Michael Corral: Vector Calculus

62 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
In practice, parametrizing a curve f(t) by arc length requires you to evaluate the
integral s= ∫ t
a ‖f′ (u)‖du in some closed form (as a function of t) so that you could then
solve for t in terms of s. If that can be done, you would then substitute the expression
for t in terms of s (which we calledα(s)) into the formula for f(t) to get f(s).
Example 1.43. Parametrize the helix f(t)=(cost,sint,t), for t in [0,2π], by arc length.
Solution: By Example 1.41 and formula (1.43), we have
s=
∫ t
0
‖f ′ (u)‖du=
∫ t
0
√
2du=
√
2t for all t in [0,2π].
So we can solve for t in terms of s: t=α(s)= √ s .
(
2
s
∴ f(s)= cos√ ,sin s s
)
√ , √ for all s in [0,2 √ 2π]. Note that‖f ′ (s)‖=1.
2 2 2
Arc length plays an important role when discussing curvature and moving frame
fields, in the field of mathematics known as differential geometry. 16 The methods involve
using an arc length parametrization, which often leads to an integral that is either
difficult or impossible to evaluate in a simple closed form. The simple integral in
Example 1.43 is the exception, not the norm. In general, arc length parametrizations
are moreusefulfor theoreticalpurposesthanfor practical computations. 17 Curvature
and moving frame fields can be defined without using arc length, which makes their
computationmucheasier,andthesedefinitionscanbeshowntobeequivalenttothose
using arc length. We will leave this to the exercises.
The arc length for curves given in other coordinate systems can also be calculated:
Theorem 1.22. Suppose that r=r(t),θ=θ(t) and z=z(t) are the cylindrical coordinates
of a curve f(t), for t in [a,b]. Then the arc length L of the curve over [a,b]
is ∫ b √
L= r ′ (t) 2 +r(t) 2 θ ′ (t) 2 +z ′ (t) 2 dt (1.44)
a
Proof: The Cartesian coordinates (x(t),y(t),z(t)) of a point on the curve are given by
x(t)=r(t)cosθ(t), y(t)=r(t)sinθ(t), z(t)=z(t)
so differentiating the above expressions for x(t) and y(t) with respect to t gives
x ′ (t)=r ′ (t)cosθ(t)−r(t)θ ′ (t)sinθ(t), y ′ (t)=r ′ (t)sinθ(t)+r(t)θ ′ (t)cosθ(t)
16 See O’NEILL for an introduction to elementary differential geometry.
17 For example, the usual parametrizations of Bézier curves, which we discussed in Section 1.8, are
polynomialfunctionsin 3 . Thismakestheircomputationrelativelysimple, which, inCAD,isdesirable.
But their arc length parametrizations are not only not polynomials, they are in fact usually impossible
to calculate at all.