Begin2.DVI

origin, which sweeps out the curve as the parameter xvaries. In figure 6-15, the

position vector
r is illustrated. This position vector has the representation

r =
r (x) = xˆe 1 +f(x)ˆe 2. (6 .60)

As the parameter xvaries, the position vector
r represents the distance and direction

of the point (x, f (x)) with respect to the origin. The derivative

d
r

dx =
r

′(x) = ˆe 1 +f′(x)ˆe 2 (6 .61)

is also illustrated in figure 6-15. Observe that the derivative represents the tangent

vector to the curve at the point (x, f (x)).There can be two tangent vectors to the

curve at (x, f (x)), namely

d
r

dx =
r

′(x)and −d
r

dx =−
r

′(x).Unless otherwise stated, the

tangent vector in the sense of increasing parameter xis to be understood.

The cross product of the unit vector ˆe 3 , out of the plane of the curve, and the

tangent vector dxd
r =
r ′(x)to the curve, gives a normal vector N
to the curve at the

point (x, f (x)).One calculates this normal vector using the cross product

N
 =ˆe 3 ×d
r

dx =

∣∣

∣∣

∣∣

eˆ 1 ˆe 2 ˆe 3

0 0 1

1 f′(x) 0

∣∣

∣∣

∣∣=−f

′(x)ˆe 1 +eˆ 2 (6 .62)

Note there can be two normal vectors to the curve at the point (x, f (x)), namely N

and −N .
To verify that N
is normal to the tangent vector at a general point (x, f (x))

one can examine the dot product of the normal and tangent vector N
·

d
r

dx and show

N
 ·d
r

dx = [−f

′(x)ˆe 1 +ˆe 2 ]·[ˆe 1 +f′(x)ˆe 2 ] = 0

which demonstrates these two vectors are perpendicular to one another. Further,

the magnitudes of the normal vector N
and the tangent vector d
rdx are equal and can

be represented

|N
|=

∣∣

∣∣d
r

dx

∣∣

∣∣=√1 + [f′(x)]^2. (6 .63)

One can use the magnitudes of the tangent and normal vectors to construct unit

vectors in the tangent and normal directions at each point (x, f (x)) on the plane

curve. One finds these unit vectors have the form

ˆet= ˆe^1 +f

′(x)ˆe 2

√

1 + [f′(x)]^2

and ˆen=

−√f′(x)eˆ 1 +ˆe 2

1 + [f′(x)]^2

. (6 .64)