10.4 MINIMAL PATH PROBLEMS
10.5 MAXIMUM AND MINIMUM PROBLEMS
IN MOTION AND ELSEWHERE
10.6 VECTOR FUNCTIONS FOR MOTION IN A PLANE
Warning:
Only some of this is review.
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.
B terminal
point
AB
A
initial
point
The length is
AB

B terminal
point
AB
A
initial
point
A vector is represented by a directed line segment.
Vectors are equal if they have the same length and
direction (same slope).

y
A vector is in standard position if the
initial point is at the origin.
 v1 , v2 
x
The component form of this vector is:
v  v1 , v2
y
A vector is in standard position if the
initial point is at the origin.
 v1 , v2 
x
The component form of this vector is:
The magnitude (length) of
v  v1 , v2
v  v1 , v2
is:
v  v12  v22

(-3,4)
P
(-5,2)
Q
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0
v -1
-2
(-2,-2)
-3
-4
-5
-6
The component form of
PQ is:
v  2, 2
1 2 3 4 5 6
v
 2   2
2
2
 8
2 2

If
0,0
v 1
Then v is a unit vector.
is the zero vector and has no direction.

Vector Operations:
Let u  u1 , u2 , v  v1 , v2 , k a scalar (real number).
u  v  u1 , u2  v1 , v2  u1  v1 , u2  v2
(Add the components.)
u  v  u1 , u2  v1 , v2  u1  v1 , u2  v2
(Subtract the components.)

Vector Operations:
Scalar Multiplication:
Negative (opposite):
ku  ku1 , ku2
u   1 u  u1 , u2

u
v
u + v is the resultant vector.
u+v
(Parallelogram law of addition)
v
u

The angle between two vectors is given by:
u1v1  u2v2
  cos
u v
1
This comes from the law of cosines.
See page 524 for the proof if you are interested.

The dot product (also called inner product) is defined as:
u  v  u v cos  u1v1  u2v2
Read “u dot v”
Example:
3, 4  5, 2   3 5   4  2   23

The dot product (also called inner product) is defined as:
u  v  u v cos  u1v1  u2v2
This could be substituted in the formula for the angle
between vectors (or solved for theta) to give:
 uv 
  cos 

u v
1

Example:
Find the angle between vectors u and v:
u  2,3 , v  2,5

2,3


2,5


u

v

1
1
  cos 
  cos 
2,3

2,5
u
v



 11 
 cos 

 13 29 
1




 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 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
u
E
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
v
60o
u
E

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
We need to find the magnitude
and direction of the resultant
vector u + v.
v
u+v
u
E

N
The component forms of u and
v are:
u  500,0
v
70
v  70 cos 60 , 70sin 60
u+v
500
v  35,35 3
Therefore:
u  v  535,35 3

u  v  535  35 3
2
and:
u
E
35 3
  tan
535
1

2
 538.4
 6.5

N
538.4
6.5o
E
The new ground speed of the airplane is about 538.4 mph,
and its new direction is about 6.5o north of east.
p
Any vector
v  a, b
can be written as a linear
combination of two standard unit vectors.
i  1,0
v  a, b
 a,0  0, b
 a 1,0  b 0,1
 ai  bj
j  0,1
The vector v is a linear combination
of the vectors i and j.
The scalar a is the horizontal
component of v and the scalar b is
the vertical component of v.

We can describe the position of a moving particle by a
vector, r(t).
r t 
f t  i
g t  j
r t   f t  i  g t  j
If we separate r(t) into horizontal and vertical components,
we can express r(t) as a linear combination of standard
unit vectors i and j.

In three dimensions the component form becomes:
r t   f t  i  g t  j  h t  k

Graph on the TI-89 using the parametric mode.
r  t    t cos t  i   t sin t  j
MODE
Graph…….
Y=
t 0
2
xt1  t  cos t
yt1  t  sin t
ENTER
ENTER
WINDOW
GRAPH
8p

Graph on the TI-89 using the parametric mode.
r  t    t cos t  i   t sin t  j
MODE
Graph…….
Y=
t 0
2
xt1  t  cos t
yt1  t  sin t
ENTER
ENTER
WINDOW
GRAPH

Most of the rules for the calculus of vectors are the same as
we have used, except:
Speed  v  t 
“Absolute value” means
“distance from the origin” so we
must use the Pythagorean
theorem.
v t 
velocity vector

Direction 
v t 
speed

Example 5:
r  t    3cos t  i   3sin t  j
a) Find the velocity and acceleration vectors.
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t  p / 4 .

Example 5:
r  t    3cos t  i   3sin t  j
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t  p / 4 .
3
3
p 
p
p  
i
j
velocity: v     3sin  i   3cos  j  
4 
4
2
2
4 
p 
p
3
3
p  
i
j
acceleration: a     3cos  i   3sin  j  
4 
4
4 
2
2

Example 5:
r  t    3cos t  i   3sin t  j
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t  p / 4 .
3
3
p 
v   
i
j
2
2
4
3
3
p 
a   
i
j
2
2
4
2
speed:
direction:
p    3   3 
v 

 

2  2

4
v p / 4 
v p / 4 
2
9 9


3
2 2
3 / 2
3/ 2   1 i  1 j

i
j
2
2
3
3


Example 6:
 

r  t   2t 3  3t 2 i  t 3  12t j
a) Write the equation of the tangent where

 
t  1.

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
At t  1 :
position:
tangent:
v  1  12i  9 j
r  1  5i  11j
 5,11
slope:
y  y1  m  x  x1 
3
y  11    x  5 
4
9
3

12
4
3
29
y  x
4
4

Example 6:

 

r  t   2t 3  3t 2 i  t 3  12t j

 

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
b) Find the coordinates of each point on the path where
the horizontal component of the velocity is 0.
2
6
t
 6t .
The horizontal component of the velocity is
6t  6t  0
2
t t  0
r  0   0i  0 j
 0, 0 
2
t  t  1  0
t  0, 1
r 1   2  3 i  1 12  j
r 1  1i  11j
 1, 11
p
…interesting how they posed this
question. They are not asking for the
acceleration at t = 1 they are asking how
much of the acceleration helped the
velocity! This is not as easy to compute.
This is not the acceleration at t = 1. This is how much of the
acceleration helped the velocity!