# A biconditional statement for arc length of a function

1. Mar 26, 2013

### stripes

1. The problem statement, all variables and given/known data

Show that $\gamma : [a, b] \rightarrow \Re^{2}$ is a parameterization of $\Gamma$ if and only if the length of the curve from $\gamma(a)$ to $\gamma(s)$ is $s - a$; i.e.,

$\int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.$

2. Relevant equations

3. The attempt at a solution

Part 1; show $\left| \gamma ' (s) \right| = 1 \Rightarrow \int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.$

We have

$\ell = \int ^{b}_{a} \left| \gamma ' (t) \right| dt = \int ^{b}_{a} \left| \gamma ' (s(t)) s'(t) \right| dt = \int ^{b}_{a} \left| \gamma ' (s(t))\right| \left| s'(t) \right| dt = \int ^{\ell}_{0} \left| \gamma ' (s) \right| ds = \int ^{\ell}_{0} 1 ds = s(\ell) - s(0) = s - a.$

My question here is: did I do the change of variables correctly? Specifically in the limits of integration?

Now how do I prove the converse; i.e., how do I show that $\int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a \Rightarrow \left| \gamma ' (s) \right| = 1 ?$

$\int ^{s}_{a} \left| \gamma ' (t) \right| dt$
$= \int ^{s}_{a} \sqrt{x'(t)^{2} + y'(t)^{2} } dt$
$= \int ^{s}_{a} \sqrt{ (x'(s(t)) s'(t) )^{2} + (y'(s(t)) s'(t) )^{2} } dt$
$= \int ^{s}_{a} \sqrt{ s'(t)^{2} (x'(s)^{2} + y'(s)^{2}) } dt$
$= \int ^{s}_{a} s'(t) \sqrt{ (x'(s)^{2} + y'(s)^{2}) } dt$
$= \int ^{s}_{t=a} \sqrt{ (x'(s)^{2} + y'(s)^{2}) } ds$
$= \int ^{s}_{t=a} \left| \gamma' (s) \right| ds = s - a$
$= \frac{d}{ds}(\int ^{s}_{t=a} \left| \gamma' (s) \right| ds)$
$= \frac{d}{ds}(s - a)$
$= 1 - 0 = 1$

so $1 = \left| \gamma' (s) \right| .$

or is that garbage?

Last edited: Mar 26, 2013
2. Mar 26, 2013

### voko

This statement is incorrect. Is it supposed to be "natural parametrization" rather than just "parametrization?

3. Mar 26, 2013

### stripes

I do believe that the statement I am to prove is correct. I have copied it directly out of my text. My instructor says my proof is complete. I'm not sure what you meant.

And it is natural parametrization.

4. Mar 26, 2013

### voko

The statement is correct only if it is about natural parametrization. It is not correct for an arbitrary parametrization.

5. Mar 26, 2013

### stripes

Oh my bad. I failed to mention that in my