Prove that any curve can be parameterized by arc length

[stripes]

Homework Statement

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

Homework 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}.

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.

 

[voko]

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

 

[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?

 

[voko]

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

 

[stripes]

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

 

[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.

 

[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.

 

[voko]

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

 

[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?

 

[voko]

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

 

[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...

 

[voko]

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

 

[stripes]

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

 

[stripes]

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

 

[voko]

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

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'?

 

[stripes]

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