Length of path examples

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

$\includegraphics[width=2.5in]{pathline.eps}$

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 $\displaystyle \le$ t $\displaystyle \le$ 1

c'(t)	= (2, - 1, - 1)
\|\|c'(t)\|\|	= $\displaystyle \sqrt{{2^2+(-1)^2 + (-1)^2}}$ = $\displaystyle \sqrt{{6}}$

$\displaystyle \int_{a}^{b}$ |c'(t)| dt = $\displaystyle \int_{0}^{1}$ $\displaystyle \sqrt{{6}}$ dt = $\displaystyle \sqrt{{6}}$

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

p(t) = (1, 2, 3) + (e^t -1)(2, -1, -1), 0 $\displaystyle \le$ t $\displaystyle \le$ 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 (e^t - 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)	= e^t(2, - 1, 1)
\|\|p'(t)\|\|	= e^t\|(2, -1, -1)\| = e^t $\displaystyle \sqrt{{6}}$ ,

$\displaystyle \int_{a}^{b}$ \|\|p'(t)\|\| dt	= $\displaystyle \int_{0}^{{\log 2}}$ e^t $\displaystyle \sqrt{{6}}$ dt
	= $\displaystyle \sqrt{{6}}$ (e^log 2 - e⁰)
	= $\displaystyle \sqrt{{6}}$ (2 - 1) = $\displaystyle \sqrt{{6}}$ ,

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.

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

c'(t)	= (- sin t, cos t, 1)
\|\|c'(t)\|\|	= $\displaystyle \sqrt{{\sin^2 t + \cos^2 t + 1^2}}$ = $\displaystyle \sqrt{{2}}$ .

$\displaystyle \int_{0}^{{6\pi}}$ $\sqrt{{2}}$ dt = $\sqrt{{2}}$ t $\displaystyle \Big\vert _{0}^{{6\pi}}$ = 6 $\sqrt{{2}}$ $\displaystyle \pi$ $\displaystyle \approx$ 26.7.