Month: January 2022

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202201260719 Problem 1.57

Let A=\{ x:x^2-3x+2=0\}, B=\{ x:x^2\le 16\}. Determine whether or not A\subset B.

Extracted from M. R. Spiegel. (1969). Schaum’s Outline of Theory and Problems of Real Variables

Set A and set B are described according to property method; if described by the roster method:

\begin{aligned} A & =\{ 1,2\} \\ B & = \{ -4,-3,-2,-1,0,1,2,3,4\} \\ \end{aligned}

Obviously set A is a subset of set B.

Roughwork.

(set A)

\begin{aligned} x^2-3x+2 & = 0 \\ (x-2)(x-1) & = 0 \\ x & = 1,2 \\ \end{aligned}

(set B)

\begin{aligned} x^2 & \le 16 \\ x^2 - 16 & \le 0 \\ (x-4)(x+4) & \le 0 \\ -4\le x & \le 4 \\ \end{aligned}

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202201251933 Circumference 001

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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202201211321 Problem 1.1

Two particles move along the x-axis uniformly with speeds v_1=8\,\mathrm{m/s} and v_2=4\,\mathrm{m/s}. At the initial moment the first point was 21\,\mathrm{m} to the left of the origin and the second 7\,\mathrm{m} to the right of the origin. When will the first point catch up with the second? Where will this take place? Plot the graph of the motion.

Extracted from A. A. Pinsky. (1980). Problems in Physics.

Set-up.

Rename the two particles by a and b. The velocity of particle a is \mathbf{v}_a=+8\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}} and that of particle b is \mathbf{v}_b=+4\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}}. The particles at time t=0 are located on the x-axis with x-coordinates x_a=-21 and x_b=+7 respectively.

Roughwork.

When t=0:

when t=1:

when t=2:

when t=3:

when t=4:

when t=5:

when t=6:

when t=7:

Solution.

The positions x_a(t), x_b(t) of particle a, b can be expressed in a function of discrete time interval

t=\{ t_i\in\mathbb{Z^{+}}\textrm{ s.t. } t_{i+1}-t_{i}=t_{i}-t_{i-1}=1\},

i.e.,

\begin{aligned} x_a(t_{i+1}) & =x_a(t_{i})+8 \\ x_b(t_{i+1}) & =x_b(t_{i})+4 \\ \end{aligned}

or simply, in continuous time intervals,

\begin{aligned} x_a(t) & =-21+8t \\ x_b(t) & = 7+4t \\ \end{aligned}

Particle a will meet particle b when x_a(t)=x_b(t) at some time t', as

\begin{aligned} x_a(t') & = x_b(t') \\ -21+8t' & = 7+4t' \\ 4t' & = 28 \\ t' & = 7 \\ \end{aligned}

so the place of meeting is

x_a(7)=-21+8(7)=\boxed{35}=7+4(7)=x_b(7).