Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #5

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

Given a chance to counter Sir Isaac Newton’s famous quote of “What goes up must come down”, do you think it is true in all scenarios?

Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #4

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

Can you explain how a privacy screen protector works on your smartphone?

Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #3

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

In designing a circuit, an engineer needs to use five 5KΩ resistors to design a resistors network of approximately 4.3KΩ. How should he place the resistors to achieve that resistance?

Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #2

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

Assuming an object is moving in a circular motion in a polar coordinate system given by $x = r \text{cos} \theta$ and $y = r \text{sin} \theta$. Can you derive the formula of the centripetal acceleration? Hint: look at one of the axes and think of the direction and what is centripetal acceleration

Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #1

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

Kepler’s Third Law states that the period of a planet’s orbit is directly proportional to the cube of the semi-major axis of its orbit. This allows scientists to calculate distances of planets in our solar system. Can you derive the equation using what you have learned in Motion in a Circle and Gravitational Field?

[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #1

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by Aaron. More of him can be found here.

Why study Physics?

I have questioned many students and getting answers ranging from “Just a subject that I have to take in A-Level” to “I do not know what else to take?”

It is not just a subject:

Physics can be seen as a ‘tool’ to enable you to be more logical and you practise and learn to question ‘Why’, which is an invaluable asset to many other jobs or careers

A student’s dad was having a good chat with me and he was a control systems engineer back in his younger years in The States for a good decade or so. Thereafter he went on to take an MBA and he is now an MD in a local bank. Even if you are not going to major in engineering, it can be a good training.

Helping out with the house:

You can of course repair and fix things in the house without studying Physics but if you know more about topics like pressure, forces, electricity and how things work, won’t it be awesome that you are more knowledgeable? It can be a general knowledge and it will help you out in life along the way when things are screwed at home.

It puts your maths to good use:

Lots of formulas? Well, they help physicists and engineers to understand the world we live in.

Becoming an Engineer:

It is a myth that only bankers are being paid well. Only the good bankers are well paid, similar to engineers. If you excel in the things you do, you will be rewarded. Engineers are not technicians. Yes, they can fix things but they innovate and SOLVE problems too.

A world without Physics:

You will not be looking at this now on your computer/laptop or mobile device literally. Engineers continuously apply Physics to push the frontiers of technology.

It is not just a subject:

Physics can be seen as a ‘tool’ to enable you to be more logical and you practise and learn to question ‘Why’, which is an invaluable asset to many other jobs or careers

A student’s dad was having a good chat with me and he was a control systems engineer back in his younger years in The States for a good decade or so. Thereafter he went on to take an MBA and he is now an MD in a local bank. Even if you are not going to major in engineering, it can be a good training.

Helping out with the house:

You can of course repair and fix things in the house without studying Physics but if you know more about topics like pressure, forces, electricity and how things work, won’t it be awesome that you are more knowledgeable? It can be a general knowledge and it will help you out in life along the way when things are screwed at home.

It puts your maths to good use:

Lots of formulas? Well, they help physicists and engineers to understand the world we live in.

Becoming an Engineer:

It is a myth that only bankers are being paid well. Only the good bankers are well paid, similar to engineers. If you excel in the things you do, you will be rewarded. Engineers are not technicians. Yes, they can fix things but they innovate and SOLVE problems too.

A world without Physics:

You will not be looking at this now on your computer/laptop or mobile device literally. Engineers continuously apply Physics to push the frontiers of technology.

Differentiation Question #1

Given that $y = \frac{8}{x^3} - \frac{6}{x^2} + \frac{5}{2x}$, find the approximate percentage change in $y$ when $x$ increases from 2 by 2%.

Probability Question #4

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears $k$ times ($k = 1, 2, 3$), he wins $$k$. If his number fails to appear, he loses$1. Calculate the gambler’s expected winnings

Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

let $\frac{dy}{dx} = 0$

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

let $\frac{dy}{dx} = 0$

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…

A little reminder to students doing Calculus now

When $\frac{dy}{dx} = 0$, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

Students should write the actual values of $\alpha^-, \alpha, \alpha^+$ and $\frac{dy}{dx}$ in the table.

We use this under these two situations:
1. $\frac{d^2y}{dx^2}$ is difficult to solve for, that is, $\frac{dy}{dx}$ is tough to be differentiated
2. $\frac{d^2y}{dx^2} = 0$

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.