# Homework Help: Prove that any curve can be parameterized by arc length

1. Mar 26, 2013

### stripes

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

Prove that any curve $\Gamma$ can be parameterized by arc length.

2. Relevant equations

Hint: If η is any parameterization (of $\Gamma$ I am guessing), let $h(s) = \int^{s}_{a} \left| \eta ' (t) \right| dt$ and consider $\gamma = \eta \circ h^{-1}.$

3. The attempt at a solution

Given the hint, we have $\gamma = \eta \circ h^{-1} (t) = (x(h^{-1} (t), y(h^{-1} (t))$ and thus $\left| \gamma ' (t) \right| = \sqrt{ [x' (h^{-1} (t))]^{2} + [y' (h^{-1} (t))]^{2}}$ but what good is this if I can't find the inverse of h(t)?

I know I have to show there exists a parameterization $\gamma : [a, b] \rightarrow \Re^{2}$ of $\Gamma$ so that $\left| \gamma ' (t) \right| = 1$ for all t. Perhaps this hint will reveal this parameterization, but I can't see how. Any help would be appreciated.

2. Mar 26, 2013

### voko

Your expression for $|\gamma'(t)|$ is incorrect. Fix that, then apply the rule for differentiating inverse functions.

3. Mar 28, 2013

### stripes

What exactly is wrong with my expression?

If gamma is a parameterization, then |γ'| = sqrt ( x'(t)^2 + y'(t)^2), is it not?

4. Mar 28, 2013

### voko

The equation is correct in #3, but not in #1.

5. Mar 28, 2013

### stripes

But the parametrization is a function of h^-1 which is a function of t. Would I need to use chain rule?

6. Mar 28, 2013

### voko

Yes, you have to use the chain rule. Which you tried, but you either did not do it correctly, or did not write the result correctly.

7. Mar 28, 2013

### stripes

Oh I forgot because h is a function of s, not t. All this time I thought it was a function of t. Alright I'm plugging away I will be back.

8. Mar 28, 2013

### voko

No, that was not the problem. The problem is that the chain rule should have resulted in a product of two derivatives.

9. Mar 28, 2013

### stripes

So

$\eta (t) = (x(t), y(t))$ is any parametrization. Then let

$h(s) = \int^{s}_{a} | \eta ' (t) | dt$

Now consider
$\gamma (t) = \eta (h^{-1} (s)) = (x(h^{-1} (s)), y(h^{-1} (s))).$

We have

$| \gamma ' (t) | = | \gamma ' (h^{-1} (s)) | = \sqrt{(x'(h^{-1} (s))([h^{-1}]' (s))^{2} + (y'(h^{-1} (s))([h^{-1}]' (s))^{2}}$

$= ... = |\eta ' (s) | \sqrt{ x'(h^{-1} (s))^{2} + y'(h^{-1} (s))^{2}}$

but now what?

10. Mar 28, 2013

### voko

That is not correct. $(h^{-1})' \ne |\eta'|$.

11. Mar 28, 2013

### stripes

Well we defined $h(s) = \int^{s}_{a} | \eta ' (t) |dt$ didn't we? So by the FTC we end up with $| \eta ' (t) | or | \eta ' (s) |$?

Chain rule is really irritating me right now....

12. Mar 28, 2013

### voko

Yes, $h'$ is indeed equal to $|\eta'|$. But you have $h^{-1}$ to differentiate.

13. Mar 28, 2013

### stripes

Right. So it'd be |n'|', but that still doesn't seem useful.

14. Mar 28, 2013

### stripes

Furthermore, how am I going to deal with the square root stuff?

15. Mar 28, 2013

### voko

Why would that be so? Say you have y = f(x), and you also have its inverse x = g(y): g(f(x)) = x. Is there an equation linking f' and g'?

16. Mar 28, 2013

### stripes

I think thats enough for me today. I'll hand in wha I've got. Thanks for your help!