So, my main motive for understanding this concept comes from a problem I had to solve.

it reads: Find an arc length parameterization of the line segment from $(1,2)$ to $(5,-2)$

In the book I'm using they don't have an explicit example of how to derive the arc length parameterization of a line segment, but I've kind of figured it out from poking around a few places, but I want to check my understanding of the concept as well as the solution I derive. Comments/critiques welcome at both the conceptual and minute detail.

Okay, so to find the arc length parameterization you determine the displacement vector

$$ \vec{v} = <4,-4> \text{ and the distance } \mid \mid \vec{v} \mid\mid = \sqrt{32} $$

Thus the parameterization of this line segment can be given of the form:

$$ \left\{ \begin{array}{l l} x=1 + \frac{4t}{\sqrt{32}} \\ y=2 - \frac{4t}{\sqrt{32}}\\ \end{array} \right. $$

So, check me here. I'm creating something like a unit vector, with the fraction component, no? Since that contains both the vector component divided by it's distance, and the starting point component which "pushes" the unitized parameter to the correct place, right?

I know the whole point of discussing this is so that we have a better understanding of a unit tangent vector, which we use to find curvature- and this is just trying to drive home the fact that we're creating a frame of reference (this portion of the curve).

Let me know if my understanding is off in some way, I'm learning this on my own, so help is appreciated.

  • $\begingroup$ I don't see anything wrong with how you've constructed your arc-length parametrization. But I think it's worth emphasizing something: While $\langle x,y \rangle$ isn't a unit vector, its $t$-derivative is. (Which, as you mention in the next paragraph, means it has a unit tangent vector.). You could in fact start there, and take a definite integral in $t$ in order to produce the parametrization you found. $\endgroup$ – Semiclassical Sep 9 '14 at 2:26
  • $\begingroup$ so I could integrate the unit vector to determine this same parameterization? $\endgroup$ – Adam Sep 9 '14 at 13:30
  • $\begingroup$ Should be able to, yes---try it out! $\endgroup$ – Semiclassical Sep 9 '14 at 13:32
  • $\begingroup$ If you do get it to work by integration of the unit tangent vector, then I'd encourage you to post that as an answer. That way we can provide feedback and you'll get a chance to build some more rep. $\endgroup$ – Semiclassical Sep 9 '14 at 20:36

Integrating each separate unit vector : x followed by y.

$$ \frac{4}{\sqrt{32}} \int_{0}^{t} ds = \frac{4t}{\sqrt{32}} $$

$$ \frac{-4}{\sqrt{32}} \int_{0}^{t} ds = \frac{-4t}{\sqrt{32}} $$

Placing this parameterization relative to the starting point provides the answer.

\begin{array}{l l} x=1 + \frac{4t}{\sqrt{32}} \\ y= 2 - \frac{4t}{\sqrt{32}} \\ \end{array}

  • $\begingroup$ Nicely done. You can make this even a bit sharper if you note that your first LHS may be rewritten as $$x(t)-x(0)=\int_0^t x'(s)\,ds$$ so that the 'starting 'point' is already contained in $x(0)$. (Same for $y$.) $\endgroup$ – Semiclassical Sep 9 '14 at 22:56
  • $\begingroup$ Very cool. Thanks for the help! $\endgroup$ – Adam Sep 9 '14 at 23:51

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