Talk:Arc Length for Parametric Equations

I would like to argue that this has to be a definition rather than a theorem. This because the expression $\dfrac{dy}{dx}$ (sorry for omitting mathrm) needn't be defined everywhere (compare $x(t)=0$, $y(t)=t$, a vertical line). Also, a very important theorem is missing stating that arc length is independent of the chosen parametrisation. Maybe this can be presented as a theorem but that would require very careful reasoning and splitting the curve into multiple parts for which the standard arc length applies. In any case, this page has merit as a confirmation that both ways of computing arc length coincide for simple graphs of functions. --Lord_Farin 06:35, 20 March 2012 (EDT)


 * Also, I've just noticed, the limits of integration are glossed over: it starts at $\alpha \to \beta$ and ends up $a \to b$ with no explanation as to why. Or even, come to that, what $\alpha$ and $\beta$ actually are in the first place. --prime mover 09:14, 20 March 2012 (EDT)


 * I switched the limits of integration, my original concern was that perhaps integrating WRT t instead of x would change the limits of integration. I remember changing the page several times back and forth unable to decide how to present this possibility, and I ended up just getting confused; that was the main reason for the proofread template.
 * Larson has this as a theorem. However, it could be that he wouldn't consider $x(t)=0$, $y(t)=t$ "smooth", though that seems like a stretch. I had a similar issue regarding Definition:Arc Length, that I had a hard time separating out what's a definition and what's a theorem. Take a look at the earliest revisions of that page and PM's comments on the talk page, there. In general I have had a hard time as to splitting concepts such as "justification of definitions", "explanations of definitions", from things that need to be proven. I was under the impression that justifications are definitions, and proofs are required only to show that definitions are well-defined/the things being defined exist. This is a particular vivid example of a situation where I'm confused about these issues. --GFauxPas 09:41, 20 March 2012 (EDT)
 * Would it help any to say something like, if $\frac{dy}{dx}$ is not defined, consider $\frac{dx}{dy}$ instead? --GFauxPas 09:51, 20 March 2012 (EDT)

That's possible; it requires careful reasoning and splitting up the integral into (possibly) a lot of parts, as there is no limit on the number of times either will not be defined. However, even then, artificial examples can be crafted that are continuously differentiable, yet have a point where neither quotient is defined (i.e., both $y' = x' =0$ for a certain time $t$); one may impose that such points do not occur, but I'm not sure if this is necessary. Also, I suspect that the 'continuously differentiable' can be weakened to 'differentiable' or at the very least 'piecewise cont. diff.' without affecting the validity of the formula. Piecewise continuity will also deal with the crafted examples I just mentioned. Keep up the good work in any case! --Lord_Farin 14:12, 20 March 2012 (EDT)
 * Thanks for the compliment! Larson has continuously differentiable, but my Calc III prof. says that most books she's seen have piecewise cont. differentiable. I just used what book I had in front of me. In any event it would evaluate to the same thing, it's just that the integral would be improper according to Larson. --GFauxPas 14:15, 20 March 2012 (EDT)


 * I hardly see what's improper about $\displaystyle \int_0^1 \sqrt x \mathrm dx$... Still, there is work here; I agree with you that care is to be taken about the limits of integration, especially when it occurs that $t_1<t_2$ does not imply $x(t_1)<x(t_2)$ (the curve backs up on itself, and is not the graph of a function anymore; more than one $y$ associated to one $x$; the major reason for passing over to parametrised equations). As you are probably discovering, this stuff is hard to deal with in full generality. --Lord_Farin 14:22, 20 March 2012 (EDT)


 * I meant improper as in e.g. Length of Arc of Cycloid for more than one arc. Re: potential problems, we did stipulate that the graph can't intersect itself, but perhaps that's not precise/accurate enough. --GFauxPas 14:33, 20 March 2012 (EDT)

Yes, I noticed that stipulation, but it's very unnecessary and in my opinion should be disposed of. It simply doesn't matter for the result of the integral (well, unless one wants to deal with doubly traversing an arc as traversing it only once (i.e., from $0$ to $1$ and back has length $1$ instead of $2$)) and is an annoying, easy to overcome restriction by splitting the integral into multiple parts. --Lord_Farin 14:39, 20 March 2012 (EDT)
 * Larson does that because he wants to define the length of the curve to not be twice as long if you traverse it twice, like you said. But yeah you can split the integral into parts, but then if you don't write it as two parts, is the integral improper? Does it matter? I dunno. Delete that part if you don't think it belongs. --GFauxPas 14:51, 20 March 2012 (EDT)