Part 1: The Unit Normal

In the last section, we explored those ideas related to velocity-namely, distance, speed, and the unit tangent vector. In this section, we do the same for acceleration by exploring the concepts of linear acceleration, curvature, and the unit normal vector.

To begin with, the unit tangent T satisfies || T( t) || = 1, which implies that T(t) ·T( t) = 1. Differentiation yields

d dt ( T( t) ·T( t))   =  d dt  1 dT dt ·T + T· dT dt  =  0 2T·  dT dt  =  0

That is, the derivative of T is orthogonal to T.  Indeed, the unit normal vector is the unit vector  

N = 1 || dT/dt|| dT dt

(2)

That is, N is a unit vector which is orthogonal to T.

EXAMPLE 1    Find the unit normal N to the helix r( t) = á 4cos( t), 4sin(t) ,3t ñ       

Solution: Since the velocity is v = á -4sin( t) ,4cos( t) ,3 ñ , the speed is

v =  16sin2(t) +16cos2(t)+9  =  16+9 = 5

and consequently the unit tangent vector is

T =  1 v v =  -4 5 sin(t) , 4 5 cos( t) , 3 5

The derivative of the unit tangent vector is

dT dt  =  d dt -4 5 sin(t) , 4 5 cos( t) , 3 5   =   -4 5 cos( t) , -4 5 sin(t) ,0

which has a length of

dT dt  =   16 25 cos2(t) +  16 25 sin2( t) +02  =  4 5

Thus, the unit normal is

N =  1 || dT/dt|| dT dt  =  1 4/5 -4 5 cos( t) ,  -4 5 sin( t) ,0

which simplifies to N = á -cos( t) ,-sin( t) ,0 ñ .  For example, if t=p /2, then N = á 0 ,-1 ,0 ñ , as is shown below along with the unit tangent T at the same point.