January 25, 2022 – Physicspupil

Draw a circle of radius $R$ around the center $(R,0)$ .

As the center of a circle is equidistant from all points on the circumference, using Pythagorean theorem, we have

$\begin{aligned} (x-R)^2+(y-0)^2 & = R^2 \\ y^2 & = R^2 - (x-R)^2 \\ y & = \pm\sqrt{2xR-x^2} \\ |y| & = \sqrt{2xR-x^2}\\ \end{aligned}$

Attempts. (reinventing the wheel)

Considering the differentials $\mathrm{d}x$ and $\mathrm{d}y$ , we have

$\mathrm{d}y =\displaystyle{\frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x}$

Suppose I do not know the circumference is $2\pi R$ long. Let its unknown length be $s$ , and let it be partitioned into infinitesimal $\mathrm{d}s$ , such that

$s=\displaystyle{\int\mathrm{d}s}$

Assume we may write

$(\mathrm{d}s)^2=(\mathrm{d}x)^2+(\mathrm{d}y)^2$ .

Expand the right hand side as follows

$\begin{aligned} &\quad \textrm{RHS} \\ & = (\mathrm{d}x)^2+(\mathrm{d}y)^2 \\ & = (\mathrm{d}x)^2 + \bigg( \frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x\bigg)^2 \\ & = \bigg(\frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2 \\ \end{aligned}$

Then

$\begin{aligned} \mathrm{d}s & = \sqrt{\bigg( \frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2} \\ s = \int\mathrm{d}s & = \int_{x=0}^{2R} \frac{R}{\sqrt{2xR-x^2}}\,\mathrm{d}x \\ \end{aligned}$

Suppose I evaluate the integral above $\textrm{\scriptsize{NOT}}$ by direct substitution $\textrm{\scriptsize{BUT}}$ by Riemann sum, so the definite integral due to Riemann is given by

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x}=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\, x$ .

For $i=0,1,2,\dots ,n$ , let $P=\{ x_i\}$ be a regular partition of $[0,2R]$ . Then

$x=\displaystyle{\frac{b-a}{n}=\frac{2R}{n}}$ .

By right-endpoint approximation for Riemann sums, for each interval $[x_{i-1},x_i]$ , we have

$\displaystyle{x_i=x_0+ix=0+i\bigg[\frac{2R}{n}\bigg]=\frac{2Ri}{n}}$ .

Let $f(x)\stackrel{\textrm{def}}{=}\displaystyle{\frac{R}{\sqrt{2xR-x^2}}}$ . Thus,

$\begin{aligned} f(x_i) & = \frac{R}{\sqrt{(2)\displaystyle{\bigg(\frac{2Ri}{n}\bigg)}(R)-\displaystyle{\bigg(\frac{2Ri}{n}\bigg)^2}}} \\ & = \dots \\ & = \frac{n}{2\sqrt{in-i^2}} \\ \end{aligned}$

Writing the Riemann sum in the form

$\begin{aligned} \sum_{i=1}^{n}f(x_i)\, x & = \sum_{i=1}^{n}\frac{n}{2\sqrt{in-i^2}}\bigg(\frac{2Ri}{n}\bigg) \\ & = R\cdot \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}} \\ & = R\cdot g(n) \\ \end{aligned}$

Inspecting $R\cdot g(n)$ where

$g(n) = \displaystyle{ \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}}}$

I guess, under correction, that $g(n)=2\pi$ .

(to be continued)

Solution. (arc-length parametrization)

Referring to the equation of locus on the very first line:

$(x-R)^2+(y-0)^2=R^2$ ,

then parameterizing $x(\theta )$ , $y(\theta )$ by $\theta$ ,

$\begin{aligned} x & = R+R\cos\theta \\ y & = R\sin\theta \\ \end{aligned}$

and computing the derivatives wrt $\theta$ :

$\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}\theta} & = -R\sin\theta \\ \frac{\mathrm{d}y}{\mathrm{d}\theta} & = R\cos\theta \\ \end{aligned}$

Note that

$\displaystyle{s=\int\mathrm{d}s}$

where

$\mathrm{d}s=\sqrt{\displaystyle{\bigg(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}}\,\mathrm{d}\theta$ .

Then,

$\begin{aligned} s & = \int \mathrm{d}s \\ & = \int \sqrt{\bigg( \frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg( \frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}\,\mathrm{d}\theta \\ & = \int \sqrt{(-R\sin\theta )^2+(R\cos\theta )^2}\,\mathrm{d}\theta \\ & = \int \sqrt{R^2}\,\mathrm{d}\theta \\ & = \int_{\theta =0}^{2\pi} R\,\mathrm{d}\theta \\ & = 2\pi R \\ \end{aligned}$

(to be continued)

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Day: January 25, 2022

202201260719 Problem 1.57

202201251933 Circumference 001