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.
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)|| | =  =  | ||
∫
ab||c'(t)||dt = ∫
01 dt =  |
Clearly, it was silly to calculate the length this way. We knew the length of the
line segment must be ||d|| = 
. 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. |
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 , |
| ∫ ab||p'(t)||dt | = ∫
0log 2et dt | ||
=  (elog 2 - e0) | |||
=  (2 - 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)|| | =  =  . |
∫
06π dt =  = 6 π ≈ 26.7. |
