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r
dR t () = dx t () ê + dy t () ê + dz t () ê (7.54)
dt dt 1 dt 2 dt 3

and the unit vector tangent to the curve is given by:

r
dR ()t
ˆ
() =
r
Tt dt (7.55)
dR () t
dt
This is, of course, the unit vector that is always in the direction of the velocity
of the particle.

r
LEMMA r dV t ()
If a vector valued function Vt() has a constant value, then its derivative is
orthogonal to it. dt
PROOF The proof of this lemma is straightforward. If the length of the vector
r r
is constant, then its dot product with itself is a constant; that is, Vt Vt()⋅ () = C.
r
dV t () r
Differentiating both sides of this equation gives ⋅ Vt() = 0 , and the
dt
orthogonality between the two vectors is thus established.
ˆ
The tangential unit vector ()Tt is, by definition, constructed to have unit
length. We construct the norm to the curve by taking the unit vector in the
direction of the time-derivative of the tangential vector; that is,

dT ˆ () t
ˆ
() =
Nt dt (7.56)
dT ˆ ()t
dt

The curvature of the curve is

ˆ
dT t ()
dt
κ= r (7.57)
dR t ()
dt

© 2001 by CRC Press LLC