Vector Valued
Functions
Velocity and Acceleration
Written by Judith McKaig
Assistant Professor of Mathematics
Tidewater Community College Norfolk,
Virginia
Definitions of Velocity and Acceleration:
If x and y are twice differentiable functions of t and r is a
vector-valued function given by r(t) = x(t)i + y(t)j, then
the velocity vector, acceleration vector, and speed at
time t are as follows:
Velocity  v(t )  r (t )  x(t )i  y(t ) j
Acceleration  a(t )  r (t )  x(t )i  y(t ) j
Speed  v(t )  r (t ) 
 x(t )   y(t )
2
2
The definitions are similar for space functions of the
form: r(t) = x(t)i + y(t)j + z(t)k
Velocity  v(t )  r (t )  x(t )i  y(t ) j  z (t )k
Acceleration  a(t )  r (t )  x(t )i  y(t ) j  z(t )k
Speed  v(t )  r (t ) 
 x(t )   y(t )   z(t )
2
2
2
r(t )  ti  t 2 j
Example 1: The position vector
describes the path of an object moving in the xy-plane.
a. Sketch a graph of the path.
b. Find the velocity, speed, and acceleration of the object at any
time, t.
c. Find and sketch the velocity and acceleration vectors at t = 2
Solution:
a. To help sketch the graph of the path, write the following
parametric equations:
x(t )  t
y(t )  t 2
The curve can then be represented by
the equation
y  x2 with the
orientation as shown in the graph.
b. Velocity  v(t )  r (t )  x(t )i  y(t ) j
Acceleration  a(t )  r (t )  x(t )i  y(t ) j
Speed  v(t )  r (t ) 
 x(t )   y(t )
2
2
So the following vector valued functions represent
velocity and acceleration and the scalar for speed:
r (t )  ti  t 2 j
v(t) = i + 2tj
a(t) = 2j
Speed  12  (2t ) 2  1  4t 2
c. At t = 2, plug into the equations
above to get:
the velocity vector v(2) = i + 4j,
the acceleration vector a(2) = 2j
To sketch the graph of the
velocity vector, start at the
initial point (2,4) and move
right 1 and up 4 to the
terminal point (3,8).
Sketch the acceleration
similarly.
v(2) = i + 4j
a(2) = 2j
Example 2: The position vector r (t )  3cos ti  2sin tj
describes the path of an object moving in the xy-plane.
a. Sketch a graph of the path.
b. Find the velocity, speed, and acceleration of the object at any
time, t.
c. Find and sketch the velocity and acceleration vectors at (3,0)
Solution:
a. To help sketch the graph of the path, write the following
parametric equations:
x
x  3cos t , so  cos t
3
y
y  2 sin t , so  sin t
2
Since
y



can be
sin 2 t  cos2 t  1 , the curve
2
2
represented by the equation
x
y

1
9
4

which is an ellipse with the orientation as
shown in the graph.

x

b.
By differentiating each component of the vector,
you can find the following vector valued functions
which represent velocity and acceleration. You
can use the formula to find the scalar for speed:
v(t) = -3sinti + 2costj
a(t) = -3costi-2sintj
r(t) = 3costi + 2sintj
y
Speed 
(3sin t ) 2  (2 cos t ) 2
Speed  9sin 2 t  4 cos 2 t

v(0)=2j
c. The point (3,0) corresponds to
t = 0. You can find this by solving:
3cos t = 3
cos t = 1
t=0
At t = 0, the velocity vector is given
by v(0) = 2j,
and the acceleration vector is
given by a(0) = -3i
a(0)=-3i




x

Example 3: The position
vector r describes the path
of an object moving in
space. Find the velocity,
acceleration and speed of
the object.
r (t )  t 2 , t , 2t
3
2
Solution: Recall, you are given r(t) in component form. It can be written
in standard form as:
3
2
r (t )  t 2i  tj  2t k
The velocity and acceleration can be found by differentiation:
1
2
v (t )  2ti  j  3t k
3  12
a (t )  2i  t k
2
The speed is found using the formula and simplifying:
Speed= v (t )  4t 2  9t  1
For comments on this presentation you
may email the author Professor Judy Gill at
jgill@tcc.edu or the publisher of the VML,
Dr. Julia Arnold at jarnold@tcc.edu.