The post The Slide Rule Watch appeared first on Ringo Mok.

]]>I was born in 1991 and when I was old enough to learn about numbers, the digital calculator has already replaced the slide rule – an analogue calculator that, as the name suggests, required the physical action of sliding markers around to do calculations such as multiplication, division, exponents and trigonometry. For many millenials like myself, the slide rule is like a cassette tape to Gen Z – most of us have no idea what it is and it really doesn’t matter because we have better technology now.

However, as a mathematician, I am can’t help but be in awe of such a device, no matter how obsolete it may be, for its brilliant and ingenious application of logarithms.

In this article, I’ll explain the basics of how it works, but first let’s talk about using two rulers for adding and subtracting:

To do a simple sum like 6 + 4, we place the origin of the yellow ruler on top of the 6 of the metal ruler, and read the answer off the metal ruler where the yellow ruler reads 4.

To do a subtraction of say 11 – 5, we place the yellow 5 above the metal ruler’s 11, and then read back at the origin of the yellow ruler to get an answer of 6.

This is pretty intuitive.

When it comes to the slide rule, the markers are not equally spaced apart, but rather they are spaced according to a logarithm scale of base 10. Which means the origin of the log scale starts at 1, not 0, since \(\log_{10}(1) = 0\), and these are followed by log 2, log 3, log 4, … log 9.

Now, using log laws notice the following:

\(\log_{10}(10) = 1\)

\(\log_{10}(20) = \log_{10}(10\times 2) = \log_{10}(10) + \log_{10}(2) = 1 + \log_{10}(2)\)

\(\log_{10}(30) = 1 + \log_{10}(3)\)

\(\ldots\)

\(\log_{10}(100) = 2\)

\(\log_{10}(200) = 2+\log_{10}(2)\)

\(\log_{10}(300) = 2+\log_{10}(3)\)

\(\ldots\)

This means every time we get to the next power of 10, the exponent of all the markers after them are increased by one as well. i.e. when you measure a \(2 + \log_{10}(5)\) marker on the ruler, it represents the number 500 and a \(3 + \log_{10}(5)\) marker represents 5000.

For more precision, the slide rule would usually exclude the numbers 1, 2, 3, through to 9 but instead mark out the logarithms of 10, 11, 12, …, 99 then move on to 100, 110, 120, 130, …, 990, 1000, 1100, 1200, so on. The more significant figures marked will mean more precision.

Consider the multiplication \( 1.5 \times 70\).

Now consider the following:

\(\log_{10}(1.5 \times 70) = \log_{10}(1.5) \times \log_{10}(70) = -1 + \log_{10}(15) + \log_{10}(70)\).

Now the calculator will have markings for \(\log_{10}(15) and \log_{10}(70)\), and it also becomes an addition problem. We saw from my example with the yellow ruler and metal ruler that it is easy to calculate the addition and subtraction of two numbers when we have their markings.

Using the same method as that, we can find the answer for \(\log_{10}(15) + \log_{10}(70)\). All we have to do is keep track of the exponent which will tell us what power of 10 to multiply the reading by.

During the second world war, increasing demand by pilots for a quick way to do calculations in the cockpit prompted the watchmaker company Breitling to meet this request. Willy Breitling asked mathematician Marcel Robert to design a watch bezel and dial to do logarithmic calculations such as those mentioned above. Thus, the Breitling Chronomat watch, which was later improved upon by the Breitling Navitimer watch, was created.

In the creation of this design, Marcel Robert would have had to apply his knowledge of radians, arc lengths and trigonometry to the logarithmic scale! I can see a lot of potential mathematics problems he would have had to solve to finally arrive at this solution.

Unfortunately, I do not have the funds to purchase such an iconic piece myself, so I make do with the Seiko SSC-009 until one day, perhaps… For calculations, the only complaint that I have with the SSC-009 is the presence of parallax error from the bezel being slightly raised above the dial.

Let’s have a look at what we can do with this.

To find \(1.5 \times 70\), we position the bezel’s marking for 15, above the origin of the inner dial, 10. Remember the arc length between 10 and 15 represents a distance of \(\log_{10}(15)\). We then add a distance of \(\log_{10}(70)\), so from the inner dial we add an arc length between 10 and 70. Reading off the outer bezel, we get the reading 10.5. Remember we need to keep track of exponents, which requires us to have some sense with numbers, and we know the answer is therefore 105.

Consider this scenario: the exchange rate in Hong Kong is 5.5 HKD for 1 AUD. If I wanted to buy an item for 74 HKD, find the equivalent price in AUD.

For this problem, I position the 74 above the number 55, then read off the answer from above the inner dial’s 10. This works because \(\log_{10}(74 \div 5.5) = \log_{10}(74) – \log_{10}(5.5)\) and it becomes a matter of subtraction. Using the technique outlined in yellow/metal ruler scenario, I read off my answer from the origin, 10.

Hence, 74 HKD is about 13.5 AUD.

The great thing about unit conversions is that they’re usually just simple multiplication or divisions by a set constant. For example, converting miles to kilometres is just a multiplication by 1.609. On the logarithmic scale, therefore, the number of miles is always the same distance away from the number of kilometres.

Positioning the arrow for km above 70, I can then read off the other distance markings for their equivalent conversions!

My brother is training to become a commercial pilot. Cadets training to become pilots still need to know how to operate a circular slide rule in their examinations but realistically, they just use GPS and automatic instruments. However, when these instruments fail, they will need to default back to these methods without a digital calculator.

Furthermore, when a plane experiences turbulence, one cannot use a digital calculator – it could fall out of your hands and buttons are difficult to push. A slide rule on your wrist? That will always work despite electrical and physical interference!

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]]>The post Statistics in the New HSC appeared first on Ringo Mok.

]]>In my formative years of high school, the topic of Statistics often involved drawing up tables, tallying up personally irrelevant data, doing mundane calculations on a calculator, and getting out a protractor and ruler to do summary statistic diagrams (which spells TREK!). I could not help but notice an adverse reaction from everyone, from both teachers and students, towards the topic. People treated Statistics as this topic that you just had to do because it was related to Mathematics but it really wasn’t Mathematics. This attitude of course had a bleeding effect onto my own attitude towards the topic. Even today in the staffroom that I work in, there are many teachers who still hold a major dislike towards Statistics.

It didn’t help that during my first year of University, I was completely turned off Statistics as my lecturer mumbled and fumbled his way through lectures with an accent and stutter. I was turned off learning anything new in what I perceived to be a completely useless and uninteresting subject. I also didn’t so well but that was probably due to my lack of interest in it. I didn’t do anymore Statistics study in my Degree after first year.

So when I heard that Statistics was being added into the HSC, I initially reacted with shock and disappointment. I could not believe it. I believe this reaction was due to the culmination of all the negative experiences I had with Statistics compounded with the fact that I had not done my own research and learning in the subject – I was a poor student of Statistics and now I have to teach it? Perhaps this was an experience that other teachers can relate to.

Such a negative and preconceived view that I had would require some eye-opening insight and mathematical maturity to overcome – and so begins my journey to gradually appreciate and nurture my interest in it.

In teaching the IB courses with Statistics, I started interacting more with senior level Statistics just out of necessity. Unfortunately, my earlier classes in my career had to suffer from the bleeding effect of a negative attitude towards the topic, emanating from their teacher. Through teaching, I found that I really didn’t know much about the topic – I thought that what I learned in high school was sufficient but there was a lot more that I didn’t really know as I hadn’t learned it for myself. I couldn’t even explain what p-value meant!

From 2017, my head of department modelled to his staff what using data to make informed decisions in teaching looked like. This started with collecting and tallying up multiple choice responses in a half-yearly or yearly assessment task. The insights drawn from this exercise were surprising. Anecdotal evidence and judgement in what students need is often not enough – cold hard facts from the data they produce is more important: *You may have taught something, but have your students learned it?*

We also had staff meetings where we looked at RAP data from HSC performance. The trends we saw such as how the state and our school did not really understand significant figures was a surprising one. We definitely taught it, we thought it was easy, but obviously the students didn’t get it. And so we did something about it in our teaching of the next cohort.

The power of Statistical Analysis was so obvious.

I had merely been blinded to do anything with it because of my stubborn preconceptions.

In doing more exercises like these, later building upon the analysis of cohort multiple choice, I also started adding more granularity to the data by examining class clusters and per-question responses. The performance insights as I looked from class to class was interesting to say the least. For example, my class did not do as well in the topic of Bits and Bytes (when I thought we nailed it). This informed my own teaching to go back to revisit that. Such insight wouldn’t have been possible if we just looked at whole cohort data.

Further conversations with my head of department revealed to me that he loves Statistics. In the school, he is also the Director of Statistics and helps other heads of departments in other faculties to understand their data. I wanted to be like him because what he was doing was so beneficial and it was kinda cool too – helping others find insight in something that they would otherwise be unable to comprehend fully so they can do something about it. Before I knew it, data analysis became something I am also passionate about. In wanting more, I started googling for courses and found that at the University of Sydney, the Graduate Certificate of Data Science offered what looks to be the perfect packaged of up-skilling my data analysis abilities. And so I applied and I am now studying part time there!

In my lectures I have learned what I should have learned in first year Statistics. It was probably due to a combination of my increased awareness of the relevance of Statistics and my own mathematical maturity that allowed me to learn this stuff properly now. Some of the material I learned throughout my undergraduate degree helped also (like Measure Theory). It also helped that my lecturer was quite audible!

So Statistics is being added into the HSC courses? Bring it on!

When teaching the new syllabus, some of the material can be cross referenced with textbooks, worksheets and resources from the old syllabus. These dot points don’t really present a difficulty in the teaching.

However, there aren’t many resources readily available in the usual channels of published textbooks or worksheets for Statistics. These resources will need time to build up. Teachers will need time to up-skill their own understanding of the topic before they teach it – if they don’t, it’ll become a classic example of the blind leading the blind.

It also doesn’t help that the syllabus doesn’t define what a random variable is very clearly.

A random variable is a variable whose possible values are outcomes of a statistical experiment or a random phenomenon. (p. 73)

Know that a random variable describes some aspect in a population from which samples can be drawn. (p. 47)

HSC Mathematics Advanced Syllabus

There’s a few problems with this.

The first definition presented in the glossary is quite a self-referential definition. A random variable is a variable. Great.

The second definition in the content outcomes is merely a qualitative description. A vague understanding of what a random variable *does* rather than what it *is*.

A random variable, usually denoted with a capital *X*, is actually a *function* that maps each element of a sample space to a real number (there’s a further generalisation in University mathematics, but for the sake of a high school understanding this is suficient).

Basically, the domain of a random variable is the sample space and the range is a subset of the real numbers. If the number of elements in the sample space is countable then it’s a discrete random variable and if it’s uncountable, then it’s continuous. (The definition of countable and uncountable is for another day).

For example, suppose we roll a 4 sided die with faces coloured blue, green, red and yellow. An example random variable that allows us to model the scenario is as follows:

\[X(blue) = 1, X(green) = 2, X(red) = 3, X(yellow) = 4\]

There are of course an infinite number of ways you can define this random variable. There’s nothing stopping you from defining them as \(X(blue) = 342, X(green) = 111, X(red)=-2313, X(yellow)=99999\) but that just makes things hard on yourself and most likely will have no benefit in modelling the situation.

Now we can ask questions like “what is the probability of getting a green or a blue?” which translates to finding \(P(X < 3)\).

So \[P(X<3) = P(X=2) + P(X=1) = P(\{green\}) + P(\{blue\})\]

Now if the dice is fair, the random variable follows a uniform distribution which means each outcome has an equally likely chance of occurring. By mapping the sample space of blue, green, red and yellow to numbers (1, 2, 3, 4 in this example), we don’t have to concern ourselves with those words but rather focus on the behaviour of the numbers under a certain distribution. Most of the time when it’s numbers we are studying, we just let \(X(x) = x\) for all \(x \in \Omega\) where \(\Omega\) is the sample space. This is why sometimes, capital *X *and lower case *x* are interchangeable in statistics questions and we still get the correct answers.

I guess I should write more (clearly) on these kind of things later in the future.

All in all, I welcome the addition of Statistics to the new syllabus and I am excited to teach it to my students!

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]]>The post L’Hopital’s Limit From A Phone Game appeared first on Ringo Mok.

]]>Turns out the solution to the game was just simply the word “answer” as the instruction was to simply enter that in. However, the integral and the limit shown there is possible to do and that is what this post is going to focus on. I don’t know what the game is called so I can’t link it. If I find out, I’ll edit this and link it in.

So here’s the problem in case you can’t see the screenshot:

\[\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x^2\ln\left(1+\frac{1}{x}\right)}\]

Spoiler

The first thing to notice is that \(\displaystyle\lim_{x\rightarrow \infty} \left(1+\frac{1}{x}\right)^x\) is actually the definition of the constant \(e\).

By using logarithm laws, we can manipulate the denominator to become just \(x\):

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x\ln\left(1+\frac{1}{x}\right)^x}\\=&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x\ln\left(e\right)}\\=&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x}\end{align*}\]

Now this is a \(\displaystyle \frac{\infty}{\infty}\) limit so we can use L’Hopital’s rule to help us evaluate the limit. If we differentiate the numerator, we apply the Fundamental Theorem of Calculus and the integral sign just goes away:

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{\left[x^2\left(e^\frac{1}{x}-1\right)-x\right]}{1}\\=&\lim_{x\rightarrow \infty}x^2\left(e^\frac{1}{x}-1\right)-x\\=&\lim_{x\rightarrow \infty}x^2e^\frac{1}{x}-x^2-x\end{align*}\]

We can algebraically manipulate this into a fraction:

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{e^\frac{1}{x}-1-\frac{1}{x}}{\frac{1}{x^2}}\end{align*}\]

This is a \(\displaystyle\frac{0}{0}\) limit so we can apply L’Hopital’s Rule:

\[\lim_{x\rightarrow \infty} \frac{\frac{-1}{x^2}e^{\frac{1}{x}}+\frac{1}{x^2}}{\frac{-2}{x^3}}\]

Simplifying this and then using L’Hopital’s Rule once more:

\[\begin{align*}&\lim_{x\rightarrow \infty} \frac{-e^{\frac{1}{x}}+1}{\frac{-2}{x}}\\=&\lim_{x\rightarrow \infty}\frac{\frac{1}{x^2}e^{\frac{1}{x}}}{\frac{2}{x^2}}\end{align*}\]

Finally, we can see that the factors of \(\displaystyle\frac{1}{x^2}\) cancel out leaving us with:

\[\lim_{x\rightarrow \infty} \frac{e^\frac{1}{x}}{2}=\frac{1}{2}\]

A way to check this answer is to just chuck it into a graphing application and check the horizontal asymptote:

Yep. The answer is definitely \(\frac{1}{2}\).

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]]>The post No, You Don’t Need Quadratic Formula appeared first on Ringo Mok.

]]>The whole point of a question that uses the sum and product of roots to solve it is to avoid finding out exactly what those roots are.

Let \(f(x) = x^3+3bx^2 +3cx +d\).

(1) Show that \(y=f(x)\) has 2 distinct turning points if and only if \(b^2>c\).

(2) If \(b^2 > c\) show that the vertical distance between the turning points is \[4(b^2-c)^\frac{3}{2}\]

Part 1 Solution

Differentiating \(f(x)\):

\[f'(x) = 3x^2 + 6bx+3c\]

There are two distinct turning points if and only if \(f'(x)=0\) has two distinct solutions, i.e. the discriminant of the quadratic is positive.

\[\begin{align*}(6b)^2-4(3)(3c) &> 0\\36b^2-36c&>0\\b^2&>c\end{align*}\]

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Part 2 Solution

Since the leading term of \(f(x)\) is \(x^3\), then we know that the graph takes a cubic shape where if \(\alpha\) and \(\beta\) are the \(x\) coordinates of the turning points and \(\alpha < \beta\), then \(f(\alpha)>f(\beta)\).

Hence, we need to calculate the value of \(f(\alpha)-f(\beta)\) to find the vertical distance between the turning points.

Firstly, using Vieta’s formulas (otherwise known as sum and product of roots):

\[\alpha + \beta = \frac{-6b}{3}=-2b\]

\[\alpha\beta = \frac{3c}{3}=c\]

As a corollary, we can also derive the following result:

\[\begin{align*}(\alpha-\beta)^2&=\alpha^2-2\alpha\beta +\beta^2\\(\alpha-\beta)^2&=(\alpha + \beta)^2 -4\alpha\beta\\(\alpha-\beta)^2&=(-2b)^2-4(c)\\\alpha -\beta&=-\sqrt{4b^2-4c}\\&=-2\sqrt{b^2-c}\end{align*}\]

Now the value of \(\alpha – \beta\) must be negative as we have already stated that \(\alpha <\beta\).

Now the vertical distance of the turning points:

\[\begin{align*}f(\alpha)-f(\beta)&=\alpha^3-\beta^3+3b\alpha^2-3b\beta^2+3c\alpha-3c\beta\\&=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)+3b(\alpha-\beta)(\alpha+\beta)+3c(\alpha-\beta)\\&=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2+3b(\alpha+\beta)+3c)\\&=(\alpha-\beta)\left((\alpha+\beta)^2-\alpha\beta+3b(\alpha+\beta)+3c\right)\\&=-2\sqrt{b^2-c}\left((-2b)^2-c+3b(-2b)+3c\right)\\&=-2\sqrt{b^2-c}(4b^2-c-6b^2+3c)\\&=-2\sqrt{b^2-c}(-2b^2+2c)\\&=-2\sqrt{b^2-c}(-2)(b^2-c)\\&=4(b^2-c)^\frac{3}{2}\end{align*}\]

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]]>The post Trigonometric Tricks in Geometry appeared first on Ringo Mok.

]]>In the triangle \(ABC\), \(AB \perp AC\), \(AE=EC\), \(AC=\sqrt{2}DE\). Find the angle \(BDE\).

Spoiler

Firstly, let \(\angle ACE\) be equal to \(\theta\).

The triangle \(ACE\) is isosceles as there are two equal sides \(AE\) and \(CE\).

\(\angle CAE\) is equal to \(\angle ACE\) since they are the base angles of the triangle \(ACE\) that is isosceles. So \(\angle CAE = \theta\).

\(\angle CAE + \angle ACE + \angle AEC = 180^\circ\) (Angle sum of a triangle)

\[\theta + \theta + \angle AEC = 180^\circ\]

\[\angle AEC = 180^\circ-2\theta\]

Now we also know that:

\(\angle DAE = 90^\circ – \theta\) (Complementary angles)

\(\angle EDA = 180^\circ – \alpha\) (Angles on a straight angle)

In Triangle \(ADE\) and using Sine Rule:

\[\begin{align*}\frac{\sin(180^\circ – \alpha)}{x}&=\frac{\sin(90^\circ – \theta)}{1}\\\frac{\sin \alpha}{x} &= \cos\theta\\\sin\alpha &=x\cos\theta\end{align*}\]

In Triangle \(ACE\) and using Sine Rule:

\[\begin{align*}\frac{\sin (180^\circ – 2\theta)}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{\sin(2\theta)}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{2\sin\theta\cos\theta}{\sqrt{2}}&=\frac{\sin\theta}{x}\\\frac{2\cos\theta}{\sqrt{2}}&=\frac{1}{x}\\x\cos\theta&=\frac{\sqrt{2}}{2}\end{align*}\]

Equating the two results for \(x\cos\theta\) together, we get:

\[\begin{align*}\sin\alpha &=\frac{\sqrt{2}}{2}\\\alpha &=45^\circ\end{align*}\]

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]]>The post Geometry Induction appeared first on Ringo Mok.

]]>Prove that an \(n\)-sided convex polygon has \(\frac{1}{6}n(n-1)(n-2)\) triangles that can be drawn inside it by connecting any three of the vertices.

Spoiler

For \(n=3\): \[\frac{1}{6}\times 3\times (3-1) \times (3-2) = 1\]

It is obvious that there is only one triangle to form with three vertices. Hence it is true for the base case of \(n=3\).

(Side note: For \(n<3\) the statement is trivially true as there is no such thing as polygons of less than 3 sides)

Assume that it is true for \(n=k\) for some integer \(k\geq 3\). That is,

\(\frac{1}{6}k(k-1)(k-2)\) triangles can be formed between \(k\) number of vertices.

We are required to prove that the number of triangles possible to be formed for \(n=k+1\) is:

\[\frac{1}{6}(k+1)(k)(k-1)\]

To construct the case for \(n=k+1\), add a vertex between the first and the \(k\)th vertex:

There are two cases when counting the number of triangles that can be formed:

- Case 1 – When \(x_{k+1}\) is not a vertex of the triangles. The number of triangles that can be formed in the polygon defined by \(x_1x_2x_3\ldots x_k\) is simply the induction hypothesis as there are \(k\) vertices. That is, in Case 1 the number of triangles possible is:

\[\frac{1}{6}k(k-1)(k-2)\] - Case 2 – When \(x_{k+1}\) is a vertex of the triangles. There then remains 2 vertices to be chosen from the other \(k\) vertices to form a triangle. Therefore, the number of triangles possible in Case 2 is:

\[\left(\matrix{k\\2}\right)=\frac{k!}{(k-2)!2!}=\frac{k(k-1)}{2}\]

Hence, the number of triangles formed for \(k+1\) vertices is found by adding Case 1 and Case 2 together:

\[\begin{align*}&\frac{1}{6}k(k-1)(k-2)+\frac{k(k-1)}{2}\\=&\frac{1}{6}k(k-1)(k-2+3)\\=&\frac{1}{6}k(k-1)(k-+1)\\=&\frac{1}{6}(k+1)(k)(k-1)\end{align*}\]

Therefore the statement is true for \(n=k+1\).

By the principle of mathematical induction, it is therefore true for integers \(n\geq 3\) that \(\frac{1}{6}n(n-1)(n-2)\) triangles can be formed from a convex polygon of \(n\) vertices.

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]]>The post The Perfect View appeared first on Ringo Mok.

]]>Here it is:

A painting 3 metres high is placed on a wall with its lower edge 1 metre above the eye level of an observer. Find how far from the wall the observer should stand in order to get the best view of the painting (i.e. so that the angle subtended at his/her eye by the painting is a maximum).

Spoiler

We can represent the question with the following diagram:

\(\theta\) is the angle subtended by the the painting and the eye, \(\alpha\) is the angle subtended by the horizontal to the eye and the bottom of the painting.

Using right angled trigonometry,

\[\tan(\theta + \alpha) =\frac{4}{x}\]

Expanding the left using the appropriate trigonometric identity:

\[\frac{\tan \theta + \tan \alpha}{1-\tan \theta \tan \alpha}=\frac{4}{x}\]

We can replace \(\tan\alpha\) with \(\frac{1}{x}\) using right angled trigonometry:

\[\frac{\tan\theta + \frac{1}{x}}{1-\tan\theta\cdot\frac{1}{x}}=\frac{4}{x}\]

Multiply the denominators across to the numerators of the other sides:

\[x\left(\tan\theta + \frac{1}{x}\right)=4\left(1-\tan\theta\cdot\frac{1}{x}\right)\]

\[x\tan\theta + 1=4-\frac{4\tan\theta}{x}\]

\[x\tan\theta + \frac{4\tan\theta}{x}=3\]

\[\left(x+\frac{4}{x}\right)\tan\theta=3\]

\[\left(\frac{x^2+4}{x}\right)\tan\theta = 3\]

\[\tan\theta = \frac{3x}{x^2+4}\]

At this stage, most people would make \(\theta\) the subject and then differentiate with respect of \(x\). This is fine, but it will lead to using a lot of chain rule.

If we just differentiate both sides with respect to \(\theta\) anyway:

\[\begin{align*}\frac{d}{d\theta}\left(\tan\theta\right) &= \frac{d}{d\theta}\left(\frac{3x}{x^2+4}\right)\\&=\frac{d}{dx}\left(\frac{3x}{x^2+4}\right)\cdot\frac{dx}{d\theta}\end{align*}\]

The second line works due to the chain rule. This is not a usual way of differentiating but it saves me a lot of work trying to differentiate \(\tan^{-1}\). Hence:

\[\begin{align*}\sec^2\theta&=\frac{3(x^2+4)-2x(3x)}{(x^2+4)^2}\cdot \frac{dx}{d\theta}\\1&=\frac{-3x^2+12}{(x^2+4)^2}\cos^2\theta\cdot\frac{dx}{d\theta}\\\frac{d\theta}{dx}&=\frac{-3x^2+12}{(x^2+4)^2}\cos^2\theta\end{align*}\]

For stationary points, \(\frac{d\theta}{dx}=0\), i.e.

\[\begin{align*}\frac{-3x^2+12}{(x^2+4)^2}&=0\mbox{ or }\cos^2\theta=0\\x&=2\mbox{ or } \theta=\frac{\pi}{2}\end{align*}\]

However, \(\theta = \frac{\pi}{2}\) is not a physically possible solution (looking up to the sky).

Hence, \(x=2\). Testing this solution to confirm that it is a maximum:

Notice that the denominator and \(\cos^2\theta\) are positive in \(\frac{d\theta}{dx}\). Hence, we only need to test if \(x=2\) is a maximum in the numerator expression:

\[-3x^2+12\]

Since the graph of this is a concave down parabola, \(x=2\) must be a maximum.

Hence, the observer should stand 2 metres away from the painting to maximise his viewing pleasure.

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]]>The post Special Integral: \(\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\) appeared first on Ringo Mok.

]]>This integral requires the special result:

\[\int_0^a f(x)\;dx = \int_0^a f(a-x)\;dx\]

It’ll also require trigonometric identities, logarithms, graphs and integration techniques.

Find the value of:

\[\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Spoiler

Using the special result:

\[\begin{align*}I=&\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\\I=&\int_0^\frac{\pi}{2}\ln\left(\sin \left(\frac{\pi}{2}-x\right)\right)\;dx\\I=&\int_0^\frac{\pi}{2}\ln(\cos x)\;dx\end{align*}\]

By adding the first and third lines above together:

\[\begin{align*}2I=&\int_0^\frac{\pi}{2}\left(\ln(\sin x)+\ln(\cos x)\right)\;dx\\=&\int_0^\frac{\pi}{2}\ln(\sin x \cos x)\;dx\\=&\int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\cdot 2\sin x\cos x\right)\;dx\\=&\int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\sin 2x\right)\;dx\end{align*}\]

We finally arrive at the following result (call it Equation *):

\[\begin{align}2I=&\int_0^\frac{\pi}{2}\left(\ln\left(\frac{1}{2}\right)+\ln(\sin 2x)\right)\;dx\end{align}\]

By looking at the graph of \(\ln(\sin 2x)\) in the domain \(0 < x < \frac{\pi}{2}\):

There is an axis of symmetry at \(x=\frac{\pi}{4}\), which means that:

\[\int_0^\frac{\pi}{2}\ln(\sin 2x)\;dx=2\int_0^\frac{\pi}{4}\ln(\sin 2x)\;dx\]

Using integration by substitution:

\[\begin{align*}u &= 2x\\du &= 2dx\end{align*}\]

When \(x=0,\;u=0\) and \(x=\frac{\pi}{4},\;u=\frac{\pi}{2}\).

\[2\int_0^\frac{\pi}{4}\ln(\sin 2x)\;dx=\int_0^\frac{\pi}{2}\ln(\sin u)\;du\]

Since the value of the definite integral doesn’t depend on what letter we use for the variable, i.e.:

\[\int_0^\frac{\pi}{2}\ln(\sin u)\;du=\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Hence:

\[\int_0^\frac{\pi}{2}\ln(\sin 2x)\;dx=\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Notice that the right hand side is actually the integral \(I\). Therefore, we can do the following in Equation *:

\[\begin{align*}2I &= \int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\;dx + I\\ I & = \int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\;dx\\&=\left. x\ln\left(\frac{1}{2}\right)\right|_0^\frac{\pi}{2}\\&=\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\end{align*}\]

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]]>The post Vector Question from Hong Kong 1990 appeared first on Ringo Mok.

]]>In the figure above, \(\vec{x}\) and \(\vec{y}\) are unit vectors, each of which makes an angle \(\theta\) with \( \vec{z}\), where \(0<\theta<\frac{\pi}{2}\).

- Show that \(\vec{x}\cdot\vec{z}=\vec{y}\cdot\vec{z}\).
- Let \(\vec{z}=m\vec{x}+n\vec{y}\). By expressing \(\vec{x}\cdot\vec{z}\) and \(\vec{y}\cdot\vec{z}\) in terms of \(m\), \(n\) and \(\theta\), show that \(m=n\).

Part 1 Solution

Since \(\vec{x}\) and \(\vec{y}\) are unit vectors, then they have a length of 1. Using this, we observe that:

\[\begin{align*}\vec{x}\cdot\vec{z}&=|\vec{x}||\vec{z}|\cos \theta\\&=|\vec{z}|\cos\theta\end{align*}\]

\[\begin{align*}\vec{y}\cdot\vec{z}&=|\vec{y}||\vec{z}|\cos \theta\\&=|\vec{z}|\cos\theta\end{align*}\]

Therefore: \[\vec{x}\cdot{z} = \vec{y}\cdot\vec{z}\]

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Part 2 Solution

Without loss of generality, we can let \(\vec{x}=\left(\begin{matrix}1\\0\end{matrix}\right)\), which is a unit vector. We can do this due to the fact that any coordinate system can be rotated and translated so that the vector \(\vec{x}\) is mapped to \(\left(\begin{matrix}1\\0\end{matrix}\right)\).

Let \(\vec{y}=\left(\begin{matrix}y_1\\y_2\end{matrix}\right)\) for some \(y_1\), \(y_2\in\mathbb{R}\).

By considering the horizontal and vertical vector components of \(\vec{y}\) as shown below:

We can see that \(y_1=|\vec{y}|\cos 2\theta=\cos 2\theta\), since \(\vec{y}\) is a unit vector.

Now as \(\vec{z}=m\vec{x}+n\vec{y}\):

\[\begin{align*}\vec{z}&=m\left(\begin{matrix}1\\0\end{matrix}\right)+n\left(\begin{matrix}y_1\\y_2\end{matrix}\right)\\&=\left(\begin{matrix}m+ny_1\\ny_2\end{matrix}\right)\end{align*}\]

\[\begin{align*}\vec{x}\cdot\vec{z}&=\left(\begin{matrix}1\\0\end{matrix}\right)\cdot\left(\begin{matrix}m+ny_1\\ny_2\end{matrix}\right)\\&=m+ny_1\\&=m+n\cos 2\theta\end{align*}\]

\[\begin{align*}\vec{y}\cdot\vec{z}&=\left(\begin{matrix}y_1\\y_2\end{matrix}\right)\cdot\left(\begin{matrix}m+ny_1\\ny_2\end{matrix}\right)\\&=my_1+ny_1^2+ny_2^2\\&=my_1+n(y_1^2+y_2^2)\\&=my_1+n|\vec{y}|^2\\&=my_1+n\\&=m\cos 2\theta + n\end{align*}\]

From part 1, since \(\vec{x}\cdot\vec{z}=\vec{y}\cdot\vec{z}\):

\[\begin{align*}m+n\cos 2 \theta &= m\cos 2\theta +n\\ m-m\cos 2\theta &= n-n\cos 2\theta \\ m(1-\cos 2\theta) & = n (1-\cos 2 \theta)\\ \mbox{therefore } m&=n\end{align*}\]

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]]>The post Things to Reconsider: The Integral of \(\frac{1}{x}\) appeared first on Ringo Mok.

]]>Consider the following integral:

\[\int \frac{1}{x}\,dx = \log_e|x| + C\]

When learning this integral, teachers often stress the need to include the absolute value signs. The reasoning behind this is often to do with the fact that the logarithm of a negative number is undefined.

However, is it *really* necessary to add the absolute value signs? The basis of this document is to discover the answer to this question and in the process, question some habits in the teaching and learning of Mathematics.

Consider the following equation:

\[ x^2=-1 \]

Before learning about the existence of the imaginary unit \(i\), the answer to this equation is simply: there is no real solution for \(x\). However, after learning about \(i\), many students can now proudly announce that the answer is \(x=\pm i\), as if they have some new esoteric knowledge that separates them from the rest of those still stuck in the real field.

For these two types of students, those who are aware of the complex numbers and those who are not, this equation is intepreted very differently. There is a solution for \(x\) if \(x\in \mathbb{C}\) but there isn’t a solution for when \(x \in \mathbb{R}\). So which answer do we really accept?

If a Mathematics Extension 2 student was asked to solve this equation in an examination and simply wrote that there is no real solution, is he correct? Similarly, if a student from Year 10 wrote the answer \(x=\pm i\), will he be penalised for learning something ahead of his course?

My argument for these cases is that the fault is not with the students at all. The fault lies with the question.

The question fails to define the variable \(x\).

In order for it to be absolutely clear, the question needs to be rewritten as follows:

\[\mbox{Solve for }x\in \mathbb{C}: x^2 = -1\]

Now it would be rather annoying to always define the variable for which we are solving for, so usually we just assume it to be \(\mathbb{C}\) or \(\mathbb{R}\) depending on which course the student is taking. I think we need to stop making this assumption and at the very least start examination papers, worksheets, textbooks and questions with disclaimers, such as: “For all questions, \(x \in \mathbb{R}\)” or “For all questions, the domain of \(f(x)\) is a subset of \(\mathbb{R}\)”. We need to clearly define the domain in which the questions lie!

When we clearly define domains in which the question’s solution is to be given in, some more interesting questions arise. Here are some examples:

- Solve for \(x\in \mathbb{N}: x^2+2x=0\)

A discussion in whether \(0\in\mathbb{N}\) will result from this question. Depending on which school of thought the student is in, this question could yield zero solutions or one solution.

- Solve for \(x\in \mathbb{Q}: 2x^2+3x+1=0\)

Would a question like this make students more aware of checking that their solution makes sense? For \(x\in\mathbb{R}\), there will two solutions but for \(x\in\mathbb{Q}\), there is only one!

- Factorise \(x^4-9\) over \(\mathbb{Q},\mathbb{R},\mathbb{C}\)

To factorise over three different types of numbers will yield three different results. Is the student who factorised over \(\mathbb{Q}\) rather than \(\mathbb{R}\) less deserving of full marks? This would not be an issue if the domain was specified.

- For \(f: \mathbb{R}\rightarrow\mathbb{C}\) such that \(f(x)= \sqrt{x}\), evaluate \(f(-4)\).

Without defining the domain and codomain of the function, it actually won’t make sense to ask the value of \(f(-4)\). The clarity of such a task is not apparent until the function’s domain and codomain is specified.

This is why I generally dislike questions that ask “what is the domain of the function…” because they inherently don’t make sense due to this issue. What they mean to ask is “what is the maximal possible domain of the function as a subset of \(\mathbb{R}\)”.

Consider the following:

\[\begin{align*}y&=\log_e(-x)\quad \mbox{ for } x<0\\\frac{dy}{dx}&=\frac{-1}{-x}\\&= \frac{1}{x}\end{align*}\]

With this argument, it will then be enough to justify the need for absolute values in the integral:

\[\int\frac{1}{x}\,dx=\log_e|x|+C\]

This seems all fine until we realise that we have made an assumption on the function \(\log_e(x)\).

If we make the assumption that \(\log_e(x)\) has a real valued range then the absolute signs are definitely required. However, without any explicit definition of this case, then the same scenario as the \(x^2=-1\) equation arises: Who is more correct – the student who works in \(\mathbb{R}\) or the one who works in \(\mathbb{C}\)?

If we define the domain and codomain/range of \(\log_e(x)\) to be \(\mathbb{C}\), then something very interesting happens.

Leonard Euler discovered that for any \(x\in\mathbb{R}\),

\[e^{ix}=\cos x + i\sin x\]

This is known as Euler’s formula (not to be confused with the Euler Characteristic Formula \(V+F=E+2\)).

If we let \(x=\pi\), then \(e^{i\pi}=\cos\pi+i\sin\pi=-1\) and this results in the famous identity: \[e^{i\pi}+1=0\]

Now let us consider the following for the function \(\mathbb{C}\rightarrow\mathbb{C}:\log_e(z)\). Also recall that for any \(z\in\mathbb{C}\) that \(z = r(\cos x + i\sin x)\) where \(x = \arg z\) and \(r=|z|\).

\[\begin{align*}\log_e(z) &=\log_e(r(\cos x + i\sin x))\\&=\log_e(re^{ix})\\&=\log_e(r)+\log_e(e^{ix})\\&=\log_e(r)+ ix\\\end{align*}\]

Therefore:

\[\log_e(z) = \log_e|z| + i\arg z\]

Now because any number \(z\in\mathbb{C}\) technically has an infinite number of arguments separated by multiples of \(2\pi\), then there are also technically an infinite number of complex logarithms. This has some interesting applications:

Solve for \(z\in\mathbb{C}: e^z=-1\)

\[\begin{align*}z&=\log_e(-1)\\&=\log_e|-1|+i\arg (-1)\\&=i\arg (-1)\\&=i(\pi + 2k\pi)\mbox{ where }k\in\mathbb{Z}\\&=(2k+1)i\pi\end{align*}\]

There are infinitely many solutions for this equation!

The fact that there are infinitely many complex logarithms provides a motivation to define the *principal* complex logarithm:

\[\mbox{Log}_e z = \log_e|z| + i\mbox{Arg} z\]

Note the capital letters to mean the principal logarithm or the principal argument of the number \(z\).

With this context of the complex logarithm now, let us reconsider the integral of \(\frac{1}{x}\) and see what happens when we exclude the absolute value signs:

\[\begin{align*}\int \frac{1}{x}\, dx &= \log_e(x) + C\mbox{ (for a constant C)}\\&=\log_e|x| + i\arg x + C\end{align*}\]

Now \(i\arg x\) is a constant as well, which means this final line can be rewritten as:

\[\int \frac{1}{x}dx = \log_e|x| + K\mbox{ where }K=i\arg x+C\]

Hence, this shows that for the complex logarithm, the absolute value signs are actually not necessary because they are already implied from the way the complex logarithm works!

This may be a nice discovery, but should we now tell students that they don’t need to write the absolute value signs anymore? Absolutely not!

Even though the absolute values are not necessary over the domain and codomain of the complex numbers, the real valued logarithm is the assumed function that students in high school are expected to work with. Although I have gripes with the fact that we assume too much as discussed earlier, we must accept that the general notion of such functions at the high school level in differential and integral calculus is that they are real valued. This is mainly due to the fact that most theorems in high school have been proven true for real numbers, but may or may not fall apart for complex numbers.

When solving \(x\in\mathbb{R}: \cos x = 2\), there is no real solution.

Now consider \(e^{ix}=\cos x + i\sin x\) and \(e^{-ix}=\cos x – i\sin x\). Adding them together we get:

\[\begin{align*}e^{ix}+e^{-ix}&=2\cos x\\\frac{e^{ix}+e^{-ix}}{2}&=\cos x\end{align*}\]

Now let’s solve \(x\in\mathbb{C}: \cos x = 2\).

\[\begin{align*}\frac{e^{ix}+e^{-ix}}{2}&=2\\e^{ix}+e^{-ix}&=4\\e^{2ix}+1&=4e^{ix}\\e^{2ix}-4e^{ix}+1&=0\\e^{ix} & = \frac{4\pm \sqrt{4^2-4(1)(1)}}{2}\\e^{ix} & = \frac{4\pm\sqrt{12}}{2}\\e^{ix} & = 2\pm \sqrt{3}\\ix & = \log_e(2\pm\sqrt{3})\\x & = \frac{1}{i}\log_e(2\pm\sqrt{3})\\x & = -i\log_e(2\pm\sqrt{3})\\x & = -i(\log_e|2\pm\sqrt{3}| + i2k\pi) \mbox{ where } k\in\mathbb{Z}\\x & = -i\log_e|2\pm\sqrt{3}| + 2k\pi\end{align*}\]

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