Blaga P. Lectures on the differential geometry of - tiera.ru

42 Chapter 1. Space curves
1.7 The curvature of a curve
Let (I, r = r(t)) be a regular parameterized curve. Let (J, ρ = ρ(s)) be a naturally parameterized
curve, equivalent to it. Then �ρ ′ (s)� = 1, while the vector ρ ′′ (s) is orthogonal
to ρ ′ (s) 6 .
One can show that ρ ′′ (s) does not depend on the choice of the naturally parameterized
curve equivalent to the given curve r = r(t). Indeed, if (J1, ρ 1 = ρ 1(˜s)) is another
equivalent naturally parameterized curve with the parameter change ˜s = λ(s), then, from
the condition
�ρ ′ (s)� = �ρ ′ 1 (λ(s))� = 1
we get |λ ′ (s)| = 1 for any s ∈ J. Thus, λ ′ = ±1 and, therefore,
˜s = ±s + s0, where s0 is a constant. It follows that
ρ ′′ (s) = ρ ′′ 1(˜s) (λ ′ (s)) 2
��������
=1
+ρ ′ 1 · λ ′′ (s) = ρ
����
=0
′′ 1(˜s).
Definition 1.7.1. The vector k = ρ ′′ (s(t)) is called the curvature vector of the parameterized
curve r = r(t) at the point t, while its norm, k(t) = �ρ ′′ (s(t))� – the curvature of
the parameterized curve at the point t.
We shall express now the curvature vector k(t) as a function of r(t) and its derivatives.
We choose as natural parameter the arc length of the curve. Then we have
r(t) = ρ(s(t)) ⇒
r ′ (t) = ρ ′ (s(t)) · s ′ (t)
where s ′ (t) = �r ′ (t)�, s ′′ (t) = �r ′ (t)� ′ = d
dt
r ′′ (t) = ρ ′′ (s(t)) · (s ′ (t)) 2 + ρ ′ (s(t)) · s ′′ (t),
� √ � r
r ′ (t) · r ′ (t) = ′ (t) · r ′′ (t)
�r ′ . Thus, we have
(t)�
k(t) = ρ ′′ (s(t)) = r′′
�r ′ � 2 − r′ · r ′′
�r ′ � 4 · r′ . (1.7.1)
Now, since the vectors ρ ′ and ρ ′′ are orthogonal, while ρ ′ has unit length, we have
k(t) = �k(t)� = �ρ ′′ � = �ρ ′ × ρ ′′ �.
6 Indeed, since ρ ′ is a unit vector, we have ρ ′2 = 1. Differentiating this relation, we obtain that ρ ′ ·ρ ′′ = 0,
which expresses the fact that the two vectors are orthogonal.