Richard Feyman: colourful life.

https://www.hindustantimes.com/more-lifestyle/feyn-balance-meet-one-of-the-world-s-most-influential-colourful-physicists/story-a6T3fQUHspE47zxYHGoHWO.html

Thanks to Hindustan Times and Rachel Lopez.

 

Remembering Richard Feynman on his 100th birth centenary

https://www.hindustantimes.com/more-lifestyle/infectious-enthusiastic-relevant-a-physicist-s-take-on-richard-feynman/story-jD3jualkHCX5WuXcCp0ndL.html

Thanks to Hindustan Times and Prof. Arnab Bhattacharya, TIFR, Mumbai.

Co-ordinate geometry practice for IITJEE Maths: Ellipses

Problem 1:

Find the locus of the point of intersection of tangents to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}=1, which are at right angles.

Solution I:

Any tangent to the ellipse is y = mx + \sqrt{a^{2}m^{2}+b^{2}}….call this equation I.

Equation of the tangent perpendicular to this tangent is y=-\frac{-1}{m}x+\sqrt{\frac{a^{2}}{m^{2}}+b^{2}}…call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}

\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})

\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.

Problem II:

A tangent to the ellipse x^{2}+4y^{2}=4 meets the ellipse x^{2}+2y^{2}=6 at P and Q. Prove that the tangents at P and Q of the ellipse x^{2}+2y^{2}=6 at right angles.

Solution II:

Let the tangent at R(2\cos {\theta}, \sin{\theta}) to the ellipse x^{2}+4y^{2}=4 meet the ellipse x^{2}+2y^{2}=6 at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point S(\alpha,\beta). Then, PQ is the chord of contact of the point S(\alpha,\beta) with respect to ellipse two, and so its equation is

\alpha x + 2\beta y=6….call this “A”.

PQ is also the tangent at R(2\cos {\theta}, \sin{\theta}) to the first ellipse and so the equation can be written as (2\cos{\theta})x+(4\sin{\theta})y=4….call this “B”.

Comparing “A” and “B”, we get \frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}

\Longrightarrow \cos{\theta}=\frac{\alpha}{3} and \sin{\theta}=\frac{\beta}{3}

\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9}=1 \Longrightarrow \alpha^{2}+\beta^{2}=9

The locus of S( \alpha, \beta) is x^{2}+y^{2}=9, or x^{2}+y^{2}=6+3, which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

Problem 3:

Let d be the perpendicular distance from the centre of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1 to the tangent drawn at a point P on the ellipse. If F_{1} and F_{}{2} are the two foci of the ellipse, then show that (PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}})

Solution 3:

Equation of the tangent at the point P(a\cos {\theta}, b\sin{\theta}) on the given ellipse is \frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1. Thus,

d= |\frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}}+\frac{\sin^{2}{\theta}}{b^{2}}}}|

\Longrightarrow d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}

We know PF_{1}+PF_{2}=2a

\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}…call this equation I.

Also, (PF_{1}.PF_{2})^{2}=[ (a\cos{\theta}-ae)^{2}+(b\sin{\theta})^ {2}].[(a\cos{\theta}+ae)^{2} + (b\sin{\theta})^{2}], which in turn equals,

[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]. [a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ], that is,

a^{4}[ (\cos^{2}{\theta}+e^{2}) -2e\cos{\theta}+\sin^{2}{\theta} - e^{2} \sin^{2}{\theta} ]. [ (\cos^{2}{\theta}+e^{2}) + 2e \cos{\theta} + \sin^{2}{\theta} - e^{2}\sin^{2}{\theta}] = a^{4}[ 1-2e\cos{\theta}+e^{2}\cos^{2}{\theta} ] [1+ 2e \cos{\theta} + e^{2}\cos^{2}{\theta} ]

that is,

a^{4}[ (1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}(1-e^{2}\cos^{2}{\theta})^{2}

\Longrightarrow PF_{1}. PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})

Now, from I, we get (PF_{1} - PF_{2})^{2} = 4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta}) = 4a^{2}e^{2}\cos^{2}{\theta},

also, 1-\frac{b^{2}}{a^{2}} = 1 - \frac{b^{2}\cos^{2}{\theta} + a^{2}\sin^{2}{\theta}}{a^{2}} = \frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}

Hence, (PF_{1} - PF_{2})^{2} = 4a^{2}(1-\frac{b^{2}}{d^{2}})

we will continue later,

Cheers,

Nalin Pithwa

B.S. in Mathematics: IIT Bombay program:

http://www.math.iitb.ac.in/Academics/bs_programme.php

Note that the admission is through IITJEE Advanced only.

Nalin Pithwa.

Co-ordinate Geometry problems for IITJEE : equations of median, area of a triangle, and circles

Problem I:

If A(x_{1}, y_{1}), B(x_{1}, y_{1}) and C(x_{3}, y_{3}) are the vertices of a triangle ABC, then prove that the equation of the median through A is given by:

\left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1  \end{array}\right | + \left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right |=0

Solution I:

If D is the mid-point of BC, its co-ordinates are ( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} )

Therefore, equation of the median AD is \left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1\\ \frac{x_{2}+x_{3}}{2} & \frac{y_{2}+y_{3}}{2} & 1 \end{array} \right|=0, which in turn, implies that,

\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2}+x_{3} & y_{2}+y_{3} & 2 \end{array}\right |=0

Now apply the row transformation R_{3} \rightarrow 2R_{3} to the previous determinant. So, we get

\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0, using the sum property of determinants.

Hence, the proof.

Problem 2:

If \triangle_{1} is the area of the triangle with vertices (0,0), (a\tan {\alpha},b\cot{\alpha}), (a\sin{\alpha}, b\cos {\alpha}), and \triangle_{2} is the area of the triangle with vertices (a,b), (a\sec^{2}{\alpha}, b\csc^{2}{\alpha}), and (a+a\sin^{2}{\alpha}, b+b\cos^{2}{\alpha}), and \triangle_{3} is the area of the triangle with vertices ( 0, 0), ( a\tan{\alpha}, -b\cos{\alpha}), (a\sin{\alpha},b\cos{\alpha}). Then, prove that there is no value of \alpha for which the areas of triangles, \triangle_{1}, \triangle_{2} and \triangle_{3} are in GP.

Solution 2:

We have \triangle_{1}=\frac{1}{2}|\left | \begin{array}{ccc}0 & 0 & 1 \\ a\tan{\alpha} & b\cot {\alpha} & 1 \\ a \sin{\alpha} & b\cos{\alpha} & 1 \end{array}\right ||=\frac{1}{2}ab|\sin{\alpha}-\cos{\alpha}|, and

\triangle_{2}=\frac{1}{2}|\left | \begin{array}{ccc}a & b & 1 \\ a\sec^{2}{\alpha} & b\csc^{2}{\alpha} & 1 \\ a + a\sin^{2}{\alpha} & b + b\cos^{2}{\alpha} & 1 \end{array} \right | |.

Applying the following column transformations to the above determinant, C_{1} \rightarrow -aC_{3} and C_{2}-bC_{3}, we get

\triangle_{2}=\frac{1}{2}ab\left | \begin{array}{ccc}0 & 0 & 1 \\ \tan^{2}{\alpha} & \cot^{2}{\alpha} & 1 \\ \sin^{2}{\alpha} & \cos^{2}{\alpha} & 1 \end{array}\right | = \frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha}) and \triangle_{3}=\frac{1}{2}|\left | \begin{array}{ccc} 0 & 0 & 1 \\ a\tan{\alpha} & -b\cot{\alpha} & 1 \\ a\sin{\alpha} & b\cos{\alpha} & 1 \end{array} \right | |=\frac{1}{2}ab |\sin {\alpha}+\cos{\alpha}|

so that \triangle_{1}\triangle_{3}=\frac{1}{2}ab\triangle_{2}.

Now, \triangle_{1}, \triangle_{2} and \triangle_{3} are in GP, if \triangle_{1}\triangle_{3}=\triangle_{2}^{2} \Longrightarrow \frac{1}{2}ab\triangle_{2}=\triangle_{2}^{2} \Longrightarrow \triangle_{2}=\frac{1}{2}ab

\Longrightarrow \triangle_{2}=\frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})=\frac{1}{2}ab \Longrightarrow (\sin^{2}{\alpha}-\cos^{2}{alpha})=1, that is,

\alpha = (2m+1)\pi/2, where m \in Z. But, for this value of \alpha, the vertices of the given triangles are not defined. Hence, \triangle_{1}, and \triangle_{2} and \triangle_{3} cannot be in GP for any value of \alpha.

Problem 3:

Two points P and Q are taken on the line joining the points A(0,0) and B(3a,0) such that AP=PQ=QB. Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to b^{2}, is

(a) x^{2}+y^{2}-3ax+2a^{2}-b^{2}=0

(b) 3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0

(c) x^{2}+y^{2}-5ax+6a^{2}-b^{2}=0

(d) x^{2}+y^{2}-ax-b^{2}=0.

Ans. b.

Solution 3:

Since AP=PQ=QB, the co-ordinates of P are (a,0) and of Q are (2a,0), equations of the circles on AP, PQ, and QB as diameters are respectively.

Please draw the diagram.

So, we get

(x-0)(x-a)+y^{2}=0

(x-a)(x-2a)+y^{2}=0

(x-2a)(x-3a)+y^{2}=0

So, if (h,k) be any point of the locus, then 3(h^{2}+k^{2})-9ah+8a^{2}=b^{2}.

So, the required locus of (h,k) is 3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0.

More later,

Nalin Pithwa.

Bill Gates returns to Harvard to talk: Math 55

https://www.thecrimson.com/article/2018/4/27/bill-gates-event/

How to solve equations: Dr. Vicky Neale: useful for Pre-RMO or even RMO training

Dr. Neale simply beautifully nudges, gently encourages mathematics olympiad students to learn to think further on their own…

Stephen Hawking in his own words: excerpts: My Brief History

I. Another early memory was getting my first train set. Toys were not manufactured during the war, at least not for the home market. But, I had a passionate interest in model trains. My father tried making me a wooden train, but that didn’t satisfy me, as I wanted something that moved on its own. So, he got a secondhand clockwork train, repaired it with a soldering iron, and gave it to me for Christmas when I was nearly three. That train didn’t work very well. But my father went to America just after the war, and when he came back on the Queen Mary he bought my mother some nylons, which were not obtainable in Britain at that time. He bought my sister Mary a doll that closed its eyes when you laid it down. And he brought me an American train, complete with a cowcatcher and a figure-eight track. I can still remember my excitement as I opened the box.

Clockwork trains, which you had to wind up, were all very well, but I really wanted were electric trains. I used to spend hours watching a model railway club layout in Crouch End, near Highgate. I dreamed about electric trains. Finally, when both my parents were away somewhere, I took the opportunity to draw out of the Post Office bank all of the very modest amount of money that people had given me on special vacations, such as my christening. I used the money to buy an electric train set, but frustratingly enough, it didn’t work very well either. I should have taken the set back and demanded that the shop or the manufacturer replace it, but in those days the attitude was that it was a privilege to buy something and it was just your bad luck, if it turned out to be faulty. So, I paid for the electric motor of the engine to be serviced, but it never worked very well, even then.

Later on, in my teens, I built model aeroplanes and boats. I was never very good with my hands, but I did this with my school friend John McClenahan, who was much better and whose father had a workshop in their house. My aim was always to build working models that I could control. I didn’t care what they looked like. I think it was the same drive that led me to invent a series of very complicated games with another school friend, Roger Ferneyhough. There was a manufacturing game, complete with factories in which units of different colours were made, roads and railways on which they were carried, and a stock market. There was a war game, played on a board of 4000 squares, and even a feudal game, in which each player was a whole dynasty, with a family tree. I think these games, as well as the trains, boats, and aeroplanes, came from an urge to know how systems worked and how to control them. Since I began my PhD, this need has been met by research into cosmology. If you understand how the universe operates, you control it, in a way.

II. When I was thirteen, my father wanted me to try for Westminister School, one of Britain’s main public schools (what in the United States are called private schools). At that time, there was a sharp division in education along class lines, and my father felt that the social graces such a school would give me would be an advantage in life. My father believed that his own lack of poise and connections had led to him being passed over in his career in favour of people of less ability. He had a bit of a chip on his shoulder because he felt that other people who were not as good but who had the right background and connections had got ahead of him. He used to warn me against such people.

Because my parents were not well-off, I would have to win a scholarship in order to attend Westminister. I was ill at the time of the scholarship examination, however, and did not take it. Instead, I remained at St. Albans School, where I got an education that was as good as, if not better than, the one I would have had at Westminister. I have never found that my lack of social graces has been a hindrance. But I think physics is a bit different from medicine. In physics it doesn’t matter what school you went to or whom you are related. It matters what you do.

I was never more than about halfway up the class (it was a very bright class.) My classwork was untidy, and my handwriting was the despair of my teachers. But, my classmates gave me the nickname Einstein, so presumably they saw signs of something better. When I was twelve, one of my friends bet another friend a bag of sweets that I would never amount to anything. I don’t know if this bet was ever settled, and if so, which way it was decided.

I had six or seven close friends, most of whom I’m still in touch with. We used to have long discussions and arguments about everything from radio-controlled models to religion and from parapsychology to physics. One of the things we talked about was the origin of the universe and whether it had required a God to create it and set it going. I had heard that light from distant galaxies was shifted towards the red end of the spectrum and that this was supposed to indicate that the universe was expanding. (A shift towards the blue would have meant it was contracting.) But, I was sure there must be some other reason for the red shift. An essentially unchanging and everlasting universe seemed so much more natural. Maybe light just got tired, and more red, on its way to us, I speculated. It was only after about two years of PhD research that I realized that I had been wrong.

I was always very interested in how things operated, and I used to take them apart to see how they worked, but I was not so good at putting them back together again. My practical abilities never matched up to my theoretical inquiries. My father encouraged my interest in science, and he even coached me in mathematics until I got to a stage beyond his knowledge. With this background and my father’s job, I took it as natural that I would go into scientific research.

When I came to the last two years of school, I wanted to specialize in mathematics and physics. There was an inspiring maths teacher, Mr. Tanta, and the school had also just built a new maths room, which the maths set had as their classroom. But my father was very much against it because he thought there wouldn’t be any jobs for mathematicians except as teachers. He would really have liked me to do medicine, but I showed no interest in biology, which seemed to me to be too descriptive and not sufficiently fundamental. It also had a rather low status at school. The brightest boys did mathematics and physics; the less bright did biology.

My father knew I wouldn’t do biology, but he made me do chemistry and only a small amount of mathematics. He felt this would keep my scientific options open. I am now professor of mathematics, but I have not had any formal instruction in mathematics since I left St. Albans School at the age of seventeen. I have had to pick up what I know as I went along. I used to supervise undergraduates at Cambridge and kept one week ahead of them in the course.

Physics was always the most boring subject at school because it was so easy and obvious. Chemistry was much more fun because unexpected things, such as explosions kept happening. But physics and astronomy offered the hope of understanding where we came from and why we are here. I wanted to fathom the depths of the universe. Maybe I have succeeded to a small extent, but there’s still plenty I want to know.

III. My early work showed that classical general relativity broke down at singularities in the Big Bang and black holes. My later work has shown how quantum theory can predict what happens at the beginning and end of time. It has been a glorious time to be alive and doing research in theoretical physics. I’m happy if I have added something to our understanding of the universe.

A humble tribute to Professor Hawking …to understand him from a layman’s viewpoint…by Nalin Pithwa.

Reference: My Brief History by Stephen Hawking, Bantam Press.

Amazon India link:

An immortal genius Professor Stephen Hawking is no more in this world

Some tributes that I can only humbly share …

 

 

 

 

Happy Pie with Pi…

via Everything you wanted to know about pi but were afraid to ask