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  

r

 ( ) Acceleration   

r

  

i

  

i



j



j

Speed  

r

       2 The definitions are similar for space functions of the form:

r

(t) = x(t)

i

+ y(t)

j

+ z(t)

k

Velocity



Acceleration

 

r



( )



r

   

i

  

i



j

  

j



k



k

Speed

 

r

         2

Example 1 : The position vector

r

t

i

t

2

j

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: 

t



t

2 The curve can then be represented by

y



x

2 orientation as shown in the graph.

b.

Velocity  Acceleration  

r

 ( )  

r

  

i

  

i



j



j

Speed  

r

       2 So the following vector valued functions represent velocity and acceleration and the scalar for speed:

v

(t) =

i

+ 2t

j a

(t) = 2

j

Speed  1 2 

t

2  1  4

t

2

c.

At t = 2, plug into the equations above to get: the velocity vector

v

(2) =

i

+ 4

j

, the acceleration vector

a

(2) = 2

j

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.

r

t

i

t

2

j

a(2) = 2j v(2) = i + 4j

Example 2 : The position vector

r



3cos

t

i



2sin

describes the path of an object moving in the xy-plane.

t

j

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

x

parametric equations: 

t

x 3



cos

t

 y

y



y



sin

t

2

sin

t

 cos

t

 1 represented by the equation

x

9 2 

y

4 2  which is an ellipse with the orientation as 1 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) = -3sint

i

+ 2cost

j r

(t) = 3cost

i

+ 2sint

j a

(t) = -3cost

i

-2sint

j

y Speed  2  2 Speed  9 sin 2

t

 4 cos 2

t

 v(0)=2j

c.

The point (3,0) corresponds to t = 0. You can find this by solving: a(0)=-3i 3cos t = 3 cos t = 1 t = 0   x  At t = 0, the velocity vector is given by

v

(0) =

2j

, and the acceleration vector is given by

a

(0) = -3

i

 

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

 2

, , 2

t

2 3

r

Solution : Recall, you are given

r

(t) in component form. It can be written in standard form as: 3 

t

2

i

t

j

2

t

2

k

The velocity and acceleration can be found by differentiation:

v

 2

t

i

3

t

1 2

k

a

2

i

3

t

 1 2

k

2 The speed is found using the formula and simplifying: Speed=

v

 4

t

2 1

For comments on this presentation you may email the author Professor Judy Gill at [email protected]

or the publisher of the VML, Dr. Julia Arnold at [email protected]

.