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I know that formula of a length of arc of a curve $y=f(x)$ from abscissae a to b is given by $$\int\limits_{a}^{b} \sqrt{1+\{f^\prime (x)\}^2}\;\mathrm{d}x$$ But according to my college professor, he says that formula to determine the length of arc is $$\lim\limits_{h\rightarrow 0} \sum\limits_{r=1}^{\frac{b-a}{h}} \sqrt{\left\{ f\Big(a+rh\Big)-f\Big(a+(r-1)h\Big)\right\}^2 +h^2}$$

He also asked us to find the length of circumference of the unit circle at origin $(x^2+y^2=1)$ using this limit formula only, but I don't know how I can compute this. So please can anyone help me to solve this in this particular way using limits.

Soham Sarkar
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1 Answers1

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The $y=f(x)$ in this case becomes $y=\pm\sqrt{1-x^2}$ and using symmetry across the $x$- and $y$-axes we can take $x\in[0,1]$, so with your limit formula we have

$$4\lim_{h\to 0} \sum_{r=1}^{\frac1h}\sqrt{\left\{\sqrt{1-(rh)^2}-\sqrt{1-((r-1)h)^2}\right\}^2 +h^2}\\ =4\lim_{h\to 0}\sum_{r=1}^{\frac 1h}\sqrt{2+2rh^2-2r^2h^2-2\sqrt{(1-r^2h^2)(1-r^2h^2+2rh^2-h^2)}}\\ =4\sqrt 2\lim_{h\to 0}\sum_{r=1}^{\frac 1h}\sqrt{1+rh^2-r^2h^2-\sqrt{(1+rh^2-r^2h^2)^2-h^2}}\\ r\in\left[1,\frac 1h\right]\to 1+rh^2-r^2h^2\in[h,1],\frac 1h\to n, \frac 1{1+rh^2-r^2h^2}\to t^2, t\in[1,\sqrt n]\\ =4\sqrt 2\lim_{n\to\infty}\frac 1n\sum_{t=1}^{\sqrt n}\sqrt{\frac 1{t^2}-\sqrt{\frac 1{t^4}-\frac 1{n^2}}}\\ =4\sqrt2\lim_{n\to\infty}\frac1{n}\sum_{t=1}^{\sqrt n}\frac {\sqrt{1-\sqrt{1-\frac{t^4}{n^2}}}}t\\ =4\sqrt2\lim_{n\to\infty}\frac 1n\sum_{t=1}^{\sqrt n}t\sqrt{\frac{\frac1{n^2}}{1+\sqrt{1-\frac{t^4}{n^2}}}}\\ =4\sqrt2\lim_{n\to\infty}\frac1{n^2}\sum_{t=1}^{\sqrt n}\frac t{\sqrt{1+\sqrt{1-\frac{t^4}{n^2}}}}$$

which is still a little ways from something reasonable.

We can use one approach as described in https://math.stackexchange.com/a/217251/86846 by considering the area of the same circle, especially if we take areas of triangles having height sides equal to the radius (i.e., $1$), and their bases as the same length calculation as used for our arc length, so each area is $\frac 12bh$, or

$$\frac 12\cdot 1\cdot\sqrt{\left(\sqrt{1-r^2h^2}-\sqrt{1-(r-1)^2h^2}\right)^2+h^2}$$

giving our total area formula (converging from overestimates) as

$$2\lim_{h\to 0} \sum_{r=1}^{\frac1h}\sqrt{\left\{\sqrt{1-(rh)^2}-\sqrt{1-((r-1)h)^2}\right\}^2 +h^2}$$

This approach only gives us that the circumference of the unit circle is twice the area, but at least it's a minor bit of progress.

For an alternate approach, consider taking a cue from https://tutorial.math.lamar.edu/classes/calcii/arclength.aspx on this formula and using one more "symmetry fold", allowing us to rewrite the initial limit as

$$8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}h\sqrt{1+\left(\frac{-rh}{\sqrt{1-r^2h^2}}\right)^2}\\ =8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}h\sqrt{1+\frac{r^2h^2}{1-r^2h^2}}\\ =8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}\frac h{\sqrt{1-r^2h^2}}\\ =8\lim_{n\to\infty}\sum_{r=1}^\frac{\sqrt 2n}2\frac 1{\sqrt{n^2-r^2}}$$

but this fails to meet the professor's strict instructions...

abiessu
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  • But doesn't this limit becomes a trivial limit when we take the limit within the summation? Also I don't know how to compute infinite series. $$\lim\limits_{h\rightarrow 0} \sum\limits_{r=1}^{\frac{2}{h}} \Rightarrow \sum\limits_{r=1}^{\infty}\lim\limits_{h\rightarrow 0}$$ – Soham Sarkar Jan 28 '21 at 18:46
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    In this case we haven't adjusted $r$ and/or our formula to remove the dependency of $r$ on $h$, so we can't swap the limit and sum just yet. – abiessu Jan 28 '21 at 18:51
  • But is this limit finally solvable and can we get $2\pi$ as the answer? – Soham Sarkar Jan 28 '21 at 18:55
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    It should be, I'm doing some more research to get closer to that result. – abiessu Jan 28 '21 at 18:59
  • I'm not having much luck pushing this further. My research has yielded the above plus https://math.stackexchange.com/a/3664317/86846 as well as https://math.stackexchange.com/a/217251/86846 but these do not follow the prescribed path to get to the intended results. – abiessu Feb 03 '21 at 16:37