Category Archives: applications of maths

Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

Question 1.

If the point on $y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}$, where $\alpha>0$, where the tangent is parallel to $y=x$ has an ordinate $\frac{u^{2}}{4a}$, then what is the value of $\alpha$?

Question 2:

Prove that the segment of the tangent to the curve $y=c/x$, which is contained between the coordinate axes is bisected at the point of tangency.

Question 3:

Find all the tangents to the curve $y = \cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$.

Question 4:

Prove that the curves $y=f(x)$, where $f(x)>0$, and $y=f(x)\sin{x}$, where $f(x)$ is a differentiable function have common tangents at common points.

Question 5:

Find the condition that the lines $x \cos{\alpha} + y \sin{\alpha} = p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$.

Question 6:

Find the equation of a straight line which is tangent to one point and normal to the point on the curve $y=8t^{3}-1$, and $x=4t^{2}+3$.

Question 7:

Three normals are drawn from the point $(c,0)$ to the curve $y^{2}=x$. Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other.

Question 8:

If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2} + p_{2}^{2}=a^{2}$.

Question 9:

Show that the curve $x=1-3t^{2}$, and $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$. If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t= \tan {\psi} +\sec {\psi}$. If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Question 10:

Find the condition that the curves $ax^{2}+by^{2}=1$ and $a^{'}x^{2} + b^{'}y^{2}=1$ intersect orthogonality and hence show that the curves $\frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1$ and $\frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1$ also intersect orthogonally.

More later,

Nalin Pithwa.

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

Another set of “easy to moderately difficult” questions:

1. The function $y = \frac{}x{1+x^{2}}$ decreases in the interval (a) $(-1,1)$ (b) $[1, \infty)$ (c) $(-\infty, -1]$ (d) $(-\infty, \infty)$. There are more than one correct choices. Which are those?
2. The function $f(x) = \arctan (x) - x$ decreases in the interval (a) $(1,\infty)$ (b) $(-1, \infty)$ (c) $(-\infty, -\infty)$ (d) $(0, \infty)$. There is more than one correct choice. Which are those?
3. For $x>1$, $y = \log(x)$ satisfies the inequality: (a) $x-1>y$ (b) $x^{2}-1>y$ (c) $y>x-1$ (d) $\frac{x-1}{x}. There is more than one correct choice. Which are those?
4. Suppose $f^{'}(x)$ exists for each x and $h(x) = f(x) - (f(x))^{2} + (f(x))^{3}$ for every real number x. Then, (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general. Find the correct choice(s).
5. If $f(x)=3x^{2}+12x-1$, when $-1 \leq x \leq 2$, and $f(x)=37-x$, when $2. Then, (a) $f(x)$ is increasing on $[-1,2]$ (b) $f(x)$ is continuous on $[-1,3]$ (c) $f^{'}(2)$ doesn’t exist (d) $f(x)$ has the maximum value at $x=2$. Find all the correct choice(s).
6. In which interval does the function $y=\frac{x}{\log(x)}$ increase?
7. Which is the larger of the functions $\sin(x) + \tan(x)$ and $f(x)=2x$ in the interval $(0?
8. Find the set of all x for which $\log {(1+x)} \leq x$.
9. Let $f(x) = |x-1| + a$, if $x \leq 1$; and, $f(x)=2x+3$, if $x>1$. If $f(x)$ has local minimum at $x=1$, then $a \leq$ ?
10. There are exactly two distinct linear functions (find them), such that they map $[-1,1]$ and $[0,2]$.

more later, cheers,

Nalin Pithwa.

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

“Easy” questions:

Question 1:

Find the slope of the tangent to the curve represented by the curve $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$ at the point $(2,-1)$.

Question 2:

Find the co-ordinates of the point P on the curve $y^{2}=2x^{3}$, the tangent at which is perpendicular to the line $4x-3y+2=0$.

Question 3:

Find the co-ordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $x^{2}/8 + y^{2}/18=1$ so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest.

Question 4:

The function $f(x) = \frac{\log (\pi+x)}{\log (e+x)}$, where $x \geq 0$ is

(a) increasing on $(-\infty, \infty)$

(b) decreasing on $[0, \infty)$

(c) increasing on $[0, \pi/e)$ and decreasing on $[\pi/e, \infty)$

(d) decreasing on $[0, \pi/e)$ and increasing on $[\pi/e, \infty)$.

Fill in the correct multiple choice. Only one of the choices is correct.

Question 5:

Find the length of a longest interval in which the function $3\sin(x) -4\sin^{3}(x)$ is increasing.

Question 6:

Let $f(x)=x e^{x(1-x)}$, then $f(x)$ is

(a) increasing on $[-1/2, 1]$

(b) decreasing on $\Re$

(c) increasing on $\Re$

(d) decreasing on $[-1/2, 1]$.

Fill in the correct choice above. Only one choice holds true.

Question 7:

Consider the following statements S and R:

S: Both $\sin(x)$ and $\cos (x)$ are decreasing functions in the interval $(\pi/2, \pi)$.

R: If a differentiable function decreases in the interval $(a,b)$, then its derivative also decreases in $(a,b)$.

Which of the following is true?

(i) Both S and R are wrong.

(ii) Both S and R are correct, but R is not the correct explanation for S.

(iii) S is correct and R is the correct explanation for S.

(iv) S is correct and R is wrong.

Indicate the correct choice. Only one choice is correct.

Question 8:

For which of the following functions on $[0,1]$, the Lagrange’s Mean Value theorem is not applicable:

(i) $f(x) = 1/2 -x$, when $x<1/2$; and $f(x) = (1/2-x)^{2}$, when $x \geq 1/2$.

(ii) $f(x) = \frac{\sin(x)}{x}$, when $x \neq 0$; and $f(x)=1$, when $x=0$.

(iii) $f(x)=x |x|$

(iv) $f(x)=|x|$.

Only one choice is correct. Which one?

Question 9:

How many real roots does the equation $e^{x-1}+x-2=0$ have?

Question 10:

What is the difference between the greatest and least values of the function $f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)$?

More later,

Nalin Pithwa.

Applications of Derivatives: A Quick Review

Section I:

The Derivative as a Rate of Change

In case of a linear function $y=mx+c$, the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from $x_{0}$ to $x_{1}$, y changes m times as much:

$y_{1}-y_{0}=m(x_{1}-x_{0})$

Thus, the slope $m=(y_{1}-y_{0})(x_{1}-x_{0})$ gives the change in y per unit change in x.

In more general case of differentiable function $y=f(x)$, the difference quotient

$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$, where $h \neq 0$

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative $dy/dx = f^{'}(x)$, which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position $x(t)$. Then, at time $t+h$ the position of the object is $x(t+h)$ and $x(t+h)-x(t)$ is the change in position that the object experienced during the time period t to $t+h$. The ratio

$\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}$

gives the average velocity of the object during this time period. If

$\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)$

exists, then $x^{'}(t)$ gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

$a(t) = v^{'}(t) = x^{''}(t)$

The speed is by definition the absolute value of the velocity: speed at time t is $|v(t)|$

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is $x(t)$, the body’s jerk at time t is

$j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}$

Differentials

Let $y = f(x)$ be a differentiable function. Let $h \neq 0$. The difference $f(x+h) - f(x)$ is called the increment of f from x to $x+h$, and is denoted by $\Delta f$.

$\Delta f = f(x+h) - f(x)$

The product $f^{'}(x)h$ is called the differential of f at x with increment h, and is denoted by $df$

$df = f^{'}(x)h$

The change in f from x to $x+h$ can be approximated by $f^{'}(x)h$:

$f(x+h) - f(x) = f^{'}(x)h$

Tangent and Normal

Let $y = f(x)$ be the equation of a curve, and let $P(x_{0}, y_{0})$ be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve $y = f(x)$ at P is given by $(\frac{dy}{dx})_{(x_{0}, y_{0})}$

Thus, the equation of the tangent to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is $y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})$

Since PM is perpendicular to PT, it follows that if $(\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0$, the slope of PN is

$- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}$

Hence, the equation of the normal to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is

$y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})$

The equation of the normal parallel to the x-axis is $y = y_{0}$, that is, when $(\frac{dy}{dx})_{(x_{0}, y_{0})} = 0$. The length of the tangent at $(x_{0}, y_{0})$ is PT, and it is equal to

$y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}$

The length of the normal is PN and it is equal to $y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}$

If the curve is represented by $x = f(t)$ and $y = g(t)$, that is, parametric equations in t, then

$\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)}$ where $g^{'}(t)= \frac{dy}{dt}$ and $f^{'}(t) = \frac{dx}{dt}$. In this case, the equations of the tangent and the normal are given by

$y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)]$ and $[y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0$ respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let $y = f(x)$ and $y=g(x)$ be two curves, and let $P(x_{0}, y_{0})$ be their point of intersection. Also, let $\psi$ and $\phi$ be the angles of inclination of the two tangents with the x-axis, and let $\theta$ be the angle between the two tangents. Then,

$\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}$

Example 1:

Write down the equations of the tangent and the normal to the curve $y = x^{3} - 3x + 2$ at the point $(2,4)$.

Solution 1:

$\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9$.

Hence, the equation of the tangent at $(2,4)$ is given by $y-4 = 9(x-2) \Longrightarrow 9x-y-14=0$ and the equation of the normal is $y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0$.

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) f(x) is continuous on $[a,b]$, (ii) f(x) is derivable on $(a,b)$, and (iii) f(a) = f(b). Then, there exists a $c \in (a,b)$ such that $f^{'}(x)=0$.

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) $f(x)$ is continuous on $[a,b]$, and (ii) $f(x)$ is derivable on $(a,b)$. Then, there exists a $c \in [a,b]$ such that

$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$

Example 2:

The function $f(x) = \log {\sin(x)}$ satisfies the conditions of Rolle’s theorem on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$, as the logarithmic function and $\sin (x)$ are continuous and differentiable functions and $\log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}$.

The conclusion of Rolle’s theorem is given at $c=\frac{\pi}{2}$, for which $f^{'}(c) = \cot (c) = \cot (\pi/2) =0$.

Rolle’s theorem for polynomials:

If $\phi(x)$ is any polynomial, then between any pair of roots of $\phi(x)=0$ lies a root of $\phi^{'}(x)=0$.

Monotonicity:

A function $f(x)$ defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any $x_{1}, x_{2} \in D$ and $x_{1} < x_{2}$, we have $f(x_{1}) \leq f(x_{2})$, $f(x_{1}) < f(x_{2})$, $f(x_{1}) \geq f(x_{2})$ and $f(x_{1}) > f(x_{2})$ respectively. The function $f(x)$ is said to be monotonic if it possesses any of these properties.

For example, $f(x) = e^{x}$ is an increasing function, and $f(x)=\frac{1}{x}$ is a decreasing function.

Testing monotonicity:

Let $f(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then,

(i) for $f(x)$ to be non-decreasing (non-increasing) on $[a,b]$ it is necessary and sufficient that $f^{'}(x) \geq 0$ ($f^{'}(x) \leq 0$) for all $x \in (a,b)$.

(ii) for $f(x)$ to be increasing (decreasing) on $[a,b]$ it is sufficient that $f^{'}(x)>0$ ($f^{'}(x)<0$) for all $x \in (a,b)$.

(iii) If $f^{'}(x)=0$ for all x in $(a,b)$, then f is constant on $[a,b]$.

Example 3:

Determine the intervals of increase and decrease for the function $f(x)=x^{3}+2x-5$.

Solution 3:

We have $f^{'}(x) = 3x^{2}+2$, and for any value of x, $3x^{2}+2>0$. Hence, f is increasing on $(-\infty, -\infty)$. QED.

The following is a simple criterion for determining the sign of $f^{'}(x)$:

If $a,b \geq 0$, then $(x-a)(x-b)>0$ iff $x > \max (a,b)$ or $x < \min(a,b)$;

$(x-a)(x-b)<0$ if and only if $\min(a,b) < x < \max(a,b)$

Maxima and Minima:

A function has a local maximum at the point $x_{0}$ if the value of the function $f(x)$ at that point is greater than its values at all points other than $x_{0}$ of a certain interval containing the point $x_{0}$. In other words, a function $f(x)$ has a maximum at $x_{0}$ if it is possible to find an interval $(\alpha, \beta)$ containing $x_{0}$, that is, with $\alpha < x_{0} < \beta$, such that for all points different from $x_{0}$ in $(\alpha, \beta)$, we have $f(x) < f(x_{0})$.

A function $f(x)$ has a local minimum at $x_{0}$ if there exists an interval $(\alpha, \beta)$ containing $x_{0}$ such that $f(x) > f(x_{0})$ for $x \in (\alpha, \beta)$ and $x \neq x_{0}$.

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point $x_{0}$ of the function $f(x)$ is that $f^{'}(x_{0})=0$, or $f^{'}(x_{0})$ does not exist. The points at which $f^{'}(x)=0$ or $f^{'}(x)$ does not exist, are called critical points.

First Derivative Test:

(i) If $f^{'}(x)$ changes sign from positive to negative at $x_{0}$, that is, $f^{'}(x)>0$ for $x < x_{0}$ and $f^{'}(x)<0$ for $x > x_{0}$, then the function attains a local maximum at $x_{0}$.

(ii) If $f^{'}(x)$ changes sign from negative to positive at $x_{0}$, that is, $f^{'}(x)<0$ for $x, and $f^{'}(x)>0$ for $x > x_{0}$, then the function attains a local minimum at $x_{0}$.

(iii) If the derivative does not change sign in moving through the point $x_{0}$, there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation $f^{'}(x)=0$. Then,

(i) c is a local maximum point if $f^{''}(c)<0$.

(ii) c is a local minimum point if $f^{''}(c)>0$.

However, if $f^{''}(c)=0$, then the following result is applicable. Let $f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0$ (where f^{r} denotes the rth derivative), but $f^{(n)}(c) \neq 0$.

(i) If n is even and $f^{(n)}(c)<0$, there is a local maximum at c, while if $f^{(n)}(c)>0$, there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point $c \in D$ if $f^(x) \leq f(c)$ for all x in D and a least value (or absolute minimum) at c, if $f(x) \geq f(c)$ for all x in D.

If f is continuous at every point of D, and $D=[a,b]$, a closed interval, the f assumes both a greatest value M and a least value m, that is, there are $x_{1}, x_{2} \in [a,b]$ such that $f(x_{1})=M$ and $f(x_{2})=m$, and $m \leq f(x) \leq M$ for every $x \in [a,b]$.

Example 4:

a) $y=x^{2}$, with domain $(-\infty, \infty)$. This has no greatest value; least value at $x=0$

b) $y=x^{2}$ with domain $[0,2]$. This has greatest value at $x=2$ and least value at $x=0$.

c) $y=x^{2}$ with domain $(0,2]$. This has greatest value at $x=2$ and no least value.

d) $y=x^{2}$ with domain $(0,2)$. This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function $f(x)$ on the interval $[a,b]$ is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval $(a,b)$, and the values $f(a), f(b)$ of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.

Nalin Pithwa.

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.

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

Some tributes that I can only humbly share …

Listening math : Gaurish Korpal way :-) :-) :-)

https://gaurish4math.wordpress.com/2018/02/17/listening-maths/

Thanks Gaurish —

From — Nalin Pithwa.