10.2 day 2 Vector Valued Functions
Everglades National Park, FL
Photo by Vickie Kelly, 2006
Greg Kelly, Hanford High School, Richland, Washington
Everglades National Park, FL
Photo by Vickie Kelly, 2006
Everglades National Park, FL
Photo by Vickie Kelly, 2006
Everglades National Park, FL
Photo by Vickie Kelly, 2006
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
8

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   / 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   / 4 .
3
3
 

  
i
j
velocity: v     3sin  i   3cos  j  
4 
4
2
2
4 
 

3
3
  
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   / 4 .
3
3
 
v   
i
j
2
2
4
3
3
 
a   
i
j
2
2
4
2
speed:
direction:
    3   3 
v 

 

2  2

4
v  / 4 
v  / 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
