2.2 Principal normal and curvature

(2.16) |

which states that is orthogonal to the tangent vector, provided it is not a null vector. This fact can be also interpreted from the definition of the second derivative

(2.17) |

As shown in Fig. 2.4, the direction of
becomes perpendicular to the tangent
vector as
. The unit vector

(2.18) |

which has the direction and sense of is called the

When
is moved from
to
, then
,
and
form an isosceles triangle (see Fig. 2.4), since
and
are unit tangent
vectors. Thus we have
as
and hence

(2.19) |

is called the

The vector
is called the *curvature vector*, and measures the rate of
change of the tangent along the curve. By definition
is
nonnegative, thus the sense of the normal vector is the same as that
of
.

The curvature for arbitrary speed (non-arc-length parametrized) curve
can be obtained as follows.
First we evaluate
and
by the chain rule

where is the parametric speed. Taking the cross product of and we obtain

For the planar curve, we can give the curvature
a sign by
defining the normal vector such that
form a right-handed screw, where
as shown in Fig.
2.5. The point where the
curvature changes sign is called an *inflection
point* (see also Fig. 8.3).

According to this definition the unit normal vector of the plane curve
is given by

and hence from (2.23) we have

For a space curve, by taking the norm of (2.23) and using (2.4), we obtain

The normal vector for the arbitrary speed curve can be obtained from
, where
is the unit *binormal vector* which will be introduced in
Sect. 2.3 (see (2.41)).

The unit principal normal vector and curvature for implicit curves
can be obtained as follows. For the planar curve the normal vector can
be deduced by combining (2.14) and
(2.24) yielding

where only the sign of was used (although it is not necessary).

We will introduce a derivative operator with respect to arc length
so that the derivation becomes simple. If we rewrite the plane
implicit curve as
where
is arc length along the
implicit curve, the total derivative with respect to the arc length
becomes

(2.28) |

Now if we replace and by using (2.5) and (2.14) (+ sign), we obtain the derivative operator with respect to arc length

By applying the operator (2.29) to (2.14) (+ sign) and equating with (using (2.20) and (2.27)), we obtain

(2.30) |

For a 3-D implicit curve, we can deduce a derivative operator [444] similar to (2.29),

where is the tangent vector of the 3-D implicit curve (see (2.15)) given by

(2.32) |

and

(2.33) | |||

(2.34) | |||

(2.35) |

By applying the derivative operator (2.31) to
**
**
we obtain

(2.36) |

which gives

Taking the cross product of

Thus,

(2.39) |

A different derivation of the curvature of a 3-D implicit curve is given in Sect. 6.3.2.