Calculus III:

Vector functions
Arc Length of a vector function

1. Arc Length of a vector function

Let consider the vector function (t) = 〈 f(t), g(t), h(t) 〉

We want to determine le length of the curve related to this vector function on the interval a ≤ t ≤ b.

If we write this vector function into the parametric form, we will have:

x = f(t) , y = g(t), z = h(t), and the arc length is given by

L = ∫_a^b √{f'(t)² + g'(t)² + h'(t)²} dt.

The integrand is the magnitude of the tangent vector, so:

L = ∫_a^b||(t)|| dt

Example

Let's consider the following vector function:

(t) = (2t, - cos t, sin t)
We want to determine le length of the curve on the interval 0≤ t ≤ 4.

So (t) = 〈 2, sin t, cos t 〉

The length of this tangent vector is:

||(t)|| = sqrt(2² + sin²t + cos²t) = √5

The length is then:

L = ∫₀⁴√5 dt = 4 √5.

2. Arc length function s(t)

We want to know the length of the arc s(t) at a certain time t, and the express the vector function (t) in termes of s.

This allow to determine the final point on the curve after traveling a distance s.

The length of the arc s(t) can be written as:

s(t) = ∫₀^t ||(ξ)|| dξ,

where ξ is a spacial variable along the curve.

Example

Let's determine the arc length function for
(t) = 〈 2t, - cos t, sin t 〉 .

After travelling a distance of 2π√5, where is then the final point on the curve ?

(t) = 〈 2t, - cos t, sin t 〉. So

(t) = 〈 2, sin t, cos t 〉

||(t)|| = sqrt(2² + sin²t + cos²t) = √5

Therefore

s(t) = ∫₀^t ||(ξ)|| dξ = ∫₀^t √5 dξ = √5 t

s(t) = √5 t → t = s/√5

In terms, of s, the original vector function becomes:

(t) = 〈 2s/√5, - cos (s/√5), sin (s/√5) 〉.

With the reparameterization we can now tell where we are on the curve after we’ve traveled a distance of s = 2π√5 = 14.04 units, along the curve. We start the measurement of distance from t = 0.

(s) = 〈 2s/√5, - cos (s/√5), sin (s/√5) 〉.

(s = 2π√5) = 〈 4π, - 1, 0 〉.

So, after raveling a distance of s = 2π√5 = 14.04 units on the curve, we are at the point:

(s = 2π√5) = 〈 4π, - 1, 0 〉