
b.) Find the velocity, acceleration, speed, and direction of motion at t = /4
4cos2,
4sin4 )4/(on Accelerati
a
2,22
2,22
2,22/4)(Speed v 10)2()22( 22
Example 7:
A particle moves in an elliptical path so that its position at any time
t ≥ 0 is given by <4 sin t, 2 cos t>.
c.) Sketch the path of the particle and show the velocity vector at the point (4, 0).
Graph parametrically
x = 4 sin t
y = 2 cos t At (4, 0):
4 = 4 sin t and 0 = 2 cos t
1 = sin t 0 = cos t
v(t) = <4 cos t, -2 sin t> = <0, -2>
Example 7:
A particle moves in an elliptical path so that its position at any time
t ≥ 0 is given by <4 sin t, 2 cos t>.
d.) Does the particle travel clockwise or counterclockwise around the origin?
The vector shows the particle travels clockwise around the origin.
Example : 3 2 32 3 12t t t t t r i j
2 26 6 3 12d
t t t tdt
r
v i j
a) Write the equation of the tangent where .1t
At :1t 1 5 11 r i j 1 12 9 v i j
position: 5,11 slope:9
12
tangent: 1 1y y m x x
311 5
4y x
3 29
4 4y x
3
4
The horizontal component of the velocity is .26 6t t
Example 6: 3 2 32 3 12t t t t t r i j
2 26 6 3 12d
t t t tdt
r
v i j
b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0.
26 6 0t t 2 0t t
1 0t t 0, 1t
0 0 0 r i j
1 2 3 1 12 r i j
1 1 11 r i j
0,0
1, 11