Length of path examples

Example 1

Write a parameterization for the straight-line path from the point (1,2,3) to the point (3,1,2). Find the path length.

PIC

Solution: The vector from (1,2,3) to (3,1,2) is d = (3, 1, 2) - (1, 2, 3) = (2,-1,-1). We can parametrized the line segment by

c(t) = (1, 2, 3) + t(2,-1,-1), 0 t 1

To find path length, we calculate

c'(t) = (2,-1,-1)
||c'(t)|| = ∘ -------------------
  22 + (- 1 )2 + (- 1)2 = √ --
  6
Therefore, the length of the path is
ab||c'(t)||dt = 01√ --
  6dt = √ --
  6

Clearly, it was silly to calculate the length this way. We knew the length of the line segment must be ||d|| = √6--. But, this simply illustrates the method of calculating path length.

Example 2

Another parameterization for the line segment of example 1 is

p(t) = (1, 2, 3) + (et - 1)(2,-1,-1), 0 t log 2.
Find the length of the line segment using this parametrization.

This might seem tricky at first. You can just take it by faith, skip the rest of this paragraph, and proceed. Or, just notice that (et - 1) is zero when t = 0, and it is 1 when t = log 2. Indeed, a particle with position p(t) at time t does move along the straight line from (1,2,3) to (3,1,2) as t goes from 0 to log 2. It just doesn’t move at a constant speed. You can read about another example where particles move along the same curve but at different speeds.

Solution: We simply use the definition of path length to find the length of the line segment using using this parameterization. We calculate

p'(t) = et(2,-1, 1)
||p'(t)|| = et||(2,-1,-1)|| = et√ --
  6,
so the path length is
ab||p'(t)||dt = 0log 2et√ --
  6dt
= √--
 6(elog 2 - e0)
= √--
 6(2 - 1) = √ --
  6,
which agrees with example 1.

Examples 1 and 2 illustrate an important principle. The length of a path does not depend on its parametrization. Of course, this makes sense, as the distance a particle travels along a particular route doesn’t depend on its speed.

Example 3

Find the length of the helix parametrized by c(t) = (cos t, sin t,t) for 0 t 6π. (This was the example used in the previous reading.)

Solution: We calculate

c'(t) = (- sin t, cos t, 1)
||c'(t)|| = ∘  ------------------
   sin2 t + cos2t + 12 = √ --
  2.
The length is
06π√ --
  2dt =     |
√ --|6π
  2t|0 = 6√ --
  2π 26.7.