Expressing the Acceleration in Terms of the unit Tangent and the Unit Normal


We have seen that given r ( t)



x( t)  i  y( t)  j then the velocity vector is v( t)


dr
dt
dx dy
i 
j
dt
dt
and the velocity vector is tangent to the curve at each point along the trajectory.

Further the acceleration vector is


a( t )
2
d r
dt
2
2
2
d x d y
i 
 j.
2
2
dt
dt
However, there are 2 ways in which a particle can accelerate: either due to a change in speed or a change in
direction. In this form of the acceleration we have we can't really see this.
We want to express the acceleration in terms of a tangential component aT and a normal component aN.
The tangential component will tell us the change in speed and the normal component will give us the change


in direction. In this discussion we concentrate on finding T and N the unit tangent and unit normal vectors.
In subsequent discussions we will concentrate on finding find aT and aN .
Let

T


dr

T
dt


dr
dt
be the unit tangent . Since we know the velocity is tangent to the curve it follows:


dT

We claim the unit normal is N
dt


dT
. It is obvious it is a unit vector so all that remains to be shown is
dt


dT

that T is perpendicular to
dt


 dT
i.e. T
dt
0.
Thm
Let

w( t) be any vector valued function.

w(t)
If
constant


 dw
then w( t) 
dt
0
Pf:




dw   dw
 w  w
dt
dt
 
d 
w w
dt
But
and
 
d 
w w
dt
d
dt


 dw
2 w
dt

2
 
w 

2
 
w  is a constant
Therefore
It follows
 
d 
w w
dt


 dT
T
dt


 dw
2 w
dt
0
0 is a corollary as

T
1
Summary

The unit tangent is T


dr


dT

dt
and the unit normal is N


dr
dt


dT
dt
dt

Now we have another problem. In 2 space there are actually 2 possibilities for N an inward
and an outward normal . (In 3space there are an infinity of possibilities)
We want

N to be the vector pointing inward and we call this the inward unit normal.


dT
Does

N
dt


dT
satisfy this at all points ?
dt


Recall if we have any unit vector u then it can be written u


cos ( ) i  sin( ) j

where θ is the angle between u and the horizontal measured counterclockwise from the horizontal.


dT


dT d

d dt
dt

sin()i  cos ()j  d
dt




Note sin() i  cos () j is a 900 counter clockwise rotation of T
Note in the diagram on the left
d
dt
0


dT
so
dt




is in the same direction as sin() i  cos () j
pointing inward along the curve.
Note in the diagram on the right
d
0
so
dt
and again pointing inward along the curve.


dT
dt
is oppositely directed to
sin()i  cos ()j 
See Animation 1
Example


r ( t)
 2
t i  t  j


dr

T( t)
dt


dr



 i  2 t j
1
1  4 t

2
dt

N is a little more difficult
The Calculation of


dT

4 t
dt
3
1  4t 
2


dT



 i  2t j 
1  4 t
2

1
dt
3
1  4t2


 2j
1
2

 
 4 t i  2 j

1
3
2
1  4t2
1
3
1  4t2
2


2

 
 4 t i  2 j
2
16 t  4
1


dT

N
 1 4 t2
dt


dT
dt
1

3
1  4t2
See Animation 2
2

3
2
2

16 t  4
4ti  2j 
1
2
2 4 t  1

 
 4 t  i  2 j
