User talk:Engeer

An intuitive derivation for the Catenary[edit]

Dear Engeer,

I would like to discuss the decision on removing my contribution of an alternative derivation for the Catenary equation. The reason for adding that derivation was that in my perspective, the wiki page does not include any logical derivation. All derivations presented require foresight in the solution. A statement such as "A reasonably straightforward approach to integrate this is to use hyperbolic substitution" make the logical connection between hyperbolic functions and the catenary equation opaque.

During the writing of this message, I realized that a far more intuitive, perhaps simplest solution is possible. I will soon add this contribution to the page. This derivation does not rely on any insight of trigonometric functions in any form. I look forward to your review.

Also, I would like to suggest removing the scaling constant 'a' by using a new variable z = x/a (making z physically dimensionless). This is effectively the free scaling parameter which was the last missing piece with the other 2 integration constants being for translation. This would also significantly simply all derivations.

I hope to hear from you.

Kind regards,

Thatyougoon

Thatyougoon (talk) 22:12, 22 May 2024 (UTC)[reply]

Okay, I'm interested in your "insight-free" version. The one you offered earlier required recognizing the sqrt(1-x^2) form as suitable for a triginometric substitution, just as the recognition of the sqrt(1+x^2) form is suitable for a hyperbolic substitution. Your solution went off on a digression involving complex numbers, which in my experience requires a more sophisticated reader than I feel is the target of this article. The relation cosh^2 - sinh^2 = 1 may be a little less well known than the relation cos^2 + sin^2 = 1, but it's not *that* obscure, and it leads to a much simpler derivation.

As to rescaling to z=x/a: I agree that for abstract mathematics this is cleaner, but I feel that the point of the catenary is a physics/engineering artifact, and so explicitly showing the scaling factor used for arbitrary length units is useful. (I, myself, have to fight the temptation to show the conclusion as "y/a = cosh(x/a)", which is sorta half-way between your z abstraction and the standard form.) Engeer (talk) 22:23, 24 May 2024 (UTC)[reply]

Actually, while I am curious about your "more intuitive" approach, I do question the foundation of your basic objection. We are not pre-supposing or guessing at the substitution, we are recognizing a common pattern in integrating certain forms of expressions.

When integrating an expression which includes a

{\sqrt {\pm a^{2}\pm x^{2}}}

component, and there is not something "helpful" such as a

x

outside of the radical, the techniques used on the Trigonometric_substitution page are bog-standard techniques. For the particular case of

{\sqrt {a^{2}+x^{2}}}

, there are two common options: use the

\sec ^{2}=\tan ^{2}+1

identity or the

\cosh ^{2}=1+\sinh ^{2}

relation. These are standard tools that should be in the toolkit of anyone who might need to integrate such equations. If one tries the

\sec ^{2}

substitution one happens to wind up with a mess that requires a fair bit of effort to integrate, whereas using the

\cosh ^{2}

substitution things turn out to work out quickly and easily.

Or, one could just consult a table of integrals and see the

\int 1/{\sqrt {a^{2}+x^{2}}}=\sinh ^{-1}{x/a}

entry, with no insight whatsoever (an earlier version of the Catenary page did this; I didn't like it, and added the "hyperbolic substitution" link to explain how the integral can be found). Engeer (talk) 00:00, 28 May 2024 (UTC)[reply]

Welcome![edit]

Hello, Engeer, and welcome to Wikipedia! I hope you like the place and decide to stay. Below are some pages you might find helpful. For a user-friendly interactive help forum, see the Wikipedia Teahouse.

Introduction
The five pillars of Wikipedia
How to edit a page
How to write a great article
Simplified Manual of Style
Your first article
Discover what's going on in the Wikimedia community
Feel free to make test edits in the sandbox
and check out the Task Center, for ideas about what to work on.

I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, please see our help pages, and if you can't find what you are looking for there, please feel free to ask me on my talk page or place {{Help me}} on this page and someone will drop by to help. Again, welcome! Nerd271 (talk) 11:39, 25 May 2024 (UTC)[reply]