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It is easy to use calculus to find that in a circular segment, if the chord is held constant, then $$\left(\frac{\partial a}{\partial s}\right)_c = R$$ Where $c$ is the chord, held constant, $a$ is area of circular segment and $s$ is arc length. Is there an elementary geometric argument for this?

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Arya
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MaudPieTheRocktorate
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    The problem is contradictory because a cord couldn’t be held constant when evaluating a limit where the cord length becomes infinitesimal. Aside of that, I think that there isn’t a geometric proof that circumvents infinitesimals, hence calculus. If it existed, it would involve operating with the exact value of $\pi$ which can not be determined exactly outside calculus. – WindSoul Jan 14 '23 at 16:32
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    Notations like $\frac{\partial a}{\partial s}$ work only if we have a very clear idea of what function(s) we are working with. In what specific way are you finding the area $a$ as a function of $s$ and other variables? It would help if you wrote the formula out explicitly; this would help people understand what you mean by $\frac{\partial a}{\partial s}$. I might be able to guess what you're thinking, but it would be more straightforward for you to say it yourself. – David K Jan 14 '23 at 17:20
  • Have you tried approximating the circular wedge as two triangles? – Alex K Jan 14 '23 at 18:09
  • @DavidK how about now – MaudPieTheRocktorate Jan 15 '23 at 12:12
  • We already had "the chord is held constant". What I'm having trouble with is writing the area as an explicit function of $c$ and $s.$ If I actually set $c$ as a constant, so that segment length and area become functions of a single variable, for example the height of the segment, I can derive your result using total derivatives, but it was a complicated, laborious derivation and I could only get to the result by knowing what I was looking for. I got no intuition from it. You say you "easily" found your result; I'm wondering how. It might yield some clues. – David K Jan 15 '23 at 16:37
  • I used $\theta$ as the variable, and wrote $a, s$ as functions of $\theta, c$, then just bruteforced $a(\theta + d\theta, c), c(\theta + d\theta, c)$. This was enough for Wolfram Alpha to verify the result. But that's not how I found the result. I found the result by solving the isoperimetric problem with Lagrangian multipliers, and interpreting one of the multipliers as a shadow price for the length of the curve. – MaudPieTheRocktorate Jan 15 '23 at 21:08

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